Class 10

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 8 Structure of Atom Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 8th Lesson Structure of Atom

10th Class Chemistry 8th Lesson Structure of Atom Textbook Questions and Answers

Improve Your Learning

Question 1.
What information does the electronic configuration of an atom provide? (AS1)
Answer:

• The distribution of electrons in shells, sub-shells and orbital in an atom is known as electronic configuration.
• It provides the information of position of an electron in the space of atom.
• The distribution of electrons in various atomic orbitals provides an understanding of the electronic behaviour of the atom and in turn its reactivity.
• The short hand notation is as shown below.

Question 2.
a) How many maximum number of electrons that can be accommodated in a principal energy shell?
Answer:
The maximum number of electrons that can be accommodated in a principal energy shell is 2n². Here n is principal quantum number.

b) How many maximum number of electrons that can be accommodated in a sub-shell?
Answer:
The maximum number of electrons that can be accommodated in a sub-shell is 2(2l +1) (where l is orbital quantum number).

c) How many maximum number of electrons can that be accommodated in an orbital?
Answer:
The maximum number of electrons that can be accommodated in an orbital is 2.

d) How many sub-shells are present in a principal energy shell?
Answer:
The number of sub-shells in a principal energy shell is n (n is principal quantum number).

e) How many spin orientations are possible for an electron in an orbital?
Answeer:
The spin orientations possible for an electron in an orbital are 2.

Question 3.
In an atom the number of electrons in M-shell is equal to the number of electrons in the K and L-shell. Answer the following questions. (AS1)
a) Which is the outermost shell?
Answer:
The outermost shell is N shell.

b) How many electrons are there in its outermost shell?
Answer:
Two electrons are there in outermost shell.

c) What is the atomic number of element?
Answer:
The atomic number of element is 22.

d) Write the electronic configuration of the element.
Answer:
The element is Ti (Titanium). Its electronic configuration is 1s²2s²2p⁶3s²3p⁶4s²3d².

Reason :

• Electrons enter M shell after completion of K and L.
• So the number of electrons in M shell is 10.
• But after completion of 3p orbital electron enters 4s before entering to 3d.
• So outermost orbit or shell is N shell.
• So the atomic number of element is 22.
• Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d².

Question 4.
Rainbow is an example for continuous spectrum – explain. (AS1)
(OR)
Which is naturally occurring continuous spectrum ? Explain.
Answer:

• Rainbow is a spectrum of different colours (VIBGYOR) with different wavelengths.
• These colours are continuously distributed.
• There is no fixed boundary for each colour.
• Hence, rainbow is a continuous spectrum.

Question 5.
How many elliptical orbits are added by Sommerfeld in third Bohr’s orbit ? What was the purpose of adding these elliptical orbits? (AS1)
Answer:
Sommerfeld added two elliptical orbits to Bohr’s third orbit.

Purpose of adding elliptical orbits :

• Bohr’s model failed to account for splitting of line spectra and line spectrum.
• In an attempt to account for the structure of line spectrum, Sommerfeld modified Bohr’s atomic model by adding elliptical orbits.

Question 6.
What is absorption spectrum?
Answer:
Absorption spectrum: The spectrum formed by the absorption of energy when electron jumps from lower energy level to higher energy level is called absorption spectrum. It contains dark lines on bright background.

Question 7.
What is an orbital? How it is different from Bohr’s orbit? (AS1)
(OR)
Comparison between orbit and orbital.
Answer:
The region of space around the nucleus where the probability of finding electron is maximum is called orbital. Whereas orbit is the path of the electron around the nucleus.

These two are differentiated like this.

Question 8.
Explain the significance of three quantum numbers in predicting the positions of an electron in an orbit. (AS1)
(OR)
How are quantum numbers helpful to understand the atomic structure?
Answer:
Significance of three quantum numbers in predicting the positions of an electron in an orbit.

1) Principal quantum number (n) :
The principal quantum number explains about the size and energy of shells (or) orbitals. It is denoted by n.

As ‘n’ increases, the orbitals become larger and the electrons in those orbitals are farther from the nucleus.

It takes values 1, 2, 3, 4, ……………. for that the shells are represented by letters K, L, M, N, ……….

The number of electrons in a shell is limited to 2n².

2) The Angular – momentum quantum number (l) :
The angular momentum quantum number defines the shape of the orbital occupied by the electron and the orbital angular momentum of the electron, is in motion.

l takes values from 0 to n – 1 for these values the orbitals are designated by letters s, p, d, f, ………….. etc.

l also governs the degree with which the electron is attached to nucleus. The larger the value of l, the smaller is the bond with which it is maintained with the nucleus.

3) Magnetic orbital quantum number (m_l) :
The orientation of orbital with external magnetic field determines magnetic orbital quantum number.

m_l has integer values between – l and l including zero.

The number of values for m, are 2l + l, which give the number of orbitals per sub-shell. The maximum number of electrons in orbitals in the sub-shell is 2 (2l + l).

Question 9.
What is nl^x method? How is it useful? (AS1)
(OR)
What is nl^x method? How is it useful in electronic configuration?
Answer:
The shorthand notation consists of the principal energy level (n value) the letter representing sub – level (l value), and the number of electrons (x) in the sub-shell is written as superscript nl^x.

It is useful in writing electron configuration of elements. For example, in Hydrogen (H), the set of quantum numbers is n = 1, l = 0, m_l = 0, m_s = ½ or – ½. The electronic configuration is

Question 10.
Following orbital diagram shows the electronic configuration of nitrogen atom. Which rule does not support this? (AS1)

(OR)
Write the correct electronic configuration of the given nitrogen atom with the help of Hund’s rule.
Answer:

• This electron configuration does not support Hund’s rule.
• According to Hund’s rule, the orbitals of equal energy are occupied with one elec-tron each before pairing of electrons starts.
• Here, pairing of electrons in 2p_x orbital was taken place without filling of an elec-tron in 2p_z orbital.
• Hence the correct electron configuration is as follows.

Question 11.
Which rule is violated in the electronic configuration 1s⁰ 2s² 2p⁴?
Answer:

• Aufbau principle is violated in this electronic configuration because according to Aufbau principle, electron enters orbital of lowest energy.
• Among 1s, 2s and 2p, Is has least energy.
• So Is orbital must be filled before the electron should enter 2s.

Question 12.
Write the four quantum numbers for the differentiating electron of sodium (Na) atom. (AS1)
Answer:
The electronic configuration of sodium (Na) is 1s² 2s² 2p⁶ 3s¹. So the differentiating electron enters 3s. Therefore the four quantum numbers are

Question 13.
What is emission spectrum?
(OR)
When radiation is emitted what is the name given to such spectrum? Explain such spectrum.
Answer:

• The spectrum produced by the emitted radiation is known as emission spectrum.
• This spectrum corresponds to liberation of energy when an excited electron returns back to ground state.

Emission spectrum is of two types :

1) Continuous spectrum :
When white light passes through a prism it dissociates into seven colours. This spectrum is called continuous spectrum.

2) Discontinuous spectrum :
Discontinuous spectrum is of two types.

a) Line spectrum :
The spectrum with sharp and distinct lines. It is given by gaseous atoms.

b) Band spectrum :
The spectrum very closely spaced lines is known as band spectrum. It is given by molecule.

Question 14.
i) An electron in an atom has the following set of four quantum numbers to which orbital it belong to : (AS2)
Answer:

This electron belongs to 2s orbital.
Spin is in clockwise direction. ⇒ 2s¹

ii) Write the fojur quantum numbers for Is1 electron. (AS1)
Answer:
The four quantum numbers for Is1 electron are

Question 15.
Which electronic shell is at a higher energy level K or L? (AS2)
Answer:
L – shell is at higher energy level, because it is far from nucleus than K shell.

Question 16.
Collect the information regarding wavelengths and corresponding frequencies of three primary colours red, blue and green. (AS4)
Answer:
The wavelengths and corresponding frequencies of three primary colours red, blue and green are given below.

Question 17.
The wavelength of a radio wave is 1.0 m. Find its frequency. (AS7)
Answer:
c = 3 × 10⁸ m/s ; λ = 1m ; c = vλ ⇒ v = \(\frac{\mathrm{c}}{\lambda}=\frac{3 \times 10^{8}}{1}\) = 3 × 10⁸ Hz.

Question 18.
Why are there exemptions in writing the electronic configurations of Chromium and Copper?
Answer:
1. Elements which have half-filled or completely filled orbitals have greater stability.

2. So in chromium and copper the electrons in 4s and 3d redistributes their energies to attain stability by acquiring half-filled and completely filled d-orbitals.

3. Hence the actual electronic configuration of chromium and copper are as follows.

Fill In The Blanks

1. If n = 1, then angular momention quantum number (l) = …………………
2. If a sub-shell is denoted as 2p, then its magnetic quantum number values are …………………, …………………, …………………
3. Maximum number of electrons that an M-shell contain is / are …………………
4. For ‘n’, the minimum value is ………………… and the maximum value is …………………
5. For?, the minimum value is ………………… and the maximum value is …………………
6. For’m/ the minimum value is ………………… and the maximum value is …………………
7. The value of ‘ms’ for an electron spinning in clockwise direction is ………………… and for anti-clockwise direction is …………………
Answer:

1. 0
2. – 1, 0, + 1
3. 18
4. 1, – ∞
5. 0, (n – 1)
6. – l, + l
7. + ½, – ½

Multiple Choice Questions

1. An emission spectrum consists of bright spectral lines on a dark back ground. Which one of the following does not correspond to the bright spectral lines?
A) Frequency of emitted radiation
B) Wavelength of emitted radiation
C) Energy of emitted radiations
D) Velocity of light
Answer:
D) Velocity of light

2. The maximum number of electrons that can be accommodated in the L-shell of an atom is
A) 2
B) 4
C) 8
D) 16
Answer:
C) 8

3. If l = 1 for an atom, then the number of orbitals in its sub-shell is
A) 1
B) 2
C) 3
D) 0
Answer:
C) 3

4. The quantum number which explains about size and energy of the orbit or shell is
A) n
B) l
C) m_l
D) m_s
Answer:
A) n

10th Class Chemistry 8th Lesson Structure of Atom InText Questions and Answers

10th Class Chemistry Textbook Page No. 112

Question 1.
How many colours are there in a rainbow? What are they?
Answer:
There are seven colours in a rainbow. They are Violet, Indigo, Blue, Green, Yellow,Orange and Red.

Question 2.
What are the characteristics of electromagnetic waves?
A.nswer:
Electromagnetic energy is characterised by wavelength (l) and frequency (u).

10th Class Chemistry Textbook Page No. 113

Question 3.
Can we apply this equation c = uA, to a sound wave?
Answer:
Yes. It is a universal relationship and applies to all waves.

10th Class Chemistry Textbook Page No. 114

Question 4.
What happens when you heat an iron rod on a flame? Do you find any change in colour while heating an iron rod?
Answer:

• When we heat an iron rod, some amount of heat energy that was absorbed by iron rod is emitted as light.
• First iron turns into red (lower energy corresponding to higher wavelength) and as the temperature rises it glows and turns into orange, yellow, blue or even white respectively (higher energy and lower wavelength).
• If we go on heating the rod, it turns into white light which includes all visible wavelengths.
• So we find some changes in colour while heating an iron rod.

Question 5.
Do you observe any other colour at the same time when one colour is emitted?
Answer:
While heating the rod if the temperature is high enough, other colours will also be emitted, but due to higher intensity of one particular emitted colour (eg.: red), others cannot be observed.

Question 6.
How do various colours come from fire works?
(OR)
Do you enjoy Deepavali fire works? Variety of colours is seen from fire works. How do these colours come from fire works?
Answer:
Yes. The electrons present in atoms of elements absorb energy and move to excited states and they return to ground state with emission of energy in visible spectrum. So the colours observed during fire works are the emitted energy by various elements in different fire works.

10th Class Chemistry Textbook Page No. 115

Question 7.
Do you observe yellow light in street lamps? Which will produce yellow light?
Answer:
Yes, sodium vapours produce yellow light in street lamps.

Question 8.
Why do different elements emit different flame colours when heated by the same non-luminous flame?
Answer:

• All the materials are made up of atoms and molecules. These atoms and molecules possess certain fixed energy.
• An atom or molecule having lowest possible energy is said to be in ground state.
• When we heat the materials the electrons of these atoms gain energy and move to excited states (higher energy state).
• An atom of molecule in excited state can emit light to lower its energy in order to get stability and come back to ground state.
• Light emitted in such process has certain fixed wavelength for one kind of atoms.
• The light emitted by different kinds of atoms is different because the excited states electrons will go are different. So different elements produce different flame colours.

Question 9.
What happens when an electron gains energy?
Answer:
The electron moves to higher energy level called the excited state.

10th Class Chemistry Textbook Page No. 116

Question 10.
Does the electron retain the energy forever?
Answer:
The electron loses the energy and comes back to its ground state. The energy emitted by the electron is seen in the form of electromagnetic energy.

Question 11.
Did Bohr’s model account for the splitting of line spectra of a hydrogen atom into finer lines?
Answer:
No, Bohr’s model failed to account for splitting of line spectra.

Question 12.
Why is the electron in an atom restricted to revolve around the nucleus at certain fixed distances?
Answer:
In order to explain the atomic spectra, Bohr-Sommerfeld model proposed that the electrons are restricted to revolve around the nucleus at certain fixed distances.

10th Class Chemistry Textbook Page No. 117

Question 13.
Do the electrons follow defined paths around the nucleus?
Answer:
No, they revolve around the nucleus in a region called orbital.

Question 14.
What is the velocity of the electron?
Answer:
It is very close to light.

Question 15.
Is it possible to find exact position of electron? How do you find the position and velocity of an electron?
Answer:
No, as the electrons are very small, light of very short wavelength is required for this task.

This short wavelength light interacts with the electron and disturbs the motion of electron. So it is not possible to find exact position and velocity of electron simultaneously. Whereas we can find the region where the probability of finding electron is more.

Question 16.
Do atoms have a definite boundary, as suggested by Bohr’s model?
Answer:
Yes, atoms have definite boundary.

Question 17.
What do we call the region of space where the electron might be, at a given time?
Answer:
The region of space around the nucleus where the probability of finding an electron is maximum, called an orbital.

10th Class Chemistry Textbook Page No. 118

Question 18.
What information do the quantum numbers provide?
Answer:
The quantum numbers describe the space around the nucleus where the electrons are found and also their energies.

Question 19.
What does each quantum number signify?
Answer:
The quantum numbers signify the probability of finding electron in the space around nucleus.

10th Class Chemistry Textbook Page No. 119

Question 20.
What is the maximum value of/for n = 4?
Answer:
The maximum value of / for n = 4 is 3.

Question 21.
How many values can l have for n = 4?
Answer:
l takes values from 0 to n – 1. So l has 4 values for n = 4. Those values are 0, 1,2, 3.

Question 22.
Do all the p-orbitals have the same energy? A. Orbitals in the sub-shell belonging to same shell possess same energy but they differ in their orientations.

10th Class Chemistry Textbook Page No. 121

Question 23.
How are two electrons in the Helium atom arranged?
Answer:
They are arranged in pair in Is orbital and the electronic configuration is 1s².

10th Class Chemistry Textbook Page No. 122

Question 24.
What are the spins of two electrons in an orbital?
Answer:
The two electrons in an orbital have opposite spins. If one is clockwise spin, then other electron has anti-clockwise spin.

Question 25.
How many electrons can occupy an orbital?
Answer:
An orbital can hold only two electrons.

10th Class Chemistry 8th Lesson Structure of Atom Activities

Activity – 1

Question 1.
Explain the wave nature of light.
(OR)
How does light behave? Explain.
Answer:

• Light is an electromagnetic wave.
• Electromagnetic waves are produced when an electric charge vibrates.
• This vibrating electric charge creates a change in the electric field. The changing electric field creates a changing magnetic field.
• This process continues with both the created fields being perpendicular to each other and at right angles to the direction of propagation of the wave.
• This electromagnetic wave is produced.

Activity – 2

Question 2.
Write an activity which shows metal produces colour in flame.
(OR)
‘Metal produces colour in a flame.’ Prove the statement by giving examples.
Answer:
A)

• Take a pinch of cupric chloride in a watch glass and make a paste with concentrated hydrochloric acid.
• Take this paste on a platinum loop and introduce it into a non-luminous flame.
• Cupric chloride produces a green colour flame.

B)

• Take a pinch of strontium chloride in a watch glass and make a paste with concentrated hydrochloric acid.
• Take this paste on a platinum loop and introduce it into a non-luminous flame.
• Strontium chloride produces a crimson red flame.

Activity – 3

Question 3.
Complete the electronic configuration of the following elements.

Answer:

AP State Board Syllabus AP SSC 10th Class Biology Important Questions Chapter 1 Nutrition.

AP State Syllabus SSC 10th Class Biology Important Questions 1st Lesson Nutrition

10th Class Biology 1st Lesson Nutrition 1 Mark Important Questions and Answers

Question 1.
Define photosynthesis. (OR) What is photosynthesis?
Answer:
Photosynthesis is the process by which plants containing the green pigment called chlorophyll which build up complex organic molecules from relatively simple inorganic ones using sun light as an energy source.

Question 2.
What is the balanced equation to show the process of photosynthesis?
Answer:

Question 3.
Bhumika told that “If there were no green plants, all life on the earth would come to an end” Comment.
Answer:
All the living organisms on the earth depends on the plants either directly or indirectly for the food and oxygen.

Question 4.
Which disease occurs in child when there is an immediate second pregnancy or repeated child births in a mother?
Answer:
Marasmus

Question 5.
Give two examples for nutritional deficiency diseases.
Answer:
Nutritional deficiency diseases: Kwashiorkor, Marasmus, Beri-Beri, Glossitis, Pellagra, Anemia, Scurvy, Rickets, etc.

Question 6.
Your neighbour’s children appear with swollen legs, hands and other body parts. They have dry skin and frequently suffering from Diarrhoea. What are the reasons for it ? What suggestions do you give to their parent ?
Answer:
The children appear with swollen legs, hands and other body parts. They have dry skin and suffering from diarrhoea.
The reason for it is malnutrition and they are suffering from Kwashiorkor and Marasmus diseases.
They are advised to take proteins and calorie rich foods like liver, meat, eggs, milk, fruits, cereals and leafy vegetables.

Question 7.
What suggestions do you give to your friend suffering from constipation?
Answer:
Constipation can often be avoided by
a) having plenty of roughages in our daily diet. Ex : Leafy vegetables, beans, cabbage, etc.
b) drink plenty of water daily.
c) avoid junk food,
d) swallow the food only after its thorough mastication.

Question 8.
Which organelle of the leaf absorbs energy from the sunlight for photosynthesis ?
Answer:
Chloroplast.

Question 9.
Which gas is evolved in the diagram experiment? How can this gas be tested for confirmation?

Answer:
i) Oxygen gas is evolved in this experiment.
ii) If the burning splinter is kept near the mouth of the test tube, it burns brightly.

Question 10.
Classify the following vitamins into water soluble and fat soluble
i) Riboflavin ii) Retinol iii) Tocoferol iv) Thiamin
Answer:
Water soluble vitamins : Riboflavin (B₂), Thiamin (B₁)
Fat soluble vitamins : Retinol (A), Tocoferol (E)

Question 11.
Prepare your own tabular column to get information about food deficiency diseases from a doctor.
Answer:

Question 12.
Why is KGH used in Mohl’s half leaf experiment? (OR)
Why do we use KOH solution in Mohl’s half-leaf experiment?
Answer:
1) We conducted Mohl’s half leaf experiment to prove CO₂ is essential for photosynthesis.
2) So KOH is used in this experiment to absorb the CO₂ present inside the bottle.

Question 13.
What questions will you ask a doctor to know about malnutrition ?
Answer:

1. What is malnutrition?
2. What are the causes for malnutrition?
3. What are the different types in malnutrition?
4. How can we overcome the malnutrition?
5. What are the reasons for malnutrition in our country?

Question 14.
Name the vitamin which is synthesized by the bacteria present in the human intestine.
Answer:
B₁₂ (Cyanocobalamine) is the vitamin which is synthesized by the bacteria present in the human intestine.

Question 15.
Which digestive juice doesn’t contain enzymes ?
Answer:
Bile juice which is produced by liver doesn’t contain enzymes. But it converts fats into small globules. This process is called emulsification.

Question 16.
One student takes high calorie food. Another student takes less calorie food. But both are affected with diseases. Name the diseases by which they are affected.
Answer:
Student takes high calorie food affected by – Obesity.
Student takes less calorie food affected with – Marasmus.

Question 17.
Mention the two chemicals which you have used in an experiment to test the presence of starch in the leaf.
Answer:
The chemicals used in an experiment to test the presence of starch in the leaf are

1. Methylated spirit
2. Iodine solution.

Question 18.
18. Write two slogans for campaign on Mal-nutrition.
Answer:

1. Be cool and say no to fast food.
2. Be smart – Eat smart
3. Eat healthy – Look healthy – Feel healthy

Question 19.
Identify the two parts A and B indicated in the given figure.

Answer:
A – Stomach
B – Large intestine

Question 20.
Doctors advise not to eat food items at the time of fever. What are the reasons for this ?
Answer:

1. At the time of fever body temparature is high.
2. At that time digestive enzymes do not properly work to digest food.
3. That’s why doctors advise not to eat food items at the time of fever.

Question 21.
If we chew the grains like wheat, jowar, rice we feel sweet. Why?
Answer:

1. The grains like wheat, jowar, rice contain carbohydrates.
2. Ptyalin acts on them and coverts them into sugars. So we feel sweet.

Question 22.
What teeth you use when you eat peas and banana? Why?
Answer:

1. Premolars and molars are used when we eat peas and banana.
2. Their function is to chew and grind the seed and food material.

Question 23.
What is autotrophic nutrition?
Answer:
It is a type of nutrition in which an organism makes its own food from the simple inorganic materials like carbon dioxide and water using light as source of energy.

Question 24.
What is nutrition ?
Answer:
Nutrition: Nutrition is the process of intake or procurement of nutrients.

Question 25.
What are the main modes of nutrition ?
Answer:
Autotrophic nutrition and heterotrophic nutrition are the main modes of nutrition.

Question 26.
What are different types of heterotrophic nutrition ?
Answer:
Heterotrophic nutrition is of three types. They are:

1. Saprophytic nutrition
2. Parasitic nutrition and
3. Holozoic nutrition.

Question 27.
What are the symptoms of disease pellagra?
Answer:
Dermatitis, diarrhoea, loss of memory and scaly skin are the symptoms of disease pellagra.

Question 28.
Which vitamin deficiency causes Rickets ? What are its symptoms?
Answer:
Vitamin D Calciferol deficiency results in Rickets. The symptoms shown are improper formation of bones, knocknees, swollen wrists, delayed dentition, weak bones, etc.

Question 29.
For proper vision which vitamin is required? What is its chemical name?
Answer:
Vitamin A is required for proper vision. The chemical name of vitamin ‘A’ is Retinol.

Question 30.
What are the complex molecules produced by plants from simple Inorganic substances?
Answer:
Carbohydrates, proteins and lipids are produced from simple inorganic substances like water and CO₂.

Question 31.
What are the essential factors required for photosynthesis?
Answer:
Carbon dioxide, water, chlorophyll and sunlight are the essential factors required for photosynthesis.

Question 32.
What is the equation for photosynthesis proposed by C.B.Van Neil in the year 1931?
Answer:

Question 33.
In which form are carbohydrates stored in plants?
Answer:
Carbohydrates are stored in plants in the form of starch.

Question 34.
What is the reagent that is used to test the presence of starch in leaves?
Answer:
The reagent that is used to test the presence of starch in leaves is Iodine solution. The colour of the leaf will turn to blue-black in colour.

Question 35.
Who found that water was essential for the increase of plant mass?
Answer:
Von Helmont found that water was essential for increase of plant mass in the year 1648.

Question 36.
Who performed series of experiments in 1770 to reveal the role of air in growth of green plants?
Answer:
Joseph Priestly performed series of experiments in 1770 to reveal the role of air in growth of green plants.

Question 37.
What is the role of potassium hydroxide solution kept inside the glass bottle in the Mohl’s half leaf experiment?
Answer:
Potassium hydroxide solution kept inside the glass bottle in the Mohl’s half leaf experiment absorbs the carbon dioxide present inside the bottle.

Question 38.
Who found that gas bubbles liberated from hydrilla plant contain the gas oxygen?
Answer:
Jan Ingenhousz in 1779 found that the gas bubbles liberated from hydrilla plant contain the gas oxygen.

Question 39.
Who coined the term chlorophyll for the extract of green coloured substance from the leaf?
Answer:
Pelletier and Caventou in the year 1817 coined the term chlorophyll for the extract of green coloured substance from the leaf.

Question 40.
Chloroplast is formed by how many membranes?
Answer:
Chloroplast is formed by 3 membranes.

Question 41.
What is grana?
Answer:
Grana : The stacked sac like structures formed by the third layer of chloroplasts is called grana.

Question 42.
What is the function of stroma?
Answer:
It is believed to be responsible for enzymatic reactions leading to the synthesis of glucose, which inturn join together to form starch.

Question 43.
What is stroma?
Answer:
Stroma : The fluid filled portion of chloroplast is called as stroma.

Question 44.
What are the two major phases found in photosynthesis?
Answer:
The two major phases found in photosynthesis are:

Question 45.
Why is light reaction phase called photochemical phase?
Answer:
A series of chemical reactions occur in a very quick succession initiated by light and therefore the phase is technically called the photochemical phase.

Question 46.
Where does the light reaction take place?
Answer:
The light reaction takes place in chlorophyll containing thylakoids called grana of Chloroplasts.

Question 47.
What are the end products of light reaction?
Answer:
The end products of light reaction are O₂, ATP and NADPH.

Question 48.
What are called the assimilatory powers?
Answer:
ATP and NADPH formed at the end of the light reaction are called assimilatory powers.

Question 49.
What are dark reactions?
Answer:
Dark reactions: The reactions that occur in both presence or absence of light are called dark reactions. The occurance of dark reaction is independent of light.

Question 50.
Who observed dark reactions?
Answer:

1. The entire series of reactions involved in the conversion of CO₂ to glucose were identified by Melvin Calvin.
2. The dark reactions are also called as Calvin Cycle.

Question 51.
What is Calvin cycle?
Answer:
Calvin cycle: The cycle of reactions in fixation of carbon dioxide to glucose is called Calvin cycle.

Question 52.
How is glucose produced during dark reaction?
Answer:
In the dark reaction, the hydrogen of the NADPH is used to combine with CO₂ by utilizing ATP energy and ultimately produce glucose.

Question 53.
Write some of the events that occur in the chloroplasts during photosynthesis.
Answer:

1. Conversion of light energy to chemical energy.
2. Splitting of water molecule.
3. Reduction of carbon dioxide to carbohydrates.

Question 54.
What is photolysis of water? Which gas is released?
Answer:
Photolysis of water:

1. Photolysis of water Splitting of water molecule by light activated chlorophyll molecule is known as photolysis of water.
2. Oxygen is released during (photosynthesis) photolysis of water.

Question 55.
In which cells of the leaves photosynthesis takes place?
Answer:
The mesophyll cells of leaf containing palisade and spongy tissue photosynthesis takesplace.

Question 56.
What is the ultimate source of energy?
Answer:
The ultimate source of energy is the Sun.

Question 57.
What are guard cells?
Answer:
Guard cells: The two kidney shaped cells which surrounds the stoma are called guard cells.

Question 58.
How does CO₂ enter into leaf?
Answer:
CO₂ present in the atmosphere enters through stomata into the cells of leaf by diffusion.

Question 59.
What is the first stable product formed in dark reaction or photosynthesis?
Answer:
Phosphoglyceric acid or PGA is the first stable product formed in dark reaction or photosynthesis.

Question 60.
Why ATP and NADPH are required in photosynthesis?
Answer:
ATP and NADPH are required for the utilization of carbon dioxide and formation of glucose during photosynthesis.

Question 61.
Why chloroplasts are green in colour?
Answer:
Chloroplasts are green in colour due to the presence of a green colouring pigment called chlorophyll.

Question 62.
What are the examples for parasitic organisms?
Answer:
The examples for parasitic organisms are cuscuta (plant), lice, leeches and tapeworms (animals).

Question 63.
How does amoeba take food into the body?
Answer:
Amoeba takes in food using temporary finger like extensions called pseudopia of the cell surface.

Question 64.
How does paramoecium take food?
Answer:
Food is moved to the cytosome by the movement of cilia which covers the entire surface of the cell where the food is ingested.

Question 65.
What is ingestion?
Answer:
Ingestion : The process of taking food into the body is called ingestion.

Question 66.
Name the three pairs of salivary glands. What is the enzyme secreted by them?
Answer:

1. The three pairs of salivary glands in mouth are :
   i) Parotid glands
   ii) Submandibular glands and
   iii) Sublingual glands.
2. The enzyme secreted by salivary glands is amylase (ptyalin).

Question 67.
What is the role of amylase in digestion of food?
Answer:
Amylase helps in the breakdown of complex carbohydrates to simple ones.

Question 68.
What is digestion?
Answer:
Digestion : The process of breaking down of complex food substances into simple substances so that they can be used by the body with the help of enzyme is called digestion.

Question 69.
How does the food from oesophagus move into the stomach?
Answer:
Food passes through oesophagus by wave like movements called peristaltic movements and reaches the stomach.

Question 70.
What is chyme?
Answer:
Chyme : Chyme is a soft slimy substance of food in which some proteins and carbohydrates have broken down.

Question 71.
What is the function of sphincter muscle present at the exit of stomach?
Answer:

1. The sphincter muscle is responsible for regulating the openings of the stomach into small intestine.
2. So that only small quantities of the food material may be passed into the small intestine from the stomach at a time.

Question 72.
What does the gastric juice contain?
Answer:
The gastric juice secreted by the walls of stomach contains Hydrochloric acid, protein digesting enzyme, pepsin and mucus.

Question 73.
What makes the internal condition of the intestine gradually to a basic or alkaline one?
Answer:
Liver and pancreatic juice make the internal condition of the intestine gradually to a basic or alkaline one.

Question 74.
What is emulsification?
Answer:
Emulsification : Fats are digested by converting them into small globule like forms by the help of bile juice secreted from liver. This process is known as emulsification.

Question 75.
For what intestinal juice secreted by small intestine is responsible?
Answer:
The enzymes present in intestinal juice finally convert the protein to amino acids, complex carbohydrates into glucose and fats into fatty acids and glycerol.

Question 76.
What are the enzymes that act on proteins?
Answer:
Pepsin and Trypsin are the enzymes that act on proteins. Both these enzymes convert proteins to peptones.

Question 77.
What is absorption?
Answer:
Absorption : Transport of the products of digestion from the walls of the intestine into blood is called absorption.

Question 78.
What is defecation?
Answer:
Defecation: The passage of undigested material from the body by the way of anus is called defecation.

Question 79.
What are roughages in the food?
Answer:
Roughages are the fibres of either carbohydrates or fats which help in constipation.

Question 80.
What are the parts of human digestive system?
Answer:
Mouth, buccal cavity, pharynx, oesophagus, stomach, small intestine, large intestine, rectum and anus.

Question 81.
What is a balanced diet?
Answer:
Balanced diet: Diet containing nutrients in required amounts is known as balanced diet.

Question 82.
What are the nutrients present in balanced diet?
Answer:
Carbohydrates, proteins, fats, vitamins and mineral salts are the nutrients present in balanced diet.

Question 83.
What is malnutrition?
Answer:
Malnutrition : Eating of food that does not have one or more than one nutrients in required amount is known as malnutrition.

Question 84.
What are different types of malnutrition?
Answer:
Malnutrition is of three types. They are:

1. Calorie malnutrition
2. Protein malnutrition and
3. Protein calorie malnutrition.

Question 85.
What are the two sources of vitamins to our body?
Answer:
The two sources of vitamins to our body are one is diet and the other is bacteria present in the intestine synthesise enzymes and supply them to the body.

Question 86.
Vitamins are classified into how many groups?
Answer:
Vitamins are classified into two groups. They are fat soluble (eg : A, D, E and K) and water soluble (eg : B complex, Vitamin C).

Question 87.
What is the chemical name of vitamin B₁₂?
Answer:
The chemical name of vitamin B₁₂ is Cyanocobalamine.

Question 88.
Which vitamin deficiency causes sterility in males?
Answer:
Vitamin E. i.e., tocopherol deficiency causes sterility in males.

Question 89.
What is the name of vitamin ‘C’? Deficiency of vitamin ‘C’ causes which disease?
Answer:
The chemical name of vitamin ‘C’ is Ascorbic acid. Its deficiency results in Scurvy disease.

Question 90.
What are saprophytes?
Answer:
Saprophytes: Saprophytes are the organisms which obtain their food from dead plants, dead and decaying animal bodies and other organic matter. Eg: Fungi, many bacteria.

Question 91.
What is peristaltic movement?
Answer:
Peristaltic movement: The muscles present in the wall of oesophagus rhythmically contracts and relaxes. This produces an wave like movement known as a peristaltic movement.

Question 92.
Name the intestinal enzymes present in small intestine and what are their functions.
Answer:
Peptidases and sucrases are the enzymes present in the intestinal juice. Peptidases converts peptides to amino acids. Sucrase converts sucrose (cane sugar) into glucose.

Question 93.
Name the parts of small intestine.
Answer:
The anterior part of small intestine is called the duodenum, the middle part is the jejunum and the posterior part is called the ileum.

Question 94.
What is the enzyme present only in children?
Answer:
Renin is the enzyme present only in children. It helps in curdling of milk.

Question 95.
How many types of teeth are present in our mouth?
Answer:
Four types of teeth are present in our mouth. They are incisors, canine, premolars and molars.

Question 96.
What is the use of tongue?
Answer:
Tongue is useful for mixing and pushing the food in between teeth and helps to push into oesophagus.

Question 97.
What was the opinion of C.B. Van Neil on the equation for photosynthesis?
Answer:
For each molecule of carbohydrate formed, one molecule of water and one molecule of oxygen is also produced.

Question 98.
“Plants are capable of surviving under a range of situations.” How do you support this statement?
Answer:

1. Plants are capable of surviving under a range of situations.
2. They survive from very hot, dry and brightly lighted conditions to wet, humid and dimly lighted ones.
3. The requirement of light and other factors varies from one plant to other.

Question 99.
From where do we get energy to do work?
Answer:
We get energy to do work from the food we eat.

Question 100.
What did Priestly hypothesize on the experiment he conducted on the role of air in the growth of plants?
Answer:
Priestly hypothesized that plants restore to the air whatever breathing animals and burning candles remove.

Question 101.
How does gaseous exchange occur in plants?
Answer:
Gaseous exchange occurs in plants through the stomata present in leaves and also through the lenticels present on stems.

Question 102.
How do the aquatic plants acquire CO₂ to manufacture food?
Answer:
Aquatic plants utilizes or absorbs the carbon dioxide dissolved in water in the form of bicarbonates to manufacture food.

Question 103.
In the experiment to prove that starch is produced during photosynthesis. Why do we boil the leaf in alcohol?
Answer:
We boil the leaf in alcohol to remove all the chlorophyll present in the leaf. The leaf turns to white in colour.

Question 104.
In the experiment to show that sunlight is necessary for photosynthesis, why do we immerse leaf in boiling water?
Answer:

1. Immersing leaf in boiling water will breakdown the cell membranes of leaf cells.
2. It makes the leaf more permeable to iodine solution, so that it can reach the starch present inside the leaf cells.

Question 105.
Why do we have to destarch the leaf before conducting experiment on photosynthesis?
Answer:
This is because if starch is present it will interfere with the result of the experiment.

Question 106.
Why is a water bath used for heating alcohol in which leaf is kept inside the beaker for testing the presence of starch?
Answer:
A water bath is used here for heating alcohol because alcohol is a highly inflammable liquid. So if alcohol is heated directly over a flame, then it will catch fire at once.

Question 107.
What are the reasons for vomiting?
Answer:
Causes for vomiting :

1. Overeating especially when the food contain a high proportion of fat.
2. When we eat something very indigestible or poisonous.

Question 108.
Why do we feel bilious or liverish?
Answer:
We feel bilious or liverish because of having eaten rich meals for several days.

10th Class Biology 1st Lesson Nutrition 2 Marks Important Questions and Answers

Question 1.
Vitamin A, D, E and K are fat soluble vitamins. Write the deficiency diseases and resources of these vitamins in a tabular form.
Answer:

Question 2.
A doctor visited your school to check up the health of school children. What kind of questions do you ask to know about the pancreas?
Answer:

1. Where is pancreas located?
2. Why pancreas is called as mixed gland?
3. What are the hormones released by the pancreas?
4. What are the enzymes released by the pancreas?
5. What are the disorders occur if pancreas does not work properly?
6. What is the role of pancreas in digestion?

Question 3.
Look at the following equation and answer the questions:
Fats + Bile → Fat globules
a) What is the name of that reaction?
b) Which gland plays major role in this reaction?
Answer:
a) The name of the reaction is Emulsification.
b) The gland LIVER plays major role in this reaction.

Question 4.

We know that food is the main source to maintain biological processes in a perfect manner. Our diet should be a balanced one which contains proper amount of carbohydrates, proteins, vitamins, mineral salts and fats. Two third of world population is affected by food related diseases. Some of them are suffering by consuming high calorific food. Most of them are facing various diseases due to lack of balanced diet. Eating of food that does not have one or more than one nutrients in required amount is known as malnutrition. Poor health, will full starvation, lack of awareness of nutritional habits, socio-economic factors are all the reasons of malnutrition.

i) Define Balanced diet.
ii) What is malnutrition and what could be the possible reasons for it?
Answer:
i) The food containing proper amount of carbohydrates, proteins, vitamins, mineral salts and fats is known as “balanced diet”.

ii) Eating of food that does not have one or more than one nutrients in required amounts is known as “malnutrition”.
Poor health, starvation, lack of awareness of nutritional habits, socio-economic factors are all reasons for malnutrition.

Question 5.
What questions do you ask your teacher to know about the obesity and its consequences?
Answer:

1. What are the reasons for Obesity?
2. How to reduce the body weight?
3. What are the consequences of obesity?
4. What type of food you suggest to people suffering from obesity?

Question 6.
Prepare four questions for the Gastro-enterologist to know the problems that occur in the digestive system.
Answer:

1. Why do the people suffer from indigestion problems?
2. Why do we get vomitings?
3. Why do we get belching?
4. Why do we get ulcers in stomach?
5. What is acidity?

Question 7.
Balance the following equation. Write what you have received through this equation.

Answer:

1. This equation represents the process of photosynthesis. For the photosynthesis to occur, four factors are required. They are carbon dioxide, water, light and chlorophyll.
2. 6 molecules of carbon dioxide and 12 molecules of water in the presence of sun light and chlorophyll forms 1 molecule of glucose, 6 molecules of water and 6 molecules of oxygen.

Question 8.
Pregnant ladies are advised to eat leafy vegetables and take folic acid pills. Why?
Answer:

1. Folic acid is the part of the B – a complex family of vitamins.
2. Dark green leafy vegetables are a good source of folic acid.
3. Folic acid is often prescribed for pregnant women because it is essential for the normal development of the brain and spinal cord of the foetus.
4. It also help in protection against birth defects.
5. It also required for the synthesis of nucleic acids DNA and RNA.
   Hence pregnant ladies are advised to eat leafy vegetables and take folic acid pills.

Question 9.
You want to know about intestinal juice from your nearby doctor. Which questions do you ask him?
Answer:

1. Which part of the small intestine secrete intestinal juice?
2. Intestinal juice contains which enzymes?
3. Intestinal juice converts peptids into?
4. Sucrose present in intestinal juice converts sucrose to?
5. Digestion of which nutrient resumes in small intestine?
6. Which digestive juice involved in the complete digestion of food?

Question 10.
Photosynthesis process provides food for all organisms. It mainly takes place by two phases. Light reaction is first phase.
H₂O → H + OH
Above equation in light reaction shows what?
Answer:

1. The first phase of photosynthesis is light reaction. It occurs in grana of the chloroplast.
2. During this phase light-activated chlorophyll molecule splits water molecule into Hydrogen (H⁺) and Hydroxyl ions (OH^–)
3. This reaction is known as photolysis of water which means splitting by light (photo means light, lysis means breaking). This was discovered by Hill. Hence it is also called Hill’s reaction.

Question 11.
Write names of the given sentences.
i) Organelle in which photosynthesis occurs.
Answer:
Chloroplast

ii) Life process in which complex food is converted into simple substances.
Answer:
Digestive system

iii) The part after large intestine in digestive tract.
Answer:
Anus

iv) Deficiency of Vitamin “K” causing disease.
Answer:
Delay in blood clotting.

Question 12.
Write about the nutrition in Amoeba. (OR)
Observe the below given organism and write its name. How nutrition occurs in this organism ?
Answer:

1. The mode of nutrition in Amoeba is holozoic nutrition.
2. Amoeba takes in food using temporary finger like extensions called pseudopodia.
3. These pseudopodia fuse over the food particle forming food vacuole.
4. Inside the food vacuole, complex substances are broken down into simple ones which then diffuse into the cytoplasm.
5. The remaining undigested material is moved to the surface of the cell and thrown out.

Question 13.
Write a short note on heterotrophic nutrition.
Answer:

1. Heterotrophic nutrition is that mode of nutrition in which an organism cannot make its own food from simple inorganic materials like CO₂ and water and depends on other organisms for food.
2. Examples are all animals, most bacteria and fungi.
3. Heterotrophs depend on others for their food requirement.

Question 14.
How is a more serious form of indigestion caused?
Answer:

1. A more serious form of indigestion is caused by stomach and duodenal ulcers.
2. These conditions occur more often in people who may be described as hurried or worried.
3. The ulcers occur more often in busy people who get into the habit of hurrying over meals and rushing from one activity to another without sufficient rest.

Question 15.
What is parasitic nutrition? Write briefly about it.
Answer:

1. Parasitic nutrition is that nutrition in which an organism derives its food from the body of another living organism without killing it.
2. The organism which obtains the food is called a parasite and the organism from whose body food is obtained is called the host.
3. A parasite usually harms the host.
4. The host may be a plant or an animal.
5. Parasitic mode of nutrition is seen in several fungi, bacteria, a few plants like cuscuta and animals like plasmodium.

Question 16.
Write about holozoic mode of nutrition.
(OR)
What is holozoic nutrition?
Answer:

1. Holozoic nutrition means feeding on solid food.
2. Holozoic nutrition is that nutrition in which an organism takes the complex organic food materials into its body by the process of ingestion.
3. The ingested food is digested and then absorbed into the body cells of the organism.
4. The undigested and unabsorbed part of the food is thrown out of the body of the
   organism by the process of ingestion.
5. The human beings and most of the animals have a holozoic mode of nutritions.

Question 17.
Briefly explain about saprophytic nutrition.
(OR)
What are saprophytes? How does nutrition occur in them?
Answer:

1. Saprophytic nutrition is that nutrition in which an organism obtain its food from dead and decaying plant and animal bodies.
2. Organisms having saprophytic mode of nutrition are called saprophytes.
3. Fungi and bacteria are saprophytes.
4. The saprophytes breakdown the complex organic molecules present in dead and decaying matter and convert them into simple substances outside their body.
5. These simple substances are then absorbed by saprophytes as their food.

Question 18.
What is malnutrition? What are the reasons for it? What are the different types of malnutrition?
Answer:

1. Eating of food that does not have one or more than one nutrients in required amount is known as malnutrition.
2. Reasons for malnutrition: Poor health, will-full starvation, lack of awareness of nutritional habits, socio economic factors are all the reasons for malnutrition.
3. Different types of Malnutrition:
   1. Calorie malnutrition,
   2. Protein malnutrition and
   3. Protein calorie malnutrition.

Question 19.
Write a short note on obesity. (OR)
What are the ill effects of obesity?
Answer:

1. Obesity is due to over eating and excess of energy intake.
2. It is a big health hazard. Obese children when grow, they will be target of many diseases.
3. Obesity children usually suffer from diabetes, cardiovascular, renal, gall bladder problems.
4. Eating junk foods and other food habits lead to obesity.

10th Class Biology 1st Lesson Nutrition 4 Marks Important Questions and Answers

Question 1.
Answer the following questions by observing the diagram showing the experiment:
a) What will you prove by this experiment?
Answer:
By this experiment we will going to prove oxygen is released during photosynthesis.

b) What apparatus do you use in this experiment?
Answer:

1. Beaker
2. Test-tube
3. Funnel
4. Hydrilla Plant
5. Burning splinter.

c) What would be the results if the experiment is done in shadow?
Answer:
If we conduct this experiment in shadow no change in the water level of the Test-tube. No photosynthesis occur. No air bubbles are form.

d) What will you do to obtain result from the experiment?
Answer:
If the burning splinter is kept near the mouth of test tube, burns brightly.

Question 2.
Answer the following questions by observing the diagram showing the experiment.
a) What will you prove by this experiment?
Answer:
Carbon dioxide (CO₂) is necessary for the photosynthesis.

b) What apparatus do you use in this experiment?
Answer:
Wide mouthed bottle, split cork, KOH solution, Iodine, Potted plant.

c) Why do we use KOH solution in this experiment?
Answer:
KOH is used for the absorption of CO₂ in the bottle.

d) Why did we study two leaves in this experiment?
Answer:
We should test two leaves of which one must be having the availability of CO₂ and other must not be having the availability of CO₂ to prove that CO₂ is essential for photosynthesis.

Question 3.
Describe an experiment conducted by Joseph Priestley which revealed the essential role of air in the growth of green plants.
(OR)
What is the role of air in the growth of green plants?
(OR)
Write the experiment of bell jar and pudina plant, performed by Priestley to prove that air plays key role in Photosynthesis.
Answer:

1. Joseph Priestley in 1770 performed a series of experiments that revealed the essential role of air in the growth of green plants.
2. Priestley discovered oxygen in 1774.
3. Priestley observed that a candle burning in a closed space, a bell jar, soon gets extinguished.
4. Similarly, a mouse would soon suffocate in a closed space of the bell jar.
5. He concluded that a burning candle or an animal, both somehow, damage air.
6. But when he placed a mint plant in the same bell jars, he found that the mouse stayed alive and the candle when lighted from outside continued burning in the presence of the mint plant.
7. Priestly hypothesized as plants restore to the air whatever breathing animals and burning candles remove.

Question 4.
B₁, B₂, B₃, A, C, D, E, K are the symbols of vitamins. Classify these vitamins based on solubility and diseases due to vitamins deficiency.

Answer:

Question 5.
During Photosynthesis, several events occurs in the chloroplast. Explain the light dependent reactions.
(OR)
Write the mechanism of light dependent reactions in Photosynthesis.
Answer:
Light reaction takes place in grana thylakoids of chloroplast.
The following events occurs in the light dependent reaction.
1. Step -1: The chlorophyll on exposure to light energy becomes activated by absorbing photons of light energy.
2. Step – II: The activated energy is used in splitting the water molecule into hydrogen (H⁺) and hydroxyl ion (OH^–). This reaction is known as photolysis or Hill’s reaction.
H₂O → H + OH
3. Step – III : Water (H₂O) and oxygen (O₂) are produced by the OH^– ions through a series of reaction.
4. Step – IV: H⁺ ions are involved in the synthesis of ATP and NADPH which are the end products of light reaction. These are called assimilatory powers.

Question 6.
You might have conducted and experiment in your school laboratory to prove that CO₂ is essential for Photosynthesis. Raju, who is in 9th class, wanted to perform the same experiment. He had some doubts regarding this experiment. Clarify them.
i) Prior to the experiment, the potted plant was kept in a dark room for a week. Why?
Answer:
To destarch the plant.

ii) KOH pellets were kept in the glass jar. Why?
Answer:
To absorb CO₂

iii) Write the apparatus used to perform this experiment.
Answer:
Potted plant, wide mouthed transparent bottle, splitted cork.

iv) What will be the result, if the same experiment is conducted in dark?
Answer:
Photosynthesis does not take place.

Question 7.
Write the differences between light and dark reactions of photosynthesis.
Answer:

Question 8.
Draw a neat labelled diagram of internal structure of leaf. Which mesophyll cells of the leaf consist of chloroplast?
Answer:
Palisade parenchyma and spongy parenchyma consist of chloroplast.

Question 9.
Keep a plant for a week in the dark. Then insert one leaf of this plant into a transparent bottle containing potassium hydroxide solution. Keep the plant in sunlight. Test the leaf in the bottle and any other leaf of the plant with Iodine, after a few hours.
i) What is the aim of this experiment?
ii) What will be observed in the leaf kept in the bottle and a leaf from the plant?
iii) Why was the plant kept in the dark first and then in the sunlight?
Answer:
i) To prove that CO₂ is necessary for photosynthesis.
ii) The leaf part kept in the bottle does not show colour change in iodine test, whereas the leaf exposed to air turns bluish-black in the iodine test.
iii) To remove starch from leaves (destarching).

Question 10.
Analyse the following information and answer the questions.

i) Which vitamins deficiency causes diseases related to bones?
Answer:
Calciferol – Vitamin – D, deficiency causes diseases related to bones,

ii) Which vitamins we get by eating fruits?
Answer:
Retinol – Vitamin – A, and Tocoferol – Vitamin – E we get them by eating fruits,

iii) Which vitamin deficiency causes Paralysis? To prevent iii), what type of food we have to eat?
Answer:
Paralysis is caused by the Vitamin (B}) Thiamine deficiency.
To prevent this disease, we have to eat cereals, oil seeds, vegetables, milk, meat, fish and eggs.

iv) To avoid vitamin deficiency diseases, what type of food we have to eat?
Answer:
To avoid vitamin deficiency diseases, we have to take proper diet.
Our diet should be a balanced one which contains proper amount of carbohydrates, proteins, vitamins, mineral salts and fats.

Question 11.
What are fat soluble vitamins? Explain the deficiency diseases due to deficiency of fat soluble vitamins and their symptoms.
Answer:
Fat soluble vitamins :

Question 12.
Read the table and answer the following questions.

i) Name the anti-sterility vitamin from the above table.
Answer:
Tocoferol (Vitamin E) is the anti – sterility vitamin.

ii) WTiich vitamin deficiency that causes the gums bleeding?
Answer:
Ascorbic acid (Vitamin C) deficiency causes the gums bleeding.

iii) Name the fat soluble vitamins from the above.
Answer:
Retinol (Vitamin A), Tocoferol (Vitamin E), Phylloquinone (Vitamin K) are fat soluble vitamins.

iv) Name the symptoms that appear due to deficiency of vitamin K.
Answer:
Delay in blood clotting, over bleeding are the symptoms appear due to deficiency of vitamin K.

Question 13.
Mention different modes of nutrition. Explain the mode of nutrition in Cuscuta.
Answer:
There are different modes of nutrition
Parasitic Nutrition in Cuscuta :

1. Cuscuta plant is the best example for parasitic nutrition.
2. It contains no chlorophyll and has root like structures called haustoria.
3. With the help of haustoria, cuscuta absorbs nutrients from it’s host plant.
4. As cuscuta grows on the host plants, its roots gradually degenerated and it establishes firmly on the host plant.
5. Meanwhile, the root rots away after stem contact has been made with a host plant.
6. As dodder grows, it sends out new haustoria and establishes itself very firmly on the host plant.
7. After growing in a few spirals around one host shoot, the dodder finds its way to another.
8. It continues to twine and branch until it resembles a fine densley tangled web of thin stems enveloping the host plant.

Question 14.
Is the malnutrition reason for diseases? Why? Write any of such disease and its characters.
Answer:

1. Yes, malnutrition is the reason for occurrance of diseases.
2. Malnutrition is eating of food that does not have one or more than one nutrients in required amount.
3. This results in scarcity of nutrients for the proper growth and health of the individual.
4. Kwashiorkor disease occurs in children due to the deficiency of proteins in the diet.
5. Characteristic features of kwashiorkor disease :
   1. Body parts becomes swollen due to accumulation of water in the intercellular spaces.
   2. Very poor muscle development, swollen legs, fluffy face, difficult to eat, diarrhoea, dry skin are the symptoms of Kwashiorkor disease.
   3. The child becomes lethargic and shows little interest in its surroundings or in playing and learning.

Question 15.
Which issues do you take into consideration to tell that plants play a key role in animals nutrition?
Answer:

1. Plants play a very important role in nutrition of animals. Actually plants are the producers whereas animals are dependent on plants for their nutrition.
2. Many plants or plants parts are eaten as food. There are around 2000 plant species which are cultivated for food. Nutrients like carbohydrates, proteins, fats, vitamins and minerals are available from these plants.
3. Seeds of plants are good sources of food for animals including humans because they contain many healthful fats.
4. Infact majority of the food consumed by human beings are seed based food. Edible seeds include cereals like wheat, rice, maize etc. and legumes like pea, groundnut and nuts.
5. Oil seeds are often pressed to produce rich oils like sunflower, groundnut, sesame etc.
6. Seeds are typically high in unsaturated fats and considered as a health food.
7. Fruits are the ripened ovaries of plants including the seeds within. Many plants and animals have coevolved such that the fruits of the former are an attractive food source to the later.
8. Vegetables are a second type of plant matter that is commonly eaten as food.
9. Hence all the nutrients required by animals are available to the animals from plants. Hence plants are playing an important role in nutrition of animals including human beings.

Question 16.
Give examples for the vitamin deficiency diseases.
Answer:

Question 17.
We know that by taking different types of food materials we will get vitamins. For this what changes shall we take in our food habits?
Answer:

1. Eating of rice polished only for one time gives us vitamin thiamine in abundant amount.
2. We should eat balanced food containing nutrients in equal quantities.
3. Fresh leafy vegetables are rich source of vitamins Riboflavin, Folic acid and Vitamin ‘C’. So our diet should contain these leafy vegetables.
4. Sufficient amount of non-vegetarian foods like meat, poultry, fish, kidney and liver should present in our food.
5. Eating fresh fruits also add number of vitamins to our body.
6. Whole cereals like wheat, rice provide number of vitamins to our body. So we shall eat required quantities of cereals in our food.

Question 18.
Describe what disaster occurs on earth, if photosynthesis life process stops.
Answer:

1. Plants are the universal food providers. If they stop photosynthesis all the animals would die due to starvation.
2. Plants release oxygen during photosynthesis. It is essential for survival of all living beings. If photosynthesis does not take place, there would be no oxygen on the earth and no chance of survival of life on this planet.
3. plants utilise CO₂ for photosynthesis. If it does not occur, CO₂ utilisation is stopped and hence increase in the levels of CO₂ This leads to global warming.

Question 19.
Describe the experiment conducted by Jan Baptist Von Helmont and discuss his results in the growth of plant body mass.
(OR)
How is water required for the growth of plant body mass?
Answer:

1. In the year 1648 a Belgian scientist Jan Baptist Von Helmont conducted an experiment that continued for five years.
2. He took a small willow tree and planted it in a large pot of soil.
3. Before he did the experiment he carefully measured the mass of the dry soil and mass of the tree.
4. He covered the soil with a lid so that nothing could fall on to the surface of the soil and add to its mass.
5. There were holes in the lid so that the tree could grow out of the soil and so that air and water could reach the roots.
6. Von Helmont left the tree for five years giving it only rain water to drink.
7. At the end of five years he measured the mass of the tree and the mass of the dry soil for a second time.
8. The results of this experiment are shown below.
   

   Mass (Kg)
   	At start	After five years	Change in mass (Kg)
   Tree	2.27	76.74	74.47
   Drv soil	90.72	90.66	0.06

9. This experiment changed the belief of hundreds of years. This is because Von Helmont arrived at a result that
   a) The substances needed for the growth of a plant do not come from the soil only.
   b) The plant grow because of the water it gets.

Question 20.
Write an account on the mechanism of dark reaction.
(OR)
Write an account on carbon fixation.
(OR)
Briefly explain the Melvin Calvin cycle.
Answer:

1. The reactions of photosynthesis which do not require light energy are called dark reactions. These reactions occur both in presence and absence of light.
2. Dark reactions occur in the stroma of the chloroplast and these were discovered by American scientist Melvin Calvin.
3. In the first reaction carbon dioxide is accepted by Ribulose 1 – 5 biphosphate, a five carbon sugar to form hexose sugar.
4. It is an unstable compound so it breaks down into two molecules of three carbon compounds called phospho glyceric acid (PGA).
5. 
6. Phospho glyceric acid undergoes a series of reactions and is converted to glyceral- dehyde – 3 phosphate. NADPH and ATP produced in light reactions are used up at this stage.
   
7. Two molecules of glyceraldehyde – 3 phosphate combine together to form glucose molecule.
8. In the end glucose is converted into starch.
   Summary reaction :
   

Question 21.
Describe the buccal cavity of human beings.
(OR)
How is food masticated in buccal cavity?
Answer:

1. The cavity or space in the mouth is called oral cavity or buccal cavity.
2. Teeth, tongue and openings of three salivary glands are present in buccal cavity
3. Incisors, canines, premolars and molars are the different types of teeth present in the mouth.
4. Tongue is muscular and pushes the food on to the teeth during mastication. It also helps in swallowing the food into oesophagus.
5. Three pairs of salivary glands present in the buccal cavity are parotid, sub-lingual and sub-maxillary glands.
6. Parotid glands are present near the ears. The other two pairs of glands open below the tongue through ducts.

Question 22.
What is indigestion? How can we avoid indigestion?
Answer:
Indigestion: Indigestion is a general term when there is difficulty in digesting food. Healthy people can usually avoid problems related to indigestion by:

1. Having simple, well balanced meals.
2. Eating them in a leisurely manner. ,
3. Thoroughly masticating the food.
4. Avoiding taking violent excercises soon after eating food.
5. Drinking plenty of water and having regular bowel movements.

Question 23.
What are vitamins? Why they are called essential nutrients? What is their role in the human body?
Answer:

1. Vitamins are organic substances.
2. They are micro – nutrients required in small quantities.
3. Actually vitamins are not synthesised in the body and therefore they are essential nutrients.
4. Though our body cannot synthesis vitamins we do not generally suffer from vitamin deficiency disease.
5. This is because of two sources of vitamins for our body. Diet is the primary source.
6. In addition, bacteria present in the intestine synthesis vitamins and supply them to us
7. By themselves vitamins cannot generate any energy or carryout any chemical reaction.
8. In the body vitamins combine with some of the enzymes and make the enzymes active.
9. In the absence of vitamins, enzymes become inactive and cannot catalyse the reactions.
10. Vitamins are normally present in all types of natural foods – milk, meat, fruits and vegetables.
11. Vitamins are of two types:
    1. Water soluble vitamins eg: B complex and Vitamin C and
    2. Fat soluble vitamins eg: A, D, E and K.

Question 24.
Give an account of water soluble vitamins, their occurrence, deficiency diseases and symptoms.
Answer:
Vitamins, available sources and deficiency diseases:

Question 25.
a) What apparatus do you use during the conduction of the experiment?
b) Before conducting the experiment, why do you keep the potted plant in dark place for a week?
c) What test do you conduct to know the formation of carbohydrates in leaves?
d) During the test, which part of the leaf turns into blue and which part doesn’t? Why?
Answer:
a) Potted plant, Light screen, Iodine solution, Petridish, Black paper.
b) To destarch the leaves.
c) Iodine Test
d) The part of the leaf exposed to light turn into blue. Unexposed part does not turn into blue.
Reason: Photosynthesis occurs only in the presence of sunlight. The unexposed part does not receive the sunlight.

Question 26.
Draw the diagram of equipment arrangement in Hydrilla experiment. You did to prove that oxygen releases in photosynthesis. Write the reasons why test tube is placed upside down on funnel.
Answer:
Reasons for placing the test tube upside down on funnel:

1. To count the number of bubbles coming out of the hydrilla twigs per minute during photosynthesis.
2. To collect sufficient amount of oxygen at the end of the test tube so that it can be tested with glowing splinter.
3. Amount of oxygen produced in the test tube can be measured by the displacement of water within the test tube.

Question 27.
Describe the internal structure of the leaf.
(OR)
How is the internal structure of leaf modified to prepare starch through the process of photosynthesis?
Answer:

1. A transverse section of a typical leaf shows that it is covered on both sides by epidermis. Epidermal layers are covered on both sides by cuticle.
2. Lower epidermis is interrupted by a large number of openings called stomata. The central opening is called stoma.
   
3. Each stoma is surrounded by two kidney shaped cells called Guard cells.
4. Stomata regulate the exchange of gases and loss of water vapour from the leaves.
5. Mesophyll is the tissue present between two epidermal layers. Upper mesophyll tissue is called palisade parenchyma and lower tissue is known as spongy parenchyma.
6. The cells in the palisade parenchyma are long elongated arranged in rows.
7. The cells in the spongy parenchyma are irregularly arranged cells with large inter cellular spaces.
8. More number of chloroplasts are present in palisade parenchyma than the spongy parenchyma.
9. In the midrib and veins vascular bundles are present with phloem located towards the lower side and xylem towards the upper side.

Question 28.
Write about the parasitic nutrition in cuscuta plant.
Answer:

1. Dodder (genus cuscuta) is a leafless, twining, parasitic plant belongs to family convolvulaceae.
2. The dodder contains no chlorophyll and instead absorbs food through haustoria.
   
3. The dodder’s seed germinates, forming an anchoring root, and then sends up a slender stem that grows in a spiral fashion until it reaches a host plant.
4. It then twines around the stem of the host plant and throws out haustoria, which penetrate it.
5. Water is drawn through the haustoria from the host plant’s stem and xylem, and nutrients are drawn from its phloem.
6. Meanwhile, the root rots away after stem contact has been made with a host plant.
7. As the dodder grows, it sends out new haustoria and establishes itself very firmly on the host plant.
8. After growing in a few spirals around one host shoot, the dodder finds its way to another and it continues to twine and branch until it resembles a fine densley tangled web of thin stems enveloping the host plant.

Question 29.
Describe the digestion of food materials in alimentary canal with the help of a diagram.
(OR)
How is the food digested in your body?
Answer:

1. The alimentary canal is basically a long tube extending from the mouth to the anus.
2. Food is masticated by our teeth in the mouth and mixed with saliva to make it slippery and wet.
3. Saliva which contains amylase helps in the breakdown of carbohydrates into simple ones.
   
4. The soft food mixed with saliva passes through oesophagus or food pipe by wave like movements called peristaltic movements to the stomach.
5. The gastric glands in the stomach secrete gastric juice that contains hydrochloric acid, protein digesting enzyme pepsin and mucus.
6. Partially digested food chyme is released in small amounts by sphincter muscles reach small intestine.
7. Bile juice secreted by liver and pancreatic juice secreted by the pancreas release into the duodenum of small intestine.
8. Emulsification of fats is done by bile juice. Pancreatic juice contains enzyme trypsin for carrying on the process of digestion of proteins and lipase for fats.
9. Complete digestion of carbohydrates, fats and proteins takes place in the small intestine by intestinal juice.
10. Finger-like projections present in the walls of small intestine are called villi. They absorb the digested food into the blood.
11. Rest of the food is sent into the large intestine where most of the water present in it is absorbed from the food.
12. This material is then expelled through the anus which is the last part of the alimentary canal.

Question 30.
Draw a flow chart of human digestive system.
Answer:

Question 31.
Write briefly about nutrition in paramoecium. (OR)
Explain the nutrition process in paramoecium with the help of diagrams.
Answer:
Paramoecium :

1. Paramoecium is a unicellular aquatic organism.
2. The paramoecium has thin, hair like cilia all over its body.
   
3. The cilia move back and forth rapidly in water.
4. When the cilia is present around, the mouth region of paramoecium move, back and forth. They sweep the food particles present in water into the mouth of paramoecium.
   
5. This is the first step in the nutrition of paramoecium which is called ingestion and is followed by digestion, absorption, assimilation and egestion.

Question 32.
Describe the digestion of food in the stomach. (OR)
How is the food digested in your stomach ?
Answer:

1. The stomach is a large organ which expands when food enters it.
2. The muscular walls of the stomach help in mixing the food thoroughly with more digestive juices.
   
3. The gastric glands present in the wall of the stomach secrete gastric juice which contains hydrochloric acid, pepsin and mucus.
4. The hydrochloric acid facilitates the action of the enzyme pepsin and also kills the germs present in the food.
5. The enzyme pepsin digests proteins and mucus protects the inner lining of the stomach from the action of the acid under normal conditions.

Question 33.
Describe the digestion of food in the small intestine. (OR)
What are the changes that you see during the digestion of food in small intenstine?
(OR)
Small intestine is a long coiled tube. How the food is digested in small intestine of human being?
Answer:

1. The small intestine is the longest part of the alimentary canal.
2. It is the site of the complete digestion of carbohydrates, proteins and fats.
   
3. It receives the secretion of liver and pancreas for this purpose.
4. These juices make the internal condition of the intestine gradually to a basic or alkaline one.
5. Bile juice secreted by liver breaks down fats into small globules like forms. This process is called emulsification.
6. Pancreatic juice contains enzymes like trypsin for carrying on the process of digestion of proteins and lipase for fats.
7. Walls of the small intestine secrete intestinal juice. This is also known as succus entericus.
8. The intestinal juice consists of enzymes like enterokinase, peptidase, lipase, sucrase, nucleotidase, nucleosidase etc.
9. Enzymes present in the intestine completely digest the partially digested food.
10. Following digestive processes take place in the intestine.
    a) Peptidases convert peptides into amino adds.
    b) Intestinal lipase completely digest fats into fatty acids and glycerol.
    c) Enzymes sucrase, maltase, lactase hydrolyse sucrose, maltose and lactose respectively converting them into glucose.
    d) Nucleotidase and nucleosidase complete the digestion of nucleic acids.
11. The end products of digestion are absorbed in the intestine.

Question 34.
What are the health aspects of the alimentary canal?
Answer:

1. The human alimentary canal usually functions remarkably well considering how badly we treat it on occasions.
2. Vomiting is the body’s method of ridding itself unwanted or harmful substances from the stomach.
3. There are many causes of vomiting, but one of the most common is over eating, especially when the food contains a high proportion of fat.
   
4. The liver is unable to cope with the excessive fat and we get a feeling of nausea and sometime headache.
5. Indigestion is a general term used when there is difficulty in digesting food.
6. A more serious form of indigestion is caused by stomach and duodenal ulcers. These conditions occur more often in people who may be described as hurried or worried.
7. Those who are able to relax, who are not continually tensed up, and who live at a slower pace, seldom get ulcers.
8. For good health it is necessary to empty the bowels regularly.
9. If the food residues remain in the colon for too long, the bacteria present have more time to produce harmful substances which may be absorbed by the blood.
10. Constipation can often be avoided by having plenty of roughage in the diet.

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 7 Human Eye and Colourful World Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 7th Lesson Human Eye and Colourful World

10th Class Physics 7th Lesson Human Eye and Colourful World Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
What is the function of lens in human eye?
Answer:
The eye lens forms a real image on retina then we can see the objects.

Question 2.
How does it help to see objects at long distances and short distances?
Answer:
Eye lens can adjust itself in shape, so that it helps to see the objects at long and short distances.

Question 3.
How is it possible to get the image at the same distance on the retina?
Answer:
When the eye is focussed at distant object, the ciliary muscles are relaxed. So the focal length of eye lens adjusted itself which is equal to the distance of object from the retina. Then we can see the object clearly.
When the eye is focussed on a closer object the ciliary muscles adjust the focal length in such a way that the image is formed on retina and we see the object clearly.

Question 4.
Are we able to see all objects in front of our eye clearly?
Answer:
Yes. By maintaining the 25 cm distance of the object from our eye, we can see the objects in front of our eye.

Question 5.
How do the lenses used in spectacles correct defects of vision?
Answer:
To form image on retina.
To correct
myopia- Concave lens
hypermetropia i- Convex lens

Improve Your Learning

Question 1.
How do you correct the eye defect Myopia? (AS1)
Answer:

• Some people cannot see objects at long distances but can see nearby objects clearly. This type of defect in vision is called ‘Myopia’ or ‘Near sightedness’.
• Myopia is corrected by using a con-cave lens of focal length equal to the distance of the far point F from the eye.
• This lens diverges the parallel rays from distant object as if they are coming from the far point.
• Finally the eye lens forms a clear im-age at the retina.

Question 2.
Explain the correction of the eye defect Hypermetropia. (AS1)
Answer:

• A person with hypermetropia can see distant objects clearly but cannot see objects at near distances. This is also known as ‘far sightedness’.
• Eye lens can form a clear image on the retina when any object is placed beyond near point.
• To correct the defect of hypermetropia, we need to use a lens which forms an image of an object beyond near point at H, when the object is between H and L. This is possible only when a double con¬vex lens is used.
• The image acts like an object for the eye lens. Hence final image due to eye is formed at retina.

Question 3.
How do you find experimentally the refractive index of material of a prism? (AS1)
(OR)
Write the experimental procedure in finding the refractive index of a prism.
(OR)
Which quantity will decide whether a given medium is denser or rarer? How do you find that quantity of prism experimentally?
Answer:
Refractive index decides whether a medium is denser or rarer.
Aim :
To find the refractive index of a prism.

Materials required :
Glass prism, white chart of size 20 x 20 cm, pencil, pins, scale and protractor.

Procedure :

1. Keep a prism on white chart.
2. Draw the triangular base of the prism with pencil.
3. Remove the prism.
4. The shape of the outline drawn prism is triangle and name its vertices as P, Q and R.
5. PQ and PR be the refracting surfaces.
6. Find the angle between PQ and PR. This is the angle of the prism (A) (or) Refracting angle.
7. Mark M on PQ and draw a perpendicular line to PQ at M.
8. Place the centre of the protractor at M, along the normal and mark an angle of 30° and then draw a line up to M. This line denotes incident ray. This angle is called angle of incidence.
9. Place the prism in its position (triangle) again.
10. Now fix two pins vertically on the line at points A and B .
11. See the images of pins through the 2nd refracting side (PR).
12. Fix another two pins at points C and D such that all the four pins appear to lie along the same line.
13. Remove the pins and prism, join the pin-holes. Draw the incident and emergent rays.
14. The angle between the normal and the emergent ray at N is the angle of emergence.
15. The line passing through the points A, B, M, N, C and D represents the path of light when it suffers refraction through prism.

The angle of deviation :

• Extend incident and emergent rays are intercept at a point ‘O’.
• The angle between these two rays is the angle of deviation (d).
• Note the emergent deviation angles for different values of i, in the given table.

• Draw the graph between angle of incidence on X – axis and the angle of deviation on Y – axis.
• We notice that the angle of deviation decreases first and then increases with increase of the angle of incidence.
• Mark points on a graph paper and join the points to obtain a graph (smooth curve).
• Draw a tangent line to the curve, which parallel to X-axis, at the lowest point of the graph.

• The point where this line cuts the Y- axis gives the angle of minimum deviation. It is denoted by D.
• From the graph, we notice that, at angle of minimum deviation, the angle of incidence is equal to the angle of emergence.
• By finding A and D we can find refractive index of prism by using formula .

Question 4.
Explain the formation of rainbow. (AS1)
(OR)
What is the natural spectrum occuring in sky? Explain the formation of that spectrum.
Answer:

• A rainbow is a natural spectrum of sunlight in the form of bows appearing in the sky when the sun shines on rain drops.
• It is combined result of reflection, refraction and dispersion of sunlight from water droplets, in atmosphere.
• It always forms in the direction opposite to the sun.
• To see a rainbow, the sun be must behind us and the water droplets falls in front of us.
• When a sunlight enters a spherical raindrop, it is refracted and dispersed. The different colours of light are bent in different angles.
• When different colours of light falls on the back inner surface of drop, it (Water drop) reflects (different colours of light) interwnally (total internal reflection).
• The water drops again refract the different colours, when it comes out from the raindrop.
• After leaving this different colours from the raindrop as rainbow, reach our eye. Thus, we see a rainbow.

Question 5.
Explain briefly the reason for the blue of the sky. (AS1)
Answer:

• The reason for blue sky is due to the molecules N₂ and O₂.
• The size of these molecules are comparable to the wavelength of blue light.
• These molecules act as scattering centres for scattering of blue light.
• So scattering of blue light by molecules of N₂ and O₂ is responsible for blue of the sky.

Question 6.
Explain two activities for the formation of artificial rainbow. (AS1)
(OR)
Give two activities for the formation of the artificial rainbow? And explain it.
(OR)
Suggest an experiment to create a rainbow in your classroom and explain the procedure.
Answer:
Activity -1 :

• Select a white coated wall on which the sun rays fdll.
• Stand in front of a wall in such a way that the sun rays fall on your back.
• Hold a tube through which water is flowing.
• Place your finger in the tube to obstruct the flow of water.
• Water comes out from small gaps between the tube and finger like a fountain.
• Observe the changes on wall while showering the water.

Activity – II : (Activity – 4)

• Take a metal tray and fill with water.
• Place a mirror in water such that it makes an angle to the water surface.
• Now focus white light on the mirror through the water.
• Keep a white card board sheet above the water surface.
• We may observe the colours VIBGYOR on the board.
• The splitting of white light into different colours (VIBGYOR) is called dispersion.
• So consider a white light is a collection of waves with different wavelengths.
• Violet has shortest wavelength, and red has longest wavelength.

Question 7.
Derive an expression for the refractive index of the material of a prism. (AS1)
Answer:
Derivation of formula for refractive index of a prism :
PQ and PR are refracting surfaces of prism, i, is an angle of incidence, i2 is an angle of emergence, is an angle of refraction of first surface and r2 is an angle of incidence on second surface.

1. From triangle OMN, we get
d = i₁ – r₁ + i₂ – r₂
∴ d = (i₁ + i₂)-(r₁+ r₂) ……… (1)
2. From triangle PMN, we have
∠A + (90° – r₁) + (90° – r₂) = 180°
⇒ r₁ + r₂ = A ……… (2)
3. From (1) and (2),
⇒ d = (i₁ + i₂) – A
⇒ A + d = i₁ + i₂
4. This is the relation between angle of incidence, angle of emergence, angle of deviation and angle of prism.

5. From Snell’s law, we know that
n₁ sin i = n₂ sin r

6. Let ‘n’ be the refractive index of the prism.
7. Using Snell’s law at M, refractive index of air n₁ = 1; i = i₁; n₂ = n; r = r₁.
⇒ sin i₁ = n sin r₁ …………. (4)
8. Similarly, at N, n₁ = n; i = r₂; n₂ = 1; i₁ = i₂
⇒ n sin r₂ = n sin i₂ ………… (5)
9. We know that at the angle of minimum deviation position (D), i.e. i₁ = i₂
10. We will notice that MN is parallel to the base of prism QR.

14. This is the formula for the refractive index of the prism.

Question 8.
Light of wavelength λ₁ enters a medium with refractive index n₂ from a medium with refractive index n₁ What is the wavelength of light in second medium? (AS1)
Answer:
Wavelength of first medium =λ₁
Refractive index of first medium = n₁
Let the wavelength of second medium = λ₂,
Refractive index of second medium = n₂
Light enters from first medium to second medium = \(\Rightarrow \frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\lambda_{1}}{\lambda_{2}}\) (∵ υ = const)
\(\Rightarrow \lambda_{2}=\frac{\lambda_{1} \mathrm{n}_{1}}{\mathrm{n}_{2}}\)

Note:
For the below questions the following options are given. Choose the correct option by making hypothesis based on given assertion and reason. Give an explanation.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true and R is not the correct explanation of A.
c) A is true but R is false.
d) Both A and R are false.
e) A is false but R is true.

Question 9.
Assertion (A) : The refractive index of a prism depends only on the kind of glass of which it is made of and the colour of light. (AS2)
Reason (R) : The refractive index of a prism depends on the refracting angle of the prism and the angle of minimum deviation.
Answer:
(b) Both A and R are true and R is not the correct explanation of A.

Question 10.
Assertion (A) : Blue colour of sky appears due to scattering of light.
Reason (R) : Blue colour has shortest wavelength among all colours of white light. (AS2)
Answer:
(C) A is true but R is false.

Question 11.
Suggest an experiment to produce a rainbow in your classroom and explain the procedure. (AS3)
Answer:
Activity to produce a rainbow in classroom :

• Take a prism and place it on the table near a vertical white wall.
• Take a thin wooden plank, make a small hole in it and fix it vertically on the table.
• Place the prism between the wooden plank and wall.
• Place a white light source behind the hole of the wooden plank. Switch on the light.
• Adjust the height of the prism such that the light falls on one of the lateral surfaces.
• Observe the changes in the emerged ray of the prism.
• Adjust the prism by slightly rotating it till you get an image on the wall.
• We observe a band of different colours on the wall.
• These colours are nearly equal to the colours of the rainbow, i.e., VIBGYOR.

Question 12.
Prisms are used in binoculars. Collect information why prisms are used in binoculars. (AS4)
Answer:

• Binoculars consists of a pair of identical or mirror symmetrical telescope mounted side by side and aligned to point accurately in the same direction, allowing the viewer to use both eyes, when viewing distant objects.
• The size of binoculars is reduced by using prisms.
• We get good image with more brightness.
• Objective size and optical quality should be increased by using prisms in binoculars.

Question 13.
Incident ray on one of the face (AB) of prism and emergent ray from the face AC are given in figure. Complete the ray diagram. (AS5)

Answer:

Question 14.
How do you appreciate the role of molecules in the atmosphere for the blue colour of the sky? (AS6)
(OR)
How do you appreciate the nature of molecules, responsible for the blue of the sky?
Answer:

• The sky appear blue due to atmospheric refraction and scattering of light through molecules.
• Molecules are scattering centres.
• The reason to blue sky is due to the molecules N2 and 02.
• The sizes of these molecules are comparable to the wavelength of blue light.
• In the absence of molecules there will be no scattering of sunlight and the sky will appear dark.
• We should appreciate the molecules which are scattering centres.

Question 15.
Eye is the only organ to visualise the colourful world around us. This is possible due to accommodation of eye lens. Prepare a six line stanza expressing your wonderful feelings. (AS6)
Answer:
“Many people simply see
The world in black and white
But through my eyes the world’s alive
So colourful and bright
Don’t close your mind to the sights
That light up the night and day
There’s so much to see here on this earth
And not a rupee do you have to pay”
The most obvious things aren’t the ones
There is beauty in the unknown
with willing eyes and an open mind
The true wonders you will be shown

Question 16.
How do you appreciate the working of ciliary muscles in the eye? (AS6)
(OR)
Which muscles are helpful in changing the focal length of eye lens? How do you appreciate those muscles?
Answer:

• Ciliary muscles are helpful in changing focal length of eye lens.
• The ciliary muscles which attached with eye lens help to change the focal length of eye lens.
• When the eye is focussed on a distant object, these are relaxed. So the focal length of eye lens increases to its maximum value.
• The parallel rays coming into the eye are focussed on the retina then we can see the object clearly.
• When the eye is focussed on a nearer object the muscles are strained so the focal length of eye-lens decreases. The ciliary muscles adjust the focal length and the image is formed on retina then we can see the object.
• Their process of adjusting focal length of eve lens is called accomodation.
• So ciliary muscles must be appreciated for its accomodation of eye lens.

Question 17.
Why does the sky sometimes appear white? (AS7)
(OR)
The sky sometimes appears white. What is the reason behind it?
Answer:

• Our atmosphere contains atoms and molecules of different sizes.
• According to their sizes, they are able to scatter different wavelengths of light.
• For example, the size of the water molecule is greater than the size of the N₂ or O₂.
• It acts as a scattering centre for light other frequencies which are lower than the frequency of blue light.
• On a hot day due to rise in the temperature, water vapour enters atmosphere which leads to abundant presence of water molecules in atmosphere. These water molecules scatter the colours of other frequencies (other than blue).
• All such colours of other frequencies reach our eye and white colour is appeared to us.

Question 18.
Glass is known to be transparent material. But ground glass is opaque and white in colour. Why? (AS7)
Answer:

• Ground glass is glass whose surface has been ground that produces a flat but rough finish.
• Ground glass has the effect of rendering the glass translucent by scattering of light during transmission thus blurring visibility while still transmitting light.
• To get more permanent frost, the glass may be ground by rubbing with some gritty substance.

Question 19.
If a white sheet of paper is stained with oil, the paper turns transparent. Why? (AS7)
(OR)
What is the reason behind the paper turns transparent when it is immersed in oil?
Answer:

• The paper fibres have higher index of refraction probably much greater than 1.5.
• The oil or fat also has a high index of refraction so that it nearly matches the index of refraction of the paper fibres and it reduces the scattering significantly.
• The fat adhering to the cellulose fibers lowers the index of refraction of the cellulose and also fills in air voids, so that visible light passes through the bag with significantly less scattering.
• The oil connects the fibres in the paper with a liquid which can transmit by refraction (rather than scatter) light that falls upon it. As a result, the paper stained with oil is turned transparent.

Question 20.
A light ray falls on one of the faces of a prism at an angle 40° so that it suffers angle of minimum deviation of 30°. Find the angle of prism and angle of refraction at the given surface. (AS7)
Answer:
Given that, incident ray on one of the prisms (i₁) = 40°
Angle of minimum deviation (D) = 30°
Angle of prism (A) = ?
A+D = 2i₁
⇒ A = 2i₁ – D
⇒ A = 2(40°) -30°
= 80° – 30°
= 50°
∴ Angle of prims (A) = 50°
Angle of refraction (r₁) = ?

Question 21.
The focal length of a lens suggested to a person with Hypermetropia is 100 cm. Find the distance of near point and power of the lens. (AS7)
Answer:
i) The distance of near point :
If the distance of near point is ‘d’ and focal length is ‘f’ then the relation between

Question 22.
A person is viewing an extended object. If a converging lens is placed in front of his eye, will he feel that the size of object has increased? Why? (AS7)
Answer:

• A simple magnifier allows us to put the object closer to the eye than we could normally focus and forms an enlarged virtual images.
• The principle behind this is angular magnification.
• The magnification is Ma = \(\frac{25}{f}\) for the close focus point, but since that causes eye strain, it is usually desirable to put the images at infinity giving Ma = \(\frac{25}{f}\)
• So he fees the size of object is increased. The reason is mentioned.
• Angular magnification: ‘The ratio of the angle substended at the eye by the image formed by an optical instrument to the angle substended at the eye by the object being viewed.”

Fill In The Blanks

1. The value of least distance of distinct vision is about ……………………… .
2. The distance between the eye lens and retina is about ……………………… .
3. The maximum focal length of the eye lens is about ……………………… .
4. The eye lens can change its focal length due to working of ……………………… muscles.
5. The power of lens is ID then focal length is ……………………… .
6. Myopia can be corrected by using ……………………… lens.
7. Hypermetropia can be corrected by using ……………………… lens.
8. In minimum deviation position of prism, the angle of incidence is equal to angle of ……………………… .
9. The splitting of white light into different colours (VIBGYOR) is called ……………………… .
10. During refraction of light, the character of light which does not change is ……………………… .
Answer:

1. 25 cm
2. 2.5 cm
3. 2.5 cm
4. ciliary
5. 100 cm
6. bi-concave
7. bi-convex
8. emergence
9. dispersion of light
10. frequency

Multiple Choice Questions

1. The size of an object as perceived by an eye depends primarily on
A) actual size of the object
B) distance of the object from the eye
C) aperture of the pupil
D) size if the image formed on the retina
Answer:
B) distance of the object from the eye

2. When objects at different distances are seen by the eye which of the following remains constant?
A) focal length of eye-lens
B) object distance from eye-lens
C) the radii of curvature of eye-lens
D) image distance from eye-lens
Answer:
D) image distance from eye-lens

3. During refraction, …………….. will not change.
A) wavelength
B) frequency
C) speed of light
D) all the above
Answer:
B) frequency

4. A ray of light falls on one of the lateral surfaces of an equilateral glass prism placed on the horizontal surface of a table as shown in figure. For minimum deviation of ray, which of the following is true?

A) PQ is horizontal
B) ‘QR’ is horizontal
C) ‘RS’ is horizontal
D) Either ‘PQ’ or ‘RS’ is horizontal
Answer:
B) ‘QR’ is horizontal

5. Far point of a person is 5 m. In order that he has normal vision what ki nd of spectacles should he use?
A) concave lens with focal length 5 m
B) concave lens with focal length 10 m
C) convex lens with focal length 5 m
D) convex lens with focal length 2.5 mm
Answer:
A) concave lens with focal length 5 m

6. The process of re-emission of absorbed light in all directions with different intensi¬ties by the atom or molecule is called ……………
A) scattering of light
B) dispersion of light
C) reflection of light
D) refraction of light
Answer:
A) scattering of light

10th Class Physics 7th Lesson Human Eye and Colourful World InText Questions and Answers

10th Class Physics Textbook Page No. 104

Question 1.
Can you imagine the shape of rainbow when observed during travel in an airplane?
Answer:

The shape of rainbow, when observed during travel in an airplane is circular as shown in the following figure.

10th Class Physics Textbook Page No. 88

Question 2.
Why do the values of least distance of distinct vision and angle of vision change with person and age?
Answer:

• The ciliary muscle helps the eye lens to change its focal length by changing radii of curvature of eye lens.
• When the eye is focussed on a distant object, the ciliary muscles are relaxed so that the focal length of eye lens has its maximum value which is equal to its distance from the retina.
• The working of ciliary muscle in eye changes from person to person.
• So, the values of least distance of distinct vision and angle of vision changes with person and age.

10th Class Physics Textbook Page No. 89

Question 3.
How can we get same image distance for various positions of objects?
Answer:
For different positions of object, the image distance remains constant only when focal length of lens changes.

Question 4.
Can you answer above question using concepts of refraction through lenses?
Answer:
The focal length of a lens depends on the material by which it has been made and radii of curvatures of lens. We need to change focal length of eye lens to get same image distance for various positions of object.

10th Class Physics Textbook Page No. 90

Question 5.
How does eye lens change its focal length?
(OR)
What is the role of ciliary muscles in the eye ? Write the answer in one or two sentences only.
Answer:
The ciliary muscles to which eye lens is attached help the eye lens to change its focal length by changing the radii of curvature of eye lens.

Question 6.
How does this change (eye lens changes its focal lengths) take place in eye ball?
Answer:
When the eye is focussed on a distant object the ciliary muscles are relaxed. So the focal length of eye lens has its maximum value which is equal to its distance from the retina. The parallel rays coming into the eye are then focussed on to the retina and we see the object clearly.

Question 7.
Does eye lens form a real image or virtual image?
Answer:
The eye lens forms a real and inverted image of an object on the retina.

Question 8.
How does the image formed on retina help us to perceive the object without change in its shape, size and colour?
Answer:

• The eye lens forms a real and inverted image of an object on the retina.
• This retina is a delicate membrane, which contains large number of receptors (125 million) called ‘rods’ and ‘cones’.
• They receive the light signals and identify the colour, and the intensity of light.
• These signals are transmitted to the brain through optic-nerve fibres.
• The brain interprets these signals and finally processes the information so that we perceive the object in terms of its shape, size and colour.

Question 9.
Is there any limit to change of focal length of eye lens?
Answer:
Yes. When the object is at infinity, the parallel rays from the object falling on the eye lens are refracted. They form a point-sized image on retina. In this situation, eye lens has a maximum focal length.

Question 10.
What are the maximum and minimum focal lengths of the eye lens?
Answer:
Maximum focal length is 2.5 cm and minimum focal length is 2.27 cm.

Question 11.
How can we find the maximum and minimum focal lengths of the eye lens?
(OR)
Calculate the maximum and minimum focal lengths of the eye lens.
Answer:’
a) When the object is at infinity,
u = – ∞ ; v = 2.5 cm (image distance which is equal to distance between eye lens and retina)

b) Object at a distance of 25 cm from eye. In this case, eye has minimum focal length.
Here u = -25 cm ; v 2.5 cm

10th Class Physics Textbook Page No. 91

Question 12.
What happens if eye lens is not able to adjust its focal length?
Answer:
In this case the person cannot see the object clearly and comfortably.

Question 13.
What happens if the focal length of eye lens is beyond the range of 2.5 cm to 2.27 cm?
Answer:
The vision (image) becomes blurred due to defects of eye lens.

10th Class Physics Textbook Page No. 92

Question 14.
What can we do to correct myopia?
Answer:
To correct myopia, we use concave lens.

10th Class Physics Textbook Page No. 93

Question 15.
How can you decide the focal length of the lens to be used to correct myopia?
Answer:
Assume that the object distance (u) is infinity and image distance (v) is equal to distance of far point.
u = – ∞ ; v = distance of far point = – D
Let ‘f be the focal length of bi-concave lens.

Here ‘f is negative showing that it is a concave lens.

Question 16.
What happens when the eye has a minimum focal length greater than 2.27 cm?
Answer:
In this case, the rays coming from the nearby object after refraction at eye lens, forms image beyond the retina.

10th Class Physics Textbook Page No. 94

Question 17.
How can you decide the focal length of convex lens to be used?
Answer:
Here object distance (u) = -25 cm
Image distance (v) = – d (distance of near point)
Let ‘f be the focal length of bi-convex lens.

If d > 25 cm, then ‘f’ becomes positive then we need to use bi-convex lens to correct hypermetropia.

10th Class Physics Textbook Page No. 95

Question 18.
Have you ever observed details in the prescription?
Answer:
A prescription that contains some information regarding type of lens to be used to correct vision.

Question 19.
What does it (“my sight is increased or decreased”) mean?
Answer:
Usually doctors, after testing the defects of vision prescribe corrective lenses indicating
their power which determines the type of lens to be used and its focal length.

Question 20.
What do you mean by power of lens ? flftorH)
Answer:
The reciprocal of focal length is called power of lens.

Question 21.
Doctor advised to use 2D lens. What is its focal length?
Answer:
Given power of lens P = 2D

∴ Focal length of lens (f) = 50 cm.

10th Class Physics Textbook Page No. 96

Question 22.
How could the white light of the sun give us various colours of the rainbow?
Answer:
Due to reflection, refraction and dispersion of sunlight.

Question 23.
What happens to a light ray when it passes through a transparent medium bounded by plane surfaces which are inclined to each other?
Answer:
When light incident on one of the plane surfaces and emerges from the other.

Question 24.
What is a prism?
Answer:
A prism is a transparent medium separated from the surrounding medium by consisting two refracting plane surfaces which are inclined.

10th Class Physics Textbook Page No. 97

Question 25.
What is the shape of the outline drawn for a prism?
Answer:
The outline drawn for a prism is in a triangle shape.

10th Class Physics Textbook Page No. 98

Question 26.
How do you find the angle of deviation?
Answer:
The angle between the extended incident and emergent rays is called angle of deviation.
(OR)
Extend both incident and emergent rays till they meet at a point ‘O’. Measure the angle between these two rays. This is the angle of deviation.

Question 27.
What do you notice from the angle of deviation?
Answer:
The angle of deviation decreases first and then increases with increase in the angle of incidence.

Question 28.
Can you draw a graph between angle of incidence and angle of deviation?
Answer:

Yes, we can draw the graph between angle of incidence and angle of deviation.

Question 29.
From the graph, can you find the minimum of the angle of deviation?
Answer:
Yes, we can. Draw a tangent line to the curve, parallel to X – axis, at the lowest point of the graph. The point where this line cuts Y – axis gives the angle of minimum deviation.

Question 30.
Is there any relation between the angle of incidence (i) angle of emergence (r) and piangle of deviation (d)?
Answer:
(i₁ + i₂)= A + D
i + r = A + D

10th Class Physics Textbook Page No. 101

Question 31.
In activity-3,we noticed that light has chosen different paths. Does this mean that the refractive index of the prism varies from colour to colour?
Answer:
Yes, refractive index of the prism varies from colour to colour.

Question 32.
Is the speed of light of each colour different?
Answer:
In vacuum – No, the speed of each colour is constant.
In medium – Speed is different for different colours.

Question 33.
Can you guess now, why light splits into different colours when it passes through a prism?
Answer:
Due to dispersion of light and different wavelength of colours in medium.

10th Class Physics Textbook Page No. 102

Question 34.
Does it (light passing through a prism) split into more colours? Why?
Answer:
We know the frequency of light is the properly of the source and it is equal to number of waves leaving the source per second. This cannot be changed by any medium. Hence frequency doesn’t change due to refraction. The coloured light passing through any transparent medium retains its colour.

Question 35.
Can you give an example in nature, where you observe colours as seen in activity 3?
Answer:
Yes, in rainbow. It is a good example of dispersion of light.

Question 36.
When do you see a rainbow in the sky?
Answer:
Dut to the refraction, reflection and dispersion of sunlight, when the sunlight passes through the rain drops, we can see the rainbow in the sky.

Question 37.
Can we create a rainbow artificially?
Answer:
Yes, we can create a rainbow artificially.

10th Class Physics Textbook Page No. 104

Question 38.
Why does the light dispersed by the raindrops appear as a bow?
Answer:

• A rainbow is not the flat two dimensional arc as it appears to us.
• The rainbow we see is actually a three dimensional cone with the tip of our eye.
• All the drops that disperse the light towards us lie in the shape of the cone – a cone of different layers.
• The drops that disperse red colour to our eye are on the outermost layer of the cone, similarly, the drops that disperse orange colour to our eye are on the layer of the cone beneath the red colour cone.
• In this way the cone responsible for yellow lies beneath orange and so on it till the violet colour cone becomes the innermost cone.

Question 39.
Why is the sky blue?
Answer:
A clear cloudless day – time sky is blue because molecules in the air scatter blue light from the sun more than thev scatter red liuht.

Question 40.
What is scattering?
A. Atoms or molecules which are exposed to absorb light energy and emit some part of the light energy in different directions is called scattering of light.

10th Class Physics Textbook Page No. 106

Question 41.
Why is that the sky appears white sometimes when you view it in certain direction on hot days?
Answer:

• On a hot day, due to rise in the temperature, water vapour enters atmosphere which leads to abundant presence of water molecules in atmosphere.
• These water molecules scatter the colours of other frequencies (other than blue).
• All such colours of other frequencies reach our eye and the sky appears white.

Question 42.
Do you know the reasons for appearance the red colour of sun during sunrise and at sunset?
(OR)
Sun appears red in colour during sunrise and sunset. Give reason.
(OR)
Why does sky appear red at Sunshine and Sunset?
Answer:

• The light rays from the sun travel more distance in atmosphere to reach our eye in morning and evening times.
• During sunrise and sunset except red light all colours scatter more and vanish before they reach us.
• Since scattering of red light is very less, it reaches us.
• As a result sun appears red in colour during sunrise and sunset.

Question 43.
Can you guess the reason why sun does not appear red during noon hours?
Answer:
During noon hours, the distance to be travelled by the sun rays in atmosphere is less than when compared to morning and evening hours. Therefore all colours reach our eye without scattering. Hence the sunlight appears white in noon hours.

10th Class Physics 7th Lesson Human Eye and Colourful World Activities

Activity – 1

Question 1.
How do you find least distance of distinct vision?
(OR)
What is the least distance a person can see an object comfortably and distinctly known as ? Write an activity to find that (least, distance of distinct vision) distance.
Answer:

• The least distance a person can see an object comfortably and distinctly is known as least distance of distinct vision.
• Hold the textbook at certain distance with your hands.
• Try to read the contents on the page.
• Gradually move the book towards eye, till it reaches very close to your eyes.
• You may see that printed letters on the page appear blurred or you feel strain to read.
• Now move the book backwards to a position where you can see clear printed letters without strain.
• Ask your friend to measure distance between your eye and textbook at this position.
• Note down its value.
• Repeat this activity with other friends and note down the distances for distinct vision in each case.
• Find the average of all these distances of clear vision.
• You will notice that to see an object comfortably and distinctly, you must hold it at a distance about 25 cm from your eyes.
• This 25 cm distance is called least distance of distinct vision.
• This value varies from person to person and with age.

Activity – 2

Question 2.
How can you find angle of vision?
(OR)
What is maximun angle a person is able to see whole object? Write an activity to find that angle.
Answer:

• The maximum angle, at which we are able to see the whole object is called angle of vision.
• Arrange a retort stand.
• Collect a few wooden sticks (or) PVC pipes that are used for electric wiring.
• Prepare sticks or pipes of 20 cm, 30 cm, 35 cm, 40 cm, 50 cm from them.
• Keep the retort stand on a table and stand near the table such that vour head is beside the vertical stand.
• Adjust the clamp on horizontal rod and fix at a distance of 25 cm from the eyes.
• Ask one of your friends to fix a wooden stick of 30 cm height to the clamp in vertical position.
• Now keeping your vision parallel to horizontal rod of the stand, try to see the top and bottom of wooden stick kept in vertical position.
• If you are not able to see both ends of stick at this distance (25 cm), adjust the vertical stick on the horizontal rod till you are able to see both ends of the stick at nearest possible distance from your eye.
• Fix the clamp to the vertical stick at this position.
• Without changing the position of the clamp on horizontal rod, replace this stick of 30 cm length.
• Try to see the top and bottom of the stick simultaneously without any change in the position of eye.
• Try the same activity with various lengths of the sticks.
• You can see the whole object AB which is at a distance of 25 cm (least distance of clear vision) because the rays coming from the ends A and B of the object will enter the eye.
• Similarly we can also see complete object CD with eye as explained above.
• Let us assume that AB is moving closer to eye to a position A B .
• You notice that you will be able to see only the part (EF) of the object A B because the rays coming from E and F enter your eye.

• The rays coming from A and B cannot enter your eye.
• The rays coming from the extreme ends of an object forms an angle at the eye.
• If this angle is below 60°, we can see whole object.
• If this angle is above 60°, then we see only the part of the object.
• This maximum angle, at which we are able to see the whole object is called angle of vision.
• The angle of vision for a healthy human being is about 60°.
• It varies from person to person and with age.

Activity – 3

Question 3.
Describe an activity for dispersion of light.
(OR)
What is the name given to process when white light passes through a prism it splits into different colours ? Explain the process with an activity.
Answer:

• The splitting of white light into different colours is called despersion of light.
• Do this experiment in the dark room.
• Take a prism and place it on the table near a vertical white wall.
• Take a thin wooden plank.
• Make a small hole in it and fix it vertically on the table.
• Place the prism between the wooden plank and wall.
• Place a white light source behind the hole of wooden plank.
• Switch on the light.
• The rays coming out of the hole of plank become a narrow beam of light.
• Adjust the height of the prism such that the light falls on one of the lateral surfaces.
• Observe the changes in emerged rays of the prism.
• Adjust the prism by slightly rotating it till you get an image on the wall.
• You will observe that white light is splitting into certain different colours.

Activity – 6

Question 4.
Describe an experiment for scattering of light.
(OR)
What is the principle involved in blue of the sky ? Explain the principle with an experiment?
Answer:

• Take the solution of sodium-thio-Sulphate (hypo) with sulphuric acid in a glass beaker.
• Place the beaker in which reaction is taking place in an open place where abundant sunlight is available.
• Watch the formation of grains of sulphur and observe changes in beaker.
• You will notice that sulphur precipitates as the reaction is in progress.
• At the beginning, the grains of sulphur are smaller in size as the reaction progresses, their size increases due to precipitation.
• Sulphur grains appear blue in colour at beginning and slowly their colour becomes white as their size increases.
• The reason for this is scattering of light.
• At the beginning, the size of grains is small and almost comparable to the wavelength of blue light.
• Hence they appear blue in the beginning.
• As the size of grains increases, their size becomes comparable to the wavelengths of other colours.
• As a result of this, they act as scattering centres for other colours.
• The combination of all these colours appears as white.

AP State Board Syllabus AP SSC 10th Class Biology Important Questions Chapter 4 Excretion.

AP State Syllabus SSC 10th Class Biology Important Questions 4th Lesson Excretion

10th Class Biology 4th Lesson Excretion 1 Mark Important Questions and Answers

Question 1.
What is meant by excretion?
Answer:
Excretion is a biological process involved in separation and removal of wastes from body.

Question 2.
When you are on a field trip, your friend collected a sticky substance oozed out by a plant called gum. What are the plants you observe which give gum ?
Answer:
Acacia, Neem, Drumstick, Eukalyptus and Sapota are the gum yielding plants present in our surroundings.

Question 3.
What are primary metabolites?
Answer:
The substances that are directly involved in normal growth and development and reproduction of a plant are called primary metabolites, e.g.: Carbohydrates, fats and proteins.

Question 4.
What are secondary metabolites?
Answer:
The inorganic substances that are not directly involved in the normal growth, development and reproduction of an organism are called secondary metabolites, e.g.: Alkaloids, tannins, resins, gums and latex, etc.

Question 5.
Write any two substances present both in blood and urine.
Answer:
Glucose, Sodium, Potassium, Chlorides, Urea, Creatinine, Uric Acid, Calcium, Phosphorous.

Question 6.
When you are on a field trip, you might have collected some plants which contain alkaloids. Name the alkaloids which are harmful to us.
Answer:
Nicotine, Morphine, Cocaine.

Question 7.
Write the names of any two excretory organs in human beings.
Answer:
Kidney, Skin, Lungs, Liver, Large Intestine.

Question 8.
Which plants in your surroundings are useful for the production of medicines?
Answer:
Neem, Datura, Tulasi, Calotropis.

Question 9.
The body of a person is filled with extra water and waste products. His hands and feet were swollen. What do we call this condition? Failure of which system causes this condition?
Answer:
This condition is called Uremia. This condition is caused due to failure of excretory system (Kidneys).

Question 10.
How do plants get rid of their waste materials?
Answer:

1. Plants can get rid of excess water by transpiration and guttation.
2. The other nitrogenous waste products are stored in leaves, bark and fruits when these dead leaves, bark and ripe fruits fall off from the tree then the waste products in them are removed.

Question 11.
Give an example of a case where organ donation can save one’s life.
Answer:

1. Kidney of a brain dead person can be donated to save the life of renal failure persons.
2. Cornea of a dead person can be donated to the blind.
3. Bone marrow can be donated to save life of persons suffering from blood related genetic disorders.

Question 12.
Why is urine yellow in color?
Answer:
Because of urochrome, urine is yellow in colour. It forms in the liver from dead RBC.

Question 13.
Write two slogans to popularize the awareness on “Organ Donation”.
Answer:
Slogans:

1. Organ donation saves lives.
2. Donate organs today for better tomorrow.
3. Donate organs – Live after death.

Question 14.
Write two healthy habits which you practice to protect your kidneys from diseases.
Answer:

1. Drink plenty of water.
2. Eat low salt diet that saves kidney life.
3. Drink more fruit juices.

Question 15.
What precautions you have to take in the observation of internal structure of mammalian kidney?
Answer:

1. Wash the kidney thorougly with water.
2. Ensure that blood is completely drained from it.
3. Wash your hands with antibacterial lotion after completing the dissection.

Question 16.
Write two secondary metabolites, which you use in your daily life.
Answer:
Gum, Rubber, Coffee are the examples for secondary metabolites which we use in our daily life.

Question 17.
Why is vasopressin not secreted when a person drinks a lot of water?
Answer:
Vasopressin is secreted only when concentrated urine is to be passed out. When a person drinks a lot of water, there will be no need to secrete concentrated urine. The excess water taken by the person will be sent out in the form of dilute urine. Hence, vasopressin is not secreted.

Question 18.
In urine excretory system much water is reabsorbed. What happens if it doesn’t occur?
Answer:
If water is not reabsorbed it leads to excessive repeated dilute urination called diabetes insipidus. If water does not reabsorbed we would dry up in few hours.

Question 19.
A substance given below consists of other three substances. What is that substance? Where is it produced? Uric Acid, Sodium, Oxalate, Urine.
Answer:
Urine contains all the other three substances. It is produced in kidney.

Question 20.
Name the hospitals where the organ transplantation facility is available in Hyderabad.
Answer:
Organ transplantation facility is available only in two government hospitals. They are NIMS (Nizam Institute of Medical Sciences) and Osmania hospital. This facility is also available in some corporate hospitals.

Question 21.
What is anabolism?
Answer:
The synthesis of complex molecules in living organisms from simpler ones together with the storage of energy.

Question 22.
What is catabolism?
Answer:
The breakdown of complex molecules in living organisms to form simpler ones, together with the release of energy.

Question 23.
What is the meaning of the word excretion in Latin?
Answer:
In Latin ex means out, crenere means shift.

Question 24.
What are the wastes produced during metabolic activities?
Answer:
Carbon dioxide, water, nitrogenous compounds like ammonia, urea, uric acid, bile pigments, excess salts, etc., are the wastes produced during metabolic activities.

Question 25.
What is the most poisonous of all waste products?
Answer:
Ammonia is the most poisonous of all waste products.

Question 26.
What are the substances present in blood?
Answer:
Substances present in blood are glucose, sodium, potassium chloride, urea, creatinine, uric acid, cholesterol, triglycerides, calcium, phosphorous, bilirubin, proteins, etc.

Question 27.
What are the substances present in urine?
Answer:
The substances present in urine are protein, creatinine, calcium, phosphorous, uric acid, etc.

Question 28.
What are the substances present in both blood and urine in common?
Answer:
The substances present in the blood and urine in common are sodium, potassium, glucose, chlorides, urea, proteins, creatinine, calcium, phosphorous and uric acid.

Question 29.
What are the substances that need to be removed from body?
Answer:
Creatinine, uric acid, urea, cholesterol and calcium.

Question 30.
In human beings excretory system consists of?
Answer:
The excretory system in human beings consists of a pair of kidneys, a pair of ureters, urinary bladder and urethra.

Question 31.
Where are the kidneys present in human body?
Answer:
In human body kidneys are present in the abdominal cavity attached to dorsal body wall one on either side of backbone.

Question 32.
What is the shape and colour of the kidney?
Answer:
The kidney is bean shaped and reddish brown in colour.

Question 33.
What is the size of the kidney?
Answer:
The size of the kidney is 10 cm in length, 5 – 6 cm in breadth and 4 cm in thickness.

Question 34.
Why is the position of right kidney lower than the left kidney?
Answer:
The position of right kidney is lower than left kidney due to the presence of liver above it.

Question 35.
Which artery brings oxygenated blood to kidney?
Answer:
Renal artery brings oxygenated blood to kidney.

Question 36.
What are the two distinct regions present inside the kidney?
Answer:
The two distinct regions present inside the kidney are

1. dark coloured outer zone called the cortex and
2. pale coloured inner zone called medulla.

Question 37.
Each kidney is made up of how many nephrons?
Answer:
Each kidney is made up of about 1.3 to 1.8 million nephrons.

Question 38.
What is the other name of Nephron?
Answer:
The other name of nephron is uriniferous tubule.

Question 39.
What are the two basic parts of nephron?
Answer:
The two basic parts of nephron are malphigian body and renal tubule.

Question 40.
What is meant by Glomerulus?
Answer:
Malphigian body consists of a blind cupshaped broader end of nephron called Bowman’s capsule and bunch of blood capillaries called Glomerulus.

Question 41.
Which blood vessel forms glomerulus in Bowman’s capsule?
Answer:
Afferent arteriole forms glomerulus in Bowman’s capsule.

Question 42.
Renal tubule consists of how many parts?
Answer:
Renal tubule consists three parts. They are Proximal convoluted tubule (PCT), Loop of Henle, and Distal Convoluted Tubule (DCT).

Question 43.
What is the major function of proximal convoluted tubule?
Answer:
Proximal convoluted tubule reabsorbs useful substances like glucose, amino acids, phosphate, potassium, urea and other organic solutes from the filtrate.

Question 44.
What are podocyte cells?
Answer:
Glomerulus is lined by a single layer of squamous epithelial cells called podocyte cells.

Question 45.
What is the function of loop of Henle?
Answer:

1. In the descending loop of Henle reabsorption of water from the filtrate takes place.
2. Ascending loop of Henle is impermeable to water and only ions diffuse out into the surrounding cells.

Question 46.
What is the function of Distal convoluted tubule?
Answer:
Distal convoluted tubule maintains proper concentration and pH of the urine.

Question 47.
How many stages are involved in formation of urine? What are they?
Answer:
Formation of urine involves four stages. They are

1. Glomerular filtration,
2. Tubular reabsorption,
3. Tubular secretion and
4. Concentration of urine.

Question 48.
The amount of water reabsorption in the tubule depends on?
Answer:
The amount of water reabsorption in the tubule depends on amount of excess water present in the body and the amount of dissolved wastes to be excreted.

Question 49.
In which region seventy five percent of water content of the nephric filtrate is reabsorbed?
Answer:
Seventy five percent of water content of the nephric filtrate is reabsorbed in the region of proximal convoluted tubule.

Question 50.
Which hormone is responsible to pass concentrated urine?
Answer:
Vasopressin is responsible to pass concentrated urine.

Question 51.
What is micturition?
Answer:
Micturition is the process of discharge of urine from the urinary bladder.

Question 52.
What are the composition of various substances in urine?
Answer:
The composition of various substances in urine are 96% of water, 2.5% of organic substances and 1.5% of inorganic solutes.

Question 53.
How much amount of urine is excreted per day?
Answer:
Total amount of urine excreted per day is about 1.6 -1.8 litres.

Question 54.
What is the storage capacity of urinary bladder?
Answer:
The storage capacity of urinary bladder is 300 – 800 ml.

Question 55.
What is uremia?
Answer:
If kidneys stop working completely, our body is filled with extra water and waste products. This condition is called uremia.

Question 56.
What is haemodialysis?
Answer:
Artificial filtering of blood to remove wastes by using dialysis machine is known as haemodialysis.

Question 57.
What is the time required for each dialysis session?
Answer:
The time required for each dialysis session is 3 to 6 hours.

Question 58.
What are the organs that can be transplanted from brain dead patients?
Answer:
The organs that can be transplanted from brain dead patients are kidney, liver, heart, lungs, pancreas, skin, bone, intestines and eyes (cornea).

Question 59.
Where is the transplanted kidney fixed in the body of kidney failure patient?
Answer:
The transplanted kidney is fixed in the body of patient inside the lower abdomen.

Question 60.
What is cadaver transplantation?
Answer:
The process of transplantation of organs from brain dead patients to another is called cadaver transplantation.

Question 61.
What are the other excretory organs present in human body in addition to kidney?
Answer:
The other excretory organs present in human body in addition to kidney are lungs, skin, liver, large intestine, salivary glands and lacrimal glands.

Question 62.
What are the waste products excreted by lungs?
Answer:
The waste products excreted by lungs are carbon dioxide and water.

Question 63.
Sebum of sebaceous glands in skin contains wastes like?
Answer:
Sebum of sebaceous glands in skin contains wastes like waxes, sterols, hydro carbons and fatty acids.

Question 64.
What are the metabolic wastes of haemoglobin of red blood cells in liver?
Answer:
The metabolic wastes of haemoglobin of red blood cells in liver are bile pigments like bilirubin, biliverdin and urochrome.

Question 65.
How is urea produced in liver?
Answer:
Urea is produced in liver by the deamination of proteins.

Question 66.
What are the wastes excreted by intestine?
Answer:
The wastes excreted by intestine are excess salts of calcium, magnesium and iron.

Question 67.
How do unicellular organisms remove waste products?
Answer:
Unicellular organisms remove waste products by diffusion from the body surface to the surrounding water.

Question 68.
What is the osmoregulatory organelle in amoeba and paramoecium?
Answer:
The osmoregulatory organelle in amoeba and paramoecium is contractile vacuole.

Question 69.
In which animal phyla water bathes almost all their cells in body of organisms?
Answer:
In porifera and coelenterates water bathes almost all their cells in body of organisms.

Question 70.
Through which processes plants get rid of excess water?
Answer:
Through transpiration and guttation plants get rid of excess water.

Question 71.
What are Raphides?
Answer:
Waste gets stored in the fruits in the form of solid bodies called Raphides.

Question 72.
What are alkaloids?
Answer:
The poisonous nitrogenous by products produced by plants are alkaloids.

Question 73.
Which alkaloid is used as antimalarial drug?
Answer:
Quinine is the alkaloid used as antimalarial drug.

Question 74.
Which flowers and fruits are used to extract sedative scopalamine?
Answer:
The sedative scopalamine is extracted from the flowers and fruits of Datura stramonium.

Question 75.
What are the alkaloids produced by plants?
Answer:
The alkaloids produced by plants are nimbin, nicotine, morphine, codeine quinine, reserpine, caffeine/scopolamine, etc.

Question 76.
What are tannins?
Answer:
Tannins are alkaloids which are carbon compounds.

Question 77.
Which group of plants secretes resin?
Answer:
Resin secretes by a group of plants Gymnosperms.

Question 78.
What is latex?
Answer:
Latex is a sticky, milky white substance secreted by plants.

Question 79.
Which plant of latex is used to prepare rubber?
Answer:
Rubber is prepared from the latex of Hevea braziliensis.

Question 80.
Bio-diesel is obtained from the seeds of?
Answer:
Bio-diesel is obtained from the seeds of Jatropa.

Question 81.
Which person’s kidney is used for a successful kidney transplantation?
Answer:
From a close relative kidney can be used for successful transplantation.

Question 82.
What may be the feeling of the patients who are waiting for suitable organ donations due to failure of vital organs?
Answer:
The patients who are waiting may think that any donor with a kind heart can give them life by donating their organs.

Question 83.
What may be the feeling of a doctor while transplanting the organs?
Answer:
Doctor is also a human being, he predicts that the transplantation becomes a success.

Question 84.
How can you appreciate the father of Yaswanth Kumar for donating the organs of his brain dead son? (From annexure)
Answer:
We can appreciate the father of Yaswanth Kumar for his humanity where no one can accept to take or think such a delicate decision of donating the organs of his brain dead son.

Question 85.
What might be the feeling of the father H.V. Shiva Kumar father of Yaswanth Kumar for donating the organs of his son? (From annexure)
Answer:
The father might have thought that his son can live even after his death in some other people who are in need for the suitable organs. We can appreciate father’s humanity.

Question 86.
What does the society need about organ donation?
Answer:
The society needs much awareness in organ donation so that we can jsave many lives who are in need of different organs from donors for their survival.

Question 87.
Which substances are present above the normal limits both in the blood and urine in the sample given?
Answer:
Substances present above the normal limits in the blood are: creatinine, uric acid and cholesterol. Substances present above the normal limits in the urine are creatinine, calcium, uric acid and urea.

Question 88.
When is more urine excreted?
Answer:
A large intake of liquids or water rich food increases the volume of water in the blood, hence more urine is excreted.

Question 89.
What are the uses of Tannins?
Answer:
Tannins are used in tanning of leather and in medicines.

Question 90.
What are the economic importance of gums?
Answer:
Gums are valuable being used as adhesives and binding agents in the preparation of the medicines, food, etc.

Question 91.
Why do we get peculiar smell when you shift the potted plant?
Answer:
Plants release some peculiar chemical substances into their surroundings into soil through roots. These substances attract useful microorganisms to the plant. Hence we get peculiar smell when we shift the potted plant.

Question 92.
What is chewing gum ? How is it made of?
Answer:
Chewing gum is a type of gum for chewing made dates back 5000 years. Modern chewing gum originally made of chicle, natural latex from plant.

Question 93.
How do pollen grains affect our health?
Answer:
Whenever pollen grains enter in our body they cause allergy due to the presence of nitrogenous substances. These allergens cause skin allergy and asthma. Ex: Parthenium.

Question 94.
What are the defensive mechanism developed by plants of your village to protect themselves from the herbivores ? Give two examples.
Answer:
Plants produce nitrogenous compounds to protect themselves from the herbivores. These compounds are called as alkaloids. For example,
Datura – It produces alkaloid known as scopolamine.
Cactus – Develop spines on the body.

10th Class Biology 4th Lesson Excretion 2 Marks Important Questions and Answers

Question 1.
Fill in the table on the basis of your obervations in field trip.

Answer:

Question 2.
Two kidneys are present in human beings as excretory organs. Haritha, whose age is 23 years, donated one kidney to her father. Now she has one kidney only. She gave birth to a female child.
a) How many kidneys are there in Haritha’s daughter?
Answer:
Harita’s daughter has 2 kidneys,

b) Support your answer.
Answer:
Bodily changes are not inherited, so Harita’s daughter has 2 kidneys.

Question 3.
What questions do you ask a nephrologist to know more about kidney related diseases?
Answer:

1. How are stones formed in the kidney?
2. What is dialysis?
3. What is the effect of smoking and alcoholism on kidneys?
4. What is ESRD?

Question 4.
Read the following lines and answer the questions.

Liver: It produces bile pigments (bilirubin, biliverdin and urochrome) which are metabolic wastes of haemoglobin of dead R.B.Cs. Urochrome, which is responsible for the amber colour of the urine, is eliminated through urine. Biliverdin and bilirubin are stored in gall bladder and later excreted through bile along with cholesterol and derivatives of steroid hormones, extra drug, vitamins and alkaline salts. Liver is also involved in urea formation.

i) Which substance adds colour to urine?
ii) The site of synthesis of urea in the body is …………..
Answer:
i) Urochrome
ii) Liver

Question 5.
Look at the picture and answer the questions, (label bottle on left as X and bottle on right as Y)
The above is a procedure of haemodialysis in a hospital.
i) What are the liquids X and Y, in the picture?
ii) What is the purpose of this procedure and for whom is it needed?
Answer:
i) X – Dialyzing fluid; Y – Urine
ii) It is a process of removal of poisonous nitrogenous wastes from blood. This process is needed by uremic or kidney failure patients.

Question 6.
Prepare four questions to find the reasons for obstructions in excretory system.
Answer:

1. What do we call the complete and irreversible kidney failure?
2. What happen if kidney stops working completely?
3. What is uremia?
4. Is there any solution to this problem?

Question 7.
Name the secondary metabolites which are useful in leather and rubber industry. From which plants we obtain them?
Answer:

1. Latex used in rubber and Tannins used in Leather Industry.
2. Latex – Hevea brasiliensis (Rubber Plant) Tannins – Cassia, Acacia.

Question 8.
Prepare four questions you will ask a nephrologist about Kidney failure.
Answer:

1. When does kidney fail?
2. What are the symptoms of kidney failure?
3. What precautions can we take to prevent failure of kidney?
4. Which alternative method can we adopt if kidney fails?

Question 9.
Observe the following table.

On the basis of above table, write answers to the following questions.

i) In above table, which living organisms contains kidneys as excretory organs like human beings?
Answer:
Reptiles and Aves.

ii) Write the excretory organs present in Earthworm and Cockroach.
Answer:
Earthworm – Nephridia Cockroach – Green glands

Question 10.
Observe the following table and answer the questions given below.

i) Which alkaloid we get from the fruit, is used as pain killer?
Answer:
Morphine

ii) From which part of the plant do we get Quinine?
Answer:
Bark of Cinchona

Question 11.
What is called structural and functional unit of kidney? Why?
Answer:

1. Each kidney is made up of more than one million microscopic and thin tubular units called nephrons or uriniferous tubules. Hence nephron is known as structural and functional unit of kidney.
2. Nephron’s chief function is to regulate the concentration of water and soluble substances, reabsorbing what is needed and excreting the rest as urine. Hence it is known as functional unit of kidney.

Question 12.
Blood is filtered in Bowman’s capsule of nephron. For the filtration of blood some pressure is needed. How does the pressure arises in Bowman’s capsule?
Answer:

1. Blood flows inside the glomerulus of Bowman’s capsule under the influence of pressure due to the large diameter of afferent arteriole.
2. This increases or rise the blood pressure in the glomerulus capillaries leading to ultrafiltration of the blood in the Bowman’s capsule.

Question 13.
Classify the substances given below.
Ptyaline, Leptin, Morphine, Riboflavin, Testosterone, Thyamin, Niacine, Sucrase, Nicotine, Amylase, Retinol, Quinine, Calciferol, Adrenaline, Tripsin.
Answer:
The above substances can be classified into Enzymes, Hormones, Alkaloids and Vitamins.
Enzymes: Ptyaline, Sucrase, Amylase, Tripsin
Hormones: Testosterone, Adrenaline, Leptin
Alkaloids: Morphine, Nicotine, Quinine
Vitamins: Riboflavin, Thyamin, Niacine, Retinol, Calciferol

Question 14.
How does excretion take place in phylum protozoa?
Answer:

1. Specific excretory organs are absent in unicellular organisms. These organisms remove waste products by simple diffusion from the body surface into the surrounding water.
2. Excess water is sent out through contractile vacuoles in fresh water protozoans. Ex: Amoeba surface.
3. The major portion of excess water is eliminated by diffusion through body surface.

Question 15.
What are the functions of excretory system?
Answer:

1. Excretory system in animals performs more than one function.
2. The primary function of excretory system is to maintain ionic and osmotic balance in the animal body.
3. Excretory system helps to maintain appropriate concentration of salts and correct amount of water in the body.
4. It also helps in maintaining the body volume.
5. The secondary function of excretory system is excretion of nitrogenous wastes and foreign substances such as drugs.

Question 16.
Mention the groups of animals where excretion fakes place through diffusion from body surface.
Answer:

1. In protozoans wastes are excreted by diffusion through body surface.
2. Contractile vacuoles present in some protozoans are mainly concerned with the maintenance of osmotic and ionic balance. Excretion is mainly by diffusion.
3. In sponges, coelenterates like Hydra, excretion takes place by diffusion.

Question 17.
People in cold countries get very less / no sweat. What changes occur in their skin and in other excretory organs?
Answer:

1. Usually human skin has a unique system keep in equilibrium state with the surrounding temperature.
2. During winter season the temperature is low and the body temperature should be heated to balance the tolerance.
3. So the water content in our body is sent out in the form of urine.
4. Hence, people in cold countries get less sweat or no sweat.

10th Class Biology 4th Lesson Excretion 4 Marks Important Questions and Answers

Question 1.
Analyse the following information and answer the questions.

i) Name the alkaloid which is used to cure malaria.
Answer:
Quininie

ii) Name the alkaloids used as insecticides.
Answer:
Pyrethroids

iii) Which system is stimulated by the alkaloid caffeine?
Answer:
Central nervous system

iv) Which parts of which plant is used as medicine for snake bite?
Answer:
The roots of Rauwolfia Serpentina (Snake root)

Question 2.
Which diagram do you draw to label these parts?
Answer:

1. Bowman’s capsule
2. Uriniferous tubule.
3. Collecting tubule.

Draw the diagram and label the parts.
Answer:

Question 3.
Fill the following table related to Secondary Metabolites in plants.

Answer:

Question 4.
The given parts belong to which system? Draw a neat labelled diagram of the system.
a) Kidneys b) Ureters c) Urinary bladder
Answer:
The given parts belong to human excretory system.

Question 5.
Read the following passage and answer the questions.

Plants can get rid of excess water by a process like transpiration and guttation. Waste products may be stored in leaves, bark, and fruits. When these dead leaves, bark, and ripe fruits fall off from the tree then waste products in them are removed. Waste gets stored in the fruits in the form of solid bodies called ‘raphides’. However several compounds are synthesized by the plants for their own use especially for defence. Many plants synthesize chemicals and store them in roots, leaves, seeds, etc. for protection against herbivores. Most of the chemicals are unpleasant to taste. Hence herbivores usually do not prefer to eat such plants. Some of the chemicals are toxic and may even kill the animals that eat them.

i) What are raphides?
ii) How do plants protect themselves against herbivores?
iii) How do plants get rid of wastes?
iv) Name the processes by which plants lose excess of water.
Answer:
i) The wastes stored in some fruits in the form of solid bodies are called “raphides”.
ii) Many plants synthesize chemicals and store them in roots, leaves, seeds, etc. for protection against herbivores.
iii) Waste products may be stored in leaves, bark and fruits. When these dead leaves, bark and ripe fruits fall off from the tree. Then waste products in them are removed.
iv) Plants lose excess of water by transpiration and guttation.

Question 6.

A) Which test is required to know bilirubin?
Answer:
Blood test is required to know bilirubin.

B) How is the sugar disease confirmed?
Answer:
In blood test if the glucose levels in blood before and after food are more than the normal, the sugar disease is confirmed.

C) By observing the above report, what would be the other problems faced by that patient?
Answer:
The blood pressure of the patient 160/90 as the normal is 120/80 mm/Hg.
The Glucose levels in the blood of the patient shows more than the normal before and after food.
So he is suffering from Hypertension and diabetes.

D) What are the organs affected by these problems?
The organsaffected by these problems are heart and kidneys.

Question 7.
Explain the temporary and permanent methods to be adopted for Kidney failure (ESRD) persons.
Answer:
Temporary method for ESRD persons is Dialysis (Artificial kidney). Permanent method is transplantation of kidney.

Dialysis:

1. Blood is taken out from the main artery, mixed with an anticoagulant, such as heparin and then pumped in to dialyzer.
2. In dialyzer blood flows through cellophane tubes and these tubes are embeded in the dialysing fluid.
3. The membrane separates the blood flowing inside the tube and dialysing fluid (same as plasma without nitrogenous waste).

Kidney transplantation:

1. A functioning kidney is used in transplantation from a donor preferably a close relative.
2. The kidney that is received by a recipient must be a good match to his body, to minimise the chances of rejection by the immune system of the recipient.
3. Nowadays the process of organ donation helps a lot for kidney failure patients.

Question 8.
What are the accessory excretory organs in human body? How does the liver carry out excretion as a secondary function? (OR)
Write about the accessory excretory organs and their excretory substances in human beings.
Answer:

Excretory functions of Liver:

1. Liver acts as detoxification centre of our body. Liver produces bile juice which contain bile pigments which are metabolic wastes of dead R.B.C.
2. Bilirubin, Biliverdin and urochrome are the pigments of bile juice.
3. Bilirubin, Biliverdin, cholesterol and derivatives of steroids, extra drugs, vitamins and alkaline salts are the wastes produced by liver.
4. Urochrome is eliminated through urine.
5. Liver also plays a role in the formation of concentrated urine.

Question 9.
Explain the formation of urine in a flow chart.
Answer:
Dissolved substances of blood like urea, glucose, amino acids, minerals, salts etc., are filtered out in Bowman’s capsule under high filtration pressure.
Glomerular filtrate is called primary urine.

Question 10.
Excreting wastes from the human body not only by kidneys but also by other organs helps you. How do you support it?
Answer:

1. In human body wastes are excreted not only by kidneys but also by other organs.
2. Kidney filters blood and eliminates nitrogenous wastes and other harmful things. Filters urea from blood.
3. Apart from kidney lungs, skin, liver, intestine, salivary glands and lacrymal glands.
4. Lungs remove carbon dioxide and water in respiration.
5. Skin excrete wastes in the form of sweat which contains water and certain salts.
6. Liver eliminates bile pigments bilirubin and biliverdin through urine.
7. Excess salts of calcium, magnesium and iron are excreted by epithellial cells of colon for elimination along with faeces by intestine.
8. Eccrine glands present on the forehead, the bottoms of the feet and the palms allow excess water to leave the body.
9. Salivary glands and lacrimal glands excrete small amounts of nitrogenous waste through saliva and tears.

Question 11.
Which plants can you get in your village? Among these by-products of which plants do you use in your real life?
Answer:

1. The plants grow or available in our village are Sapota, Coconut, Cassia, Mango, Guava, Borassus plantain, Tobacco, Rauwolfia, Coffee, Neem, Datura, Chrysanthe¬mum, Acacia, Pinus, Vallisneria, Teak, etc.,
2. Out of these plants alkaloids are available from the plants like Tobacco, Rauwolfia, Coffee, Neem, Datura and Chrysanthemum. The by-products from these plants are utilised in my real life.
   Plant – Use
   Tobacco – Insecticide
   Rauwolfia serpentina – Medicine for snake bite
   Coffee – Central nervous system stimulant
   Neem – Antiseptic
   Datura – Sedative
   Chrysanthemum – Insecticides
3. Tannins are the by-products of cassia, acacia. These are used in tanning of leather and in medicines.
4. Resin the by-product of Pinus is used in varnishes.
5. Gums are extracted from neem and acacia.
   They are used as adhesives and binding agents in the preparation of food, medicines.

Question 12.
Write an essay stating the advantages of by-products of plants in our real life.
(OR)
What are secondary metabolites? Briefly explain their uses.
(OR)
Not only the food of plants but also their wastes are useful to us. What evidences do you give for it?
Answer:

1. The materials which do not require for normal growth and development are called secondary metabolites. These are the by – products of plants, eg: Alkaloids, Tannins, Resins, Gums and Latex, etc. Though plants produce these chemicals for their own use man found the usage of these chemicals for own benefits. They are generally coloured and fragrant.
2. Alkaloids: These are nitrogenous by-products and poisonous. These are stored in different parts of the plants. Common alkaloids in plants and their uses are given in the table.
   

   Alkaloid	Plant	Part	Uses
   Quinine	Cinchona officinalis (Cinchona)	Bark	Antimalarial drug
   Nicotine	Nicotiana tobacum (Tobacco)	Leaves	Insecticide
   Morphine, Cocaine	Papaver somniferum (Opium)	Fruit	Pain killer
   Reserpine	Rauwolfia serpentiana (Snake bite)	Root	Medicine for snake bite
   Caffeine	Coffea Arabica (Coffee plant)	Seed	Central nervous system stimulant
   Nimbin	Azadirachta indica (Neem)	Seeds, Barks, Leaves	Antiseptic
   Scopolamine	Datura stramonium	Fruit, flower	Sedative
   Pyrethroids	Chrysanthemum sps	Flower	Insecticides

3. Tannins: Tannins are carbon compounds. These are stored in different parts of the plant and are deep brown in colour. Tannins are used in tanning of leather and in medicines, e.g. Cassia, Acacia.
4. Resin: Occur mostly in Gymnosperms in specialized passages called resin passages. These are used in varnishes, e.g. Pinus.
5. Gums: Plants like Neem, Acacia oozes out a sticky substance called gum. When branches are cut. The gum swells by absorbing water and helps in the healing of damaged parts of a plant. Gums are economically valuable and used as adhesives and binding agents in the preparation of the medicines, food, etc.
6. Latex: Latex is a sticky, milky white substance secreted by plants. Latex is stored in latex cells or latex vessels. From the latex of Hevea braziliensis (Rubber plant) rubber is prepared. Latex from Jatropa is the source of bio-diesel.
7. Modern chewing gum originally made of chick natural latex from plant.

Question 13.
Blood is purified in kidneys. So many wastes are removed from the blood in nephron of the kidney. Which issue make you surprise in excretory system?
Answer:

1. Kidneys remove nitrogenous waste from our body. They also work towards balancing the amount of vitamins , minerals, fat and protein that are found in the blood. They do this so that our body can easily perform day to day activities.
2. Our intestine makes solid waste materials and is excreted through digestive tract.
3. Each day our body eliminates around 1.6 to 1.8 liters of urine which contains liquids, minerals and vitamins that are of no use to the body.
4. The bladder of a human body is nearly the same size as the average of human brain.
5. In one individuals life span the liver can produce around 184.275 kgs of bile (6500 ounces).
6. In our lifetime an individual could urinate close to 7,850,000,000 gallons of fluid.
7. A really extraordinary fact regarding the excretory system is that upto 400 ml of urine can be held in human bladder.
8. Urine contains a high amount of urea which can be used by plants as a source of nitrogen. Because of this diluted urine can be used in gardens and potted plants.
9. It is amazing to see that each kidney is made up of approximately more than one million microscopic tubular functional units called nephrons or uriniferous tubules.

Question 14.
What is the structural and functional unit of the excretory system? Draw a neat labelled diagram of that unit.
Answer:
Nephron is the structural and functional unit of the kidney.

Question 15.
Give an account of excretory system found In different phyla of animal kingdom.
Answer:

1. Different organisms have various excretory systems and organs.
2. Following are the excretory system found in various organisms.
   

   Name of the phylum / organism	Excretory system / organ
   Protozoa	Simple diffusion from the body surface into the surrounding water.
   Porifera and coelenterates	Water bathes almost all their cells
   Platyhelminthes	Flame cells
   Nematoda	Renette cells
   Annelids	Nephridia
   Arthropoda	Green glands, Maiphigian tubules
   Mollusca	Meta nephridia
   Echinodermata	Water vascular system
   Reptiles, Birds and Mammals	Kidneys

Question 16.
How do plants manage/send out waste products from their body?
Answer:

1. Plants can get rid of excess water by a process like transpiration and guttation.
2. Waste products may be stored in leaves, bark and fruits.
3. When these dead leaves, bark and ripe fruits fall off from the tree then waste products in them are got rid of.
4. Plants get rid of carbon dioxide and oxygen through diffusion.
5. Plants release some waste products through roots also.
6. Some waste products are deposited near bark as resins or gums.
7. In many plants waste products are stored in vacuoles of the cells. Plant cells have comparatively large vacuoles.

Question 17.
In recent days many people are coming forward to donate organs of brain dead people, who met with accidents. How will you appreciate the family members of organ donor?
Answer:

1. In recent days many families are willing to donate organs of brain dead person’s. This is truely significant change in attitude of people.
2. Most of the people are burning or burying body after death. Very a few people are would like to see their very dear one’s in other people by donating organs like heart, liver, kidneys, cornea, spleen and bone marrow etc.
3. The family members of brain dead are already in great sorrow. They need great courage and so much kindness toward needy patients.
4. With nobel decision of that family giving new life to 5 – 7 persons, who has no alternate treatment, other than organ transplantation. In society every one must appreciate their courage, kindness and sympathy. They stood role model to others.

Question 18.
How is the amount of urine produced regulated?
Answer:

1. The amount of urine produced largely depends on the amount of water reabsorbed in the renal tubule.
2. The amount of water reabsorbed by the renal tubule largely depends on
   the amount of water present in excess in the body need to be removed. If water is abundantly present in the body tissues large quantities of dilute urine is send out of the body. When water is less in quantities in the body tissues a small quantity of concentrated urine is excreted.
3. When there is more quantity of dissolved wastes in the body more quantity of water is required to excrete them. So the amount of urine produced increases.
4. Deficiency of vasopressin causes excessive, repeated, dilute urination called diabetes insipidus.

Question 19.
Observe the below flow chart. Fill the boxes. Explain to which system this belongs to.

Answer:

1. Kidney
2. Loop of Henle
3. Pelvis
4. Urethra

This flow chart belongs to excretory system in human beings. It shows the way how the blood in the kidney moves filtered and urine is excreted.

Question 20.
Explain the external features of kidney in human beings.
Answer:

1. In human beings, there are a pair of bean-shaped, reddish-brown structures in the abdominal cavity attached to dorsal body wall one on either side of the backbone.
2. The size of the kidney is 10cm in length, 5 – 6 cm in breadth, and 4cm in thickness.
3. Each kidney is convex on the outer side and concave on the inner side.
   
4. The inner side of each kidney has a fissure or hilus for the entry of a renal artery, exit of a renal vein and an ureter.
5. Renal artery brings oxygenated blood loaded with waste products and renal vein carries deoxygenated blood.

Question 21.
Describe the excretory system of man.
Answer:

1. The excretory system of man consists of
   a) a pair of kidneys
   b) ureters and
   c) urinary bladder and d) urethra.
2. Kidneys are bean shaped and are located in the abdominal region on either side of vertebral column.
   
3. From hilus of each kidney there are a pair of whitish, narrow tubular structures arise. They are known as ureters.
4. The ureter travels downwards and open, in the sac like structure called the urinary bladder, which stores urine.
5. Urethra is a tube that takes urine from urinary bladder to outside.
6. The opening of urinary bladder into urethra is guarded by a ring of muscles or sphincter.

Question 22.
Describe the internal structure of kidney with the help of diagrams. (OR) Describe the structure of nephron with the help of a diagram.
Answer:

1. L.S. of kidney show two distinct regions. Dark coloured outer zone called cortex and pale coloured inner zone called medulla.
2. Each kidney is made up of approximately more than one million microscopic and thin tubular functional units called nephrons or uniferous tubules.
   
3. Each nephron has basically two parts. One is malphigian body and other is renal tubule.
4. Malphigian body consists of a blind cup shaped broader end of nephron called Bowman s capsule and bunch of fine blood capillaries called glomerulus.
5. Glomerulus develops from afferent arteriole and it gives to rise to an efferent arteriole.
6. Glomerulus functions as a filtration unit.
7. Renal tubule has three parts. They are
   1. Proximal Convoluted Tubule (PCT)
   2. Loop of Henle which is ‘U’ shaped and
   3. Distal Convoluted Tubule (DCT).
8. In tubule part reabsorption and secretion takes place. Urine is formed in the renal tubule part of nephron.
9. Distal convoluted tubules open into a collecting tube.
10. Collecting tube forms pyramids and calyces which open into the pelvis.
11. Pelvis leads into the ureter.

Question 23.
Describe the structure of renal tubule with neatly labelled diagram.
Answer:

1. Renal tubule is a specialised tubular structure made up of proximal convoluted tubule, a ‘U’ shaped tube called loop of Henle, and distal convoluted tubule.
   
2. The three tubular components are selectively permeable and only allow specific molecules to pass through them.
3. The renal tubule is surrounded by capillaries called peritubular capillaries that arise from the efferent arterioles.
4. The substances essential for the body are reabsorbed from the tubules into the peritubular capillaries and the unwanted or toxic molecules are secreted into the lumen of the renal tubule.

AP State Board Syllabus AP SSC 10th Class Biology Important Questions Chapter 2 Respiration.

AP State Syllabus SSC 10th Class Biology Important Questions 2nd Lesson Respiration

10th Class Biology 2nd Lesson Respiration 1 Mark Important Questions and Answers

Question 1.
What are the end products of Aerobic and Anaerobic Respirations?
Answer:
End products of aerobic respiration: Carbon dioxide, Water, Energy
End products of anaerobic respiration: Ethanol / Lactic acid, Carbon dioxide, Energy

Question 2.
In which organisms, blood does not supply the Oxygen?
Answer:
Arthropoda organisms (or) Insects (OR) Tracheal respiratory Organisms.

Question 3.
Hari said that stem also respires along with leaves. How do you support him?
Answer:
Lenticels on stem also help in gaseous exchange in some woody plants along with stomata.

Question 4.
Arrange the apparatus as above and heat the glucose. What will happen to lime water when glucose burns?
Answer:
Lime water turns milky due to carbon dioxide (CO₂).

Question 5.
What is the role of mitochondria in anaerobic respiration?
Answer:
The release of energy from glucose in the presence of oxygen occurs in mitochondria. In anaerobic respiration, as oxygen is absent, mitochondria have no role in respiration.

Question 6.
Fermented idli, dosa produce smell. Name the microorganism responsible for producing such smell.
Answer:
Yeast is responsible for producing such smell in fermented idli, dosa.

Question 7.
In what compound, the energy released during the breakdown of glucose is stored?
Answer:
“ATP” (Adenosine Triphosphate).

Question 8.
Label a and b in the given diagram.

Answer:
(a) Matrix, (b) Cristae.

Question 9.
Name chemical substance produced in human muscles during Anaerobic respiration.
Answer:
Lactic acid is produced in human muscles during Anaerobic respiration.

Question 10.
Why is Diazene Green solution added to the Glucose solution in anaerobic respiration experiment?
Answer:
Diazene Green solution is added to the Glucose solution in anaerobic respiration experiment to check the presence of oxygen in glucose solution.

Question 11.
Name the food material on which trypsin acts and name the end products.
Answer:
i) protein ii) end products – peptones.

Question 12.
“Respiration is the energy releasing process.” Write your opinion on this statement.
Answer:
The given statement is absolutely correct. We respire to use the oxygen to oxidise our food and release energy. This is similar like burning but a slower process. With the help of respiratory enzymes, energy released can be stored in the form of ATP for later use.

Question 13.
Identify the figure.

Answer:
Aerial roots in Mangrove plants.

Question 14.
Can we say that combustion and respiration are almost same actions? What evidences do you have for this?
Answer:

1. In both these processes sugar is converted to carbon dioxide and water.
2. Both these processes require oxygen.
3. Both combustion and respiration releases energy.

Question 15.
What is the role of epiglottis in respiration and swallowing food?
Answer:
The epiglottis is a flexible flap at the superior end of the pharynx in the throat. Epiglot¬tis acts as a lid over glottis and prevents food from entering into larynx. Air from pharynx enters the larynx while food enters into oesophagus.

Question 16.
What is the function of haemoglobin?
Answer:
During respiration haemoglobin carries oxygen to the cells and CO, from cells to lungs.

Question 17.
What is respiration?
Answer:
Respiration is the process by which food is broken down to release energy.

Question 18.
What does the word respiration mean in Latin?
Answer:
In Latin the word respiration means “to breathe”.

Question 19.
Who did comprehensive work on properties of gases, their exchange and respiration?
Answer:
Lavoisier and Priestly.

Question 20.
What was the gas liberated on heating powdered charcoal in a bell jar?
Answer:
It was fixed air. In those days carbon dioxide was known as fixed air.

Question 21.
What is oxygen debt?
Answer:
It is the inadequate supply of oxygen when we undertake strenuous exercise.

Question 22.
What is vitiated air?
Answer:
It is the term used then to show air from which the component needed for burning had been removed.

Question 23.
What is the total lung capacity of human being?
Answer:
The total lung capacity of human being is nearly 5800 ml.

Question 24.
Who was the renowned chemist who wrote a textbook of Human Physiology?
Answer:
John Daper was the renowned chemist who wrote a textbook of Human Physiology.

Question 25.
What happens when air passes through nasal cavities?
Answer:

1. Air is filtered in nasal cavity by mucus lining and the hairs growing from its sides, remove some of the tiny particles of dirt in the air.
2. The temperature of the air is brought close to that of the body.

Question 26.
What is the function of epiglottis?
Answer:
Epiglottis controls the movement of air and food towards their respective passages.

Question 27.
What is breathing?
Answer:

1. Breathing is the process of inhaling and exhaling.
2. The mechanism by which organisms obtain oxygen from the environment and release CO₂ is called breathing.

Question 28.
What are pleura?
Answer:
Pleura are the two membranes that protect lungs from injury.

Question 29.
What is the concentration of oxygen at a height of 13 km from the sea level?
Answer:
At a height of 13 km above sea level the concentration of oxygen is much lower about one-fifth as great as at sea level.

Question 30.
What is cellular respiration?
Answer:
Oxidation of glucose or fatty acids takes place in the cells releasing energy. Hence this process is known as cellular respiration.

Question 31.
Where does aerobic respiration occur in eukaryotic cells?
Answer:
Aerobic respiration occur in cytoplasm and mitochondria of eukaryotic cells.

Question 32.
What is Glycolysis?
Answer:
It is the first stage of respiration. In this breakdown of glucose molecule into two molecules of 3 carbon compound called pyruvic acid or pyruvate releasing energy.

Question 33.
What is the fate of pyruvate in the absence of oxygen in animals?
Answer:
In the absence of oxygen pyruvate will be converted to lactic acid and release small amount of energy in animals.

Question 34.
In which type of respiration pyruvate is converted into carbon dioxide and water?
Answer:
In aerobic respiration pyruvate is converted into carbon dioxide and water.

Question 35.
What is the main reason for feeling pain in muscles after strenuous exercise?
Answer:
Due to the anaerobic respiration in muscles large amounts of lactic acid is accumulated and this results in muscular pains.

Question 36.
What is fermentation?
Answer:
In the absence of oxygen, yeast cells convert pyruvic acid to ethanol. This process is called fermentation.

Question 37.
What is the method used to separate ethanol from the yeast glucose mixture in anaerobic respiration?
Answer:
The method used to separate ethanol from the yeast glucose mixture in anaerobic respiration is fractional distillation.

Question 38.
In which organisms does exchange of gases take place through diffusion?
Answer:
In Amoeba, hydra and planarians exchange of gases takes place through diffusion.

Question 39.
In tracheal respiratory system which carry air directly to the cells in the tissues?
Answer:
Trachioles, the fine branches of trachea carry air directly to the cells in the tissues.

Question 40.
What are the respiratory organs in fishes?
Answer:
Gills or bronchiae are the respiratory organs in fishes.

Question 41.
What is cutaneous respiration?
Answer:
If the respiration occurs through skin, it is known as cutaneous respiration, e.g : Leech, Earthworm and Frog.

Question 42.
What are the other areas on the plant body through which gaseous exchange take place?
Answer:
The areas on the plant body through which geseous exchange take place are the surface of roots, lenticels on the stem.

Question 43.
What is the full form of ATP? How is it formed?
Answer:
I) ATP stands for Adenosine triphosphate.
2) ATP is used to supply energy in the cells for the carrying all the metabolic processes.

Question 44.
What are the factors that control respiration?
Answer:
Oxygen and temperature are the two important factors that control the process of respiration.

Question 45.
What are the substances that are used for the production of energy in all living organisms?
Answer:
Glucose and fatty acids are used for the production of energy in all living organisms.

Question 46.
How many types of respiration are present? What are they?
Answer:
There are two types of respiration. They are :

1. Aerobic respiration and
2. Anaerobic respiration.

Question 47.
Where is energy stored in ATP?
Answer:
Energy is stored in the terminal phosphate bond in ATP which is having three phosphates attached to a molecule of Adenosine.

Question 48.
What are the power houses of the cell?
Answer:
Mitochondria are the power houses of the cell.

Question 49.
What is the main difference between respiration and combustion?
Answer:
In respiration several intermediates are produced and in combustion, there are no such intermediates are produced.

Question 50.
What is the equation that represents respiration?
Answer:
The equation that represents respiration is

Question 51.

.
What are the sites of cellular respiration?
Answer:
Mitochondria are the sites of cellular respiration.

Question 52.
What are cristae in mitochondria?
Answer:
The inner membrane of mitochondria is thrown into several folds called cristae.

Question 53.
What is the net gain of ATP molecules in Glycolysis?
Answer:

1. Four ATP molecules are produced when one molecule of glucose is converted to two molecules of pyruvate but two are consumed.
2. The remaining two ATP molecules are net gain in glycolysis.

Question 54.
How many ATP molecules are produced when one glucose molecule is completely oxidised?
Answer:
A net gain of 38 ATP molecules are formed from the total oxidation of one glucose molecule.

Question 55.
What are the three stages present in complete oxidation of glucose molecule?
Answer:
The three stages present in complete oxidation of glucose molecule are

1. Glycolysis
2. Kreb’s cycle and
3. Electron transport.

Question 56.
Why does oxidation of fatty acids give more energy?
Answer:
Oxidation of fatty acids give more energy due to the presence of more carbon atoms in them.

Question 57.
What are aquatic and terrestrial animals?
Answer:
Animals that live in water are called aquatic animals and that live on land are known as terrestrial animals.

Question 58.
Why is the rate of breathing in aquatic organisms much faster than terrestrial organisms?
Answer:

1. The amount of oxygen dissolved in water is low when compared to the amount of oxygen present in air.
2. Therefore the rate of breathing in aquatic animals is much faster than in terrestrial animals.

Question 59.
Which part of the roots is involved in the exchange of respiratory gases?
Answer:
The part of roots that are involved in the exchange of respiratory gases are root hairs.

Question 60.
What is the average breathing rate in an adult mem at rest?
Answer:
The average breathing rate in an adult man at rest is about 15 to 18 times per minute.

Question 61.
Why is the trachea prevented from collapsing?
Answer:
The walls of the trachea are supported by several ‘C’ shaped cartillagenous rings. They prevent the trachea from collapsing and closing.

Question 62.
Why deos the percentage of carbon dioxide increase in exhaled air?
Answer:
During oxidation of glucose carbon dioxide is produced as waste product. Hence the concentration of carbon dioxide increases in exhaled air.

Question 63.
How does breathing take place in mangrove plants?
Answer:
In mangrove plants breathing takes place through specialised structures called breath¬ing roots or pneumatophores.

Question 64.
How does respiration take place in plants where roots are present in wet places?
Answer:
The plants which have their roots in very wet places have much larger air spaces, connect the stems with the roots, making diffusion from upper parts.

Question 65.
Which form a continuous network all over the plant?
Answer:
The stomatal openings lead to a series of spaces between the cells inside the plant which form a continuous network all over the plant.

Question 66.
What are the reasons for the animals to develop different types of respiratory organs?
Answer:
Body size, availability of water, habitat in which they live and the type of circulatory system are some of the reasons for the animals to develop different types of respiratory organs.

Question 67.
Why do fishes die when taken out of water?
Answer:
Fishes do not have lungs to utilise oxygen for breathing. They have gills which can utilize only dissolved oxygen from water.

Question 68.
What would be the consequences of deficiency of haemoglobin in our bodies?
Answer:
Deficiency of haemoglobin in blood can affect the oxygen supplying capacity of blood to body cells. It can also lead to a disease called Anaemia.

Question 69.
What are the stages of respiration in man?
Answer:
Respiration in man occurs in two stages 1) Inhalation (or) Inspiration 2) Exhalation (or) Expiration.

Question 70.
Which part plays major role in respiration of man?
Answer:
Diaphragm plays a major role in respiration in man.

Question 71.
Which part plays major role in respiration of woman?
Answer:
In woman ribs play a major role in respiration.

Question 72.
How are lungs protected?
Answer:
Lungs are protected by two membranes called pleura. A fluid between these membranes protects the lungs from injury.

Question 73.
What is the composition of exhaled air?
Answer:
Exhaled air contains 16% of oxygen, 4% of carbon dioxide and 79% of nitrogen.

Question 74.
Why are red blood cells red in colour?
Answer:
Red blood cells are red in colour due to the presence of haemoglobin in their cytoplasm.

Question 75.
How is haemoglobin made up of?
Answer:
Haemoglobin is made up of a protein called globin, Iron (Hearn) and organic molecule called porphyrin.

10th Class Biology 2nd Lesson Respiration 2 Marks Important Questions and Answers

Question 1.
(a) Which gas turns lime water milky in this experiment?
Answer:
Carbondioxide (or) CO₂

(b) Which gas do you think might be present in less quantities in the air we breath out as compared to air around us?
Answer:
Oxygen (or) O₂

Question 2.
Balu said that, “Plants perform Photosynthesis during day time. They respire during night time”.
Do you agree with Bain? Why? Why not?
Answer:

1. I do not agree with Balu’s statement.
2. Photosynthesis depends on light for energy but respiration does not depend on light.
3. Hence, photosynthesis takes place during day time only whereas respiration takes place both day and night.

Question 3.
The sportsman who participated in 100 mtr race get more muscle pains. But the sportsman who participates in 5 km’s race get less muscle pains. What is the reason?
Answer:

1. Accumulation of lactic acid results in muscular pain.
2. During 100 m race a well trained athlete can hold his breath and afterwards he pants.
3. In this case, the muscles are using energy released during the anaerobic break down of glucose, lactic acid is produced.
4. The presence of lactic acid in the blood is the main cause of muscle fatigue. Whether it is 100 mtr race or 5 km race.
5. If the body is rested long enough the tiredness goes.

Question 4.
What happens if there is no epiglottis in human beings?
Answer:

1. Food may enters into the larynx.
2. Food may enters into the lungs leading to the death.
3. May not speak properly.
4. Entry of food and air may not be regulated properly.

Question 5.
Write two chemicals and two materials required to conduct the experiment “Heat and Carbon dioxide are evolved during anaerobic respiration”.
Materials required: Thermosflask, splitted corks, thermometer, wash bottle, glass tubes.
Chemicals required: Liquid paraffin, glucose solution, bicarbonate solution, Janus green B and Yeast cells.

Question 6.
Observe the below diagram.
A) To which biosystem is this picture related?
Answer:
Respiratory system.

B) Write the names of the parts of A, B.
Answer:
A – alveolus; B – blood capillary network

C) To which system are they linked with?
Answer:
Respiratory system; circulatory system.

D) Which process is happening here? What happens as a result of it?
Answer:
Gaseous exchange between alveolus of lungs and blood capillaries. Due to this the CO₂, present in blood capillaries enter alveolus and oxygen present in alveolus en¬ter blood capillaries.

Question 7.
A person reached a specific distance once on foot and once by running. In which situation his legs pain? Why?
Answer:

1. When a person runs to reach a specific distance gets pain in his legs.
2. This is due to the production of lactic acid in the muscles.
3. Due to the Anaerobic respiration glucose in muscles converts into lactic Acid.
4. Accumulation of lactic acid causes pain in leg muscles.

Question 8.
What is the advantage of the wet and warm passage of air from the nostrils to capillaries?
Answer:
When the air passes in nasal cavity and in the pharynx some changes take place.

1. The mucus layer and hair in the nasal cavity removes the dust particles in the air.
2. The temperature of the air brought to the body temperature.
3. Moistening the air.

Question 9.
In the experiment of anaerobic respiration with yeast
i) Why was liquid paraffin poured on glucose?
ii) What did you understood about anaerobic respiration?
Answer:
i) The supply of oxygen from the air can be stopped by pouring liquid paraffin on glucose.
ii) Anaerobic respiration takes place in the absence of oxygen. In this glucose molecule is incompletely oxidised. The end products of anaerobic respiration are ethyl alcohol or lactic acid and CO₂.
During anaerobic respiration small amount of energy is liberated (2ATP). Anaero¬bic respiration occurs in many anaerobic bacteria and human muscles cells. The anaero¬bic respiration can be represented as:
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂+ 56 K.Cal.

Question 10.
See the below table. Write what you know from it.

Answer:

1. The inhaled air consists of 21% of oxygen whereas the exhaled air contains 16% of oxygen only. This is due to utlilisation of oxygen during cellular respiration in the body. Hence the difference occurs.
2. Inhaled air contains 0.04% of carbondioxide whereas exhale air contains 4% of carbondioxide.
   The concentration of CO₂ is increased a lot due to the release of CO₂ during cellular respiration in the body.
3. Both inhale and exhale air contains 79% of nitrogen because nitrogen has no role to play in cellular respiration.

Question 11.
What is the pathway of air from nostril to alveolus?
Answer:
Draw a flow chart of Respiratory passage of Humans.

Question 12.
What happens when a baker prepares a dough by mixing yeast in it?
Answer:

1. The yeast is commonly used for fermenting bread is saccharomyces cerevisiae.
2. Baker’s yeast has the advantage of producing uniform, quick, and reliable results because it is obtained from pure culture.
3. Water is mixed with flour, salt and the fermenting agent.
4. The mixed dough is then allowed to rise one or more times.
5. Then loaves are formed and the bread is baked in air oven.

Question 13.
How does respiration in amoeba and hydra occur through diffusion? (OR)
What are the similarities in respiration of amoeba and hydra?
Answer:

1. Amoeba and hydra are aquatic organisms.
2. Respiration in them occurs through diffusion.
3. As oxygen is used by these organisms in respiration, its concentration is reduced in cytoplasm. Hence oxygen diffuses into cytoplasm from surrounding water.
4. During respiration CO₂ is continuously produced, its concentration increases in the cytoplasm, hence it diffuses into surrounding water.

Question 14.
Write a short note on ATP. (OR) Expand ATP.
Answer:

1. From the break down of glucose the energy is released and stored up in a special compound known as ATP (Adenosine Triphosphate).
2. It is a small parcel of chemical energy. The energy currency of these cells is ATP an energy rich compound that is capable of supplying energy whenever needed within the cell.
3. Each ATP molecule gives 7200 calories of energy. This energy is stored in the form of phosphate bonds.
4. If the bond is broken, the stored energy is released.

Question 15.
How do Dolphin and Crocodile respire?
Answer:

1. The aquatic animals like dolphin and crocodile respire with the help of lungs.
2. They come out of the water for air.
3. These two animals were lived on land initially.
4. Later they lived in water and developed several adaptations to live in water.

Question 16.
Why are Mitochondria called “Power houses of cell”? (QR)
What is the energy producing organ in a cell? How does it produce energy?
Answer:

1. Cellular respiration in prokaryotic cells like that of bacteria occurs within the cytoplasm.
2. In eukaryotic cells cytoplasm and mitochondria are the sites of reaction.
3. The produced energy is stored in mitochandria in the form of ATP.
4. Hence, mitochondria are called “Power houses of cell”.

Question 17.
Write the rate of respiration in different age groups of human beings.
Answer:

1. Newborn child: 32 times per minute
2. Children of 5 years: 26 times per minute
3. Man of 25 years: 15 times per minute
4. Man of 50 years: 18 times per minute

10th Class Biology 2nd Lesson Respiration 4 Marks Important Questions and Answers

Question 1.
Write about respiration in mangroves that grow in marshy lands.
Answer:

1. Mangroves grown near the marshy places respire through aerial roots or respiratory roots.
2. The root hairs exchange the gases from their surface.
3. They obtain oxygen from the airspaces present between the soil particles.
4. The plants grown in marshy places are adapted to develop aerial roots above the soil surface which helps in gaseous exchange.

Question 2.
a) What is the aim of this experiment?
Answer:
Heat is liberated during respiration.

b) What change do you observe in thermometer readings?
Answer:
Reading increases in the thermometer.

c) In your opinion, where did this heat come from?
Answer:
The heat comes from the germinating seeds which respire and releasing heat.

d) What precaution should we take, while doing this experiment?
Answer:
The bulb of the thermometer should be dip in the germinating seeds (or) sprouts.

Question 3.
You have conducted this experiment in your classroom. Now answer the following questions.
a) What do you prove by conducting this experiment?
Answer:
To test the production of heat and carbon dioxide during anaerobic respiration.

b) Why do you heat glucose solution?
Answer:
To remove the dissolved oxygen in the glucose solution.

c) How do you confirm that glucose solution is free from oxygen after heating it?
Answer:
By adding diazine green (Janus green B) solution to glucose solution, it turns to pink.

d) What are the changes you notice in the lime water?
Answer:
Lime water turns milky white.

Question 4.

i) What change did you observe in the thermometer in the given experiment?
Answer:
Raise in the temperature

ii) Where does the heat come from?
Answer:
From the germinating seeds during respiration

iii) What result you will get, if you perform this experiment with dry seeds?
Answer:
No change of temperature in thermometre.

iv) What are the apparatus used in this experiment?
Answer:
Glass jar, germinating seeds, cork, thermometer.

Question 5.
Observe the set of apparatus and answer the following questions.
i) Which process do we know with the help of this experiment?
Answer:
Combustion.

ii) How does this process differ with respiration?
Answer:
Respiration occurs in the presence of water.
Combustion occurs in the absence of water.

iii) What are the similarities between this process and respiration?
Answer:
In both processes energy is released.

iv) Which gas turns lime – water milky?
Answer:
Carbon-di-oxide (CO₂)

Question 6.
Look at the following experiment. Answer the questions.

a) What is the aim of the experiment?
Answer:
The aim of the experiment is CO₂ is released during anaerobic respiration.

b) Which agent is used to find the presence of oxygen?
What changes do you observe when oxygen is present in Glucose solution?
Answer:
To find the presence of oxygen diazine green (Janus Green B) solution is used. The blue diazine green solution turns pink when oxygen is present in the glucose solution.

c) Why is liquid paraffin poured on glucose solution?
Answer:
By pouring liquid paraffin on glucose solution, the supply of oxygen from the air can be cut off.

d) Which gas is released during the experiment? How can you prove it?
Answer:
Carbon dioxide is released.
The released CO₂ passes into lime water it turns milky.

Question 7.
Observe the following diagram and answer the following questions.

1. What do we call the membranes that cover the lungs?
2. What is the functional unit of lungs ?
3. Which part produces the sound ?
4. What does ‘X’ denote ?

Answer:

1. Pleura
2. Alveoli
3. Larynx
4. Trachea

Question 8.
Observe the diagram and answer the following questions.

a) What does the given diagram indicate?
b) What is the part ‘X’ in the diagram?
c) What is the function of the given picture?
d) To which system the given picture belongs to?
Answer:
a) The given diagram indicates mitochondria.
b) Matrix
c) Performing cellular respiration and releasing energy in the form of ATP.
d) Respiratory system.

Question 9.
Observe the experimental setup and answer the given questions.
A) What is the aim of this experiment?
B) What are the apparatus required for this experiment?
C) What changes do you observe in thermometer during this WKm experiment?
D) What will happen, if dry seeds are taken instead of germinating seeds in this experiment?
Answer:
A) Heat is liberated during respiration.
B) Glass jar, Germinating seeds, Cork and Thermometer.
C) We can notice the raise in temperature after observing the thermometer readings.
D) There will be no change of temperature in the thermometer. We can’t prove the aim of the experiment.

Question 10.
Observe the below diagram and answer the following questions:
i) What does the above setting (diagram) indicate?
Answer:
The above setting (diagram) indicates to prove that carbon dioxide and heat are liberated during anaerobic respiration by yeast cells.

ii) Why is boiled and cooled glucose covered with paraffin?
Answer:
To prevent supply of air, boiled and cooled glucose is covered with paraffin.

iii) What is the use of adding diazine green to glucose solution? What change you notice in glucose solution?
Answer:
Diazine green is added to glucose solution to know whether oxygen is present or not in glucose solution. When the availability of oxygen is less the diazine green changes to pink colour.

iv) Why is lime water used in this experiment?
Answer:
To know whether carbon dioxide is released or not in this experiment lime water is used. Carbon dioxide changes lime water to milky white.

v) Why is bulb of thermometer dipped in the glucose water?
Answer:
To know the rise in temperature of glucose solution when heated, the bulb of thermometer is dipped in the glucose water.

Question 11.
Explain with the help of a flow chart, the path way of air in humans.
Answer:

Question 12.
Study the graph and answer the following questions :
Graph showing effects of vigorous excercise on the concentration of lactic acid in blood.
i) What was the concentration of lactic acid in blood to start with?
ii) What was the greatest concentration of lactic acid reached during the experiment?
iii) What is the concentration of lactic acid after 25 minutes of exercise?
iv) What is the relationship between lactic acid and muscle pain?
Answer:
i) 20 mg/cm³
ii) 20 minutes (Or) at “B” point,
iii) 101 mg/cm³
iv) If concentration of lactic acid increases, muscle pains also increases.

Question 13.
Observe the following :

Write the answers to the following questions:
i) How many Pyruvic acid molecules form from one Glucose?
Answer:
2 Pyruvic acid molecules.

ii) What conditions influence Pyruvic acid to participate in Aerobic and Anaerobic respiration?
Answer:
Presence of oxygen

iii) In which we get more energy in both Aerobic and Anaerobic respirations?
Answer:
Aerobic respiration

iv) The chemical that is formed in human muscles during Anaerobic respiration.
Answer:
Lactic acid

Question 14.
Why does the exchange of gases happen only in alveoli, though arteries are present in pharynx, trachea and bronchus?
Answer:

1. Alveoli are tiny air sacs in the lungs surrounded by capillaries
2. They are numerous and only single cell thickness
3. They increase the efficiency of gas exchange.
4. Due to the difference in a gradient of O₂ oxygen diffuse from alveoli to blood capillaries.

Question 15.
What are the events or steps in respiration?
Answer:
The following are the events or steps in respiration.

1. Breathing: Air moves into lungs and out of lungs.
2. Gaseous exchange in lungs: Exchange of gases between alveoli and blood.
3. Gas transport by blood: Transport of oxygen from blood capillaries of alveoli to body cells and return of carbon dioxide.
4. Gaseous exchange in cells: Exchanging oxygen from blood into the cells and carbon dioxide from cells into the blood.
5. Cellular respiration: Using oxygen in cell processes to produce carbon dioxide and water, releasing energy to be used for life processes.

Question 16.
What will happen if the respiratory tract is not moist? (OR)
Why respiratory tract should be moist?
Answer:

1. If the respiratory tract is not moist the dirt particles in the inhaled air will not be removed from air in the nasal cavities and reaches lungs and creates problems to lungs.
2. The temperature of the inhaled air is brought close to that of the body for the smooth passage in the respiratory tract. If it is dry, it is not possible.
3. If the surface dries out, gas exchange will happen at a very reduced rate since fast moving gaseous oxygen molecules do not efficiently cross the alveoli membrane.
4. The reduced gas exchange is most likely not enough to support blood oxygenation for vital functions.
5. Hence respiratory tract should be moist for smooth exchange of gases.

Question 17.
Explain the process of transportation of gases through the blood.
Answer:

1. The relative amount of gases and their combining capacity with haemoglobin and other substances in blood determine their transport via blood in the body.
2. When oxygen present in the air is within normal limits (around 21%) then almost all of it is carried in the blood by binding to haemoglobin, a protein present in the red blood cells.
3. As oxygen is diffused in the blood, it rapidly combines with the haemoglobin to form oxyhaemoglobin.
4. Not only can haemoglobin combine with oxygen, but it can easily broken into haemoglobin and oxygen.
5. Carbon dioxide is usually transported as bicarbonate, while some amount of it combines with haemoglobin and rest is dissolved in blood plasma.
   

Question 18.
Why is human life impossible at higher altitudes without a supplementary supply of oxygen? (OR)
The concentration of oxygen in air decreases as we go up from sea level. Explain briefly.
Answer:

1. If haemoglobin is exposed to air at sea level, every molecule in air combines with oxygen to form oxyhaemoglobin.
2. At a height of 13 km above sea level, the concentration of oxygen is much lower about l/5th of a sea level.
3. Under these conditions about half as many molecules of oxygen combine with haemoglobin to form oxyhaemoglobin.
4. Blood cannot carry enough oxygen to the tissues.
5. Hence human life is impossible at such a high altitude without a supplementary supply of oxygen.
6. Provision for such a supply is built into modern aircraft which have pressurized cabins that maintain an enriched air supply.

Question 19.
What are the different ways in which glucose is oxidised to provide energy in various organisms? Give one example of each.
How does oxidation of glucose occur in various organisms?
Answer:

1. Glucose is the most commonly used sugar for deriving energy in plants, animals and in microorganisms.
2. In all these organisms glucose is oxidized in two stages.
3. The first stage is known as Glycolysis. It occurs in cytoplasm.
4. During glycolysis glucose is converted to two molecules of pyruvic acid.
5. In the second stage if oxygen is available pyruvic acid is converted to C02 and water, large amount of energy is released. This is known as aerobic respiration. It occurs in most of the plant and animal cells.
6. If oxygen is inadequate or not available, pyruvic acid is converted into ethanol and carbon dioxide. This is anaerobic respiration taking place in yeast cells that is called fermentation.
7. If oxygen is not available in muscle cells, the pyruvic acid is converted into lactic acid.
   

Question 20.
Write the adaptations seen in plants living in water logged conditions.
(OR)
What are the adaptations seen in magrove plants?
Answer:

1. Most plants can aerate their roots by taking in the oxygen through lenticels or through the surface of their root hairs.
2. But plants which have their roots in very wet places, are unable to do this.
3. They are adapted to these water logged conditions by having much larger air spaces which connect the stems with the roots, making diffusion from the upper parts much more efficiently.
4. The problem of air transportion is more difficult for trees and may not survive with their roots permanently in water.
5. To overcome this problem the mangrove tree of the tropics which raise up aerial roots above the surface and takes in oxygen.

Question 21.
Describe the mechanism of branchial or gill respiration in fishes.
(OR)
Briefly explain the process of exchange of gases in fishes during respiration.
Answer:

1. Some aquatic animals like fishes have developed special organs for respiration which are known as gills or branchiae.
2. Blood is supplied to gills through capillaries which have thin walls where gases are exchanged. Gills are present in the gill pouches or branchial pouches.
   
3. Gills are provided with leaf-like folds called gill lamellae.
4. Fish keeps its mouth open and lowers the floor of the oral cavity. As a result water from outside will be drawn into the oral cavity.
5. Now the mouth is closed and the floor of the oral cavity is raised.
6. Water is pushed into the pharynx and is forced to gill pouches through internal branchia apertures.
7. When water passes through gill lamellae exchange of gases takes place, that is oxygen diffuses from water to blood and CO₂ from blood into water.
8. Then water flows through external branchia aperture.

Question 22.
Explain briefly about Pranayama- the art of breathing. (OR)
How can the capacity of lungs be improved by yoga?
Answer:

1. To improve breathing capacity the saint Patanjali developed Yogabyasa.
2. The art of breathing in Yogabyasa is called Pranayama. Prana means gas, ayama means journey.
3. In Pranayama practice air is allowed to enter three lobes of lungs in order to in¬crease the amount of oxygen to diffuse into blood.
4. More amount of oxygen available to brain and tissues the body will be more active.
5. It is very important to practise Pranayama regularly to make our life healthy and active.
6. All people irrespective of age and sex should practise Pranayama under the guidance of well trained Yoga Teacher to improve the working capacity of lungs.

Question 23.
What are the experiments carried out by Lavoisier to understand the property of gases?
Answer:

1. In his early experiments Lavoisier thought that the gas liberated on heating powdered charcoal in a bell jar kept over water in a trough was like fixed air i.e., carbon dioxide.
2. The next series of experiments deals with the combustion of phosphorous in a bell jar. From this he showed that whatever it was in the atmospheric air which combined with the phosphorous was not water vapour.
3. This was respirable air, a component of air that also helped in burning.
4. The air that we breathe out precipitated lime water while that after heating metal did not.
5. From this, he concluded that there were two processes involved in respiration.
6. Lavoisier carried out another experiment by which he showed that about one sixth of the volume of ‘vitiated air’ consists of chalky acid gas (fixed air).
7. Either eminently respirable air is changed in the lungs to chalky acid air; or an exchange takes place, the eminently respirable air being absorbed, and an almost equal volume of chalky acid air being given up to the air from the lungs.
8. Lavoisier had to admit that there were strong grounds for believing that eminently respirable air did combine with the blood to produce the red colour.

Question 24.
Explain the evolutionary changes in energy-releasing system.
(OR)
What are the different respiratory systems in animal groups?
Answer:
Exchange of gases is a common life process in all living organisms, but it is not same in all.

1. Diffusion:
   1. Single-celled organisms like amoeba or multicellular organisms like hydra and planarians obtain oxygen and expel carbon dioxide directly from the body by the process of diffusion.
   2. In multicellular animals special organs are evolved.
   3. Body size, availability of water and the type of circulatory system are some of the reasons for the animals to develop different types of respiratory organs.
2. Tracheal respiratory system : In insects tracheal respiratory system is present in which small branches of trachea called trachioles carry air directly to the cells in the tissues.
3. Bronchial respiration : In fishes gills are utilised for the exchanges of gases. Blood is supplied to gills through capillaries which have thin walls for exchange of gases. This is called bronchial respiration.
4. Cutaneous respiration: 0 Respiration through skin is called cutaneous respiration.
   Eg: i) Earth worms and leeches.
   ii) Frog, an amphibian can respire through lungs and skin.
5. Pulmonary respiration : Most of the higher animals respire with the help of lungs. This type of respiration is known as pulmonary respiration. Eg: Mammals.

Question 25.
Describe the structure of mitochondria with the help of a diagram. (OR)
Which cell organelle is called energy currency or power house of cell? What do you know about its construction?
Answer:
Mitochondria is known as energy currency or power house of cell.
Structure of mitochondria:

1. Mitochondria are sac-like structures present in the cytoplasm of the cells.
2. Mitochondria have two compartments-an inner compartment and an outer compartment. The substance in the inner compartment is called matrix.
3. The matrix is surrounded by a membrane called inner membrane of mitochondria.
4. The inner membrane is thrown into several folds called cristae. The cristae extended into the matrix.
5. The space between the folds is continuous with the outer compartment.
6. On the inner membrane, projecting into the matrix are a large number of particles called elementary particles.
7. These particles have a spherical head and a stalk. They are attached to the inner membrane by their stalk and the head portion of the particle is in the matrix.
8. The outer compartment is surrounded by another membrane – the outer membrane. The outer membrane is smooth and has no projections.
9. The inner membrane, the matrix and the elementary particles in the mitochondria have large number of enzymes and other required proteins for the respiration and energy production.

Question 26.
Draw and label mitochondria. Why should we call it cell of power ?
Answer:

Oxidation of glucose molecule occurs in the mitochondria, ot cell. This is known as cellular respiration. The energy produced during cellular respiration stored in the form of ATP molecule. Energy producing cellular respiration occurs in mitochondria hence we call it cell of power or power house of the ceil.

Question 27.
Describe how oxygen enters the blood in lungs with the help of block diagram.
(OR)
How does gaseous exchange occur in lungs?
Answer:

1. Gaseous exchange takes place within the lungs by diffusion from the alveoli to blood capillaries and vice versa. Alveoli in lungs are numerous and only one cell thick.
   
2. Alveoli are surrounded by capillaries that are also one cell thick.
3. Blood, dark red in colour flows from the heart through these capillaries and collects oxygen from the alveoli.
   At the same time, carbon dioxide passes out of the capillaries and into the alveoli.
4. When we breathe out, we get rid of carbon dioxide.
5. The bright red, oxygen rich blood is returned to the heart and pumped out to all parts of the body.

Question 28.
What is the role of diaphragm and ribs in respiration? Are both active in man and woman?
Answer:
Diaphragm:

1. Diaphragm is a muscular dome shaped tissue present at the floor of the chest cavity separating abdomen from respiratory system.
2. Diaphragm expands downwards into the abdomen thus increasing chest cavity. This allows the lungs to expand as we inhale.
3. As the diaphragm contracts upwards thus decreasing the chest cavity, it allows the air to expel from the lungs.
   Ribs:
4. The ribs protect the lungs and expand as we inhale to facilitate space for the lungs to expand. The ribs then contract expelling the air from the lungs.
5. The intercostal muscles present between the ribs help in contraction and relaxation of ribs.
6. In man, diaphragm plays a major role in the respiration, while in woman, the ribs play a major role.

Question 29.
Why are alveoli so small and uncountable in number? (OR)
How do alveoli increase the area for exchange of gases?
Answer:

1. The pouch-like air sacs at the ends of the smallest branchioles are called alveoli.
2. The walls of the alveolus are very thin and they are surrounded by very thin blood capillaries.
3. It is in the alveoli that gaseous exchange takes place.
4. There are millions of alveoli in the lungs. The presence of millions of alveoli in the lungs provides a very large area for the exchange of gases.
5. And the availability of large surface area maximises the exchanges of gases.

Question 30.
Write a brief note on respiration in plants. (OR)
Does respiration occur in plants? Explain briefly about it.
Answer:

1. In most plants exchange of gases takes place through stomata.
2. There are other areas on the plant body like surface of roots, lenticels on stem, etc. the gaseous exchange takes place.
   
3. Some plants have specialized structures like breathing roots of mangrove plants as well as the tissue in orchids.
4. Breathing roots and tissue in orchids help plants to take oxygen to produce energy and release carbon dioxide.
5. Inside the plants openings lead to a series of spaces between the cells which form a continuous network all over the plant.
6. The whole system works by diffusion.
7. As the oxygen is used up by the cells a gradient develops between the cells and the air in the spaces.
8. So oxygen passes in between the air spaces and the air outside stomata and lenticels.
9. In the same way, as more carbon dioxide is given out by the cells, a gradient occurs in the reverse direction and it passes out.

Question 31.
Write a brief note on tracheal respiration in insects.
Answer:

1. In insects blood do not contain haemoglobin, and blood is white in colour. Hence it cannot carry oxygen.
2. For respiration insects adopt a special system called tracheal system.
   
3. This system consists of a series of tubes called trachea.
4. These trachea open out through small apertures called spiracles on either side of the body.
5. All tracheal tubes of each side join and form a longitudinal tracheal trunk.
6. Trachea divide into a number of branches called tracheoles which carry air directly to the tissues.
7. As the air moves in and out of the trachea, oxygen present in the air diffuses into the cells and CO₂ diffuses into the air from the cells.

Question 32.
Write about the mechanism of respiration in human beings. (OR)
How does exchange of gases take place in human beings?
Answer:

1. Respiration in man occurs in two stages. They are inspiration and expiration.
2. During inspiration air from outside enters into the lungs by increasing the chest cavity.
   
3. Increase in the chest cavity is made by pulling the diaphragm down and pushing the ribs forward.
4. As the air pressure in the lungs is reduced, air from outside enters the lungs through external nostrils, nasal cavities, internal nares, pharynx, epiglottis, larynx, trachea, bronchi and branchioles and finally reach the alveoli where exchange of gases takes place.
5. During expiration the diaphragm and ribs come back to original positions.
   
6. This reduces the volume of chest cavity.
7. So the volume of lungs is decreased and air under pressure comes out of the lungs.

Question 33.
Study the graph given below and analyse the reasons for accumulation of lactic acid in blood after strenuous exercise.

Answer:

1. This graph shows the relation between time accumulation of lactic acid in the muscles.
2. At the beginning, the amount of lactic acid in the blood is very less.
3. Gradually it is increased by vigorous exercise.
4. Within 15 minutes it goes to maximum level which causes muscle pain.
5. Then the lactic acid is removed from muscles in an hour.
6. Muscles produce energy by anaerobic respiration.
   C₆H₁₂O₆ → lactic acid + CO₂ + energy
7. In the vigorous exercise, muscle work rapidly and produce more lactic acid.
8. That’s why lactic acid concentration is increased in muscle after strenuous exercise.

Question 34.
Observe the above graph of lactic acid accumulation in the muscles of an athlete and answer the following questions.
a) What was the concentration of lactic acid in the blood to start with?
Answer:
It is 20 mg/km³.

b) What was the greatest concentration reached during the experiment?
Answer:
101 mg/cm³.

c) If the trend between points C and D were to continue at the same rate, how long might it take for the original lactic acid level to be reached once again?
Answer:
55 minutes.

d) What does high level of lactic acid indicate about the condition of respiration?
Answer:
It indicates the accumulation of lactic acid in muscles through anaerobic respiration. The presence of lactic acid in the blood is the main cause of muscular pain and fatigue.

Question 35.
Describe the structure of human lungs with the help of a diagram.

Answer:

1. A pair of lungs is present in the chest cavity one on either side of the heart.
2. Lungs are spongy and elastic. They are enclosed by two membranes called pleura.
3. Space between the two membranes of pleura is filled with fluid. Pleura protects the lungs from injury.
4. Right lung is larger than the left lung.
5. Right lung is made of three lobes while the left lung has only two lobes.
6. Lung has several thousands of alveoli which are supplied with blood capillaries.
7. Pulmonary artery brings deoxygenated blood from heart to lungs.
8. After entering the lung, this artery divides into several arterioles and capillaries and supplies deoxygenated blood to alveoli.
9. Gas exchange occurs in the alveoli.
10. Oxygenated blood is carried from the lung to heart by the pulmonary vein.

Project work
Question 1.
Observe and analyse the questions in the table given below.

A) In which age group rate of heart beat is more?
B) In which age group rate of heart beat is less?
C) Why heart beat in Athletics is less?
D) What are reasons for more rate of heart beats differences between the newly born and children?
Answer:
A) In newly borned babies which are in 0 – 3 months of age group rate of heart beat is more i.e., 100 to 150 times.
B) In athletics the rate of heart beat is less i.e., 40 – 60 times / minute.
C) The heart of athlete pump more blood per beat due to increased cardio-vascular fitness in the structure of the heart. The muscles in the heart wall thicken and the heart pumps more blood with each beat.
D)

1. Mothers who have special medical conditions such as thyroid diseases or diabetes may give birth to new borns who are temporarily tachscardic from altered hormone and glucose levels. Tachycardia is a medical term for a very rapid heart beat.
2. Some infants are born with accessory electrical tissue in the heart causes epi¬sodes of rapid heart rate.
3. In wolf – parkinson syndrome – white syndrome there are extra cells and an ac-cessory path way, causing additional heart beats.

Question 2.
Observe the table given below and analyse the questions.

A) Why heart beat is less in animals with more weight?
B) Why heart beat is more in animals with less weight?
C) What is the relationship between weight of the body and rate of heart beat?
D) Why the weight of heart is less than body weight?
Answer:
A) The animals with more weight usually have weighted hearts. In one heart beat the large-sized hearts sends high amounts of blood to circulatory system. It takes time for the fulfilment of heart. Hence heart beat is less in animals with more body weight.
B) Usually the heart is very small in less weight animals. When the animal shrinks or contracts , its heart actually decrease the volume of blood proportionately. It can compensate for the reduced volume by increasing the rate at which it can supply blood to all body parts.
C) As the weight of the body of the animal increases the rate of heart beat per minute decreases. And also as the weight of the body decrease the rate of heart beat increases.
D) Usually the body of an organism is made by number of organs which makes the body functional. As all the body parts constitute the whole organism, the heart one of the organ is usually has less weight than body weight of an animal.

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 5th Lesson Refraction of Light at Plane Surfaces

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Textbook Questions and Answers

Improve Your Learning

Question 1.
Why is it difficult to shoot a fish swimming in water? (AS1)
(OR)
If the fish is swimming in water, why it is difficult to shoot?
(OR)
A shooter finds it difficult to shoot a fish swimming in water. Why?
Answer:
Due to refraction of light, it is difficult to shoot a fish swimming in water.

Reason :
The light rays coming from the fish towards shooter, bend at water-air interface. So, shooter sees only image of the fish, but not actual fish.

Question 2.
The speed of light in a diamond is 1,24,000 km/s. Find the refractive index of diamond if the speed of light in air is 3,00,000 km/s. (AS1)
Answer:
Speed of light in diamond = 1,24,000 km/s
Speed of light in vacuum = 3,00,000 km/s

Question 3.
Refractive index of glass relative to water is 9/8. What is the refractive index of water relative to glass? (AS1)
Answer:
Refractive index of glass relative to water = \(\frac{n_{g}}{n_{w}}=\frac{9}{8}\)
∴ Refractive index of water relative to glass = \(\frac{\mathrm{n}_{\mathrm{w}}}{\mathrm{n}_{\mathrm{g}}}=\frac{8}{9} \cdot\left[\because \mathrm{n}_{12}=\frac{1}{\mathrm{n}_{21}}\right]\)

Question 4.
The absolute refractive index of water is 4/3. What is the critical angle ? (AS1)
Answer:
Absolute refractive index of water = 4/3

Critical angle of water = C = 48°5′ = 48.5°.

Question 5.
Determine the refractive index of benzene if the critical angle is 42°. (AS1)
Answer:
Critical angle of benzene = 42°.

Question 6.
Explain the formation of mirage. (AS1)
(OR)
How is the mirage formed? Explain.
(OR)
A person walking on a road observed some water being present on the road but when he went there actually he did not find water. What is that actually formed called? Explain that process.
(OR)
Sometimes during the hot summer at noon time on tar roads, it appears that there is water on the road, but there would really be no water. What do you call this phenomenon? Explain why it happens.
(OR)
Why do you see a mirage the road on a hot summer day?
Answer:

• During hot summer day, air just above the road surface is very hot and the air at higher altitudes is cool.
• We know that refractive index of air increases with density.
• So, the cooler air at the top has greater refractive index than hotter air just above the road.

• Light travels faster through the thinner hot air than the denser cool air above it.
• On hot days, the temperature decreases with height.
• Thus the refractive index of air increases with height.
• When the light from a tall object such as tree or from the sky passes, through a medium just above the road whose refractive index decreases towards ground, suffers refraction and takes a curved path because of total internal reflection.
• This refracted light reaches the observer in a direction shown as in second figure.
• This appears to the observer that the ray is reflected from ground.
• Hence we will see water on road, which is the virtual image of sky and an inverted image of tree on the road.
• Such virtual images of distant high objects cause the optical illusion called ‘mirage’.

Question 7.
How do you verify experimentally that \(\frac{\sin i}{\sin r}\) is a constant? (AS1)
(OR)
Explain the experiment that shows the relation between angle of incidence and angle of refraction through figure.
(OR)
Write an experiment to obtain the relation between angle of incidence and angle of refraction.
Answer:
Aim:
Identifying relation between angle of incidence and angle of refraction.

Materials required :
A plank, white chart, protractor, scale, small black painted plank, a semi-circular glass disc of thickness nearly 2 cm pencil and laser light.

Procedure :

1. Take a wooden plank which is covered with white chart.
2. Draw two perpendicular lines, passing through the middle of the paper as shown in the figure (a).
3. Let the intersecting point be O.
4. Mark one line as NN which is normal to the another line marked as MM.
5. Here MM represents the line drawn along the interface of two media and NN represents the normal drawn to this line at ‘O’.
6. Take a protractor and place it along NN in such a way that its centre coincides with ‘O’ as shown in figure (b).
7. Then mark the angles from 0° to 90° on both sides of the line NN.
8. Repeat the same on the other side of the line NN.
9. The angles should be represented on circular line.
10. Now place semi circular glass disc so that its diameter coincides with the interface line (MM) and its centre coincides with the point O.
11. Take the laser light and send it along NN in such a way that the laser propagates from air to glass through the interface at point O and observe the way of laser light coming from other side of disc.
12. There is no deviation.
13. Send Laser light along a line which makes 15° (angle of incidence) with NN and see that it must pass through point O.
14. Measure its corresponding angle of refraction.
15. Repeat the experiment with angle of incidences of 20°, 30°, 40°, 50° and 60° and note the corresponding angles of refraction.

Observation :

• Find sin i, sin r for every i and r note down the values in table.

• Evaluate \(\frac{\sin i}{\sin r}\) for every incident angle i.
• We get \(\frac{\sin i}{\sin r}\) as constant.
• That is the relationship between angle of incidence and angle of refraction.
• The ratio of sin i and sin r is called refractive index.

Question 8.
Explain the phenomenon of total internal reflection with one or two activities. (AS1)
Answer:
Procedure :

1. Place the semi-circular glass disc in such a way that its diameter coincides with interlace line MM and its center coincides with point O’.
2. Now send light from the curved side of the semicircular glass disc.
3. The light travels from denser medium to rarer medium.
4. Start with angle of incidence (i), equals to 0° and observe for refracted on other side of the disc.
5. It does not deviate into its path when entering rarer medium.
6. Send laser light along angles of incidence 5°, 10°, 15°, etc. and measure the angle of refraction.
7. And tabulate the results in the given table.

Observation :

• Make a table shown below and note the values ‘i’ and ‘r’.

• At particular angle of incidence, the refracted ray does not come out but grazes the interface separating the air and glass. This angle is called critical angle.
• When the angle of incidence is greater than critical angle, the light ray gets reflected into denser medium at the interface, i.e. light never enters rarer medium. This phenomenon is called total internal reflection.

Question 9.
How do you verify experimentally that the angle of refraction is more than angle of incidence when light rays travel from denser to rarer medium? (AS1)
(OR)
When the light rays travel from denser to rarer medium, how can you explain, the angle of refraction is more than angle of incidence experimentally?
Answer:
Procedure :

• Take a metal disc. Use a protractor and mark angles along its edge as shown in the figure.
• Arrange two straws at the centre of the disc, in such a way that they can be rotated freely about the centre of the disc.
• Adjust one of the straws to make an angle 10°.
• Immerse half of the disc vertically into the water, filled in a transparent vessel. While dipping, verify that the straw at 10° must be inside the water.
• From the top of the vessel, try to view the straw which is inside the water as shown in the figure.
• Then adjust the other straw which is outside of the water until both straws look like they are in a single straight line.
• Then take the disc out of the water and observe the two straws on it. You will find that they are not in a single straight line.
• Measure the angle between the normal and second straw. Note the values in the following table.

• Do the same for various angles and find the corresponding angles of refraction and note them in the table.

Observation :
We will find the angle of refraction is more than angle of incidence.
i. e., r > i.

Conclusion :
When light travels from denser (water) to rarer (air) it bends away from the normal.

Question 10.
Take a bright metal ball and make it black with soot in a candle flame. Immerse it in water. How does it appear and why? (Make hypothesis and do the above experiment) (AS2)
Answer:

• The black metallic .ball appears to be raised up in the water because the path of the ray changes its direction at the interface, separating the two media, i.e., water and air.
• This path is chosen by light ray so as to minimize time of travel between ball and eye.
• This can be possible only when the speed of light changes at interface of two media.
• In another way the speed of light is different in different media.

Hypothesis :
Speed of light changes when it travels from one medium to another medium.

Question 11.
Take a glass vessel and pour some glycerine into it and then pour water up to the brim. Take a quartz glass rod. Keep it in the vessel. Observe the glass rod from the sides of the glass vessel.
1) What changes do you notice?
2) What could be the reasons for these changes? (AS2)
Answer:

1. We cannot see the glass rod in glycerine but we can see the rod in water.
2. We can also observe an apparent image of glass rod in water.
3. Reasons:
   i) Glycerine has essentially same refractive index as glass.
   ii) So, any light passing through these is bent equally.
   iii) Since both are transparent, it is not possible for our eye to distinguish the boundary by a change in the angle of reflection, and the glass seems to vanish.
   iv) But, the refractive index of glass and water are different.
   v) So the glass rod is visible to our eye in water. .

Question 12.
Do Activity-7 again. How can you find critical angle of water? Explain your steps briefly. (AS3)
Answer:

Procedure:

1. Take a cylindrical transparent vessel.
2. Place a coin at the bottom of the vessel.
3. Now pour water until you get the image of the coin on the water surface.
4. This is due to total internal reflection.

Critical angle of water :

1. Refractive index of water = 1.33
2. The sine of critical angle of water = \(\frac{1}{\text { Refractive index }}\)
3. Sin C = \(\frac{1}{\text { 1.33 }}\) ⇒ sin C = 0.7518.
   ∴ C = 8.7°
4. ∴ The critical angle of water = 48.7°.

Question 13.
Collect the values of refractive index of the following media. (AS4)

Answer:

Question 14.
Collect information on working of optical fibres. Prepare a report about various uses of optical fibres in our daily life. (AS4)
(OR)
What do you know about the working of optical fibres and make a report of various uses of optical fibres in our daily life?
(OR)
How are the optical fibres working? What are the various uses of optical fibres in our daily life?
Answer:
1) An optical fibre is very thin fibre made of glass or plastic having radius about a micrometer (10-6 m).
2) A bunch of such thin fibres form a light pipe.

Working :
1. Optical fibre having three parts-namely core (n = 1, 7), clading (n = 1, 6) and shielding.
2. The ray of light AB gets refracted at point ‘B’ into core and incident at ‘C’ with angle of incidence i (i > c).

3. The angle of incidence is greater than the critical angle and hence total internal reflection takes place.
4. The light is thus transmitted along the fibre.
5. The optical fibre is also based on ‘Fermat’s principle.

Uses :

1. Optical fibres are used in ‘endoscopy’ to see the internal organs like throat, stomach, intestines, etc.
2. Optical fibres are used in transmitting communication signals through light pipes.
3. Optical fibres are used in international telephone cables laid under the sea, in large computer networks, etc.
4. In optical fibre about 2000 telephone signals appropriately mixed with light waves may be simultaneously transmitted through a typical optical fibre.

Question 15.
Take a thin thermocol sheet. Cut it in circular discs of different radii like 2 cm, 3 cm, 4 cm, 4.5 cm, 5 cm etc. and mark centers with sketch pen. Now take needles of length nearly 6 cm. Pin a needle to each disc at its centre vertically. Take water in a large opaque tray and place the disc with 2 cm radius in such a way that the needle is inside the water as shown in figure. Now try to view the free end (head) of the needle from surface of the water.

1) Are you able to see the head of the needle?
Now do the same with other discs of different radii. Try to see the head of the needle, each time.
Note : The position of your eye and the position of the disc on water surface should not be changed while repeating the Activity with’other discs.
2) At what maximum radius of disc, were you not able to see the free end of the needle ?
3) Why were you not able to view the head of the nail for certain radii of the discs ?
4) Does this Activity help you to find the critical angle of the medium (water) ?
5) Draw a diagram to show the passage of light ray from the head of the nail in different situations. (AS4)
Answer:

1. Yes, we can see head of the needle.

2. Height of the pin = 6 cm
Radius of the disc = x cm
Critical angle of water = C = 48.7°
Tan C = 48.7°
\(\frac{x}{6}\) = 1.138 ⇒ x = 6.828 cm
So, at radius of 6.8 cm we cannot see the free end of the needle.

3. Because the light rays coming from object undergoing total internal reflection by touching the surface of disc.

4. Yes, we can find critical angle.
Refractive index of air (n₂) = 1.003 ; Refractive index of water (n₁) = 1.33
Sin C = \(\frac{n_{2}}{n_{1}}=\frac{1.003}{1.33}\) = 0.7541 ⇒ C = 48.7°

5.

Question 16.
Explain the refraction of light through the glass slab with a neat ray diagram. (AS5)
(OR)
Draw a glass slab diagram and explain the refraction of light through glass slab.
(OR)
Write the procedure of a lab Activity to understand lateral shift of light rays through a glass slab.
(OR)
How can you find lateral shift using glass slab?
Answer:
Aim :
A) Determination of position and nature of image formed by a glass slab.
B) Understanding lateral and vertical shift.
C) Determination of refractive index of given glass slab.

Materials required :
Plank, chart paper, clamps, scale, pencil, thin glass slab and pins.

Procedure :

1. Place a piece of chart on a plank. Clamp it. Place a glass slab in the middle of the paper.
2. Draw border line along the edges of the slab by using a pencil. Remove it. You will get a figure of a rectangle.
3. Name the vertices of the rectangle as A, B, C and D.
4. Draw a perpendicular at a point on the longer sides (AB) of the rectangle.
5. Now draw a line, from the point of intersection where side AB of rectangle and perpendicular meet, in such a way that it makes 30° angle with the normal.
6. This line represents the incident ray falling on the slab and the angle it makes with normal represents angle of incidence.
7. Now place the slab on the paper in such a way that it fits in the rectangle drawn. Fix two identical pins on the line making 30° angle with normal, such that they stand vertically with same height.
8. By looking at the two pins from the other side of the slab, fix two pins in such a way that all pins appear to be along a straight line.
9. Remove the slab and take out pins. Draw a straight line by joining the dots formed by the pins up to the edge CD of the rectangle.
10. This line represents emergent ray of the light.
11. Draw a perpendicular to the line CD where our last line drawn meets the line CD.
12. Measure the angle between emergent ray and normal.
13. This is called angle of emergence.
14. The angle of incidence and angle of emergence are equal.
15. Incident emergent rays are parallel.
16. The distance between the parallel rays is called shift.

Question 17.
Place an object on the table. Look at the object through the transparent glass slab. You will observe that it will appear closer to you. Draw a ray diagram to show the passage of light ray in this situation. (AS5)
Answer:

Question 18.
What is the reason behind the shining of diamond and how do you appreciate it? (AS6)
(OR)
For which reason is the diamond shining and how is it appreciable?
Answer:

• The critical angle of diamonds is very low, i.e., 24.4°.
• So if a light ray enters diamond, it undergoes total internal reflection.
• It makes the diamond shine brilliant.
• So total internal reflection is main cause of brilliance of diamonds.
• Majority of people are attracted towards diamonds due to this property.
• So we have to thoroughly appreciate total internal reflection for brilliance of diamonds.

Question 19.
How do you appreciate the role of Fermat’s principle in drawing ray diagrams? (AS6)
(OR)
How do you admire the role of Fermat’s principle in drawing ray diagrams? Fermat’s principle: The light ray always travels in a path which needs shortest possible time to cover distance between two points.
This principle has lot of importance on optics. This is used in

1. Laws of reflection (i.e., angle of incidence = angle of reflection)
2. Laws of refraction (Snell’s law)
3. To derive refractive index of a medium.
4. To derive refractive index of glass slab.
   So, I appreciate the Fermat’s principle.

Question 20.
A light ray is incident on air-liquid interface at 45° and is refracted at 30°. What is the refractive index of the liquid? For what angle of incidence will the angle between reflected ray and refracted ray be 90°? (AS7)
Answer:
i) Given that angle of incidence (i) = 4S°
angle of refraction (r) = 30°

ii) Given that angle between reflected and refracted ray is 90°.
We know angle of incidence = angle of reflection
∴ Angle of refraction (r) = 90 – angle of incidence
= 90 – i

Critical angle = 54.7°. This angle is also known as polarising angle.

Question 21.
Explain why a test tube immersed at a certain angle in a tumbler of water appears to have a mirror surface from a certain viewing position. (AS7)
Answer:
When a test tube is immersed at a certain angle in a tumbler of water appears to have
a mirror surface from a certain viewing positions due to total internal reflection.

Explanation :

• The critical angle for glass is 42°.
• The glass and air in test tube works as denser and rarer mediums.
• The rays of light while travelling through water strike glass – air interface of test tube at an angle of more than 42° (i > c) they get totally internal reflected as shown figure.
• When these reflected rays reach the eye, they appear to come from the surface of test tube itself.
• Now the test tube appears like silvary.

Question 23.
In what cases does a light ray not deviate at the interface of two media? (AS7)
Answer:

1. When a light ray incident is perpendicular to the interface of surface, it does not undergo deviation.
2. When a light ray incident is more than critical angle, it does not undergo deviation (refraction) but it undergoes reflection to come back into the original medium.

Question 25.
When we sit at camp fire, objects beyond the fire seen swaying. Give the reason for it. (AS7)
(OR)
What are the reasons for the objects beyond the fire seen swaying, when we sit at camp fire?
Answer:

• The temperature of the surrounding air changes due to convection of heat by the camp fire.
• This leads to chang in density and refractive index of air, continuously.
• The continuous change in refractive index of air changes the refracted path of the light ray.
• This is the cause for swaying of an object.

Question 26.
Why do stars appear twinkling? (AS7)
(OR)
What is the reason for the appearance of stars like twinkling?
Answer:

• The twinkling of a star is due to atmospheric refraction of star light.
• The atmosphere consists of a number of layers of varying densities.
• When light rays coming from a star pass through this layers and undergo refraction for several times.
• Thats why stars appear twinkling.

Question 27.
Why does a diamond shine more than a glass piece cut to the same shape? (AS7)
(OR)
What is the reason for shining of diamond brightly as compared to glass piece cut?
Answer:

• The critical angle of a diamond is very low (i.e., 24.4°).
• So if a light ray enters a diamond it definitely undergoes total internal reflection.
• Whereas it is not possible with glass piece cut to the same shape.
• So diamond shines more than a glass piece.

Fill In The Blanks

1. At critical angle of incidence, the angle of refraction is ……………… .
2. n₁ sin i = n₂ sin r, is called ……………… .
3. Speed of light in vacuum is ……………… .
4. Total internal reflection takes place when a light ray propagates from …………. to …………… medium.
5. The refractive index of a transparent material is 3/2. The speed of the light in that medium is …………… .
6. Mirage is an example of ……………… .
Answer:

1. 90°
2. Snell’s law
3. 3 × 10⁸ m/s
4. denser, rarer
5. 2 × 10⁸ m/s
6. optical illusion / total internal reflection

Multiple Choice Questions

1. Which of the following is Snell’s law?

Answer:
B)

2. The refractive index of glass with respect to air is 2. Then the critical angle of glass air interface is ………………….
A) 0°
B) 45°
C) 30°
D) 60°
Answer:
C) 30°

3. Total internal reflection takes place when the light ray travels from …………….. .
A) rarer to denser medium
B) rarer to rarer medium
C) denser to rarer medium
D) denser to denser medium
Answer:
C) denser to rarer medium

4. The angle of deviation produced by the glass slab is …………… .
A) 0°
B) 20°
C) 90°
D) depends on the angle formed by the light ray and normal to the slab
Answer:
D) depends on the angle formed by the light ray and normal to the slab

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Additional Questions and Answers

Question 1.
Derive Snell’s law.
(OR)
Prove n₁ sin i = n₂ sin r.
(OR)
Derive the Snell’s formula from Fermat’s principle.
(OR)
Derive the formula in realtion with and of incidence and angle of refraction.
Answer:
Let X be the path and A be the point above X and B be the point below X.
Now, we have to find the way from A to B.

1. Let us try to calculate how long it would take to go from A to B by the two paths through point D and another through point C.
2. If we draw a perpendicular DE, between two paths at D, we see that the path on line is shortened by the amount EC.
3. On the other hand, in the water, by drawing corresponding perpendicular CF we find that we have to go to the extra distance DF in water. These times must be equal since we assumed there was no change in time between two paths.
4. Let the time taken by the man to travel from E to C and D to F be ∆t and v₁ and v₂ be the speeds of the running and swimming. From figure we get,
   EC = v₁ ∆t and DF = v₂ ∆t
   ⇒ \(\frac{\mathrm{EC}}{\mathrm{DF}}=\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}\) ………….. (1)
5. Let i and r be the angles measured between the path ACB and normal NN, perpendicular to shore line X.

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces InText Questions and Answers

10th Class Physics Textbook Page No. 56

Question 1.
Why should you see a mirage as a flowing water?
Answer:

• A mirage is a naturally occuring optical phenomenon, in which light rays are bent to produce a displaced image of distant objects or the sky.
• As light passes from colder air (higher place) to warmer air (lower place), the light ray bends away from the direction of the temperature gradient.
• Once the rays reach the viewer’s eye, the visual cortex interprets it as if it traces back along a perfectly straight “line of sight”. However this line is at a tangent to the path the ray takes at the point it reaches the eye.
• The result is that an “inferior image ” of the sky above appears on the ground.
• The viewer may incorrectly interprets this sight as water that is reflecting the sky, which is to the brain, a more reasonable and common occurrence.

Question 2.
Can you take a photo of a mirage?
Answer:

• Yes, I can take a photo of a mirage.
• Our eye can catches the total internal reflected rays.
• So, camera lens also catches the same.

10th Class Physics Textbook Page No. 46

Question 3.
What difference do you notice in fig 2(a) and Fig 2(b) with the respect to refracted rays?
(OR)
Draw the ray diagram of refraction in between denser and rarer medium.

Answer:
In figure 2(a) the light ray bends towards normal whereas in 2(b) the light ray bends away from the normal.

Question 4.
Is there any relation between behaviour of refracted rays and speed of the light?
Answer:
Yes. The speed of light changes when it travels from one medium to another medium. So the light may bend towards normal or away from normal.

10th Class Physics Textbook Page No. 47

Question 5.
Why db different material media possess different values of refractive indices?
Answer:
Refractive index depends on nature of material. So different media have different values of refractive indices.

10th Class Physics Textbook Page No. 48

Question 6.
On what factors does the refractive index of a medium depend?
Answer:
Refractive index depends on (1) Nature of material and (2) Wavelength of light used.

10th Class Physics Textbook Page No. 49

Question 7.
Can we derive the relation between the angle of incidence and the angles of refraction theoretically?
Answer:
Yes, we can derive the relation between angle of incidence and angles of refraction theoretically. We can get nt sin i = n2 sin r.

10th Class Physics Textbook Page No. 53

Question 8.
Is there any chance that angle of refraction is equal to 90° ? When does this happen?
Answer:
Yes, when angle of incidence is equal to critical angle then angle of refraction is equal to 90°.

10th Class Physics Textbook Page No. 54

Question 9.
What happens to light when the angle of incidence is greater than critical angle?
Answer:
When the angle of incidence is greater than critical angle, the light ray gets reflected into denser medium at the interface, i.e., light never enters rarer medium. This phenomenon is called total internal reflection.

10th Class Physics Textbook Page No. 57

Question 10.
How does light behave when a glass slab is introduced in its path?
Answer:
The light ray undergoes refraction two times.

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Activities

Activity – 1

Question 1.
Procedure :
Take some water in a glass tumbler. Keep a pencil in it. See the pencil from one side of glass and also from the top of the glass.
Observation:

1. How does it look?
Answer:
From the side it appears to be bent. From the top it appears as it is raising.

2. Do you find any difference between two views?
Answer:
Yes, the position of pencil is different.

Activity – 2

Question 2.
Procedure :

1. Go to a long wall (of length of 30 feet) facing the Sun. Go to the one end of a wall and ask someone to bring a bright metal object near the other end of the wall.
2. When the object is few inches from the wall, it will distort and we will see a reflected image on the wall as though the wall were a mirror.

Observation:

Why is there an image of the object on the wall?
Answer:
The image is due to refraction of light.

Activity-3 Refraction

Question 3.
Procedure: –

1. Take a shallow vessel with opaque walls such as a mug, a tin or a pan.
2. Place a coin at the bottom of the vessel.
3. Move away from the vessel until we cannot see the coin (fig. 2). Ask someone to fill the vessel with water. When the vessel is filled with water the coin comes back into view (fig. 3).

1. Why are you able to see the coin when the vessel is filled with water?
Answer:
The ray of light originated from the coin does not reach your eye when the vessel is empty. Hence you are not able to see the coin. But the coin becomes visible after the vessel is filled with water.

2. How is it possible? Do you think that the ray reaches your eye when the vessel is filled with water?
Answer:
Yes, it reaches the second instance.

3. What happens to the light ray at interface between water and air?
Answer:
It bends towards the normal.

4. What could be the reason for this bending of the light ray in the second instance?
Answer:
It is based on Fermat’s principle, which states that the light ray always travels in a path which needs shortest possible time to cover the distance between the two points.

Activity – 4

Question 4.
Prove that when light ray travels from denser to rarer medium it bends away from the normal.
Answer:


Procedure :

1. Take a metal disc. Use protractor and mark angles along its edge as shown in the figure.
2. Arrange two straws from the centre of the disk.
3. Adjust one of the straws to the angle 10°.
4. Immerse half of the disc vertically into the water, filled in a transparent vessel.
5. Inside the water the angle of straw should be at 10°.
6. From the top of the vessel try to view the straw which is inside the water.
7. Then adjust the other straw which is outside the water until both straws are in a single straight line.
8. Then take the disc out of the water and observe the two straws on it.
9. We will find that they are not in a single straight line.
10. It could be seen from the side view while half of the disc is inside the water.
11. Measure the angle between the normal and second straw. Draw table for various angles and corresponding angles of refraction.

Observation :
We observe that ‘r’ is greater than ‘i’ in all cases and when light travels from
denser to rarer medium it bends away from the normal.

Activity – 6

Question 5.
Why can we not see a coin placed in water from the side of glass?
Answer:
Procedure :

1. Take a transparent glass tumbler and coin.
2. Place a coin on a table and place glass on the coin.
3. Observe the coin from the side of the glass. We can see the coin.
4. Now fill the glass with water and observe the coin from the side of the glass tumber.
5. Now we cannot see the coin because the coin rises up due to refraction.

Activity – 7

Question 7.
Why can we see the coin in water from top? What is the phenomenon behind that?
Answer:

Procedure :

1. Take a cylindrical transparent vessel. Place a coin at the bottom of the vessel.
2. Now pour water until we will get the image of the coin on the water surface.
3. This is due to total internal reflection.

Activity – 8

Question 8.
Write an Activity to find refractive index of glass slab by calculating vertical shift.
(OR)
Explain the experiment with glass slab in determination of refraction through vertical shift.
Answer:

Procedure :

1. Take a glass slab and measure the thickness of the slab.
2. Take a white chart and fix it on the table.
3. Place the slab in the middle of the chart.
4. Draw line around it.
5. Remove the slab from its place.
6. The lines form a rectangle. Name the vertices of it as A. B, C and D. ‘
7. Draw a perpendicular to the longer line AB of the rectangle at any point on it.
8. Place slab again in the rectangle ABCD.
9. Take a needle. Place at a point P in such a way that its length is parallel to the AB on the perpendicular line at a distance of 20 cm from the slab.
10. Now take another needle and by seeing at the first needle from the other side of the slab, try to keep the needle so that it forms a straight line with the first needle. 1 1)
11. Remove the slab and observe the positions of the needles.
12. They are not in same line.
13. Draw a perpendicular line from the second needle to the line on which the first needle is placed.
14. Take the intersection point as Q.
15. The distance between P and Q is vertical shift.
16. We will get the same vertical shift placing needle at different distances.
17. 

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.1

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.1 Textbook Questions and Answers

Question 1.
By comparing the ratios \(\frac{a_{1}}{a_{2}}\), \(\frac{b_{1}}{b_{2}}\), \(\frac{c_{1}}{c_{2}}\) K find out whether the lines represented by the following pairs of linear equations intersect at a point, are parallel or are coincident.
a) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
Answer:
Given: 5x – 4y + 8 = 0
7x + 6y – 9 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{5}{7}\); \(\frac{b_{1}}{b_{2}}\) = \(\frac{-4}{6}\); \(\frac{c_{1}}{c_{2}}\) = \(\frac{8}{-9}\)
∴ \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\)
Hence the given pair of linear equations represents a pair of intersecting lines.

b) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
Answer:
Given : 9x + 3y + 12 = 0
18x + 6y + 24= 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{9}{18}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{12}{24}\) = \(\frac{1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
The lines are coincident.

c) 6x – 3y + 10 = 0
2x – y + 9 = 0
Answer:
Given: 6x – 3y + 10 = 0
2x – y + 9 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{6}{2}\) = \(\frac{3}{1}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{-1}\) = \(\frac{3}{1}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{10}{9}\)
Here \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
∴ The lines are parallel.

Question 2.
Check whether the following equations are consistent or inconsistent. Solve them graphically. (AS2, AS5)
a) 3x + 2y = 8
2x – 3y = 1
Answer:
Given equaions are 3x + 2y = 8 and 2x – 3y = 1
\(\frac{a_{1}}{a_{2}}\) = \(\frac{3}{2}\);
\(\frac{b_{2}}{b_{-3}}\) = \(\frac{-4}{6}\);
\(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\)
Hence the linear equations are consistent.


The lines intersect at (2, 1), so the solution is (2, 1).

b) 2x – 3y = 8
4x – 6y = 9
Answer:
Given: 2x – 3y = 8 and 4x – 6y = 9
\(\frac{a_{1}}{a_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{-6}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{8}{9}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
Lines are inconsistent and have no solution.

Lines are parallel.

The lines are parallel and no solution exists.

c) \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7
9x – 10y = 12
Answer:
Given pair of equations \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7 and 9x – 10y = 12
Now take \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7 ⇒ \(\frac{9x+10y}{6}\) = 7 ⇒ 9x + 10y = 42
and 9x – 10y =12
\(\frac{a_{1}}{a_{2}}\) = \(\frac{9}{9}\) = \(\frac{1}{1}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{10}{-10}\) = \(\frac{1}{-1}\) and
\(\frac{c_{1}}{c_{2}}\) = \(\frac{-42}{-12}\) = \(\frac{7}{2}\)
Since \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\) they are intersecting lines and hence consistent pair of linear equations.


Solution: The unique solution of given pair of equations is (3.1, 1.4)

d) 5x – 3y = 11
-10x + 6y = -22
Answer:
Given pair of equations 5x – 3y = 11 and -10x + 6y = -22
\(\frac{a_{1}}{a_{2}}\) = \(\frac{5}{-10}\) = \(\frac{-1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{6}\) = \(\frac{-1}{2}\) and
\(\frac{c_{1}}{c_{2}}\) = \(\frac{11}{-22}\) = \(\frac{-1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
∴ The lines are consistent.
∴ The given linear equations represent coincident lines.
Thus they have infinitely many solutions.

e) \(\frac{4}{3}\)x + 2y = 8
2x + 3y = 12
Answer:
Given pair of equations \(\frac{4}{3}\)x + 2y = 8 ⇒ \(\frac{4x+6y}{3}\) = 8 ⇒ 4x + 6y = 24 ⇒ 2x + 3y = 12
\(\frac{a_{1}}{a_{2}}\) = \(\frac{4}{2}\) = 2;
\(\frac{b_{1}}{b_{2}}\) = \(\frac{6}{3}\) = 2;
\(\frac{c_{1}}{c_{2}}\) = \(\frac{24}{12}\) = 2
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
Thus the equations are consistent.
∴ The given equations have infinitely many solutions.

f) x + y = 5
2x + 2y = 10
Answer:
Given pair of equations x + y = 5 and 2x + 2y = 10
\(\frac{a_{1}}{a_{2}}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
Thus the equations are consistent and have infinitely many solutions.

g) x – y = 8
3x – 3y = 16
Answer:
Given pair of equations x – y = 8 and 3x – 3y = 16
\(\frac{a_{1}}{a_{2}}\) = \(\frac{1}{3}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-1}{-3}\) = \(\frac{1}{3}\) and
\(\frac{c_{1}}{c_{2}}\) = \(\frac{8}{16}\) = \(\frac{1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
Thus the equations are inconsistent.
∴ They represent parallel lines and have no solution.

h) 2x + y – 6 = 0 and 4x – 2y – 4 = 0
Answer:
Given pair of equations 2x + y – 6 = 0 and 4x – 2y – 4 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{1}{-2}\) = \(\frac{-1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{-6}{-4}\) = \(\frac{3}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\)
The equations are consistent.
∴ They intersect at one point giving only one solution.


The solution is x = 2 and y = 2

i) 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
Answer:
Given pair of equations 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
\(\frac{a_{1}}{a_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{-2}{-4}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{-2}{-5}\) = \(\frac{2}{5}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
Thus the equations are inconsistent.
∴ They represent parallel lines and have no solution.

Question 3.
Neha went to a ‘sale’ to purchase some pants and skirts. When her friend asked her how many of each she had bought, she answered “The number of skirts are two less than twice the number of pants purchased. Also the number of skirts is four less than four times the number of pants purchased.”
Help her friend to find how many pants and skirts Neha bought.
Answer:
Let the number of pants = x and the number of skirts = y
By problem y = 2x – 2 ⇒ 2x – y = 2
y = 4x – 4 ⇒ 4x – y = 4


The two lines are intersecting at the point (1,0)
∴ x = 1; y = 0 is the required solution of the pair of linear equations.
i.e., pants =1
She did not buy any skirt.

Question 4.
10 students of Class-X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys then, find the number of boys and the number of girls who took part in the quiz.
Answer:
Let the number of boys be x.
Then the number of girls = x + 4
By problem, x + x + 4 = 10
∴ 2x + 4 = 10
2x = 10-4
x = \(\frac{6}{2}\) = 3
∴ Boys = 3 Girls = 3 + 4 = 7 (or)
Boys = x, Girls = y
By problem x + y = 10 (total)
and y = x + 4 (girls)
⇒ x + y = 10 and x – y = – 4


∴ Number of boys = 3 and the number of girls = 7

Question 5.
5 pencils and 7 pens together cost Rs. 50 whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Answer:
Let the cost of each pencil be Rs. x
and the cost of each pen be Rs. y.
By problem 5x + 7y = 50
7x + 5y = 46


The lines are intersecting at the point (3, 5).
x = 3 and y = 5 is the solution of given equations.
∴ Cost of one pencil = Rs. 3 and pen = Rs. 5

Question 6.
Half the perimeter of a rectangular garden, whose length is 4 m more than its width is 36 m. Find the dimensions of the garden.
Answer:
Let the width of the garden = x cm
then its length = x + 4 cm
Half the perimeter = \(\frac{1}{2}\) × 2(7+ b) = l + b
By problem, x + x + 4 = 36
2x + 4 = 36
2x = 36 – 4 = 32
∴ x = 16 and x + 4 = 16 + 4 = 20
i.e., length = 20 cm and breadth = 16 cm.
(or)
Let the breadth be x and length = y
then x + y = 36 ⇒ x + y = 36
y = x + 4 ⇒ x – y = -4


The two lines intersect at the point (16, 20)
i.e., length = 20 cm and the breadth = 16 cm.

Question 7.
We have a linear equation 2x + 3y – 8 = 0. Write another linear equation in two variables such that the geometrical representation of the pair so formed is intersect¬ing lines. Now, write two more linear equations so that one forms a pair of parallel lines and the second forms coincident line with the given equation.
Answer:
i) Given: 2x + 3y – 8 = 0
The lines are intersecting lines.
Let the other linear equation be ax + by + c = 0
∴ \(\frac{a_{1}}{a_{2}}\) ≠ \(\frac{b_{1}}{b_{2}}\); we have to choose appropriate values satisfying the condition above.
Thus the other equation may be 3x + 5y – 6 =0

ii) Parallel line \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) ≠ \(\frac{c_{1}}{c_{2}}\)
⇒ 2x + 3y – 8 = 0
4x + 6y – 10 = 0

iii) Coincident lines \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
⇒ 2x + 3y – 8 = 0 ⇒ 8x + 12y – 32 = 0

Question 8.
The area of a rectangle gets reduced by 80 sq. units if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area will increase by 50 sq. units. Find the length and breadth of the rectangle.
Answer:
Let the length of the rectangle = x units
breadth = y units Area = l . b = xy sq. units
By problem, (x – 5) (y + 2) = xy – 80 and          (x + 10) (y – 5) = xy + 50
⇒ xy + 2x – 5y – 10 = xy – 80 and                    xy – 5x + 10y – 50 = xy + 50
⇒ 2x – 5y = xy – 80 – xy + 10 and                   -5x + 10y = xy + 50 – xy + 50
⇒ 2x – 5y = – 70 and                                       -5x + 10y = 100


The two lines intersect at the point (40, 30)
∴ The solution is x = 40 and y = 30
i.e., length = 40 units; breadth = 30 units.

Question 9.
In X class, if three students sit on each bench, one student will be left. If four students sit on each bench, one bench will be left. Find the number of students and the number of benches in that class.
Answer:
Let the number of benches = x say and the number of students = y
By problem
y = 3x + 1 ⇒ 3x – y + 1 = 0
and y = 4(x – 1) ⇒ 4x – y – 4 = 0


The two lines intersect at (5, 16)
∴ The solution of the equation is x = 5 and y = 16
i.e., Number of benches = 5 and the number of students = 16

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.2

10th Class Maths 10th Lesson Mensuration Ex 10.2 Textbook Questions and Answers

Question 1.
A toy is in the form of a cone mounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm respectively. Determine the surface area of the toy. (Use π = 3.14)
Answer:

Diameter of the base of the cone d = 6 cm.
∴ Radius of the base of the cone
r = \(\frac{d}{2}\) = \(\frac{6}{2}\) = 3 cm
Height of the cone = h = 4 cm
Slant height of the cone l = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{3^{2}+4^{2}}\)
= \(\sqrt{9+16}\)
= √25
= 5 cm
∴ C.S.A of the cone = πrl
= \(\frac{22}{7}\) × 3 × 5
= \(\frac{330}{7}\) cm²
Radius of the hemisphere = \(\frac{d}{2}\) = \(\frac{6}{2}\) = 3 cm
C.S.A. of the hemisphere = 2πr²
= 2 × \(\frac{22}{7}\) × 3 × 3
= \(\frac{396}{7}\)
Hence the surface area of the toy = C.S.A. of cone + C.S.A. of hemisphere
= \(\frac{330}{7}\) + \(\frac{396}{7}\)
= \(\frac{726}{7}\) ≃ 103.71 cm².

Question 2.
A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. The radius of the common base is 8 cm and the heights of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the total surface area of the solid. [Use π = 3.14]
Answer:
Total surface area = C.S.A. of the cone + C.S.A. of cylinder + C.S.A of the hemisphere.

Cone:
Radius (r) = 8 cm
Height (h) = 6 cm
Slant height l = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{8^{2}+6^{2}}\)
= \(\sqrt{64+36}\)
= √100
= 10 cm
C.S.A. = πrl
= \(\frac{22}{7}\) × 8 × 10
= \(\frac{1760}{7}\) cm²
Cylinder:
Radius (r) = 8 cm;
Height (h) = 10 cm
C.S.A. = 2πrh
= 2 × \(\frac{22}{7}\) × 8 × 10
= \(\frac{3520}{7}\) cm²
Hemisphere:
Radius (r) = 8 cm
C.S.A. = 2πr²
= 2 × \(\frac{22}{7}\) × 8 × 8
= \(\frac{2816}{7}\) cm²
∴ Total surface area of the given solid
= \(\frac{1760}{7}\) + \(\frac{3520}{7}\) + \(\frac{2816}{7}\)
T.S.A. = \(\frac{8096}{7}\) = 1156.57 cm².

Question 3.
A medicine capsule is ih the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the capsule is 14 mm. and the width is 5 mm. Find its surface area.
Answer:

Surface area of the capsule = C.S.A. of 2 hemispheres + C.S.A. of the cylinder
i) Now for Hemisphere:
Radius (r) = \(\frac{d}{2}\) = \(\frac{5}{2}\) = 2.5 mm
C.S.A of each hemisphere = 2πr²
C.S.A of two hemispheres
= 2 × 2πr² = 4πr²
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\) × \(\frac{5}{2}\)
= \(\frac{550}{7}\)
= 78.57 mm².

ii) Now for Cylinder:
Length of capsule = AB =14 mm
Then height (length) cylinder part = 14 – 2(2.5)
h = 14 – 5 = 9 mm
Radius of cylinder part (r) = \(\frac{5}{2}\)
Now C.S.A of cylinder part = 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\) × 9
= \(\frac{900}{7}\)
= 141.428 mm²
Now total surface area of capsule
= 78.57 + 141.43 = 220 mm²

Question 4.
Two cubes each of volume 64 cm³ are joined end to end together. Find the surface area of the resulting cuboid.
Answer:
Given, volume of the cube.
V = a³ = 64 cm³
∴ a³ = 4 × 4 × 4 = 4³ , Hence a = 4 cm
When two cubes are added, the length of cuboid = 2a = 2 × 4 = 8 cm,
breadth = a = 4 cm.
height = a = 4 cm is formed.

∴ T.S.A. of the cuboid
= 2 (lb + bh + lh)
= 2(8 × 4 + 4 × 4 + 8 × 4)
= 2(32 + 16 + 32)
= 2 × 80
= 160 cm²
∴ The surface area of resulting cuboid is 160 cm².

Question 5.
A storage tank consists of a circular cylinder with a hemisphere stuck on either end. If the external diameter of the cylinder be 1.4 m. and its length be 8 m. Find the cost of painting it on the outside at rate of Rs. 20 per m².
Answer:
Total surface area of the tank = 2 × C.S.A. of hemisphere + C.S.A. of cylinder.

Hemisphere:
Radius (r) = \(\frac{d}{2}\) = \(\frac{1.4}{2}\) = 0.7 m
C.S.A. of hemisphere = 2πr²
= 2 × \(\frac{22}{7}\) × 0.7 × 0.7
= 3.08 m².
2 × C.S.A. = 2 × 3.08 m² = 6.16 m²
Cylinder:
Radius (r) = \(\frac{d}{2}\) = \(\frac{1.4}{2}\) = 0.7 m
Height (h) = 8 m
C.S.A. of the cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × 0.7 × 8
= 35.2 m²
∴ Total surface area of the storage tank = 35.2 + 6.16 = 41.36 m²
Cost of painting its surface area @ Rs. 20 per sq.m, is
= 41.36 × 20 = Rs. 827.2.

Question 6.
A hemisphere is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the length of the cube. Determine the surface area of the remaining solid.
Answer:
Let the length of the edge of the cube = a units

T.S.A. of the given solid = 5 × Area of each surface + Area of hemisphere
Square surface:
Side = a units
Area = a² sq. units
5 × square surface = 5a² sq. units
Hemisphere:
Diameter = a units;
Radius = \(\frac{a}{2}\)
C.S.A. = 2πr²
= 2π\(\left(\frac{a}{2}\right)^{2}\)
= 2π\(\frac{a^{2}}{4}\) = \(\frac{\pi \mathrm{a}^{2}}{2}\) sq. units
Total surface area = 5a² + \(\frac{\pi \mathrm{a}^{2}}{2}\) = a²\(\left(5+\frac{\pi}{2}\right)\) sq. units.

Question 7.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm and its base radius is of 3.5 cm, find the total surface area of the article.

Answer:
Surface area of the given solid = C.S.A. of the cylinder + 2 × C.S.A. of hemisphere.
If we take base = radius
Cylinder:
Radius (r) = 3.5 cm
Height (h) = 10 cm
C.S.A. = 2πrh
= 2 × \(\frac{22}{7}\) × 3.5 × 10
= 220 cm²
Hemisphere:
Radius (r) = 3.5 cm
C.S.A. = 2πr²
= 2 × \(\frac{22}{7}\) × 3.5 × 3.5
= 77 cm²
2 × C.S.A. = 2 × 77 = 154 cm²
∴ T.S.A. = 220 + 154 = 374 cm².

AP State Board Syllabus AP SSC 10th Class Social Studies Important Questions Chapter 19 Emerging Political Trends 1977 to 2000.

AP State Syllabus SSC 10th Class Social Studies Important Questions 19th Lesson Emerging Political Trends 1977 to 2000

10th Class Social 19th Lesson Emerging Political Trends 1977 to 2000 1 Mark Important Questions and Answers

Question 1.
Expand the term AIADMK.
Answer:
All India Anna Dravida Munnetra Kazagam.

Question 2.
Give any two examples for Regional Political parties.
Answer:
TDP, YSRCP, JANA SENA, TRS, AIADMK, DMK, etc.

Question 3.
Which welfare schemes initiated by N.T. Rama Rao are still continuing with some changes in Andhra Pradesh?
Answer:

1. Mid-day meal scheme in government schools.
2. Sale of rice at subsidy rates to the poor.

Question 4.
Identify at least any two states presently ruled by regional parties in India on the given Indian political map.
Answer:

Question 5.
What was the contribution of Telecom revolution?
Answer:
The contribution of Telecom Revolution:
A network of telephonic communication in the country using satellite technology increased.

Question 6.
Mention any two initiations of N.T. Rama Rao.
Answer:

1. Sale of rice at Rs. 2/- kg
2. Mid day meal scheme in government schools.
3. Liquor prohibition

Question 7.
Write about the 73rd amendment of the constitution.
Answer:
73rd amendment: The 73rd constitutional amendment created institutions of local self government at the village level and so Gram Panchayat, Mandal Parishad and Zilla Parishad are formed.

Observe the table given below and answer the questions 8 & 9.
Results of Telangana State Assembly and Parliament Elections – 2014

Question 8.
Name the two parties that secured more than 15 Assembly seats.
Answer:
Parties that secured more than 15 Assembly seats.

1. TRS
2. Congress Party
3. T.D.P

Question 9.
Why did TRS secure more seats in 2014 elections?
Answer:
TRS secured more seats in 2014 elections because it played a key role in Telangana agitation.

Question 10.
What is meant by the Coalition government?
Answer:
During the time of General Election to the Assembly and Lok Sabha, no party gain the majority to form the government at the centre or state at that time. Two or more than two political parties come together to form a single government.
(OR)
A number of national and regional parties had to come together to form governments at the centre.

Question 11.
Name some non-political movements.
Answer:
Environmental movements, the feminist movement, civil liberties movement, literacy movements.

Question 12.
Which became powerful motors of social change?
Answer:
A number of non-political movements emerged and became powerful motors of social change.

Question 13.
Which parties decided to merge together and form the Janata Party?
Answer:
The Congress, Swatantra Party, Bharatiya Jan Sangh, the Bharatiya Lok Dal and the Socia¬list Party decided to merge together and form the Janata Party.

Question 14.
Who supported the Janata Party?
Answer:
The DMK, the SAD and the CPI (M) chose to maintain their separate identities but supported the Janata Party in a common front against the Congress.

Question 15.
Who played an important role in bringing together all the anti-Congress and anti-Emergency parties?
Answer:
Senior leaders like Jayaprakash Narayan and Acharya JB Kriplani played an important role in bringing together all the anti-Congress and anti-Emergency parties to fight the elections.

Question 16.
What was the argument of the Janata Party regarding the dismiss of nine state governments?
Answer:
The Janata Party argued that the Congress party had lost its mandate to rule in the States as it had been defeated.

Question 17.
Which created a bad state in A.P.?
Answer:
In Andhra Pradesh, the frequent change of Chief Ministers by the central Congress leadership and the imposition of leaders from above created a bad taste.

Question 18.
Who moved to Assom and Bengal?
Answer:
The Bangladeshis moved to Assom and Bengal.

Question 19.
Name some communities of Assom.
Answer:
Bodos, Khasis, Mizos and Karbis.

Question 20.
Who was Bhindtanwale and what was his demand?
Answer:
Bhindranwale, the leader of the group of militant Sikhs began to preach separatism and also demanded the formation of a Sikh State- Khalistan.

Question 21.
What did the militants try?
Answer:
The militants tried to impose an orthodox life code on all Sikhs and even non-Sikhs of Punjab.

Question 22.
Who made a declaration in April 1986?
Answer:
In April 1986, an assembly at the Akal Takht, made a declaration of an independent state of Khalistan.

Question 23.
Where were the militants engaged in?
Answer:
The militants were also engaged in large scale kidnapping and extortion to raise funds for their work.

Question 24.
How were the methods used by the govern¬ment for the suppression of militancy in Punjab?
Answer:
The Government used very harsh methods for the suppression of militancy in Punjab, many of which were seen as a violation of. Constitutional rights of citizens.

Question 25.
What did Rajiv Gandhi begin?
Answer:
Rajiv Gandhi began a peace initiative in Punjab, Assam and Mizoram and also in the neighbouring country of Sri Lanka.

Question 26.
What is called the telecom revolution?
Answer:
Rajiv Gandhi initiated what is called the ‘telecom revolution’ in India which speeded up and spread the network of telephonic communication in the country using satellite technology.

Question 27.
What had been under dispute for some time regarding Babri Masjid?
Answer:
Some sections of the Hindus had begun a campaign for building a temple for Lord Rama in Ayodhya in the place of Babri Masjid.

Question 28.
What is the speciality of Elections held in 1989?
Answer:
The issue of corruption in administration and in political circles became the main plank of the election campaign for non-Congress political forces in the next elections held in 1989.

Question 29.
What is Policy Paralysis?
Answer:
Policy Paralysis means the coalition could not implement any policy which called for serious change for fear of withdrawal of support by one or the other partners.

Question 30.
Which was the first coalition to be re-elected?
Answer:
The UPA was the first coalition to be re-elected.

Question 31.
Who led the Left Front Government in West Bengal in 1977?
Answer:
Jyoti Basu of CPM led the Left Front Government in West Bengal in 1977.

Question 32.
On what did the Operation Barga depend?
Answer:
Operation Barga depended heavily on collective action by the share croppers and Panchayati Raj Institutions thus avoiding bureaucratic delays and domination of the landowning classes.

10th Class Social 19th Lesson Emerging Political Trends 1977 to 2000 2 Marks Important Questions and Answers

Question 1.
Read the following paragraph and answer the questions.

The Government used very harsh methods for the suppression of militancy in Punjab, many of which were seen as a violation of the constitutional rights of citizens. Many observers felt that such violations of constitutional rights and human rights were justified.as the constitutional machinery was on the edge of collapse due to militant activity.
Express your views on the information given above.

Answer:
There was a threat to the integration of the Indian nation due to the militancy in Punjab. If the government had not taken such actions, the map of India would be different today. So I think the government was correct.

Question 2.
Read the given data to answer the questions.

A) Which were the parties that participated in the governments of the National Front and United Front and supported the government from the outside?
Answer:
To National Front: CPM, CPI, and BJP.
To United Front: CPM.

B) Mention the name of the party that participated in the above three governments.
Answer:
J.K.N.C.

Question 3.
Based on the information given below, answer the following questions.

a) Name the first non-Congress party which formed the government at the centre.
Answer:
Janata Party is the first non-Congress party which formed the government at the centre.

b) Who is the founder of Telugu Desam Party?
Answer:
Nandamuri Taraka Rama Rao (NTR) is the founder of Telugu Desam Party.

Question 4.
Which are the newest states of India, when they created?
Answer:

Question 5.
Read the table and answer the given equations.

a) Which party won in 1996 elections and formed government?
Answer:
National Front.

b) Name the Coalition Governments mentioned in the above table.
Answer:
National Front and NDA Governments.

Question 6.
Write about people’s welfare schemes started by present Governments.
Answer:

1. Supply of rice at the cost of Rs. 1 per Kg to the white ration cardholders.
2. Pensions for the old age people and widows.
3. Free textbooks, uniforms and Midday meal scheme in government schools.
4. Housing schemes for the poor people.
5. Health scheme for the poor people.
6. Fees reimbursement to the poor for higher education, etc.

Question 7.
Read the following text and answer the questions given below.

The Congress returned to power in 1980. The Congress immediately paid back the Janatb in the same coin by dismissing the Janata and non-Congress governments in nine States. The Congress was victorious in all the States except Tamil Nadu and West Bengal.

A) Which party ruled before 1980s?
Answer:
Janata Party.

B) In which two states, the Congress party was defeated?
Answer:
Tamilnadu and West Bengal.

Question 8.
Prepare a table by classifying the given political parties into National and Regional Parties. “BJP, YSRCP, TDP, CPM, CPI, DMK, Congress-1, AGP”.

Answer:

Question 9.
Based on the information given below, answer the following questions.

i) Who was the Prime Minister at the time of demolition of Babri Masjid?
Answer:
P.V. Narasimha Rao.

ii) Give two examples of the Coalition government.
Answer:

1. Janata Dal government.
2. National Front government.
3. National Democratic Alliance (NDA).

Question 10.
Sometimes coalition governments cause ‘Policy Paralysis’. Do you agree with this statement?
Write your opinion.
Answer:
Yes. I agree with this statement. The coalition could not implement any policy which called for serious change for fear of withdrawal of support by one or the other partners.

Question 11.
“Coalition Governments cause political instability.” Comment.
Answer:

1. Sometimes no single party wins a majority of seats to form a government of its own. In such the situation, a number of political parties come together and form coalition governments.
2. A common agreement between these parties has to be arrived at, but this is not so easy.
3. Different parties put pressure on the government for their different interests.
4. The government cannot implement any policy for fear of withdrawal of support by one or the other partners. The governments become instable.
5. This is called policy paralise which is frequent in the coalition government.

Question 12.
What are the situations that paved to strengthen the regional parties in present days?
Answer:
The situations that paved to strengthen the regional parties

1. Regional aspirations – regional movements.
2. Intermediate castes strengthening – gaining political power.
3. To gain political power.
4. Defections and corruption.

Question 13.
Write the main reasons for Assam movement.
Answer:

1. Demand for autonomy.
2. Protest against the domination of Bangladesh.
3. Migration from Bangladesh.
4. Fear of losing their cultural roots.
5. Trade and other establishments were in the hands of outsiders.
6. No preference in employment for locals.

Question 14.
Observe the following table and analyse it.
Table: Seat share of various Political parties in 2014 (lok Sabha)

Answer:

1. In 2014 General elections BjP got with 282 seats and form the largest party and form the government also.
2. Indian National Congress got only 45 seats.
3. Left parties CPI + CPI (M) joined together got 10 seats.
4. The Regional parties like TDP 16 seats 8i TRS 11 seats gained In Lok’Sabha elections.

Question 15.
What are the important changes that occured in India between 1975-85?
Answer:
Many changes occurred In India between 1975-85. Some of them are:

1. Emergency was declared by smt. Indira Gandhi as she was asked to quit her Prime Minister post by Allahabad high court.
2. Janatha Government came into power in 1979.
3. Congress Party came to power in the elections after Janatha govt, failure.
4. Non-political movements like environment movements, feminist movements, civil liberties movement and literacy movements came up.

Question 16.
At present, what is the necessity of coalition politics?
Answer:
In the present multiparty system in India it is impossible for any single party to win a majority of seats and form a government of its own but in 2019 elections BJP has won the election as single party. It went as coalition.

Question 17.
Read the following paragraph and comment on it.

In Andhra Pradesh, the frequent change of Chief Ministers by the central Congress leadership and the imposition of leaders from above created a bad taste. There was a feeling that the Andhra Pradesh leadership was not getting respect from the national Congress leadership. This was perceived as an insult to the pride of the Telugu people. N.T. Rama Rao(NTR), popular film actor, chose to take up this cause. He began the Telugu Desam Party (TDP) on his 60th birthday in 1982. He said that the TDP stood for the honour and self respect of the Telugu speaking people (Teluguvari atma gauravam). He argued that the state could not be treated as a lower office of the Congress party.

Answer:

1. The Congress government frequently changed the Chief Ministers.
2. The Congress was not giving respect to Andhra Pradesh leadership.
3. The TDP was formed for the honour and self-respect of the Telugu speaking people.
4. He introduced welfare schemes like midday meals to government schools, liquor prohibition and the sale of rice for Rs. 2/- per kg.
5. These populist measures helped the TDP sweep the 1982 elections.
6. TDP emerged as a strong regional party, and challenged the Congress domination.

Question 18.
What are the effects of changes of the Telecom Revolution on the Human lifestyles.
Answer:

1. Telecom Revolution is the result of privatization of Telecommunications.
2. Number of industries invested in telecommunications.
3. “Mobiles” and Smart phones have created sensation.
4. They reduced the distance between the buyers and sellers.
5. Every family has a mobile in India.
6. Telemarketing is a creative innovation.
7. Smartphones have internet access and due to that internet facility is accessible to villagers through telephones.

Question 19.
What was Operation Blue Star?
Answer:

1. Sikhs became militant in Punjab under Bhindranwale.
2. People belonging to non-Sikhs were subjected to communal attack.
3. Sikh separatist groups hid in the Golden Temple.
4. Army had to intervene to vacate the campus.
5. This was called ‘Operation Blue Star’.

Question 20.
What factors influenced central government to use armed forces to reduce tensions in Assam?
Answer:

1. Three factors influenced the use of armed forces in the North Eastern Region.
2. Firstly, it was a sensitive border area adjacent to China, Mynmar and Bangladesh.
3. Secondly, rebel groups demanding separation from India, procured arms from outside.
4. Thirdly, they indulged in large-scale ethnic violence against minority communities.
5. The government thought this was the only way to bring about peace in the area.

Question 21.
Read the given information.

Now answer the following questions.
a) Which party was included in “Governing parties” in all the above coalition governments?
Answer:
Jammu & Kashmir National Conference (JKNC)

b) Which party gave support to NDA government?
Answer:
TDP.

c) Which party gave support to National Front and United Front from outside?
Answer:
CPM.

Question 22.
What did the emergence of competitive alternatives ensured?
Answer:
The emergence of competitive alternatives ensured that Indian voters could always exercise a reasonable choice. This also allowed many different political viewpoints and sectional interests to become active in state level and national politics.

Question 23.
How was the rule of the first non-Congress government?
Answer:
The Janata Party had come to power promising a restoration of democracy and freedom from authoritarian rule. However, the disunity among the partners had a serious effect on the governance and its rule is most often remembered for internal squabbles and defections. The factional struggle in the party soon culminated in the fall of the government within three years leading to fresh elections in 1980.

Question 24.
What happened whenever there was any political instability?
Answer:
Whenever there was any political instability or natural calamity in the neighbouring country, thousands of people moved into the State creating huge discomfort for the locals. The local people felt that they would lose their cultural roots and soon be outnumbered by the ‘outsiders’.

Question 25.
What was there besides culture and demographics?
Answer:
Besides culture and demographics, there was also an economic dimension. Trade and other establishments were in the hands of non-Assamese communities. The major resources of the State, including tea and oil were again not benefitting the locals.

Question 26.
What was the dominant thrust of the movement?
Answer:
The dominant thrust of the movement was that Assam was being treated as an “internal colony” and this had to stop. The main demands were that the local people should be given greater preference in employment, the “outsiders” should be removed and the resources should be used for the benefit of the locals.

Question 27.
Which has led to violent attempts of ethnic cleansing in Assam?
Answer:
Too much emphasis on ethnic identities had a negative impact on other communities of Assam like the Bodos, Khasis, Mizos and Karbis. Many of them too demanded autonomous status. They began to assert themselves and wanted to drive out people of other communities from their areas.

Question 28.
What did Punjab claim?
Answer:
It laid claims to the new capital city of Chandigarh which remained a union territory directly administered by the Centre. Punjab also claimed more water from Bhakra Nangal dam and greater recruitment of Sikhs in the army.

Question 29.
Write about the resolution of Akali Dal.
Answer:
The Akali Dal had passed a set of resolutions in 1978 during the Janata Party rule in the centre, calling upon the central government to implement them. Its most significant demand was to amend the Constitution to give more powers to the states and ensure greater decentralisation of powers.

Question 30.
What happened after Rajiv Gandhi’s entrance?
Answer:
After Rajiv Gandhi became the Prime Minister, he held talks with SAD and entered into an agreement with Sant Langowal, the SAD president. Though fresh elections were held in Punjab and SAD won them, the peace was short-lived as Langowal was assassinated by the militants.

Question 31.
What did Rajiv Gandhi say in his speech?
Answer:
In a famous speech Rajiv Gandhi said that out of every Rupee spent on the poor barely 15 paise reaches them I It highlighted the fact that despite huge increases in development expenditure, the story of the poor remained the same.

Question 32.
Which factors influenced the central government to use armed forces to reduce tensions in Assam?
Answer:

1. Three factors influenced the use of armed forces in the Northeastern region.
2. Firstly, it was a sensitive border area adjacent to China, Myanmar and Bangladesh.
3. Secondly, rebel groups demanding separation from India, procured arms from outside.
4. Thirdly, they indulged in large scale ethnic violence against minority communities.
5. The government thought this was the only way to bring about peace in the area.

Question 33.
What was meant by liberalization?
Answer:

1. It meant a lot of things put together like the drastic reduction of government expenditure, reducing restrictions and taxes on imports, etc.
2. It proved for reducing restrictions on foreign investments in India and allowed foreign countries to set up companies in India.
3. It is required to the opening of many sectors of the economy to private investors.
4. It brought in foreign goods and Indian businessmen were forced to compete with them.
5. It had many positive and negative impacts on India.

Question 34.
“One of the greatest weakness was undoubtedly the low priority given to primary education and public health”. Comment on it.
Answer:

1. The post-Independence era is marked with less priority to education and health.
2. The optimum development of country depends mostly on the education and health levels of the population of it.
3. It further forms part of Human Development Indicators also.
4. So, I suggest more priority should be given to education and health now.

Question 35.
Read the given information.

In 1992 government led by P.V. Narasimha Rao passed an important amendment to the Constitution to provide local self-governments a Constitutional Status. The 73rd Constitutional Amendment created institutions of local self government at the village level while the 74th Constitutional Amendment did the same in towns and cities. These were path-breaking amendments. They sought to usher in for the first time, office bearers at the local level elected on the basis of universal adult franchise. One-third of the seats were to be reserved for women. Seats were also reserved for scheduled castes and tribes.

Answer the following questions.
a) Which constitutional amendment created institution of local self-government?
b) According to which amendment general elections were conducted in towns and cities?
c) How many seats are reserved for women in local bodies?
Answer:
a) 73rd Constitutional Amendment created institutions of local self-governments for villages.
b) According to 74th Constitutional Amendment general elections were conducted in towns/cities.
c) 1/3 seats were reserved for women in local self-government elections.

Question 36.
“Do you think that the reservations will promote the social development” ? Express your ideas.
Answer:

1. Reservations will definitely promote social development.
2. Scheduled castes and tribes were drowtodden and suffered in the social stature for centuries.
3. To develop themselves and to question the injustice they meted out, reservations will of great help.
4. Reservations both in education, jobs, and legislature help them.

Question 37.
Imagine and write the main reasons for the continuation of anti Hindi movement in Tamil Nadu till today.
Answer:

1. DMK in Tamil Nadu believed the passage of Official Languages Act, 1963 was an attempt to first Hindi on the rest of the country.
2. They started a statewide campaign protesting the imposition of Hindi.
3. They organised strikes, dharnas, burning effigies, Hindi books as well as pages of constitution.
4. Still there is same feeling in Tamil Nadu.

Question 38.
“Some people think that Social Welfare Schemes do not reach eligible persons”. Express your suggestions.
Answer:

1. Despite all the attention to development in the country, much of it did not reach the real beneficiaries.
2. Despite huge increases in development expenditure the story of the poor remained the same.
3. The main reasons were political and beaurocratic corruption.
4. Enlistment of various beneficiaries also plagued by officialdom and political pressures.

Question 39.
“India needed to adapt itself to the new technologies emerging in the world, especially computer and telecommunication technologies”. Comment.
Answer:

1. Technologies like computer and communication technology are thursting the world.
2. It is believed that we should also adopt them without fail.
3. With initiatives of Rajiv Gandhi now called ‘Telecom Revolution1 was introduced in India.
4. With the help of satellite technology communications spread widely and extensively.
5. Everyone has access to mobile phones, the internet, email, facebook, Twitter, etc.

10th Class Social 19th Lesson Emerging Political Trends 1977 to 2000 4 Marks Important Questions and Answers

Question 1.
Read the text given and answer the questions.

Panchayati Raj & 73rd, 74th Amendment

In 1992, Government led by P.V. Narasimha Rao passed an Important amendment to the Constitution to provide Local Self Governments a Constitutional status. The 73rd Constitutional Amendment created Institutions of local self-government at the village level, while the 74th Constitutional Amendment did the same in towns and cities. These were pathbreaking amendments. They sought to usher in for the first time, office bearers at the local level elected on the basis of Universal Adult Franchise. One-third of the seats were to be reserved for women. Seats were also reserved for scheduled castes and tribes. The concerns of the State governments were taken into account and it was left to the States to decide on what functions and powers were to be developed to their respective local self-governments. Consequently, the powers of local self-governments vary across the country.

i) What is Local Self Government?
Answer:
The Government that formed by the people at the village, town and city level to solve the local needs is Local Self Government.

ii) Which government recognised the Constitutional status of Local self Government?
Answer:
P.V. Narasimha Rao or Congress Government.

iii) What does the 73rd Constitutional Amendment say?
Answer:
Creation of Local Self government at the village level.

iv) 1/3 of seats were to be reserved for women in Local Self Governments. Comment.
Answer:
Women need political equality and they should Involve actively In the Local Governments.

Question 2.
Read the following paragraph and answer the given questions.

Liberalization measures brought in foreign goods and forced Indian business to compete with global manufacturers. It also led to the setting up of industries and business by foreign companies in India. However, it also meant a lot of hardship for the common people as the government was forced to cut subsidies to the people and as many factories closed down due to Influx of cheap foreign goods. This also led to privatization of many public amenities tike education, health, and transport and people had to pay high prices to private service providers.

Write your opinion on the consequences of liberalization.
(OR)
What are the consequences of economic Liberalization?
Answer:

1. India was drawn into the world market.
2. Liberalization paved the way to telecom revolution.
3. Liberalization forced Indian business to compete with global manufactures.
4. The government was forced to cut subsidies which results a great loss to people and local industries.
5. It also led to the privatization of many public amenities like education, health and transport.
6. It led to globalization.
7. The policies of liberalisation have been of advantage particularly to well of sections only.

Question 3.
Explain the importance of regional parties in Democracy.
Answer:

1. Multi-party system which includes national parties and regional parties strengthens the democracy.
2. Regional parties reflect the spirit of the federalism.
3. Regional parties have good understanding of the problems and needs of the respective states.
4. They focus mainly on the development of their states.

Question 4.
Telecom revolution has brought several changes in human life nowadays. Explain them.
Answer:
Changes brought by the telecom revolution:

1. Saves time
2. Fast communication
3. Online services
4. Prosperous life
5. Addiction
6. Obesity
7. Cost of living increased
8. Affected human relations

Question 5.
Read the paragraph given below and interpret.

India was forced to open up and ‘liberalise’ its economy by allowing free flow of foreign capital and goods Into India. On the other hand, new social groups asserted themselves politically for the first time, and finally, religious nationalism and communal political mobilisation became Important features of our political life. All this put the Indian society into great turmoil, we are still coming to grips with these changes and adapting ourselves to them.

Answer:

1. Liberalisation means relaxation of previous government restrictions usually in areas of social and economic policy.
2. The twentieth century ended with India’s drawing into the world free market.
3. India was forced to open up and liberalise its economy. It allowed free flow of foreign capital and goods into India.
4. On the other hand, India seemed to have a thriving democracy in which voices of different sections of the population were making themselves heard and in which divisive and communal political mobilisation was threatening to destroy social peace.
5. It had stood the test of time for over fifty years and had built a relatively stable economy and deeply rooted democratic politics.
6. It still had not managed to solve the problem of acute poverty and gross inequality between castes, communities, regions and gender.

Question 6.
Observe the following table and write a paragraph analyzing it.
Summary of the 2014 -Indian General Elections

Answer:
The given table describes the summary of the 2014 general elections in India. In the given table two parties that is Bharatiya Janata Party and the Indian National Congress are compared. It is not only the party comparison but their alliances are also mentioned. The Bharatiya Janata Party alliance is National Democratic Alliance whereas the United Progressive Alliance is related to Indian National Congress. In these elections the NDAgot 31% of the votes whereas the UPAgot 19.31%. If we observe the seats, the BJP with its alliance won 282 whereas the INC won only 44. These elections are very crucial because the voter strongly rejected the pre-independence party which ruled India since 1947. For a long time it was a single largest party to win the seats in Lok sabha. The voters cleverly gave mandate to the Bharatiya Janata Party with the hopes that their future may be changed. The BJP announced the Prime Ministerial candidate, Narendra Modi in advance. He achieved and succeeded in Gujarat as Chief Minister. So the voters accepted him as Prime Minister also. They believed him. Congress lost faith of the people because of its failures. During Congress period there was a lot of corruption, scams and nepotism, etc. Many of the Congress members of Parliament were in court cases. Rajiv Gandhi himself declared that corruption is highly established in India. If the Bharatiya Janata Party with its alliance work for the development of the country, definitely they will win the next coming 2019 elections. So the party should keep this in mind and work in that direction.

Question 7.
Explain the effects along with the reasons for the emerging era of coalition politics.
Answer:
Reasons for the emergence of coalition era of politcs:

1. Multi party system
2. No single party securing required majority.
3. Significance of regional parties increased.
4. Congress party gradually lost people’s mandate after 1960s.

Effects:

1. No political stability
2. Isolating the ideologies
3. Giving importance to party’s interest at the cost of nation’s interest.
4. Coming to power in spite of securing less mandate.

Question 8.
Read the following paragraph and write your opinion.

The twentieth century closed with India which was drawn into the world market, India which seemed to have a thriving democracy in which voices of different sections of the population were making themselves heard and in which, divisive and communal political mobilisation were threatening to destroy social peace. It had stood the test of time for over fifty years and had built a relatively stable economy and deeply rooted democratic politics. It still had not managed to solve the problem of acute poverty and gross inequality between castes, communities, regions and gender.

Answer:
The given paragraph depicts about divisive and communal politics. These may destroy the social peace. After independence in India, stable government continued for 30, 40 years and unstability began. Main problem of solving poverty and inequalities with regard to caste, region is not yet solved.

My opinion is that the politics are only vote bank based. Sometimes the political leaders are there behind the communal riots. To throw out some Chief Minister of the same party, their party leaders encourage these riots. Caste based politics are shown at the time of tickets given to party candidates. Caste unions, and the caste group heads are distributed money to lure them to get their votes. Some constituencies are fixed for some religion because of their dominance in number. It is really a threat to democracy. Holy places of worship are also in some cases used to spread communal message. That destroys social peace.
My suggestion is that people should get awarness about this and act accordingly.

Question 9.
Observe the following table and analyse it.
The trend of Coalition Governments, 1989 – 2004

Answer:

1. The given table is about the trend of Coalition Governments during the period of the years from 1989 to 2004.
2. The details of three coalition governments and their duration, etc. are given in the table.
3. During 1989-1990 Janata Dal-led National Front formed the government. The governing parties in this government were JD, DMK, AGP, TDP, JKNC. CPM, CPI and BJP supported this government.
4. United Front formed the coalition government during 1996-1998. JKNC, TDP, TMC, CPI, AGP, DMK, MGP were the governing parties in this government. CPM supported this government.
5. During 1998-2004 BJP-led National Democratic Alliance formed the government. The governing parties in this government were JDU, SAD, AIADMK, JKNC, TMC, BJD and Shiva Sena. TDP rendered support to the NDA government.
6. The 1990s were years of very significant change in the post-Independence India.
7. With the transformation to a competitive multi-party system, it became near impossible for any single party to win a majority of seats and form a government of its own.
8. Since 1989, all governments that had formed at the national level have been either coalition or minority governments.

Question 10.
Explain about Assam movement in detail.
Answer:

Assam movement:

It is the struggle between Assamese and non-Assamese. These non-Assamese were none other than the people of Bangladesh. The youth of Assam formed All Assam Students Union (AASU) and was in the forefront of agitation. It led a number of strikes, agitations and marches to remove the so called outsiders. The problem of outsiders is not a cultural one but of economic issue. Every country or state wants to protect their cultural roots. The Assamese were most of them, Hindus and the outsiders were Muslims. The local people were afraid of their cultural roots.
Now they affect the trade and so the livelihoods of the locals had been in trouble. It is not only the problem of Assam, it happens at many states. Outsiders dominate a few areas of business and so the locals lose opportunities. In Assam the locals were not given priority or preference in employment. This was the demand of the Assamese. Gradually these demands led to communal polarisation as most of the outsiders are from Bangladesh Muslims. The movement between the Assamese and outsider Muslim led to form an idea of anti Indian stand.

Central Government took initiation and went on for talks for three years. An agreement was signed by the central government and the students union. In the next elections Assam Gana Parishad (an offshoot of AASU) came to power.

In conclusion, the formation of Bangladesh erstwhile Pakistan was taken place on the basis of religion. One’s religion can be given respect by all but it led to many disturbances. The Muslims, the outsiders of Assam occupied most of the areas of trade and business and there was distress and disappointment among the Assamese. The outsiders would have settled in Bangladesh only. They wanted their country to be separated and still they are coming to India illegally. Recently both the Prime Ministers of India and Bangladesh sat together and solved a few problems. If any problem arises, they should sit together and problems can be solved.

Question 11.
Prepare an album by collecting the photos of Prime Ministers of India and write their specialities.
Answer:

Question 12.
You may notice how simple and genuine demands of the people of Punjab were hijacked by religious and anti-national extremists. What steps do you think would have prevented this?
Answer:

1. The simple and genuine demands of Punjab were
   a) the contribution of state was ignored
   b) received unfair bargain when it was created
   c) capital remain UT
   d) more water from Bhakra Nangal and
   e) greater recruitment of Sikhs in the army.
2. Akali Dal government was dismissed by Congress.
3. A series of untoward incidents increased distance between Sikhs and the central government.
4. Militant Sikhs demanded separate state.
5. They occupied Golden Temple, then Congress used army to vacate.
6. A fallout led to the assassination of Indira.
7. Rioting in Delhi against Sikhs was followed.
8. Later Langowal made an agreement with centre but was killed by militants.
9. Militants engaged in extortion and kidnapping and lost faith of the people

Question 13.
Understand the table and answer the following questions.
Some opposition parties of 1970’s

a) Which political party fiercely opposes Hindi in the state ?
Answer:
DMK is the party which opposes Hindi in the state.

b) What is the place of activity for Jan Sangh?
Answer:
Jan Sangh is active in Northern States.

c) What is the ideology of CPI (M)?
Answer:
The ideologies of CPI (M) are radical land reforms, trade unionism and socialist policies.

d) Where is the political party which shows special attention to farmers, active ?
Answer:
The political party, which shows special attention to farmers is active in Uttar Pradesh.

e) Which party is of semireligious nature?
Answer:
SAD – Shiromani Akali Dal is of semireligious nature.

Question 14.
What were the implications of 1977 general elections?
Answer:

1. It was a historical election for democracy.
2. The Congress party was defeated at the national level for the first time.
3. Janata Party became victorious and tried to consolidate itself.
4. It dismissed nine Congress governments in states.
5. It argued that Congress had lost its mandate to rule in the states as it had been defeated.
6. Its stand somewhat proved correct by the results.
7. Except Tamil Nadu and West Bengal, Janata Party came to power in states.
8. The disunity among the partners had a serious effect on governance.
9. The government fell within three years.
10. It led to fresh elections in 1980.

Question 15.
Why was the public sympathy to Punjab militant Sikhs declined?
Answer:

1. They formed armed attachments and engaged in terrorist activities.
2. They clashed with police and other religious groups.
3. Those who were not confirmed to militant approved behaviour were killed.
4. There were civil casualities in derailing trains, exploding bombs, etc.
5. They were engaged in kidnapping, extortion to raise funds.
6. All this gradually alienated them from masses and even Sikhs.
7. Over a period, public sympathy declined rapidly.
8. Peace was finally returned to Punjab by the end of 1990s.

Question 16.
‘Coalition governments induce political instability’ – Elucidate.
Answer:

1. Since 1989, all governments at national level were coalition/minority governments.
2. A number of national and regional parties had come together.
3. So political ideologies and programmes of all parties had to be accommodated.
4. A common agreement had to be arrived at.
5. No party could pursue extreme agendas.
6. They needed to tone down their approaches.
7. It caused considerable instability.
8. Many coalitions did not last their full time.

Question 17.
How do political parties reap on communal polarisation? Provide an example.
Answer:

1. The Hindus are led by Bharatiya Janata Party.
2. In the year 1984 LokSabha elections they won only 2 seats.
3. It made great strides when it took up the Ayodhya issue.
4. It decided to campaign for the building of a temple at the site of mosque.
5. It claimed that was the birthplace of Lord Rama.
6. L.K. Advani in 1990, led a ‘Rath Yatra’ from Somanath to Ayodhya.
7. This campaign was accompanied by intense communal polarisation.
8. It caused a large number of communal conflicts.
9. In 1991 General elections BJP’s strength went up to 120.
10. It was then Rajiv was killed and sympathy wave followed the Congress, still, BJP withstood it.

Question 18.
What is meant by liberalisation?
Answer:

1. It means a lot of things put together.
2. It proposes drastic reduction of government expenditure.
3. It asks for reducing restrictions and taxes on import of foreign goods.
4. It provides for reducing restrictions on foreign investments in India.
5. It is required to the opening of many sectors of the economy to private investors.
6. It brought in foreign goods and forced Indian business to compete with them.
7. It allowed foreign countries to set up companies in India.
8. Common people suffered with cut of subsidies.
9. Many factories were closed down due to influx of cheap foreign goods.

Question 19.
Study the timeline given below and answer the following questions.
a) Who was the Prime Minister that initiated peace agreements with Sri Lanka?
Answer:
Rajiv Gandhi initiated peace agreements with Sri Lanka.

b) Which government tried to implement Mandal Commissions report?
Answer:
Janata Dal government tried to implement Mandal Commissions report.

c) Name two important incidents occurred during the period of P.V. Narasimha Rao.
Answer:
Economic liberalization and the demolition of the Babri Masjid took place during the period of P.V. Narasimha Rao.

d) When was Indira Gandhi assassinated?
Answer:
Indira Gandhi was assassinated in 1984.

e) Who were the Prime Ministers of National Front Government?
Answer:
Deve Gowda and I.K. Gujral were the Prime Ministers of the National Front Government.

f) Who were the Prime Ministers of Janata Dal Government?
Answer:
V.P. Singh and Chandrasekhar were the Prime Ministers of Janata Dal Government.

g) Who led the Congress party after the assassination of Rajiv Gandhi?
Answer:
P.V. Narasimha Rao led the Congress party after the assassination of Rajiv Gandhi.

h) Who led the NDA government?
Answer:
A.B. Vajpayee led the NDA government.

i) When was the NDA Government formed?
Answer:
NDA formed the government in 1998.

Question 20.

Read the following information and answer the questions.

Some opposition parties of 1970s

BLD – Bharatiya Lok Dal – A party which was formed of socialists who called for special attention to Indian farmers, based mainly in Uttar Pradesh.

Congress (O) – The conservative section of the Congress which had opposed the policies of Indira Gandhi.

CPI (M) – Communist Party of India (Marxist)-a party with a national presence, which strove for radical land reforms, trade unionism and socialist policies.

DMK – Dravida Munnetra Kazagam – a party based mainly in Tamil Nadu which sought greater autonomy and powers for the state.

Jan Sangh – A Hindu nationalist party largely confined to the northern States.

SAD – Shiromani Akali Dal – a party based in Punjab catering specially to the Sikhs and organised around Gurudwaras. It therefore had a semi-religious character. It was also in favour of greater autonomy to the States.

a) Which party fought for autonomy in Tamil Nadu?
Answer:
Dravida Munnetra Kazagam fought for greater autonomy in Tamil Nadu.

b) Which party showed special attention to Indian farmers mainly in UP?
Answer:
Bharatiya Lok Dal showed special attention to farmers mainly in U.P.

c) Name the regional party of Punjab.
Answer:
Shiromani Akali Dal is the regional party of Punjab.

d) Name one Hindu nationalist party.
Answer:
“Jan Sangh” is one Hindu nationalistic party.

e) Which opposed the policies of Indira Gandhi?
Answer:
Congress (O) – The conservative section of the Congress opposed the policies of Indira Gandhi.

f) What was the main aim of SAD?
Answer:
It sought for greater autonomy to Punjab.

g) Which party was confined to North India only?
Answer:
Jan Sangh was confined to North India only.

Question 21.

Read the following passage and interpret it.

Panchayati Raj & 73rd, 74th amendment

In 1992 government led by P.V. Narasimha Rao passed an important amendment to the Constitution to provide local self-governments a Constitutional Status. The 73rd Constitutional Amendment created institutions of local self government at the village level while the 74th Constitutional Amendment did the same in towns and cities. These were path-breaking amendments. They sought to usher in for the first time, office bearers at the local level elected on the basis of universal adult franchise.

One-third of the seats were to be reserved for women. Seats were also reserved for , scheduled castes and tribes. The concerns of the State governments were taken into j account and it was left to the States to decide on what functions and powers were to be devolved to their respective local self governments. Consequently, the powers of local self governments vary across the country.

Answer:

1. In 1992 P.V. Narasimha Rao s government passed the important amendments of 73rd and 74th.
2. The 73rd amendment created institutions of local self governments at the village levels.
3. The 74th amendment created institutions of local self-governments at the town and city levels.
4. They are path-breaking as the office bearers at the local level are elected on the basis of universal adult franchise.
5. Seats are reserved for women and Scheduled Castes and Tribes too.
6. Powers were devolved to their respective local self-governments.
7. Hence we can say that these two amendments were path-breaking.

Question 22.
On the outline map of India locate the On the outline map of India locate the following.

1. Andhra Pradesh
2. Assom
3. Punjab
4. Tamil Nadu
5. West Bengal
6. Uttar Pradesh
7. Nagaland
8. Mizoram
9. Bihar
10. Gujarat
11. Maharashtra
12. Ayodhya

Answer:

Question 23.
Locate the following in the given map of World.

1. Madagascar Island
2. Nigeria
3. Holland
4. Amsterdam
5. Brazil
6. Jordan
7. Israel
8. Spain
9. Palestine
10. Bangladesh

Answer:

AP State Board Syllabus AP SSC 10th Class Biology Solutions Chapter 8 Heredity Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Biology Solutions 8th Lesson Heredity

10th Class Biology 8th Lesson Heredity Textbook Questions and Answers

Improve your learning

Question 1.
What are variations? How do they help organisms?
Answer:

1. Differences in characters within very closely related groups of organisms are referred to as variations.
2. Variations develop during reproduction in organisms.
3. Variations are passed from parent to offspring through heredity.
4. Beneficial variations are selected by the nature in evolution.
5. Variations increase the survival chance of the organisms.
6. These variations help the organisms to adapt to their environments.
7. For example, green colour in the beetles is a variation that gave a survival advantage to the beetles as they cannot be seen by the crows.
8. Some variations do not help organisms to survive. For example, colour variation occurs in red beetles and some blue beetles are produced instead of red beetles as they are eaten by crows easily.

Question 2.
One student (researcher) wants to cross pure tall plant (TT) with pure dwarf (tt) plant, what would be the Fj and F2 generations? Explain.
Answer:

1. Pure tall plant has both the factors of the same type ‘TT’.
2. Pure dwarf plant has both the factors of the same type ‘tt’.
3. When a pure tall plant (TT) is crossed with pure dwarf plant (tt), all the offsprings in Fj generation are tall (Tt).
4. So all the plants are heterozygous tall, as ‘T’ is the dominating factor.
   
5. On self pollination of these F₁ generation plants the new breed can have any combination of T and t like TT, Tt, Tt or tt.
   
6. So in F₂ generation 75% of plants are tall and 25% of plants are dwarf. Thus the phenotype ratio is 3 : 1.
7. Among 75% of tall plants 25% are pure tall (TT) or homozygous tall, remaining 50% are heterozygous tall (Tt, tT).
8. The remaining 25% dwarf plants are pure or homozygous dwarf (tt).
9. So the genotype ratio is 1 : 2 : 1.

Question 3.
One experimenter cut the tails of parent rats, what could be the traits in offsprings? Do the daughter rats contain tails or not? Explain your argument.
Answer:

1. If the tails of parent rats were cut, their offsprings will have normal tails.
2. Daughter rats do not contain tails because the bodily changes are not inherited.
3. So the change would not be passed to their offsprings.
4. This was proved experimentally by Augustus Weisemann and rejected the theory ‘inheritance of acquired characters’ proposed by Lamarck.

Question 4.
In a mango garden a farmer saw one mango tree with full of mango fruits but with a lot of pests. He also saw another mango tree without pests but with few mangoes. But the farmer wants the mango tree with full of mango fruits and pests free. Is it possible to create new mango tree which the farmer wants? Can you explain how it is possible?
Answer:

1. Yes, it is possible to create new mango tree which one the former wants with full of mango fruits and pests free.
2. The former can cross two plants one with full of mangoes and pests and another plant with less mango fruits and without pests,
3. In F₁ generation he may get plants with full of mango fruits and without pests. Such plants are called hybrid plants.
4. The F₁ generation plants can be self pollinated and desired plants can be selected from the mixed population of F₂ generation.
5. The plant with desirable characters can be vegetatively propagated to get required number of plants.

Question 5.
ExplaIn monohybrid experiment with an example. Which law of inheritance can we understand? Explain.
Answer:

1. We can understand the law of inheritance with an example of monohybrid cross between pure yellow pea seeds with pure green pea seeds.
2. A pure breed (parental) yellow will have both the factors which denote them by ‘YY’ and pure breed (parental) green seed will have both the factors denote them by ‘yy’.
3. During reproduction one factor (genes) from each parent is taken to form a new pair in the progeny (off spring).
4. In F₁ generation all pea plants are Yellow.
   
5. F₁ generation pea plants are self pollinated.
6. In F₂ generation 75% of the plants produced were Yellow seeds and the remaining 25% produced were green seed. It can be represented as
   
   In F₂ generation the produced plants are YY, Yy; yY or yy.
   From the above example of monohybrid cross we can understand the following laws of inheritance.

1) When pure breed Yellow (YY) and green (yy) seeds were crossed, only Yellow seeds were expressed phenotypically in the F₁ generation. It indicates that Yellow seed character is dominent over green seed characters.
This is “LAW OF DOMINANCE”.

2) When F₁ plants are self pollinated each parent passes randomly selected allele (Y or y) of one of these factor to offsprings. This is segregation of alleles or genes during production of gametes.
This is “LAW OF SEGREGATION”.

Question 6.
What is the law of independent assortment? Explain with an example.
Answer:

1. In the inheritance of more than one pair of characters (traits), the factors for each pair of characters assorts independently of the other pairs. This is known as “Law of independent assortment”.
2. If pea plants with two different pairs of characteristics (eg. : Round / yellow and green wrinkled) are breed with each other, the F₁ progeny plants would have all round and yellow seeds.
3. This implies that round and yellow seeds are dominant characters over green and wrinkled seeds.
4. In F₂ progeny there would be some plants with round and yellow seeds and some with green and wrinkled seeds.
5. However, there would be some plants with mixed characters – yellow and wrinkled seeds and green and round seeds.
6. This depicts that round/wrinkled trait and yellow / green trait are inherited independent of each other (law of independent assortment).
   The following punnet square explains this.
   
7. The different combination of characters resulted from dihybrid cross.
   a) RRYY, RRYy, RrYy, RrYY, RRYy, RrYy, RrYy, RrYY, RrYy are having round and yellow seeds.
   b) RRyy, Rryy, Rryy have round and green seeds.
   c) rrYy, rrYy, rrYY have wrinkled and yellow seeds.
   d) rryy have wrinkled and green seeds.
8. From the result, it can be concluded that the factors for each character or trait remains separate and maintain its identity in the gametes. This is known as “Law of independent assortment”.

Question 7.
How does sex determination take place in human?
(OR)
Explain sex determination in humans with the help of flow chart.
Answer:

1. Each human cell contains 23 pairs (46) of chromosomes.
2. Out of 23 pairs, 22 pairs of chromosomes are called autosomes.
3. Remaining one pair is called allosomes or sex chromosomes.
4. There are two types of sex chromosomes – one is ‘X’ and the other is ‘Y’.
5. These two chromosomes determine the sex of an individual.
6. Females have two ‘X’ chromosomes in their cells (XX).
7. Males have one ‘X’ and one ‘Y’ chromosomes in their cells (XY).
   
8. All the gametes produced by women (ovum) will be with only X chromosomes.
9. The gametes produced by man (sperm) will be of two types, one with X chromosomes and other with Y chromosomes.
10. If the sperm carries X chromosome and fertilizes with the ovum, the resultant baby will have XX condition. So the baby will be a girl.
11. If the sperm carries Y chromosome and fertilises with the ovum, the resultant baby will have XY condition. So the baby will be a boy.

Question 8.
Explain Darwin’s theory of ‘Natural selection’ with an example.
(OR)
What do you understand by the term Natural selection? Write Darwin’s theory of evolution.
Answer:

1. Darwin proposed the theory of Natural selection.
2. Nature only selects or decides which organism should survive or perish in nature.
3. The organism with useful traits will survive and the organisms having harmful traits are going to perished or eliminated from its environment.
4. For example, a group of twelve red beetles live in a bush of green leaves.
5. They will grow their population by sexual reproduction.
6. So they generate variations in their population. Let us assume crows eat the red beetles more the population of red beetles slowly reduced.
7. Crows eat these red beetles and their population slowly reduces.
8. During this time a colour variation arises by the sexual reproduction.
9. So that there appears one beetle that is green in colour instead of red.
10. Moreover this green colour beetle passes its colour to its offsprings; so that all its progeny are green.
11. Crows cannot see the green coloured beetles on green leaves of the bushes and therefore crows cannot eat them.
12. The crows can see the red beetles and eat them as a result, there are more and more green beetles than red ones which decrease in their number.
13. The variation of green colour beetle gave a survival advantage to green beetles’ than red beetles. They were naturally selected.

Question 9.
What are variations? Explain with a suitable example.
Answer:

1. Differences in characters within very closely related groups of organisms are referred to as variations.
2. Often a new character in a group may lead to variations that are inherited.
3. If we observe parents and offsprings, there will be some similar features in the offspring of the parents.
4. At the same time we find differences between parents and offspring in their features.
5. These differences are an example of variations.
6. Variations are quite apparent among closely related groups of organisms.
7. If we take roses as an another example, we observe number of varieties in them.
8. But we can still find some characters similar to all plants.
9. Thus rose plants have similar physical features, at the same time they have differences in characters like flower colour, number of petals, leaf size, stem, spines, etc.
10. These differences in features are variations.

Question 10.
What variations generally have you observed in the species of cow?
Answer:
In the species of cow the following contrasting variations can be observed:

1. White coloured – spotted
2. Longhorns – short horns
3. Height – dwarf
4. Long-tail – short tail
5. Elongated face – stunted face
6. More milk giving – less milk giving, etc.

Question 11.
What are the characters that Mendel selected for his experiments on pea plants
(OR)
Write the seven pairs of contrasting characters in pea plant identified by Mendel and mention their traits.
Answer:
Mendel selected the following characters on pea plants for his experiment. They are:

Question 12.
In what way Mendel used the word ‘Traits’? Explain with an example.
Answer:

1. Trait is a separate variant of an organism.
2. Mendel hypothesized that characters were carried as traits.
3. An organism always carried a pair of factors for a character.
4. He also hypothesized that distinguishing traits of the same character were present in the population of an organism.
5. He assumed that the traits shown by the pea plants must be in the seeds that produce them.
6. The seeds must have obtained these traits from the parent plants.
7. The factors which are responsible for the character or trait of an organism, are now named as “genes”.
8. By all these we can assume that Mendel used the word ‘traits’ for indicating the variant of an organism expressed by a pair of factors or genes.
9. For example, height is a character of pea plant while the tallness is a trait expressed by a pair of factors either TT or Tt and dwarfness is another trait expressed by a pair of factors tt.

Question 13.
What are the differences that Mendel observed between parent and F₂ generation?
Answer:
Mendel identified the following differences between parent and F₂ generation.

Question 14.
Male is responsible for sex determination of baby – do you agree? If so write your answer with a flow chart.
Answer:

1. Yes, I agree with the statement that male is responsible for sex determination of baby.
2. There are two types of sex chromosomes in human beings, one is ‘X’ and other is ‘Y’.
   
3. Females have two ‘X’ chromosomes in their cells (XX) whereas males have one ‘X’ and one ‘Y’ chromosomes in their cells (XY).
4. All the gametes produced by woman (ovum) will be with only X chromosomes.
5. The gametes produced by man (sperm) will be of two types one with X chromosomes and other Y chromosomes.
6. If the sperm carrying X chromosome fertilizes the ovum, the resultant baby will have XX condition. So the baby will be a girl.
7. If the sperm carrying Y chromosome fertilizes the ovum, the resultant baby will have XY condition. So the baby will be a boy.
8. So the gamete produced by the male is the deciding factor for sex determination of the baby.

Question 15.
Write a brief note on analogous organs.
(OR)
What are analogous organs?
Answer:

1. The organs which are structurally different but functionally similar are known as ‘Analogous organs’.
2. Wings of birds and bats is the example for analogous organs.
3. The wings of bats are skin folds stretched mainly between elongated fingers.
4. But the wings of birds are a feathery covering all along the arm.
5. The designs of the two wings, their structure and components are different.
6. They look similar because they have common use for flying, but their origins are not common.
7. This makes the ‘analogous’ characteristics.
8. This type of evolution is called convergent evolution.

Question 16.
How do scientists utilise information about fossils?
(OR)
“Fossils are valuable material that nature had preserved to know about ancient organisms.” Write the information you have collected about fossils.
Answer:

1. Fossils are evidence of ancient life forms or ancient habitats which have been preserved by natural processes.
2. The scientific study of fossils is called ‘Palaeontology’.
3. Scientists utilise information about fossils to understand the evolutionary history of life.
4. This information is also useful to study ecology and environmental history, such as ancient climates.
5. This also helps to find out how old that certain layer of earth is.
6. This information is also utilized as indicators of possible fossil fuel deposits which are of great interest to humanity.
7. Thus scientists utilize the information on fossils to learn more about the earth’s past.

Question 17.
Mendel selected a pea plant for his experiments. Mention the reasons for the selection as these plants.
(OR)
Why did Mendel select the pea plant for his experiment? (OR)
Which characters in the pea plant are selected by Mendel, for his experiments?
What are the reasons for selecting pea plant by Mendel to conduct his experiments?
Answer:
Mendel chose the pea plant (Pisum sativam) for his breeding experiments for the following reasons.

1. It is sexually reproducing.
2. Flowers are bisexual.
3. Predominantly self-pollinated.
4. Predominantly self-fertilization.
5. Well developed characters.
6. Early hybridization.
7. It is an annual plant.
8. These plants have short maturity and can produce large number of seeds in a single generation.
9. Pea plants have short life cycle.
10. These plants can easy to grow either on the ground or in pots.

Question 18.
If the theory of inheritance of acquired characters proposed by Lamarck was true, how will the world be?
Answer:
If the theory of inheritance of acquired characters proposed by Lamarck was correct,

1. All the organisms which lost some of their body parts should give birth to the offsprings without the lost parts.
2. Rat which lost their tail should give birth to tail less rats.
3. A handicapped who lost their legs in an accident should give birth to babies v without legs.
4. A body builder’s children should be body builders.
5. But all these are not happening because bodily changes won’t be passed to its offspring.

Question 19.
Collect information on the inherited traits in your family members and write a note on it.
Answer:

1. My grandfather and father- had curling hair. I too have curling hair. So it’s an inherited trait in family.
2. My mother and I both have long noses which appear similar. It’s another inherited trait.
3. Eyes of my grandmother, my brother and mine are similar. It’s another inherited trait.
4. Ear lobes of my father, brother and mine are similar. This is another inherited trait.