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AP State Syllabus AP Board 9th Class Physical Science Important Questions Chapter 5 What is inside the Atom?

AP State Syllabus 9th Class Physical Science Important Questions 5th Lesson What is inside the Atom?

9th Class Physical Science 5th Lesson What is inside the Atom? 1 Mark Important Questions and Answers

Question 1.
Write any two limitations of Rutherford’s atomic model.
Answer:

• The revolving electron would lose energy continuously and get directed towards positively charged nucleus and eventually crash into the nucleus.
• If this is true, the atoms would become highly unstable and the matter would not exist, but matter exists.

Question 2.
In Rutherford’s experiment, when the alpha particles hit the foil, Rutherford expected that all the alpha particles would be deflected by small angles. Why did Rutherford expect the above observation?
Answer:
Rutherford thought that positive charge is distributed throughout the atom.

Question 3.
Write an application of isotopes in the field of medicine.
Answer:

• The isotope of iodine is used in the treatment of goitre.
• The istope of cobalt is used in the treatment of cancer.

9th Class Physical Science 5th Lesson What is inside the Atom? 2 Marks Important Questions and Answers

Question 1.
What is valency? Write valency of hydrogen.
Answer:

• The number of electrons present in outer most orbit of an atom is called its valency.
• Valency of hydrogen is 1.

Question 2.
Observe the table given below.

Answer the following questions.
i) Which Shell has highest number of electrons?
ii) Write the general form of the formula to find maximum number of electrons in each shell?
Answer:
i) 0 Shell ‘N’.
ii) 2n² (n = 1, 2, 3, ……..)

Question 3.
Convert 36g of water into mole?
Answer:
Water molecular formula = H₂O
At.wt. of water = (2 × 1) + 16 = 18
1 mole, water = 18 gr.
36g. of water = \(\frac{36}{18}\) = 2 moles

9th Class Physical Science 5th Lesson What is inside the Atom? 4 Marks Important Questions and Answers

Question 1.
Fill the blanks in the table using the given information. (Isotopes are not included).

Answer:

Question 2.
Fill in the missing information in the table given below.

Answer:

Question 3.
Draw the figures showing arrangement of electrons for the given elements.
1. Helium, Oxygen, Argon.
Answer:
1) Helium
2) Oxygen
3) Argon

2. How many neutrons are present in the nucleus of Sodium?
Answer:
Sodium = ²³₁₁Na
Neutrons = 23-11 = 12

Question 4.
a) Draw neat diagrams indicating the nucleus and arragement of electrons in different shell for the following elements?
i) Helium
ii) Carbon
iii) Argon
Answer:

b) Which of the above element is unstable? Why?
Answer:
Carbon is unstable. The nudes of carbon-14 atoms are unstable because they have too many neutrons relative to protons, so they gradually decay.

9th Class Physical Science 5th Lesson What is inside the Atom? Important Questions and Answers

9th Class Physical Science 5th Lesson What is inside the Atom? 1 Mark Important Questions and Answers

Question 1.
Write the names of isotopes of Hydrogen.
Answer:
Hydrogen, deuterium, tritium.

Question 2.
Which element has maximum number of isotopes?
Answer:
Cesium and Helium are the elements having maximum number of isotopes.

Question 3.
What is maximum number of electrons present in M-shell?
Answer:
Maximum number of electrons present in M-shell is 2 × 32 = 2 × 9 = 18.

Question 4.
Which atom doesn’t contain neutron in its nuclear?
Answer:
Hydrogen.

Question 5.
Show the arrangement of electrons in phosphorus through a diagram.
Answer:
Atomic number of phosphorus is 15.
Distribution of electrons : 2, 8, 5.

Question 6.
Who proposed rules for electron distribution in an atom?
Answer:
Bohr and Bury.

Question 7.
What is the first rule of Bohr and Bury for electron distribution in an atom?
Answer:
The maximum number of electrons present in a shell is given by the formula 2n², where ‘n’ is the shell number, which takes values 1, 2, 3, 4,

Question 8.
Write the second principle of Bohr – Bury.
Answer:
Each energy level or electron shell is further divided into sub-shells. The maximum number of electrons that can be accommodated in each shell is 8.

Question 9.
Write the third law of Bohr – Bury.
Answer:
Electrons cannot be filled in a given shell unless the inner shells are completely filled i.e., shells are filled in step-wise manner.

Question 10.
What is an octet?
Answer:
An outermost shell which has 8 electrons is said to possess an octet.

Question 11.
Define atomic number.
Answer:
Atomic number :
Atomic number is the number of protons in the nucleus of an atom, denoted by ‘Z’.

Question 12.
Define atomic mass number.
Answer:
Atomic mass number:
Atomic mass number is the number of protons plus the number of neutrons, denoted by A.
∵ A = Z + N

Question 13.
On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.
Answer:
The negative and positive charges are equal in magnitude are present in atom. So, the atom as a whole is electrically neutral as whole according to Thomson model.

Question 14.
On the basis of Rutherford model of an atom, which sub-atomic particle is present in the nuclear of an atom?
Answer:
α – particles are repelled by the nucleus. So it contains positive charged particle that is proton.

9th Class Physical Science 5th Lesson What is inside the Atom? 2 Marks Important Questions and Answers

Question 1.
What are the postulates of Thomson’s model of the atom?
Answer:
J.J. Thomson proposed a model of atom in 1898. The main postulates are

1. An atom is considered to be a sphere of uniform positive charge and electrons are embedded into it.
2. The total mass of the atom is considered to be uniformly distributed throughout the atom.
3. The negative and the positive charges are supposed to cancel out the atom as a whole is electrically neutral.

This model is commonly known as plum pudding model or watermelon model.

Question 2.
What are the features of Rutherford’s model of atom?
Answer:
In 1909, Rutherford proposed a model of atom based on the alpha particle scattering experiment. The main features are

1. All the positively charged material in an atom formed a small dense centre, called the nucleus of the atom. The electrons were not a part of nucleus.
2. The negatively charged electrons revolve around the nucleus in well-defined orbits.
3. The size of the nucleus is very small as compared to the size of the atom.

This model is known as planetary model because the electrons revolve around the nucleus as planets revolve around the sun.

This model failed to explain the stability of atom.

Question 3.
Write the rules proposed by Bohr – Bury for electron distribution.
Answer:
Bohr-Bury proposed the following rules for electron distribution.
Rule – 1 :
The maximum number of electrons that can be accommodated in each shell is given by a formula 2n2. Where n is the shell number, which takes the values 1, 2, 3,….
Ex : For K shell, n = 1

∴ Maximum number of electrons in K shell = 2(1)2 = 2 × 1=2

Rule – 2 : Each energy level or electron shell is further divided into subshell. The maximum number of electrons that can be accommodated in each subshell is 8.

Rule – 3: Electron cannot be filled in a given shell unless the inner shells are completely filled.

Question 4.
Explain the distribution of electrons in oxygen atom using Bohr-Bury principle.
Answer:

1. Atomic number of oxygen is 8.
2. Hence it has 8 protons and 8 electrons.

Step -1 :
The K-shell can accommodate maximum 2 electrons, so the first 2 electrons fill the shell of n = 1.

Step – 2 :
The other 6 electrons will fill the higher shell n = 2 or the L-shell.
Step – 3 :
Then the electronic structure for oxygen atom is 2, 6.

Question 5.
Fluorine atom contains 7 electrons in the outermost shell. But its valency is ‘1’. Explain.
Answer:

• Valency is the number of electrons present in the outer most orbit of an atom.
• The distribution of electrons in fluorine (atomic number 9) is 2, 7.
• Hence the valency of fluorine could be 7.
• But it is easier to fluorine to gain one electron for becoming octet.
• Hence its valency is determined by subtracting seven electrons from 8 and which gives you a valency ‘1’ for fluorine.

Question 6.
What is the importance of valency?
Answer:
1) An atom with 8 electrons or an octet in their outer most shell is chemically stable or does not combine other atoms.
Ex : Ne, Ar, etc.

2) An atom with duplet or 2 electrons in its outer most shell is also more stable when there is only one shell present in it.
Ex : He

3) Atoms of an element thus react with other atoms. So as to achieve an octet in their shell.

4) When elements reacts to form compounds, their atoms must be combining in such a way that they can attain the stable electron distribution of noble gases or inert gases.

Question 7.
How can an atom achieve octet?
Answer:
An atom can achieve an octet by two ways.

1. One by transfer of electrons.
2. Other by sharing of electrons.
3. Both the processes results in the formation of bonds between atoms.

Question 8.
Explain the notation : \({ }_{9}^{19} \mathrm{F}\)
Answer:

• F is the symbol of element (Fluorine).
• 9 is the atomic number.
• 19 is the mass number.
• Hence fluorine has 9 protons and (19 – 9 = 10) 10 neutrons in its nucleus.
• 9 electrons are revolving around the nucleus.

Question 9.
Define isotope and give two examples.
Answer:
Isotope :
The atoms of the same element which have the same number of protons but have different number of neutrons are called isotopes.

Question 10.
“Sulphur shows multiple valency.” Explain this.
Answer:
a) Atomic number of sulphur is 16.
b) The distribution of electrons is 2, 8, 6.
c) Hence the valency should be ’6′.
d) But it is easier to sulphur to gain 2 electrons than loosing 6 electrons to become octet. Hence its valency would become 2.
e) So, sulphur shows multiple valency, i.e., ‘2’ or ‘6’.

Question 11.
Show the arrangement of electrons in first 18 elements schematically.
Answer:

Question 12.
What are the applications of isotopes in our daily life?
Answer:
Applications of isotopes :

1. Some isotopes are used for solving chemical and medical mysteries.
2. Isotopes are also commonly used in the laboratory to investigate the steps of a chemical reaction.
3. The isotope of uranium is used as a fuel in nuclear reactors.
4. The isotope of iodine is used in the treatment of goitre.
5. The isotope of cobalt is used in the treatment of cancer.

Question 13.
If an atom contains one electron and one proton, will it carry any charge or not?
Answer:
It will not carry any charge because proton is positively charged particle and electron is negatively charged particle. So they would neutralise each other.

Question 14.
What do you think would be the observation if the a – particle scattering experiment is carried out using a foil of metal other than gold?
Answer:
Gold has highest malleability and ductility. Extremely very thin foil can be prepared by using Gold, which is not possible with other metals. So, we will not get same type of results using other metals.

Question 15.
Helium atom has an atomic mass of 4u and two protons in its nucleus. How many neutrons does it have?
Answer:
Atomic mass due to protons = 2 × 1 = 2 u
Neutrons have almost same mass as protons.
Mass of neutrons = 4 – 2 = 2u
∴ Number of neutrons = 2

Question 16.
Write the distribution of electrons in carbon and sodium atoms.
Answer:
Distribution of electrons in carbon is 2, 4.
Distribution of electrons in sodium is 2, 8, 1.

Question 17.
If K and L shells of an atom are full, then what would be the total number of electrons in the atom?
Answer:
Number of electrons present in K shell = 2(1)² = 2 × 1=2
Number of electrons present in L shell = 2(2)² = 2 × 4 = 8
Total number of electrons in the atom = 2 + 8 = 10

Question 18.
If number of electrons in an atom is 8 and number of protons is also 8, then
i) what is the atomic number of atom?
ii) what is the charge on the atom?
Answer:
i) The atomic number of atom is 8.
ii) The electronic configuration of atom is 2, 6. By gaining two electrons it will get nearest inert gas configuration Neon. So, the charge on the atom is – 2.

Question 19.
Chlorine occurs in two isotopic forms that is \({ }_{17}^{35} \mathrm{Cl}\) and \({ }_{17}^{37} \mathrm{Cl}\). The percentage of these forms are 75% and 25% respectively. Find atomic weight of chlorine atom?
Answer:
The atomic mass of an element is taken as an average mass of all the naturally occurring atoms of the sample element.
The average atomic mass of chlorine atom on the bases of above data

Question 20.
For the symbol H, D and T tabulate three sub atomic particles found in each of them.
Answer:

Question 21.
Na⁺ has completely filled K and L shells. Explain.
Electron configuration of Na+ is 2, 8.
So, sodium has completed filled K and L shells because the maximum number of electrons filled in K and L shells are 2 and 8 electrons respectively.

Question 22.
If Bromine atom is available in the form of say, two isotopes \({ }_{35}^{79} \mathrm{Br}\)(49.7%) and \({ }_{35}^{81} \mathrm{Br}\) (50.3%), calculate the average atomic mass of Bromine atom.
Answer:

Question 23.
The average atomic mass of sample of an element X is 16.2 u. What are the percentages of isotopes \({ }_{8}^{16} \mathrm{X}\) and \({ }_{8}^{18} \mathrm{X}\)?
Answer:

∴ The percentage of first isotope is 90 and percentage of second isotope is 10.

Question 24.
If Z = 3, what would be the valency of the elements? Also, name the element?
Answer:
Electronic configuration of element is 2, 1 and the element is Lithium.
The element will get nearest inert gas Helium configuration by loosing one electron. So its valency is 1.

Question 25.
Composition of nuclei of two atomic species X and Y are given as under.

Give the mass number of X and Y. What is the relation between the two species?
Answer:
Mass number of X = 6 + 6 = 12
Mass number of Y = 6 + 8 = 14
So these two species have same atomic number (same number of protons) and different mass numbers. Therefore these two species are isotopes.

9th Class Physical Science 5th Lesson What is inside the Atom? 4 Marks Important Questions and Answers

Question 1.
How did the idea of sub-atomic particles evolve?
Answer:

• According to Dalton, atom is indivisible.
• But Michael Faraday’s experiments on electrolysis proved that atoms were acquiring negative charge during electrolysis.
• This is contradiction to Dalton’s theory.
• This lead to an idea that there must exist some tiny particles in atom which are responsible for atom to behave sometimes as charged particles.
• As atom is considered as electrically neutral, it probably had equal number of positive and negative constituents to maintain electrical neutrality.
• This gave scope to think about sub-atomic particles.

Question 2.
How do we determine the atomic mass of an element with its isotopes?
Answer:

• In nature, most elements occur as a mixture of two or more isotopes, each isotope has a certain percentage of natural occurrence.
• The atomic mass of an element is taken as an average mass of all the naturally occuring atoms of the sample element.
  Ex : Let us calculate the atomic mass of chlorine.
• Isotope of Cl occurs in nature, in two isotopic forms, with masses 35 units and 37 units.
• The isotope with mass 35 is present in 75% in nature.
• The isotope with mass 37 is present in 25% in nature.
• The average mass of chlorine is
  

Question 3.
Describe the Rutherford’s alpha particle scattering experiment. What are the conclu¬sions of this experiment.
Answer:

Rutherford conducted alpha particle scattering experiment in 1909 to study the atom.

1. The stream of alpha particles from a source having considerable energy is directed towards a very thin gold foil.
2. The gold foil was placed inside a detector in such a way that the detector would show a flash of light when an alpha particle struck it.
3. The entire arrangement was kept in a vacuum chamber.
4. Rutherford did not expect to see large deflections of alpha particles based on Thomson’s model.
5. But Rutherford observed the deflection of alpha particles through very large angles and a few alpha particles were reflected right back.
6. From this observation, Rutherford concluded as follows.

Conclusions :

1. Most of the space inside the atom is empty.
2. All the positive charge must be concentrated in a very small space within the atom and he named it as ‘nucleus’.

Question 4.
What information do you know from \({ }_{17}^{35} \mathrm{X}\)?
Answer:
Given that \({ }_{17}^{35} \mathrm{X}\)
a) The atomic number of element is 17.
b) Hence the element is chlorine, symbol is ‘Cl’.
c) Number of protons = 17.
d) Number of electrons = 17.
e) Mass number = 35.
f) Number of neutrons = 35-17 = 18
g) Distribution of electrons in shells

h) Valency is ‘l’.
i) It gains one electron to become octet.

AP State Syllabus AP Board 9th Class Physical Science Important Questions Chapter 6 Chemical Reactions and Equations

AP State Syllabus 9th Class Physical Science Important Questions 6th Lesson Chemical Reactions and Equations

9th Class Physical Science6th Lesson Chemical Reactions and Equations 1 Mark Important Questions and Answers

Question 1.
Fe₂O₃ + 2 Al → Al₂O₃ + 2Fe
Name the compound which is oxidized in the above reaction.
Answer:
In the reaction Fe₂O₃ + 2Al → Al₂O₃ + 2Fe,
Aluminium (Al) is oxidised and formed as Al₂O₃.

Question 2.
Give an example for displacement reaction.
Answer:
1) In displacement reaction one element displaces another element from its compound and takes its place there in.

2) Eg : Zinc pieces react with dilute hydrochloric acid and liberate Hydrogen gas.
Zn_(s)+ 2HCl_(aq) → ZnCl_2(aq) + H_2(g)

Question 3.
A shiny brown coloured element ‘X’ on heating in air becomes black in colour. Assume and write the name of the element ‘X’ and also predict the substance formed black in colour.
Answer:
Element X – Copper.
Black substance – Copper Oxide

Question 4.
Iron gets rust but Gold doesn’t, why?
Answer:
Gold does not oxidized.
(OR)
Gold is least reactive metal.

Question 5.
What happens if iron articles are exposed to moist air? Write the chemical equation to represent that reaction.
Answer:
Rusting takes place on iron articles when they are exposed to moist air.
2Fe_(s) + O_2(g) + 4H⁺_(aq) → 2Fe²⁺_(aq) + 2H₂O_(l)

Question 6.
On adding dilute hydrochloric acid to copper oxide powder, the solution formed is blue green. Write the new compound formed.
Answer:
Copper oxide reacts with hydrochloric acid and forms copper chloride and water. Copper chloride is in blue green colour.
CuO + 2 HCl → CuCl₂ + H₂O

Question 7.
Write the equation for the chemical decomposition reaction of silver chloride in the presence of sunlight.
Answer:

Question 8.
Balance the following chemical equation.
Na + H₂O → NaOH + H₂
Answer:
2Na + 2H₂O → 2NaOH + H₂

Question 9.
If you keep an iron piece in solid state CuSO₄ crystals, does it get any reaction? Guess the reason.
Answer:
Reaction will not takes place if an iron piece is placed in solid state CuSO₄ crystals because there will not exist separate Cu⁺², SO₄^-2 ions in CuSO₄ Crystals. In aqueous solution, they exists. So iron can not displace copper.

Question 10.
What is a chemical equation?
Answer:
Chemical Equation :
Describing a chemical reaction using least possible words or symbols is called a chemical equation.
Ex : CaO + H₂O → Ca(OH)₂

Question 11.
What are “Reactants” and “Products”?
Answer:
1) Reactants :
The substances which undergo chemical change in the reaction are called ‘Reactants’.

2) Products:
The new substances formed in a chemical reaction are called ‘Products’.
Ex : Zn + 2HCl → ZnCl₂ + H₂ ↑
3) In above reaction Zn and 2HCl are called reactants, ZnCl₂ and H₂ are called products.

Question 12.
What is a balanced chemical equation?
Answer:
Balanced Chemical Equation :
A chemical reaction in which the number of atoms of different elements on the reactants side (left side) are same as those on product side (right side) is called a balanced chemical equation.
Ex : Mg + O₂ → MgO (Unbalanced chemical equation)
2 Mg + O₂ → MgO (Balanced chemical equation)

Question 13.
What is chemical change?
Answer:
Chemical Change :
The process in which the change is affecting the identity of a molecule by a change in chemical composition is called a “chemical change”.

Question 14.
What is physical change?
Answer:
Physical Change: The process where a change is occurring only in physical properties, without affecting the identity of the molecules is called a “physical change”.

Question 15.
What is the meaning of “Exo”, “Endo” and “Thermo”?
Answer:
The term ‘Exo’ means outside, ‘Endo’ means inside and ‘Thermo’ means heat.

Question 16.
What is oxidation? Give examples.
Answer:
Oxidation is a reaction that involves the addition of oxygen or loss of hydrogen or electrons.

Question 17.
What is a reduction? Give examples.
Answer:
The process in which a substance loses oxygen or gains hydrogen or electrons is known as reduction.

Question 18.
What is corrosion?
Answer:
Corrosion :
When some metals are exposed to moisture, acids etc. they tarnish due to the formation of respective metal oxide on their surface. This process is called “corrosion”.

Question 19.
What is “galvanizing”?
Answer:
It is a method of protecting iron from rusting by coating them a thin layer of zinc.

Question 20.
What is combustion?
Answer:
The process of burning a substance in the presence of oxygen is called “combustion”.

Question 21.
What do you mean by “rancidity”?
Answer:
Food materials containing fat / oil are exposed to air, for a long time they react with atmospheric oxygen and it is responsible for spoiling of food. This process is called “rancidity”.

Question 22.
How do you know that respiration is an exothermic reaction?
Answer:
During the respiration reaction energy releases, so it is an exothermic reaction.

Question 23.
What will happen in chemical reaction?
Answer:
New substances are formed in chemical reaction.

Question 24.
What is precipitate?
Answer:
A precipitate is a solid product’ which separates out from the solution during a chemical reaction.

Question 25.
How does precipitation reaction occur? Explain with an example.
Answer:

1. Prepare lead nitrate and potassium iodide solutions in separate test tubes.
2. Mix the two solutions.
3. A yellow colour substance which is insoluble in water, is formed. This insoluble substance is known as precipitate. The precipitate here in this reaction is lead iodide.
   Pb(NO₃)₂ + 2KI → Pbl₂ + 2KNO₃

Question 26.
What is a balanced chemical equation?
Answer:
A balanced chemical equation has an equal number of atoms of different elements in the reactants and products.

Question 27.
What is an unbalanced equation?
Answer:
An unbalanced chemical equation has an unequal number of atoms of one or more elements in the reactants and products.

Question 28.
Why do the smell and taste of food items change?
Answer:
When fats and oils are oxidized they become rancid. So their smell and taste change.

Question 29.
“Freshly cut apple turning brown, the iron articles shiny when new, but gradually become reddish brown when left for sometime ………”. How do these changes occur?
Answer:
Oxygen molecules interact with different substance from metal to living tissue which may come into contact with it. The above changes occur. These are all the examples of the process of oxidation.

Question 30.
What are antioxidants?
Answer:
Antioxidants :
The spoilage of food can be prevented by adding preservatives which prevent oxidation. The substances which prevent oxidation are called antioxidants.
(OR)
The substances which prevent oxidation added to food containing fats and oil are called antioxidants.

Question 31.
Complete the following reaction.
Pb(NO₃)₂ + 2 KI → …………….. + …………………
Answer:
Pb(NO₃)₂ + 2 KI → Pbl₂ + 2KNO₃

Question 32.
If iron nail is dipped in copper sulphate solution, after sometime copper will be formed. Write the chemical equation for this reaction.
Answer:
Fe + CuSO₄ → FeSO₄ + Cu

Question 33.
Which metal is used in the manufacture of Diwali crackers?
Answer:
The metal used in manufacture of Diwali crackers is Magnesium.

Question 34.
What are new substances formed due to decomposition of lead nitrate?
Answer:
The new substances formed are lead oxide, nitrogen dioxide and oxygen.

Question 35.
Balance the following chemical equation. C₂H₆ + O₂ → CO₂ + H₂O
Answer:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Question 36.
NH₄Cl → NH₃ + HCl. Which type of reaction is this?
Answer:
This is decomposition reaction.

Question 37.
Which chemical reaction is involved in the corrosion of iron?
Answer:
The chemical reaction involved in corrosion of iron is oxidation reaction.

Question 38.
Which metal is used for wrapping food material? Why?
Answer:
Aluminium is used to wrap food material because it reacts with oxygen and forms a protective layer of aluminium oxide which prevents further oxidation.

Question 39.
Write states of the reactants and products and also write conditions required for the completion of reaction.
Answer:
2H₂O → 2H₂ + O₂
H₂O is in liquid state. H₂ and 0₂ are in gaseous state.
The reaction is carried out in the presence of electricity.

Question 40.
Give an example for chemical combination reaction where only elements take part.
Answer:
2 H₂ + O₂ → 2H₂O
Here hydrogen and oxygen both are elements.

Question 41.
Give an example for chemical combination where element and compound take part in the reaction.
Answer:
2 NaN02 + O₂ → 2 NaNO₃
Here O₂ is element and NaNO₂ is compound.

Question 42.
Give an example where two compounds combine together to form a compound.
Answer:
CaO + H₂O → Ca(OH)₂
Here both calcium oxide, water are compounds which combine together to form calcium hydroxide.

Question 43.
Before burning in air, why is Mg ribbon cleaned by rubbing with a sand paper?
Answer:
To remove the protection layer of basic magnesium carbonate from the surface of magnesium

Question 44.
X Pb(NO₃)₂ → Y PbO + Z NO₂ + W O₂.
If the equation is a balanced equation, what are values of coefficients X, Y, Z and W?
Answer:
The balanced equation is 2 Pb(NO₃)₂ → 2 PbO + 4 NO₂ + O₂.
∴ The values are X = 2,Y = 2, Z = 4 and W = 1.

Question 45.
Why does a layer of zinc prevent rusting of iron?
Answer:
Zinc reacts with oxygen and forms layer of zinc oxide which prevents further oxidation. So a layer of zinc protects iron from rusting.

Question 46.
Why does jewellery made of gold not rust?
Answer:
Gold does not react with air because it has least reactivity. So gold jewellery does not rust.

Question 47.

What type of reaction is this? Which element between A and C is more reactive? Why?
Answer:
This reaction is chemical displacement. ‘C’ is more reactive than ‘A’. So C displaces ‘A’ from AB and occupies its place.

Question 48.
Given AB + CD → AD + CB. What type of chemical reaction is this? What is the condition which makes the reaction possible?
Answer:
This is double displacement reaction. This reaction can be possible if ionisation of both the salt solutions are possible.
i.e., AB → A⁺ + B^–
CD → C⁺ + D^–

Question 49.
An iron nail is completely immersed in a test tube containing oil. Do you observe any rust on the iron nail? If not, why?
Answer:
No, I do not observe any rust because both air and moisture are required for rusting of iron.

Question 50.
How can we prevent rusting of iron?
Answer:
Rusting of iron can be prevented or at least minimised by shielding the metal surface from oxygen and moisture. It can be prevented by painting, oiling, greasing, galvanizing, chrome plating or making alloys.

Question 51.
Suggest few methods to avoid corrosion.
Answer:
Some methods to avoid corrosion

1. Painting, oiling, greasing on the surfaces of the metals.
2. Galvanizing the metal with thin layer of zinc.
3. Making alloy metals prevents corrosion,
   eg : brass, bronze and steel.

Question 52.
Some metals react with oxygen to form their oxides. It is serious problem. Give some examples for oxidation of metals and write balanced equations.
Answer:
1) Iron reacts with oxygen and forms iron oxide.
4Fe + 3O₂ → 2Fe₂O₃.

2) Copper reacts with oxygen and forms copper oxide.
2Cu + O₂ → 2CuO

3) Silver tarnishes when reacts with hydrogen sulphide and oxygen
4Ag + 2H₂S + O₂ → 2Ag₂S + 2H₂O

Question 53.
Which pipes are suggestable/suitable for water supply? Justify your answer.
Answer:
Steel, PVC and CPVC pipes are suitable/suggestable for water supply.
Reasons:

1. They are not oxidised in moisture.
2. They do not form oxides.
3. They are not corroded.
4. So, they are durable and safe to use in supply of water.

Question 54.
Which pipes are used by you for water supply to your house?
Answer:

1. PVC, CPVC and steel pipes are used for water supply in my house.
2. Because they do not corrode.

Question 55.
List of metals are given below. Classify them into corroded and non-corroded metals. Aluminium, Silver, Iroh, Copper, Gold, Tin, Tungsten, Platinum.
Answer:
Corroded metals :
Aluminium, Silver, Iron, Copper

Non-corroded metals :
Gold, Tin, Tungsten, Platinum.

9th Class Physical Science6th Lesson Chemical Reactions and Equations 2 Marks Important Questions and Answers

Question 1.
A light yellow coloured compound ‘X’ is exposed to sunlight for some time. It is turned into gray coloured material. What is the name of ‘X’? Predict the type of chemical reaction occured in it.
Answer:
1) The name of the compound ‘X’ is Silver Bromide (AgBr).

This is a decomposition reaction and also it occurs in the presence of sunlight. Hence, it is a photo chemical reaction.

Question 2.
Oil and fat containing food item packets are flushed with Nitrogen gas. Why?
Answer:
By stopping the oxidation it prevents the rancidity of food material.

Question 3.
N_2(g) + O_2(g) + heat → 2NO_(g)
What information do you get from the above equation? Comment.
Answer:

• Nitrogen gas reacts with oxygen gas in the presence of heat energy and forms nitric oxide gas.
• It is an endothermic reaction.
• This is an example for chemical combination reaction.
• In this reaction, the reactants are gases and products are also gases.

Question 4.
Write an activity about how you conduct an experiment to show that more reactive metals replace less reactive metals from their compounds.
Answer:

• Take two iron nails and clean then by rubbing with sand paper.
• Take two test tubes and mark them ‘A’ and ‘B’.
• Pour copper sulphate solution in the test tube ‘A’ and Zinc Chloride solution in the test tube ‘B’.
• Dip iron nails in both test tubes.
• Keep them without disturbing for 20 min.

Observation :

1. The nail which is dipped in Copper Sulphate solution becomes brownish colour.
2. The nail which is dipped in Zinc Chloride solution doesn’t change.

Reaction :
In test tube ‘A’: CuSO₄ + Fe → FeSO₄ + Cu
In test tube ‘B’: ZnCl₂ + Fe → No reaction.

Conclusion :

1. Iron is more reactive than copper. So it displaces copper from Copper Sulphate solution.
2. Iron is less reactive than zinc. So, it doesn’t displace zinc from ZnClr

Question 5.
i) CaCO_3(s) → CaO_(s) + CO_3(g)
ii) 2Ag Br_(s) → 2Ag_(s) + Br_2(g)
Mention the types of reactions to which the above equations belong. Also mention which of them is a photochemical reaction.
Answer:
1) i) CaCO_3(s) → CaO_(s) + CO_2(g). It is a chemical decomposition reaction.
ii) 2Ag Br_(s) → 2Ag_(s) + Br_2(g). It is also a chemical decomposition reaction.

2) 2Ag Br_(s) → 2Ag_(s) + Br_2(g). It is a photochemical reaction. Because, this reaction
takes place in the presence of sunlight only.

Question 6.
Write the products of given reactions, if any. Give reason.
FeCl₂ + Zn →
ZnCl₂ + Fe →
Answer:
FeCl₂ + Zn → ZnCl₂ + Fe (Displacement reaction)
ZnCl₂ + Fe → No reaction. (Low reactive metals cannot displace high reactive metals)

Question 7.
Balance the following chemical equations:
i) Na + H₂O → NaOH + H₂
ii) K₂CO₃ + HCl → KCl + H₂O + CO₂
Answer:
i) 2Na + 2H₂O → 2NaOH + H₂
ii) K₂CO₃ + 2HCl → 2KCl + H₂O + CO₂

Question 8.
Observe the following balanced chemical equation and answer the given questions.
C₃H_8(g) + 5O_2(g) → 3CO_2(g) + 4H₂O_(g)
i) How many molecules of Oxygen are involved in this chemical reaction?
ii) How many moles of Propane are required to get 20 moles of Water?
Answer:
i) In this chemical reaction five molecules of oxygen are involved.
ii) Five moles of propane are required to get 20 moles of water.

Question 9.
What do you do to prevent rusting of copper and silver articles?
Answer:
I can follow some rules given below to prevent rusting of copper and silver articles.

1. Shielding the metal surface from oxygen and moisture.
2. By painting.
3. By oiling, greasing.
4. By galvanizing, chrome plating.
5. By making alloys.

Question 10.
What are the important characteristics of chemical reactions?
Answer:
The important characteristics of chemical reactions are

1. Evolution of a gas
2. Formation of a precipitate
3. Change in colour
4. Change in temperature
5. Change in state

Question 11.
What symbols do we use to indicate the physical state of reactants and products in an equation?
Answer:

1. Solid state is indicated by the symbol (s)
2. Liquid state is indicated by the symbol (l)
3. Gaseous state is indicated by the symbol (g)
4. Aqueous solution is indicated by the symbol (aq)

Question 12.
What can we do to make a chemical equation more informative?
(OR)
How can chemical equation be made more informative by knowing?
Answer:

1. Physical state,
2. Heat changes (exothermic and endothermic reactions),
3. Gas evolved,
4. Precipitate formed.

Question 13.
Commemt on “C_(s) + O_2(g) → CO_2(g) + Heat” equation.
Answer:

• The burning of carbon in oxygen is an exothermic reaction because heat is evolved in this reaction.
• An exothermic reaction is indicated by writing + Heat or + Heat energy or just + Energy on the products side of an equation.

Question 14.
Comment on “N_2(g) + O_2(g) + Heat → 2 NO_(g)” equation.
Answer:

• The reaction between nitrogen and oxygen to form nitric oxide is an endothermic reaction because heat is absorbed in this reaction.
• An endotheumic reaction is usually indicated by writing + Heat or + Heat energy or just”+ Energy” on the reactants side of an equation.

Question 15.
Balance the following equations.
1) Na + O₂ → Na₂O
2) H₂O₂ → H₂O + O₂
3) Mg(OH)₂ + HCl → MgCl₂ + H₂O
4) Fe + O₂ → Fe₂O₂
Answer:

1. 4 Na + O₂ → 2 Na₂O
2. 2 H₂O₂ → 2 H₂O + O₂
3. Mg(OH)₂ + 2 HCl → MgCl₂ + 2 H₂O
4. 4 Fe + 3 O₂ → 2 Fe₂O₃

Question 16.
2 Cu + O₂ → 2 CuO
What information do you get from above equation?
Answer:
The above equation tells us that,

1. Copper reacts with oxygen to form copper oxide.
2. The formula of copper oxide is CuO and that of oxygen is O₂.
3. 2 moles of copper atoms react with 1 mole of oxygen molecules (O₂) to produce 2 moles of copper oxide (CuO).

Question 17.
Write examples for oxidation reaction.
Answer:
Oxidation :
It is a reaction involving addition of oxygen or removal of hydrogen from a substance.
1) Magnesium reacts with oxygen to form magnesium oxide.
2 Mg + O₂ → 2 MgO

2) Copper reacts with oxygen to form copper oxide.
2 Cu + O₂ → 2 CuO

3) Iron reacts with oxygen to form ferric oxide.
4 Fe + 3 O₂ → 2Fe₂O₃

Question 18.
Write examples for reduction reaction.
Answer:
Reduction :
It is a reaction involving addition of hydrogen or removal of oxygen from a substance.
1) Nitrogen gas reacts with hydrogen gas to produce Ammonia.
N₂ + 3H₂ → 2NH₃

2) Oil reacts with hydrogen to form fat.
Oil + H₂ → Fat

Question 19.
Write the examples for corrosion reaction.
Answer:
1) The black coatings on silver.
4 Ag + 2H₂S + O₂ → 2 Ag₂S + 2H₂O

2) Green coating on copper.
2 Cu + O₂ → 2 CuO

Question 20.
Name the reactants and products in the following chemical equations.
Na₂SO₄ + BaCl₂ → BaSO₄ + NaCl
Answer:
Reactants are Sodium sulphate and Barium chloride. Products are Barium sulphate and Sodium chloride.

Question 21.
Balance the following chemical equation and follow the steps involved in balancing a chemical equation.
Cu₂S + O₂ → Cu₂O + SO₂
Answer:
Step – 1 : Write the unbalanced equation using correct chemical formula for all substances.
Cu₂S + O₂ → Cu₂O + SO₂

Step – 2 : Compare number of atoms of each element on both sides.

Balancing Cu, S, O atoms both sides
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
The equation is balanced.

Step – 3 : Write the coefficient of smallest ratio.
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂

Step – 4 : Verify above equation for balancing of atoms of each element on both sides.
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂

Question 22.
Write the difference between oxidation and reduction. Give example.
Answer:

Question 23.
Give two examples for chemical reactions in which precipitate is formed.
Answer:
The reactions in which a substance insoluble in water is formed are called precipitation reactions.
e.g.:
1) When lead nitrate solution reacts with potassium iodide solution it forms a yellow precipitate of lead iodide.
Pb(NO₃)_2(aq) + 2KI_(aq) → PbI_2(s) + 2KNO_3(aq)

2) When sodium sulphate solution reacts with barium chloride solution it forms a white precipitate of barium sulphate.
Na₂SO4_(aq) + BaCl_2(aq) → BaSO_4(s) + 2NaCl_(aq)

Question 24.
What are exothermic and endothermic reactions?
Answer:
Exothermic reaction :
A chemical reaction in which heat is released is called exothermic reaction.
Ex :
C + O₂ → CO₂ + Heat,
CaO + H₂O → Ca(OH)₂ + Heat

Endothermic reaction :
A chemical reaction in which heat is absorbed is called endothermic reaction.
Ex :
2NaHCO₃ + Heat → Na₂CO₃ + H₂O + CO₂

Question 25.
What is an alloy? Give one example.
Answer:
A metallic substance made by mixing and fusing two or more metals or a metal and a non-metal, to obtain desirable qualities such as hardness, lightness and strength is called alloy.
Ex : Brass, bronze, steel.

Question 26.
What is photochemical reaction? Give example.
Answer:
The decomposition reaction occurs in the presence of sunlight is called photochemical.. reaction.
Ex : 2AgBr_(s) → 2Ag_(s) + Br_2(g)

Question 27.
Why is power supply to our home from the electrical pole interrupted?
Answer:

• Sometimes during rainy season the power supply to our home from the electric pole will be interrupted due to the formation of the metal oxide layer on the electric wire.
• This metal oxide is an electrical insulator.

Question 28.
Name the reactions involved in the following reactions with reasons.

Answer:
a) It is a decomposition reaction carried out in the presence of heat. So it is thermal decomposition reaction.
b) It is a decomposition reaction carried out in the presence of light. So it is photochemical reaction.
c) It is a decomposition taking place in the presence of electricity. So it is electrolysis reaction.
d) Lead is more reactive than copper. So lead displaces copper from salt solution. So the reaction is chemical displacement reaction.

Question 29.
Ramu told Ravi that all material made of iron and its alloys are rusted when exposed to air. How do you correct him by asking suitable questions?
Answer:

1. What are the material we will generally use in our cooking utensils?
2.  What are the material used for surgical equipments?
3. What happens when stainless steel vessels are exposed to air?

Question 30.
What are the gases released when lead nitrate is heated and how do you identify those gases?
Answer:
When lead nitrate is heated it decomposes into lead oxide, nitrogen dioxide and oxygen.

Nitrogen diox-ide is identified by its reddish brown co|our. Oxygen makes the burning splinter to burn brightly when it is placed in test tube containing oxygen.

Question 31.
Write some chemical reactions occurring in our daily life.
Answer:

1. Souring of milk
2. Formation of curd from milk
3. Cooking of food
4. Digestion of food in our body
5. Fermentation of grapes
6. Rusting of iron
7. Burning of fuels
8. Burning of candle wax
9. Ripening of fruits.

Question 32.
If 40 gm of methane is burnt, then how much amount of CO₂ is released ?
Answer:

When 16 g of methane is burnt it releases 44 g of CO₂. The amount of CO₂ released when 40 g methane burnt 44
= \(\frac{44}{16}\) × 40=110g

Question 33.
Calculate the amount of calcium oxide formed when 2 kg of calcium carbonate is decomposed. (The atomic masses of Ca = 40 U, C = 12 U, O = 16 U).
Answer:
The balanced equation is
CaCO_3(s) → CaO_(s) + CO_2(g)
(40 + 12 + 3 x 16) U → (40 + 16) U + (12 + 2 x 16) U
100 U → 56 U + 44 U
100 g → 56g + 44g
100 g calcium carbonate on decomposition produces 56 g of calcium oxide.
Amount of calcium oxide formed due to decomposition of 2 kg of calcium carbonate

Question 34.
A solution of common salt when added to silver nitrate solution yields a precipitate of silver chloride (0.28 g). Find the mass of sodium chloride in the solution and also the mass of sodium nitrate formed.
Answer:

Question 35.
0.29 g of hydrocarbon when burnt completely in oxygen produces 448 ml of carbon dioxide at STP. From the information, calculate the
i) mass of carbon dioxide formed
ii) mass of element carbon in carbon dioxide
iii) mass of hydrogen in hydrocarbon.
Answer:
i) Gram molecular weight of carbon dioxide = 12 + 2 × 16g = 44 g.
22.4 litres of CO₂ at STP weighs 44 g.

The weight of 448 ml of CO₂ at STP = \(\frac{44}{22400}\) × 448 = 0.88 g

ii) 44g of CO₂ contains 12 g of carbon.
The amount of carbon present in 0.88 g of CO₂ = \(\frac{12}{44}\) × 0.88 = 0.24 g

iii) Weight of hydrogen = Weight of hydrocarbon – Weight of carbon
= 0.29 – 0.24 = 0.05 g

Question 36.
Calculate the weights of carbon dioxide and water that will be obtained by completely burning 0.25 g of an organic compound having molecular formula C₄H₄O₄.
Answer:

Question 37.
What weight of sulphuric acid will be required to completely dissolve 3g of magnesium carbonate? Calculate the volume of carbon dioxide evolved at STP.
Answer:

Question 38.
If 16.4 g of calcium nitrate is heated : (a) Calculate the volume of Nitrogen dioxide obtained at STP and (b) The mass of calcium oxide obtained.
Answer:

Question 39.
Give some examples for corroded and non-corroded metals and give the reasons for non-corrosion of metals.
Answer:
Examples to corroded metals :

1. Iron
2. Copper
3. Silver

Reactions:
4Fe + 3O₂ → 2Fe₂O₃
2Cu + O₂ → 2CuO
4Ag + 2H₂S + O₂ → 2Ag₂S + 2H₂O

Exmples to non-corroded metals :

1. Gold
2. Platinum
3. Brass
4. Bronze
5. Steel

Reasons:

1. Gold and platinum metals do not react with oxygen and have resistance to corrosion.
2. Brass, bronze and steel are alloys. So they have a quality of resistance to corrosion.

Question 40.
Iron is a corroded metal. Through alloying we can prevent corrosion. Justify.
Answer:

• Alloying is a very good method of improving properties of metal.
• Generally pure form of iron is very soft and stretches easily when hot.
• When iron is mixed with carbon, nickel and chromium, an alloy stainless steel is obtained.
• The stainless steel is hard and does not rust.
• So, through alloying iron we can prevent corrosion.

Question 41.
“Through alloying corrosion can be prevented.” For the justification pose some questions.
Answer:

1. What is alloying?
2. How are alloys prepared?
3. What are the qualities of alloys?
4. Which quantity prevents the corrosion in alloys?
5. Can we prevent corrosion by alloying a metal ?

9th Class Physical Science6th Lesson Chemical Reactions and Equations 4 Marks Important Questions and Answers

Question 1.
Write the balanced chemical reaction for the following and identify the type of reaction in each case.
A) Magnesium_(s) + Iodine_(g) → Magnesium iodide_(s)
B) Zinc_(s) + Hydrochloric acid_(aq) → Zinc chloride_(aq) + Hydrogen_(g)
Answer:
A) Magnesium_(s) + Iodine_(g) → Magnesium iodide_(s)
Mg + I₂ → Mgl₂. This reaction is chemical combination.

B) Zinc_(s) + Hydrochloric acid_(aq) → Zinc chloride_(aq) + Hydrogen_(g)
Zn + 2HCl → ZnCl₂ + H₂ ↑
This reaction is chemical displacement.

Question 2.
Write an activity to each of the following chemical reaction.
A) Photo chemical reaction
B) Chemical displacement reaction.
Answer:
A) Photo chemical reaction :
1) Take a pinch of Silver Bromide in a watch glass and it in the presence of sunlight.
2) Silver Bromide decomposes to silver and Bromine in sunlight.

B) Chemical displacement:

1. Take a small quantity of zinc dust in a conical flask and add some drops of dilute hydrochloric acid slowly.
2. Immediately we can observe the gas bubbles coming out from the solution.
3. This is because zinc reacts with dilute hydrochloric acid and liberates hydrogen gas.
   Zn + 2HCl → ZnCl₂ + H₂ ↑
4. In this reaction the element zinc has displaced hydrogen from hydrochloric acid.

Question 3.
Why should we balance a chemical equation ? Take any one chemical equation and explain the procedure of balancing it.
Answer:
Chemical reactions obey law of conservation of mass. So the total number of atoms of each elements in the reactants must be equal to the total number of atoms of each element in the

products. So we should have to balance chemical equation.
Eg : H₂ + O₂ → H₂O

Step -1 : Unbalanced equation = H₂ + O₂ → H₂O
Step – 2 : Compare no.of atoms of each element on both sides.

No.of 0 atoms balancing – H₂ + O₂ → 2H₂O
No.of H atoms balancing – 2H₂ + O₂ → 2H₂O

Step – 3 : The above equation is balanced and write the coefficients in the smallest ratio. 2H₂ + O₂ → 2H₂O

Step – 4 : Verify above equation for balancing of atoms each element on both sides. Hence the equation is balanced.
∴ 2H₂ + O₂ → 2H₂O

Question 4.
Balance the following chemical equations.
i) Zn_(s) + Ag NO_3(aq) → Zn (NO₃)_2(aq) + Ag_(s)
ii) Fe₂O_3(s) + C_(s) → Fe_(s) + CO_2(g)
iii) Ag_(s) + H₂S_(g) → Ag₂S_(s) + H₂O_(l)
iv) Cu_(s) + O_2(g) → CuO_(g)
Answer:
i) Zn_(s) + Ag NO_3(aq) → Zn (NO₃)_2(aq) + Ag_(s)
Balanced equation : Zn_(s) + 2 AgNO_3(aq) → Zn (NO₃)_2(aq) + 2Ag_(s)

ii) Fe₂O_3(s) + C_(s) → Fe_(s) + CO_2(g)
Balanced equation : 2Fe₂O_3(s) + 3C_(s) → 4Fe_(s) + 3CO_2(g)

iii) Ag_(s) + H₂S_(g) → Ag₂S_(s) + H₂O_(l)
Balanced equation :
We cannot balance the equation because atom ‘O’ does not exist in the reactants.

iv) Cu_(s) + O_2(g) → CuO_(s)
Balanced equation :
2Cu_(s) + O_2(g) → 2CuO_(s)

Question 5.
Write the equation for the reaction of zinc with hydrochloric acid and balance the equation. Find out the number of molecules of hydrogen gas produced in this reaction, when 1 mole of HCl completely reacts at S.T.P.
[Gram molar volume is 22.4 liters at S.T.P., Avogadro’s number is 6.023 × 10²³]
Answer:
Zn + 2HCl → ZnCl₂ + H₂

• In the reaction 2 moles of HC/ produce 1 mole of H₂.
• If 1 mole of HCl participate in reaction, Vi mole of hydrogen will be produced.
• 1 mole of H₂ gas contains 6.023 × 10²³ molecules at STP.
  Number of molecules in ½ mole of H₂ gas = 6.023 × 10²³ × ½ = 3.011 × 10²³

Question 6.
Write the examples for chemical combination.
Answer:
1) Magnesium burns in oxygen to form magnesium oxide.
2 Mg + O₂ → 2 MgO

2) When coal is burnt in oxygen, carbon dioxide is produced.
C + O₂ → CO₂ + Q (cheat energy)

3) Slaked lime is prepared by adding water to quick lime.
CaO + H₂O → Ca(OH)₂ + Q (heat energy)

4) When hydrogen reacts with oxygen, it gives water.
H₂ + O₂ → 2 H₂O

5) Hydrochloric acid is obtained by adding hydrogen to chlorine.
H₂ + Cl₂ → 2 HCl

6) Magnesium reacts with iodine to magnesium iodide.
Mg + l₂ → Mgl₂

7) Sodium reacts with chlorine to form sodium chloride.
2 Na +Cl₂ → 2 NaCl

8) Iron reacts with oxygen to form haematite.
Fe + 3O₂ → 2 Fe₂O₃

Question 7.
Write examples for chemical decomposition reaction.
Answer:
1) Calcium carbonate on heating decomposes to calcium oxide and carbon dioxide.

2) On electrolysis, water decomposes to water and hydrogen.
2 H₂O → 2 H₂ + O₂

3) Silver bromide decomposes to silver and bromine in sunlight.
2 AgBr → 2 Ag + Br₂

4) Silver chloride decomposes to silver and chlorine.
2 AgCl → 2 Ag + Cl₂

5) Glucose decomposes to ethanol and carbon dioxide.
C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂

6) Sodium bicarbonate decomposes to sodium carbonate, water and carbon dioxide.
2 NaHCO₃ + Heat → Na₂CO₃ + H₂O + CO₂

7) On heating lead nitrate decomposes to lead oxide, oxygen and nitrogen dioxide.

Question 8.
Write the examples for chemical displacement reaction.
Answer:
1) Zinc pieces react with dilute hydrochloric acid and liberate hydrogen gas
Zn + 2 HCl → ZnCl₂ + H₂

2) Iron reacts copper sulphate to form iron sulphate and copper.
Fe + CuSO₄ → FeSO₄ + Cu

3) Zinc is mixed with silver nitrate to form zinc nitrate and silver.
Zn + 2 AgNO₃ → Zn(NO₃)₂ + 2Ag

4) Lead reacts with copper chloride to form lead chloride and copper.
Pb + CuCl₂ → PbCl₂ + Cu

5) Sodium reacts with water to form sodium hydroxide and hydrogen.
2 Na + 2H₂O → 2 NaOH + H₂

6) Aluminium reacts with copper chloride to form aluminium chloride and copper.
2 Al + 3 CuCl₂ → 2 AlCl₃ + 3 Cu

7) Zinc reacts with sulphuric acid to form zinc sulphate and hydrogen.
Zn + H₂SO₄ → ZnSO₄ + H₂

Question 9.
Write examples for chemical double displacement reaction.
Answer:
1) Sodium sulphate solution on mixing with barium chloride solution forms a white precipitate of barium sulphate and soluble sodium chloride.
Na₂SO₄ + BaCl₂ → BaSO₄ + 2 NaCl

2) Sodium hydroxide reacts with hydrochloric acid to form sodium chloride and water.
NaOH + HCl → NaCl + H₂O

3) Sodium chloride spontaneously combines with silver nitrate in solution giving silver chloride precipitate.
NaCl + AgNO₃ → AgCl + NaNO₃

4) Mix lead nitrate solution and potassium iodide solution to form a yellow precipitate of lead iodide and potassium nitrate.
Pb(NO₃)₂ + 2Kl → PbI₂ + 2KNO₃

5) Calcium hydroxide reacts with nitric acid to form water and calcium nitrate.
Ca(OH)₂ + 2 HNO₃ → 2H₂O + Ca(NO₃)₂

6) Magnesium chloride reacts with potassium hydroxide to form magnesium hydroxide and potassium chloride.
MgCl₂ + 2 KOH → Mg(OH)₂ + 2 KCl

Question 10.
Balance the following equations.
1) Al(OH)₃ → Al₂O₃ + H₂O
2) NH₃ + CuO → Cu + N₂ + H₂O
3) Al₂(SO₄)₃ + NaOH → Al(OH)₃ + Na₂SO₄
4) HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + H₂O
5) NaOH + H₂SO₄ → Na₂SO₄ + H₂O
6) BaCl₂ + H₂SO₄ → BaSO₄ + HCl
Answer:

1. 2 Al(OH)₃ → Al₂O₃ + 3 H₂O
2. 2 NH₃ + 3 CuO → 3 Cu + N₂ + 3 H₂O
3. Al₂(SO)₃ + 6 NaOH → 2 Al(OH)₃ + 3 Na₂SO₄
4. 2 HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2 H₂O
5. 2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O
6. BaCl₂ + H₂SO₄ → BaSO₄ + 2 HCl

Question 11.
How many types of chemical reactions are there? Explain with examples.
Answer:
There are four types of chemical reactions. They are :
i) Chemical combination :
A chemical reaction in which two or more substances combine together to form a new substance is called chemical combination.
e.g.: 2H₂ + O₂ → 2H₂O
CaO + H₂O → Ca(OH)₂

ii) Chemical decomposition :
The reaction in which a compound breaks up into two or more simpler substances are known as decomposition reaction. These reactions are generally carried out by means of heat, light, electricity or catalysts.
e.g.: CaCO₃ → CaO + CO₂
2Pb(NO₃)₂ → 2 PbO + 4NO₂ + O₂

iii) Chemical displacement:
The chemical reaction in which one element takes the place of another element in a compound is called displacement reaction.
In these reactions, an atom or group of atoms in a molecule is replaced by another atom or a group of atoms.
e.g.: Zn + 2HCl → ZnCl₂ + H₂ ↑
Zn + CuSO₄ → ZnSO₄ + Cu

iv) Chemical double displacement: The reaction in which two compounds react to form two other compounds by mutual exchange of their ions is called double displacement reaction.
e.g.: NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)
NaNO_3(aq) + AgCl_(aq) → AgNO_3(s) + NaCl_(aq)

Question 12.
Balance the following chemical equations.
a) Na₂SO₄ + BaCl₂ → BaSO₄ + NaCl
b) Al₄C₃ + H₂O → CH₄ + Al(OH)₃
c) Pb(NO₃)₂ → PbO + NO₂ + O₂
d) Fe₂O₃ + Al → Al₂O₃ + Fe
Answer:
a) Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl
b) Al₄C₃ + 12H₂O → 3CH₄ + 4Al(OH)₃
c) 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂
d) Fe₂O₃ + 2Al → Al₂O₃ + 2Fe

Question 13.
How can we make a chemical equation information?
Chemical equations can be made more informative by expressing following characteristics of the reactants and products.
1. Expressing the physical state :
The different states, i.e. gaseous, liquid and solid states are represented by the notations (g), (l) and (s) respectively. If the substance is present as a solution in water the word aqueous is written.

2. Expressing the heat changes :
Q is heat energy which is shown with plus ’+’ sign on product side for exothermic reactions and minus sign on product side for endothermic reactions.
e.g.: a) C_(s) + O_2(g) → CO_2(g) + Q (exothermic reaction)
N_2(g) + O_2(g) → 2NO_(g) ” Q (endothermic reaction)

3. Expressing the gas evolved :
If a gas evolved in a reaction, it is denoted by an upward arrow ↑ or (g).
Zn_(s) + H₂SO_4(aq) → ZnSO_4(aq) + H_2(g)↑

4. Expressing precipitate formed: If a precipitate is formed in the reaction, it is denoted by downward arrow ↓.
AgNO_3(aq) + NaCl_(aq) → AgCl_(s) ↓ + NaNO_3(aq)

Question 14.
Give daily life examples of oxidation.
Answer:
Daily life examples :

1. Combustion of fuels.
2. Corrosion of metals.
3. Change of colour of fruits like apples, bananas, when they are cut.
4. Burning of crackers.
5. Rancidity of food material.
6. During rainy season the power supply to our home from the electric pole will be interrupted due to formation of metal oxide layer on the electric wire.
7. Rising of dough with yeast depends on oxidation of sugars to carbon dioxide and water.
8. Bleaching of coloured objects using moist chlorine.
9. Respiration.

Question 15.
What is the information giyen by balanced chemical equation?
Answer:

• A chemical equation gives information about the reactants and products through their symbols and formulae.
• It gives the ratio of molecules of reactants and products.
• As molecular masses are expressed in unified masses, the relative masses of reactants and products are known from the equation.
• If the masses are expressed in grams, then the equation also gives the molar ratios of reactants and products.
• If gases are involved, we can equate the masses to their volumes.
• Using molar mass and Avagadro’s number we can calculate the number of molecules and atoms of different substances from the equation.

Question 16.
Write the balanced equation and identify the type of reaction.
1) Magnesium Hydroxide_(aq) + Nitric Acid_(aq) → Magnesium Nitrate_(aq) + Water_(l)
2) Magnesium_(s) + Carbon Monoxide_(g) → Magnesium Oxide_(s) + Carbon_(g)
3) Barium Chloride_(aq) + Sodium Sulphate_(aq) → Barium Sulphate_(s) + Sodium Chloride_(aq)
4) Sodium Nitrate_(s) → Sodium Nitrite_(s) + Oxygen_(g)
Answer:
1) Mg(OH)_2(aq) + 2 HNO_3(aq) → Mg(NO₃)_2(aq) + 2 H₂O_(l)
It is both double displacement and neutralisation reaction.

2) Mg_(s) + CO_(g) → MgO_(s) + C_(s)
It is a redox reaction in which magnesium is oxidised and carbon monoxide is reduced.

3) BaCl_2(aq) + Na₂SO_4(aq) → BaSO_4(s) ↓ + 2 NaCl_(aq)
It is both precipitation and double displacement reaction.

4) 2 NaNO_3(s) → 2 NaNO_2(s) + O_2(g)
It is both endothermic and decomposition reaction.

Question 17.
We write symbol of water as H₂O. State why we should not write it as HO₂.
Answer:
1) The hydrogen atom has only T electron in its outermost shell, so it needs 1 more electron to achieve the stability. ‘2’ electrons are required to get inert gas electronic configuration.

2) The oxygen atom has ‘6’ electrons in its outermost shell and it needs ‘2’ more electrons to compare the stability. ‘8’ electron arrangement of inert gas is neon.

In the water molecule, central oxygen atom has two pairs of unshared electrons which have not been utilised in the formation of bonds.

So, we write symbol of water as H20 and we should not write it as HO₂.

Question 18.
Latha took some quantity of powder of a substance in a test tube. Heated it with spirit lamp. A gas was liberated. She sent the gas into another test tube. The colour of solution in the second test tube turned into milk white.
Answer the following questions :
a) Which substance was heated?
b) Which gas was liberated?
c) What was the solution taken in second test tube?
d) Which type of chemical reactions involved the experiment?
Answer:
a) The substance is calcium carbonate.
b) The gas liberated is carbon dioxide.
c) The solution taken in the second test tube was solution of slaked lime.
d) Two types of reactions took place in this experiment, i.e. decomposition and double displacement.
CaCO_3(s) → CaO_(s) . CO₂ ↑
Ca(OH)_2(aq) + CO_2(g) → CaCO_3(s) + H₂O

Question 19.
A light yellow colour substance (some quantity) on a watch glass is put in the sunlight. It changes into grey colour substance.
a) What is the light yellow colour substance?
b) What is the grey colour substance?
c) Which type of chemical reaction it is?
d) Write the chemical equation for the reaction.
Answer:
a) The light yellow colour substance is silver bromide.
b) The grey colour substance is silver.
c) The type of chemical reaction is photochemical reaction.

Question 20.
Heat is liberated in the reactions where water is added to calcium oxide and hydrochloric acid added to zinc pieces.
Rarnu says that they are same type of chemical reactions. Eswar’s opinion is that the reactions are not similar. What is the basis of Eswar’s opinion ? Write equations for the above reaction.
Answer:
Eswar’s thinking is correct. Although both are exothermic reactions, the type of chemical reactions is different.
Case (i) : When water is added to calcium oxide it forms calcium hydroxide. It is an example for combination reaction.
CaO_(s) + H₂O → Ca(OH)_2(aq)

Case (ii) : When hydrochloric acid is added to zinc pieces it would liberate hydrogen gas which is an example for displacement reaction.
Zn_(s) + HCl_(aq) → ZnCl_2(aq) + H_2(g) ↑

Question 21.
We see many combustion and oxidation reactions in our daily life. Among them every combustion reaction is an oxidation reaction. But not all the oxidation reactions are combustion reactions. Do you agree or disagree with the statement? Explain with proper reasons.
Answer:
1) Yes, I agree with the statement because combustion reaction is nothing but burning of a substance in the presence of oxygen so it is an oxidation reaction. So every combustion reaction is an oxidation reaction.

2) Whereas some reactions which do not require burning still they are oxidation reactions.
Eg :

1. Corrosion of metals.
2. Change of colour of fruits like apples, bananas when they are cut.
3. Rancidity of food materials.
4. Respiration.
5. Bleaching of coloured objects using moist chlorine.
   So all oxidation reactions are not combustion reactions.

Question 22.
Explain the following :
1) What happens when iron filings are added to zinc sulphate solution?
2) What happens when solid silver nitrate is added to solid sodium chloride?
Answer:
1) No reaction takes place because zinc is more reactive than iron. So iron cannot displace zinc from its salt solution.

2) No reaction takes place because in solid state silver nitrate as well as sodium chloride is unable to dissociate into constituent ions. So exchange of ions is not possible.

Question 23.
Why is steel not used for surgical equipment and what is the material used for surgical equipment? Why is that material used for surgical equipment?
Answer:
1) Steel undergoes rusting when It exposed to air. So it is not useful for preparation of surgical equipment. If we use it in surgical equipment it may cause septic of wound.

2) The material used for surgical equipment is stainless steel which is an alloy of iron, carbon, nickel and chromium. Chromium does not easily react with oxygen. So, addition of chromium makes the stainless steel free from rusting. Therefore it is used in surgical equipment.

Question 24.
Give reasons for the following.
1) Why do we add salt to water during electrolysis of water?
2) Why does hydrogen gas put off the burning splinter and it burns with blue flame?
3) Why does carbon dioxide turn lime water milky?
Answer:
1) Pure water is a bad conductor of electricity. By adding a small amount of salt the solution becomes conductor of electricity.

2) Hydrogen does not support combustion. So it puts off the burning splinter and also it is combustible. So burns with blue flame.

3) When we pass carbon dioxide through lime water it turns into milky because lime water (calcium hydroxide) reacts with carbon dioxide and forms a white milky substance, i.e. calcium carbonate.

Question 25.
A student was given the following substances and was asked to show types of chemical reactions through experiment. Write how he would have done that. Copper sulphate solution, barium chloride solution, ferrous sulphate crystals, iron nails, calcium oxide, water.
Answer:
Given chemicals are CuSO₄ solution, BaSO₄ solution, iron nails (Fe), Ferrous sulphate (FeSO₄), Calcium oxide (CaO) and water (H₂O).

i) Chemical combination :
Chemicals chosen : CaO, H₂O
CaO + H₂O → Ca(OH)₂

When we add water to calcium oxide it produces Calcium hydroxide. This is an example for chemical combination.

ii) Chemical decomposition:
Chemicals chosen : Ferrous sulphate

When we heat FeSO₄ it dissociates into Ferric oxide, Sulphur dioxide and Sulphur trioxide respectively.

iii) Chemical displacement :
Chemicals chosen : Iron nail – CuSO₄
Fe + CuSO₄ → FeSO₄ + Cu ↓

When iron nail is placed in CuSO₄ solution, the solution turns into light green due to formation of FeSO₄ and reddish brown deposit of Copper.

iv) Double displacement:
Chemicals chosen : CuSO₄ solution and BaCl2 solution.
CuSO_4(aq) + BaCl_2(aq) → BaSO_4(s) ↓ + CuCl_2(s)

When CuSO₄ is mixed with BaCl₂ solution it forms white precipitate of BaSO₄ and aqueous solution of Copper (II) Chloride.

This is an example for double displacement reaction.

Question 26.
Take two beakers and prepare lead nitrate aqueous solution and potassium iodide aqueous solutions. What are the colours of the solutions? Now mix them in another beaker. What happens? What type of chemical reaction it is? What are products?
Answer:
The colours of the solutions are white or colourless. When lead nitrate solution is mixed with potassium iodide solution we will get yellow precipitate of lead iodide. This is double displacement reaction. The products are lead iodide and potassium nitrate.
Pb(NO₃)_2(aq) + 2Kl_(aq) → Pbl_2(s) + 2KNO_3(aq)

Question 27.
Observe the following equation which shows the action of heat on Calcium Nitrate
2 Ca(NO₃)₂ → 2 CaO + 4 NO₂ + O₂
a) How many moles of NO₂ are formed when a mole of 2 Ca(NO₃)₂ is decomposed?
b) What is the volume of NO₂ produced when 164 gm of Ca(NO₃)₂ is heated at constant temperature and pressure?
c) Calculate the mass of Calcium Oxide formed when 82 gm of Ca(NO₃)₂ is heated.
d) What is the quantity of Ca(NO₃)₂ required to produce 5 moles of gaseous products?
Answer:
Given balanced equation is
a) From the balanced equation 2 moles of Ca(NO₃)₂ releases 4 moles of NO₂.
b) Molecular weight of Ca(NO₃)₂ and CaO respectively are 164 and 56.
From the equation at STP 2 × 164 g. of Ca(NO₃)₂ releases 4 × 22.4 litres of NO₂.
At similar conditions 164 g. of Ca(NO₃)₂ releasing NO₂ in litres is
= \(\frac{164}{2 \times 164}\) × 22.4 × 4 = 2 × 22.4 = 44.8 liters.

c) From the balanced equation 164 g. of Ca(NO₃)₂ decomposes and forms 112 g. of CaO. Similarly 82 g. of Ca(NO₃)₂ decomposes and forms
\(\frac{82}{164}\) × 112 = 56 g. of CaO.

d) From the above balanced equation 2 moles of Ca(NO₃)₂ releases 5 moles of gaseous products.
∴ The mass of Ca(NO₃)₂ required is 2 × 164 = 328 g.

Question 28.
Zn + HCl → ZnCl₂ + H₂
Calculate amount of zinc required to release 500g of hydrogen. (Zn = 65 U, H = 1 U, Cl = 35.5 U are the atomic masses).
Answer:
The balanced chemical equation is
Zn + 2 HCl → ZnCl₂ + H₂
65 U + (2 × 36.5) U → (65 + 2 × 35.5) U + 2 U
65 g + 73 g → 136 g + 2g
As per the balanced equation
65 g Zinc is reacting with hydrochloric acid to produce 2g of Hydrogen.

Question 29.
Calculate the volume, mass and number of molecules of carbon dioxide when 104 g of acetylene (C₂H₂) burnt in air. (Atomic masses of C = 12 U, H = 1 U, O = 16 U).
Answer:
The balanced chemical equation is
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
2 × (2 × 12U + 2 × lU) + 5 × (2 × 16U) → 4 × (12U + 2 × 16U) + 2(2 × 1U + 16U)
52 g + 160g → 176 g + 36 g
52 g of acetylene when burnt in air produces 176 g carbon dioxide.
The amount of carbon dioxide produced when 104 g acetylene burnt

At S.T.P. 1 gram molar mass of any gas occupies 22.4 litres.
So 44g carbon dioxide occupies 22.4 litres volume.
The volume occupied by 352 g carbon dioxide = \(\frac{352}{44}\) × 22.4 = 8 × 22.4 = 179.2 litres.
44 g of carbon dioxide i.e., 1 mole of CO₂ contains 6.02 × 10²³ molecules. So the number of molecules present in 352 g of carbon dioxide
352
= \(\frac{352}{44}\) × 6.02 × 10²³ = 8 × 6.02 × 10²³
= 4.816 × 10²⁴ molecules.

AP State Syllabus AP Board 9th Class Physical Science Important Questions Chapter 7 Reflection of Light at Curved Surfaces

AP State Syllabus 9th Class Physical Science Important Questions 7th Lesson Reflection of Light at Curved Surfaces

9th Class Physical Science 7th Lesson Reflection of Light at Curved Surfaces 1 Mark Important Questions and Answers

Question 1.
Which mirror is used as rear-view mirror in the vehicles?
Answer:
Convex mirror is used as rear view mirror in the vehicles.

Question 2.
What is the relation between focal length (f) and radius of curvature (R)?
Answer:
The radius of curvature of a spherical mirror is twice to its focal length.
⇒ R = 2f (or) f = \(\frac{R}{2}\).

Question 3.
Can a virtual image be photographed by a camera?
Answer:
Yes, virtual image can be photographed by a camera.

Question 4.
Complete the diagram and draw the image.

Answer:

Question 5.
Predict and write the reason, why the value of the distance of the object (u) is always negative in the mirror equation.
Answer:
i) Direction of the incident rays is taken as positive (+ve).
ii) Object distance is measured from the pole to the object in the opposite direction of incident rays.

Question 6.
Which property of concave mirror is used in making the solar cooker?
Answer:
Rays coming parallel to the principal axis of a concave mirror is focused at focal point. Based on this property solar cooker is made.

Question 7.
Draw the ray diagram to show the formation of image for the object of height 1 cm. placed at 5 cm. distance, in front of a convex mirror having the radius of curvature R = 5 cm.
Answer:

Question 8.
What is reflection?
Answer:
The light rays falling on a surface are returned into the original medium. This phenomenon is called reflection.

Question 9.
What is the relation between focal length and radius of curvature?
Answer:
Radius of curvature = 2 x focal length
∴ R = 2f (or) f = \(\frac{R}{2}\)

Question 10.
What is the mirror formula for spherical mirrors?
Answer:
The mirror formula is \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}\)
f = focal length of mirror ; u = object distance ; v = image distance

Question 11.
What is a real image? What is a virtual image?
Answer:
Real image :
The image formed due to convergence of light rays. The real image can be caught on the screen.

Virtual image :
The image that we get by extending the rays backwards is called a virtual image. A virtual image cannot be caught on the screen.

Question 12.
What is focal length?
Answer:
The distance between focus and vertex.

Question 13.
What is radius of curvature?
Answer:
The distance between vertex and centre of curvature.

Question 14.
What is magnification?
Answer:
The ratio of size of image to size of object is called magnification.

(OR)

The ratio of image distance to object distance is called magnification.

Question 15.
Why are concave and convex mirrors called spherical mirrors?
Anwer:
The reflecting surface of convex and concave mirror is considered to form a part of the surface of a sphere. So they are called spherical mirrors.

Question 16.
What is a reflecting surface?
Answer:
The surface used for reflection is called reflecting surface.

Question 17.
What is principal axis?
Answer:
The horizontal line which passes through the centre of curvature is called principal axis.

Question 18.
What is meant by converging of light rays?
Answer:
If light rays after reflection meet at a point, then we say the light rays are converging.

Question 19.
When do you say light rays are diverging?
Answer:
If light rays appear as if they are coming from a point after reflection, then we say light rays are diverging.

Question 20.
When does a ray reflect in the same path from a concave mirror?
Answer:
When it passes through centre of curvature.

Question 21.
When a light ray travelling from parallel to principal axis falls on concave mirror, then what is the path of reflected ray?
Answer:
The reflected ray passes through focal point.

Question 22.
Where do you place the vessel in solar cooker?
Answer:
We place the vessel in solar cooker at the focal point.

Question 23.
Name a mirror that can give an erect and enlarged image of an object.
Answer:
Concave mirror can give an erect and enlarged image of an object.

Question 24.
Can a convex mirror burn a paper? If not, why?
Answer:
The rays coming parallel to principal axis after reflection diverge from the mirror. So we cannot burn a paper by using a convex mirror.

Question 25.
Which mirror has wider field of view?
Answer:
A convex mirror has wider field of view, that’s why they are used as rear view mirrors in vehicles.

Question 26.
Why does our image appear thin or bulged?
Answer:
Due to converging or diverging of light rays from the mirror.

Question 27.
Why is angle of incidence equal to angle of reflection when a light ray reflects from a surface?
Answer:
Because light selects the path that takes least time to cover a distance.

Question 28.
Are angle of reflection and angle of incidence also equal for curved surface?
Answer:
Yes, it is equal for curved surfaces like spherical mirrors.

Question 29.
What is a spherical mirror? Give different types of spherical mirrors.
Answer:
If the reflecting surface of mirror is considered to form a part of the surface of sphere, then it is called spherical mirror. Spherical mirrors are of two types :

1. Concave mirror
2. Convex mirror

Question 30.
Write about various distances related to mirrors.
Answer:
The various distances related to mirrors are
1) Focal length (f) :
The distance between vertex and focus is called focal length.

2) Radius of curvature (R) :
The distance between vertex and centre of curvature is called radius of curvature.

Question 31.
We wish to obtain an erect image of an object using a concave mirror of focal length of 15 cm. What should be range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object?
Answer:
The range of distance of object is between 0 and 15 cm.
The image is virtual and erect.
The image is larger than the object.

Question 32.
Name some apparatus which can work on the principle of reflection of light.
Answer:
Plane mirror, spherical mirrors, periscope, kaleidoscope.

Question 33.
If you want to get parallel beam by using concave mirror, then where do you keep the source?
Answer:
The object should be kept at focus because the light rays coming from focus after reflection from mirror travel parallel to principal axis.

Question 34.
If you want to form the image of an object at infinity, then where do you keep the object?
Answer:
The object should be kept at focus; then the image would be formed at infinity.

Question 35.
How do you get a virtual image with a concave mirror?
Answer:
When we place the object between vertex and focus then we will get a virtual image.

Question 36.
Why do dentists use concave mirror?
Answer:
If the object is between mirror and its focus we get enlarged virtual and straight image by using concave mirror. So dentists use this principle to see inner parts of mouth.

Question 37.
A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it. Where is the image located?
Answer:
> 10 cm.

Question 38.
What is your opinion on elevating buildings with mirrors?
Answer:
The mirrors used in elevating buildings are reinforced, tough and laminated glasses. These mirrors provide safety and make the buildings attractive.

Question 39.
Identify the mirror having focal length +15 cm.
Answer:
Convex mirror (since the focal length of convex mirror is taken as positive).

Question 40.
If the focal length of mirror is 10 cm, what is that mirror?
Answer:
The mirror is concave (since the focal length of concave mirror is taken as negative).

Question 41.
Can we focus a sunlight at a point using a mirror instead of magnifying glass?
Answer:
Yes, by using concave mirror we can focus sunlight at a point.

Question 42.
To reduce glaze of surroundings the windows of some department stores, rather than being vertical, slant inward at the bottom. How does this reduce glaze?
Answer:
This slant reflects the sunlight further down towards the ground, then it would happen as if they are vertical.

Question 43.
Why do we prefer a convex mirror as a rear-view mirror in the vehicles?
Answer:
It gives erect and small image and covers large distance.

Question 44.
An object is placed at a distance 8 cm from a concave mirror of radius of curvature 16 cm. What are the characteristics of image?
Answer:
The image is real, inverted, and same size.

Question 45.
What happens when light falls on an opaque object?
Answer:
Some part of light is reflected back and remaining part is absorbed.

Question 46.
What happens when light is reflected from transparent object?
Answer:
Some part of light is reflected and remaining part is partly transmitted or partly absorbed.

Question 47.
Which objects at your home act as spherical mirrors?
Answer:
Objects at home that act as spherical mirrors are :

1. Spoons
2. Spectacles
3. Sink
4. Cooking vessel

Question 48.
Complete the following ray diagram.

Answer:

The light ray passing through centre of curvature falls normal to the concave mirror. So it retraces the same path.

Question 49.
If focal length is 20 cm, then what is radius of curvature of mirror?
Answer:
f = 20 cm
R = 2f = 2 × 20 = 40 cm.

Question 50.
The radius of curvature of a spherical mirror is 20 cm. What is the focal length?
Answer:
Radius of curvature (R) = 20 cm
R 20
Focal length (f) = \(\frac{\mathrm{R}}{2}=\frac{20}{2}\) = 10 cm.

Question 51.
The focal length of convex mirror is 16 cm. What is its radius of curvature?
Answer:
f = 16 cm
R = 2f = 2 × 16 = 32 cm

Question 52.
Write any two uses of concave mirror in our daily life.
Answer:
Uses of concave mirror :

1. Concave mirrors are used by dentists to see enlarged image of tooth.
2. Concave mirrors are used in car head lights.

Question 53.
Write any two uses of convex mirror in our daily life.
Answer:
Uses of convex mirror :

1. Convex mirrors are used as rear view mirrors in vehicles because convex mirrors increase field of view.
2. Convex mirrors are used in street light reflectors as they spread light over greater

Question 54.
Suggest a new use with a spherical mirror.
Answer:
Spherical mirrors are newly adapted in ATMs.

Question 55.
Focal length of a concave mirror is x. Find the sum of focal length and radius of curvature.
Answer:
Focal length = x; Radius of curvature = 2 x focal length = 2x.
The sum of focal length and radius of curvature = x + 2x = 3x.

Question 56.
If the angle between the mirror and incident ray is 40°, then find the angle of reflection.
Answer:
Given that angle between incident ray and mirror = 40°.
Suppose angle of incidence = x.
∴ 40 + x = 90
x = 90 – 40 = 50°.

But we know angle of incidence = angle of reflection
∴ Angle of reflection = 50°.

9th Class Physical Science 7th Lesson Reflection of Light at Curved Surfaces 2 Marks Important Questions and Answers

Question 1.
Your friend has a doubt that whether a concave mirror or a convex mirror is used as a rear view mirror in the vehicles. What questions will you ask to clarify his doubts?
Answer:

• Is the image in a rear-view mirror smaller or larger when compared to real object?
• Which mirror forms smaller image than the object in the given mirrors?

Question 2.
The focal length of a huge concave mirror is 120 cm. A man is standing in front of it at a distance of 40 cm. What are the characteristics of his image in that mirror?
Answer:
i) Image form in the mirror
ii) Virtual image
iii)Erected image
iv) Enlarged image

Question 3.
How can you find out the focal length of concave mirror experimentally when there is no sunlight?
Answer:
Place the object / candle in front of the mirror and adjust the screen to get image on it. Measure the object distance, image distance. Substitute the values (as per sign connection) in mirror formula \(\left(\frac{1}{f}\right)=\left(\frac{1}{u}\right)+\left(\frac{1}{v}\right)\). We get the focal length of mirror.
(OR)
Place the object / candle and the screen at same point in front of the mirror. Adjust this set of material to get sharp image on the screen.

Measure the distance from mirror to object/screen. This distance is the radius of curvature and make it half, it gives focal length of the mirror.

Question 4.
The magnification of the image by the concave mirror is – 1. Mention the four char-acteristics of image from the above information.
Answer:

1. Image will be formed at the centre of curvature (C).
2. Image size is equal to that of the object size. ‘
3. Inverted image.
4. Real image.

Question 5.
Write about different points related to mirrors.
Answer:
The different points related to mirrors are
1) Vertex (P) :
The point where the central axis touches the mirror is called vertex.

2) Focus or focal point (F):
The light rays coming from distinct object appear to meet at point in case of concave mirror and tend to meet at point when drawn backward in case of convex mirror. That point is known as focus or focal point.

3) Centre of curvature (C) :
It is centre of the sphere to which the mirror belongs.

Question 6.
What happens if light rays parallel to principal axis fall on the concave mirror, and draw ray diagram?
Answer:

The light rays that are parallel to the principal axis get reflected such that they pass through the focal point of the mirror. R₁ is such ray in figure.

Question 7.
What happens to a ray which passes through focal point and falls on the concave mirror, and also draw the ray diagram?
Answer:

The light ray which goes through the focal point of the mirror travels parallel to principal axis. R₂ is such ray in figure.

Question 8.
How does an image form due to convex mirror? Draw the ray diagram.
Answer:

• The parallel rays coming from distance object tend to diverge after reflection.
• If we extend the reflected rays backwards they meet at ‘F’, i.e. focal point of the convex mirror.

Question 9.
Which light ray after reflection will travel along the same path in opposite direction? What can be such a ray for a spherical mirror? Draw the ray diagram.
Answer:

1. Any ray that is normal to the surface, on reflection, will travel along the same path but in opposite direction.
2. The line drawn from the centre of curvature of mirror is perpendicular to the tangent at the point, the line meets the curve.
3. So if we draw a ray starting from the tip of the object going through the centre of curvature to meet the mirror, it will get reflected along the same line. This ray is shown as R₃ in the figure.

Question 10.
What happens if an object is placed at centre of curvature of a mirror? Draw the ray diagram.
Answer:

From the ray diagram we conclude that the image of the object will be formed at the same distance as the object and it will be inverted and of the same size. The image is real because it forms on a screen.

Question 11.
Draw the ray diagrams with convex mirror and write rules of ray diagram of convex mirror.
Answer:
Rule -1 :

A ray running parallel to main axis, on meeting the convex mirror will get reflected so as to appear as if it is coming from the focal point.

Rule – 2 :

This is converse of rule 1. A ray going in the direction of focal point after reflection will become parallel to main axis.

Rule – 3 :

A ray going in the direction of the centre of curvature will get reflected back in opposite direction, and looks like that is coming from the centre of curvature.

Question 12.
Why do we use parabolic mirror instead of concave mirror?
Answer:

1. We use parabolic mirror instead of concave mirror because with the concave mirror all the rays coming parallel in it may not be focused at focal point (F).
2. Those rays which are very nearer to principal axis will only be focused at focal point.
3. It is very effective to make the mirror parabolic in order to make all the rays to converge at focus.

Question 13.
The magnification of mirror is given as – 3. What is the inference do you get from this information?
Answer:
Magnification – ve indicates it is an inverted image. So it is a real image.
Magnification 3 indicates the image size is three times the object size. So the image is enlarged. Since it is forming real image the mirror is concave.

Question 14.
Why are we able to see various objects around us?
Answer:
We are able to see various objects around us due to the diffused light reflected from these objects reaches to our eye which gives sense of vision to those objects.

Question 15.
Which type of mirror is used as a reflector in street lamp?
Answer:
The reflectors of the street lamp are made in convex in shape so that reflected rays diverge over the larger area on ground. Therefore convex mirror acts as a reflector in street lamp.

Question 16.
Which type of mirror is used in doctor’s head lamp?
Answer:
Doctors use head lamp to examine nose, throat, teeth, etc. of patients. In this lamp a parallel beam of light is allowed to fall on the concave mirror. The reflected light concentrates on focus on the mirror on a smaller area to be examined. So the concave mirror is used in doctor head lamp.

Question 17.
Would you able to burn a paper using concave mirror?
Answer:

1. Concave mirror focuses the parallel sun rays at focal point of the mirror.
2. So with a small concave mirror we can heat up and burn a paper.

Question 18.
How do you find the focal length of concave mirror?
Answer:

1. Hold a concave mirror such that sunlight falls on it.
2. Take a small paper and slowly move it in front of the mirror until we will get smallest and brightest spot of the Sun.
3. Find the distance between mirror and image of the Sun that will give the focal length of mirror.

Question 19.
See the table and identify the mirrors in each case.

Answer:
Magnification negative indicates that it is inverted image and also real. Magnification -1 means same size. So the mirror which gives real image of same size is concave mirror. So ‘X’ is concave.
The magnification +1 means the image is virtual, erect and same size. So the mirror Y is plane.
The magnification + 0.5 means the image is virtual, erect and diminished. So the mirror Z is convex.

Question 20.
An object is placed at various positions in front of concave mirror of focal length 10 cm. Complete the table by using given information without actually doing the problem.

Answer:

Question 21.
Draw a normal at any point of a concave mirror.
Answer:

Question 22.
Identify the following in a ray diagram showing the reflection of light in a concave mirror.
a) Pole of the mirror
b) Principal axis
c) Centre of curvature
d) Focal point
e) Focal length
Answer:

Question 23.
Complete the ray diagram.

Answer:
First we have to draw normal at point of contact to the concave mirror and then we have to use laws of reflection to draw the reflecting ray.

Question 24.
Figure shows two parallel light rays falling on a convex mirror. Complete the ray diagram.

Answer:

Question 25.
See the belog figure and complete the ray diagram.

Answer:

Question 26.
Assume that an object is kept at a distance of 20 cm in front of a concave mirror. If its focal length is 30 cm, then
a) what is the image distance?
b) what the magnification of mirror in this case?
Answer:
Object distance = u = 20 cm

Question 27.
There is an object in front of convex mirror at a distance of 5 cm. If its focal length is 10 cm, then
a) what is the image distance?
b) what is its magnification?
Answer:

9th Class Physical Science 7th Lesson Reflection of Light at Curved Surfaces 4 Marks Important Questions and Answers

Question 1.
Sudheer wants to find focal length of a concave mirror experimentally.
a) What apparatus does he need?
b) Is the screen required or not? Explain.
c) Draw the table required to tabulate the values found in his experiment.
d) What is the formula used by him to find focal length?
Answer:
a) Apparatus required to Sudheer are

1. Concave mirror,
2. White paper or screen,
3. Scale,
4. V – stand,
5. Candle.

b) Yes, screen is required.
To catch and measure the image distance screen is required.

c) Table for observation and calculation of ‘f’.

d) Focal length \(\Rightarrow \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}} \text { (or) } \mathrm{f}=\frac{\mathrm{uv}}{\mathrm{u}+\mathrm{v}}\)
This is the formula used by him to find a focal length.

Question 2.
Show the formation of image with a ray diagram when an object is placed on the principal axis of a Concave mirror between focus and centre of curvature of the mirror.
Answer:

Question 3.
An object of 6 cm height is placed at a distance of 30 cm in front of a concave mirror of focal length 10 cm. At what distance from the mirror, will the image be formed? What are the characteristics of the image?
Answer:

Question 4.
List the materials required for conducting an experiment to find the focal length of a concave mirror. Explain the experimental process also.
Answer:
a) Material required for conducting an experiment to find the focal length of a concave mirror are

1. concave mirror
   a piece of paper
2. meter scale.

b) Procedure of the experiment:

1. Hold a concave mirror such that sunlight falls on it.
2. Take a small paper and slowly move it in front of the mirror and find out the point where we get the smallest and brightest spot, which will be the image of the sun.
3. Measure the distance of this spot from the pole of the mirror.
4. This distance is the focal length (f) of the mirror.

Another Experiment:
a) Material required :
A candle, paper / screen, concave mirror, V-stand, measuring tape or meter scale.

b) Procedure :
1) Place the concave mirror on V-stand, a candle and meter scale as shown in the figure.

2) Keep the candle at different length from the mirror (10 cm to 80 cm) along the axis and by moving the paper find the position where the sharp image is got on the paper.

3) Measure the image distance (o) and note in the given table.

4) Find the average of focal lengths (f) obtained in the experiment.

5) The average T is the focal length of the given mirror.

Question 5.
An object of height 5 cm is placed at 30 cm distance on the principal axis in front of a concave mirror of focal length 20 cm. Find the image distance and size of the image.
Answer:
Object distance (u) = – 30 cm, Focal length (f) = – 20 cm, Height of object (h_o) = 5 cm, Image distance (v) = ?, Height of image (h_i) = ?

∴ The image is real, inverted with a height of 10 cm.

Question 6.
A student conducted an experiment to observe characteristics of images formed by spherical mirrors and recorded his observations as follows. Observe the table and answer the questions.

i) Above said information belongs to which spherical mirror?
ii) In which situation, magnification is less than 1.
iii) An object of height 8 cm placed at centre of curvature on principal axis, then where do you get the image and what is its height?
iv) “All real images are inverted”. Justify the statement by using above table.
Answer:
i) It is a Concave mirror.
ii) When object is kept beyond ‘C’ then magnification is less than 1.
iii) Image formed at ‘C’. The height of the image is 8 cm.
iv) According to the table if the image is erected image, it is a real image. In all the other cases every real image is virtual image.

Question 7.
In the following cases calculate the magnification values for a concave mirror. Give reason.
a) When the object is at the focal point of the mirror.
b) When the object is between focal point and the pole.
Answer:
In the case of concave mirror
a) When the object is at the focal point of the mirror, then its magnification value is -1.

Reason :
In this case size of the image is large, compared with the object. It is called virtual image. Image is formed behind the mirror so magnification has negative sign.
Nature of the object: It is real, inverted, enlarged and forms at infinity.

b) When the object is between focal point (F) and the pole (P) of the mirror, then its magnification value is +1.

Reason :
In this case image is formed on the same side of the object and it is also virtual image so, the sign of the magnification is positive.

Nature of the object:
It is virtual, erect, enlarged and on the same side of the object.

Question 8.
Write the derivation of mirror formula.
(OR)
Derive \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\).
(OR)
A student wants to find the image distance for a given object distance of a mirror. Then derive a formula for the mirror.
Answer:
Derivation of mirror formula :
In the figure P = pole, C = centre of curvature and F= focus of the concave miror. Object AB is placed beyond C. Image AB’ is formed in between F and C.
From the diagram triangles A’B’C and ABC are similar triangles.
\(\frac{\mathrm{AB}}{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{B}^{\prime} \mathrm{C}}\) ………………… (1)

Question 9.
Where is the base of the candle going to be in the image when the object is placed on the axis of the mirror beyond ‘C’?
Answer:

1. Any ray starting from a point on the axis and travelling along the axis will reflect on the axis itself.
2. So the base of the image is going to be the axis.
3. If the object is placed vertically on the axis, the image is going to be vertical.
4. Draw perpendicular from point A to axis.
5. The intersection point is the point where the base of the image of the candle is going to be formed

Question 10.
What happens if an object placed at a distance less than the focal length of the concave mirror? Draw the ray diagram.
(OR)
When do you get a virtual image by using a concave mirror and draw the ray diagram?
Answer:

1. The candle object (0) is placed at a distance less than the focal length of the mirror.
2. The first ray (R₁) will start from tip of the object and run parallel to axis to get reflected so as to pass through focal length.
3. The second ray (R₂) is the ray starting from the tip of the object and going through the focal point but it is not possible as such a ray will not meet the mirror.
4. The third ray (R₃), starting from the tip of the object goes to the centre of curvature but that also seem not to be possible.
5. Now consider a ray (R₄) that starts from the tip and goes in such a direction that it would go through the centre of curvature if extended backwards.
6. This ray is normal to surface and so will be reflected along the same line in opposite direction and will go through centre of curvature.
7. The two reflected rays diverge and will not meet.
8. When we extend these rays backward they appear to be coming from one point.
9. As seen from the figure (2) the image will be erect and enlarged and virtual.

Question 11.
A person in a dark room looking through a window can clearly see a person outside in the daylight, whereas the person outside cannot see the person inside. Why?
Answer:

• There is usually some reflection that occurs at an interface between the two materials but most often of light passing through.
• Imagine you are inside in the dark. A person outside in bright sunlight is sending out (reflection) lots of light, most of which would come through the window to you, so you see them clearly.
• Since it is so bright outside, there is also a good amount of light which reflects back towards them.
• This can distract them from little bit of light from you that is going towards them, so they have much harder time seeing you.

Question 12.
What is magnification? Derive an expression for magnification.
Answer:
Magnification :
The ratio of height of image to height of object is called magnification.

Question 13.
What are the rules to be followed while drawing ray diagrams?
Answer:
Various rules to be followed for drawing ray diagrams.
1) A ray parallel to the principal axis, after reflection will pass through the principal focus in case of a concave mirror or appear to diverge from the principal focus in case of convex mirror.

2) A ray passing through the principal focus of a concave mirror or a ray which is directed towards the principal focus of convex mirror, after reflection will emerge parallel to the principal axis.

3) A ray passing through the centre of curvature of a concave mirror or directed in the direction of centre of curvature of a convex mirror, after reflection, is reflected back along the same path.

4) A ray incident obliquely to the principal axis, towards a point P on the concave mirror or a convex mirror, is reflected obliquely. The incident ray and reflected rays follow laws of reflection.

Question 14.
How do you make a parallel beam with an experiment?
Answer:
Aim :
Making a beam of parallel lines.

Material used :
Two pins, thermocol block, candle.

Procedure:

1. Stick two pins on a thermocol block.
2. The pins are exactly parallel to each other.
3. As we can see in the figure, when a source of light is kept very near, we see the shadows diverging (from the base of the pins).
4. As we move the source away from the pins, the divergent angle starts reducing.
5. If we move the source far away, we will get parallel shadows. Thus we get a beam of parallel lines.

Question 15.
Write a table which shows the image formed by a concave mirror for different positions and also give size and nature of image.
Answer:

Question 16.
Ray diagrams of concave mirror.
Answer:
Object is placed ‘Infinitely’ :

Object is placed between ‘P’ and ‘F :

Question 17.
Complete the following ray diagrams and give reasons.

Answer:
a)

Reason :
The light appears to be passing through centre of curvature after reflection from convex mirror retraces the same path.

b)

Reason :
The light ray making certain angle of incidence ‘q’ with principal axis follows laws of reflection and reflects with same angle q.

c)

Reason :
The light ray which travels parallel to principal axis after reflection from convex mirror diverges from mirror and if we extended the ray backwards it passes through principal focus.

d)

Reason :
The light ray which travels parallel to principal axis after reflection from convex mirror diverges from mirror when the light ray drawn backwards passes through focus and second light ray follows laws of reflection (i.e., ∠i = ∠r).

These extended backward light rays meet and form a virtual, diminished and erect image between pole and focus inside the mirror.

Question 18.
Complete the following diagram to obtain image of object AB.

Answer:

1. The light ray which appears to coming from focus after reflection from convex mirror travels parallel to principal axis.
2. The light ray which is incident with certain angle ’x’ at pole ‘P’ reflects with same angle from convex mirror.
3. These two extended light rays meet at B. So AB’ is the image of the object AB.

Question 19.
An object 4 cm in size is placed at 25 cm in front of a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and size of image.
Answer:
Given that f = – 15 cm ; u = – 25 cm ; h0 = 4 cm ; v = ?

So the image is enlarged and inverted.

Question 20.
Focal length of a concave mirror is f. The distance from its focal point to the object is P. Find the ratio of heights of image.
Answer:
Concave mirror is a part of spherical mirror.

Question 21.
When do we get a blurred image from a distant object by using concave mirror?
Answer:

1. The diagram shows a few rays starting from the tip of the flame.
2. The reflected rays intersect A. So the reflected image of tip of flame will be at intersection point A.
3. If we hold the paper at any point before or beyond point A (for example at point B), we see that the rays will meet the paper at different points,
4. So the image of the tip of the flame will be formed at different points due to these rays.
5. If we draw more rays emanating from the same tip we will see that point A they will meet but at point ‘B’ they won’t.
6. So the image of the tip of the flame will be sharp if we hold the paper at A and become blurred (due to mixing of multiple images) when we move the paper slightly in any direction (forward or backward).

Question 22.
An object of size 7 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed? Find size and nature of the image.
Answer:
Given, h₀ = 7 cm ; u = – 27 cm ; f = – 18 cm ; v = ?

Question 23.
An object 3 cm high is placed at a distance of 15 cm from a concave mirror, the radius curvature is 20 cm. Find the nature, position and size of the image. (V = -30 cm, m = -2, h₂ = -6 cm)
Answer:
h₀ = 3 cm; u = -15 cm; r = -20 cm; f = \(\frac{r}{2}\) = – 10 cm

Question 24.
Focal length of a concave mirror is 15 cm. An object of length 5 cm is placed in front of this mirror. Draw neat diagrams to find the length and position of image when object is at (I) 5 cm, 0i) 12 cm, (iif) 20 cm, (jv) 35 cm away from the mirror.
Answer:
i) An object is kept at a distance of 5 cm from the mirror.

Andhra Pradesh AP Board 5th Class Maths Solutions 8th Lesson Fractions Textbook Exercise Questions and Answers.

AP State Syllabus 5th Class Maths Solutions Chapter 8 Fractions

I. Observe the following table.

Answer:

What do you observe from the above table?
Answer:
Here, in all fractions the numerator is less than the denominator.

II. If Hema and Gopi got 7 and 9 biscuits then complete this table.

Answer:

III. Observe the fractions : \(\frac{3}{4}, \frac{4}{5}, \frac{6}{7}, \frac{12}{13}, \frac{25}{28}\)

Answer:

What do you say?
Answer:
Here is all fractions have numerator is greater than or equal to denominator. These types of fractions are called improper fractions.

Do this: (TextBook Page No.129)

Question 1.
Write any 5 proper fractions.
Answer:
Proper fractions: \(\frac{3}{4}, \frac{4}{5}, \frac{6}{7}, \frac{12}{13}, \frac{25}{28}\)

Question 2.
Write any 5 1m proper fractions.
Answer:
Improper fractions: \(\frac{7}{6}, \frac{26}{22}, \frac{21}{20}, \frac{28}{25}, \frac{13}{12}\)

Question 3.
Write any 5 mixed fractions.
Answer:
Mixed fractions: \(3 \frac{2}{3}, 7 \frac{1}{2}, 9 \frac{3}{5}, 8 \frac{2}{3}, 6 \frac{5}{7}\)

Question 4.
Convert these fraction into mixed fraction \(\frac{5}{2}, \frac{7}{3}, \frac{9}{4}, \frac{11}{2}\)
Answer:
Conversion of fraction into mixed fraction.

Question 5.
Convert these fractions into improper fraction \(4 \frac{2}{3}, 5 \frac{3}{4}, 6 \frac{2}{5}, 3 \frac{1}{2}\).
Answer:
Conversion of Mixed fractions into improper fractions

\(3 \frac{1}{2}=\frac{2 \times 3+1}{2}=\frac{7}{2}\).

Do this: (TextBook Page No. 132)

Question 1.
Write any three equivalent fractions to the given fractions.
a) \(\frac{4}{8}\)
b) \(\frac{1}{3}\)
c) \(\frac{3}{7}\)
d) \(\frac{20}{24}\)
Answer:
a) Equivalent fractions to \(\frac{4}{8}=\frac{8}{16}=\frac{12}{24}=\frac{16}{32}\)
b) Equivalent fractions to \(\frac{1}{3}=\frac{3}{9}=\frac{2}{6}=\frac{4}{12}\)
c) Equivalent fractions to \(\frac{3}{7}=\frac{9}{21}=\frac{6}{14}=\frac{12}{28}\)
d) Equivalent fractions to \(\frac{20}{24}=\frac{40}{48}=\frac{60}{72}=\frac{80}{96}\)

Exercise 1:

Question 1.
Simplify the following fractions. (by cancellation method).
(i) \(\frac{105}{15}\)
(ii) \(\frac{200}{20}\)
(iii) \(\frac{7}{10}\)
(iv) \(\frac{666}{66}\)
(v) \(\frac{125}{1000}\)
(vi) \(\frac{120}{200}\)
Answer:

Question 2.
Simplify the following fractions. (by H.C.F. method)
(i) \(\frac{12}{18}\)
(ii) \(\frac{14}{35}\)
(iii) \(\frac{22}{55}\)
(iv) \(\frac{27}{36}\)
(v) \(\frac{128}{164}\)
(vi) \(\frac{210}{427}\)
Answer:
(i) HCF of 12 and 18 is 6.
\(\frac{12 \div 6}{18 \div 6}=\frac{2}{3}\)
So, \(\frac{2}{3}\) is the simplest form of \(\frac{12}{18}\).

(ii) HCF of 14 and 35 is 7.
\(\frac{14 \div 7}{35 \div 7}=\frac{2}{5}\)
So, \(\frac{2}{5}\) is the simplest form of \(\frac{14}{35}\).

(iii) HCF of 22 and 55 is 11.
\(\frac{22 \div 11}{55 \div 11}=\frac{2}{5}\)
So, \(\frac{2}{5}\) is the simplest form of \(\frac{22}{55}\).

(iv) HCF of 27 and 36 is 9.
\(\frac{27 \div 9}{36 \div 9}=\frac{3}{4}\)
So, \(\frac{3}{4}\) is the simplest form of \(\frac{27}{36}\).

(v) HCF of 128 and 164 is 4.
\(\frac{128 \div 4}{164 \div 4}=\frac{32}{41}\)
So, \(\frac{32}{41}\) is the simplest form of \(\frac{128}{164}\).

(vi) HCF of 210 and 427 is 7.
\(\frac{210 \div 7}{427 \div 7}=\frac{30}{61}\)
So, \(\frac{30}{61}\) is the simplest form of \(\frac{210}{427}\).

Question 3.
Convert the following fractions into the simplest form by both the methods.
(i) \(\frac{16}{64}\)
(ii) \(\frac{12}{18}\)
(iii) \(\frac{30}{50}\)
(iv) \(\frac{40}{25}\)
(v) \(\frac{16}{32}\)
(vi) \(\frac{8}{40}\)
Answer:
(i) 1s tMethod: HCF of 16 and 64 is 4.
\(\frac{16 \div 4}{64 \div 4}=\frac{4}{16}=\frac{1}{4}\)
So, \(\frac{1}{4}\) is the simplest form of \(\frac{16}{64}\).

2nd Method =

(ii) 1st Method HCF of 12 and 28 is 4.
\(\frac{12 \div 4}{28 \div 4}=\frac{3}{7}\),
So \(\frac{3}{5}\) is simplest form of \(\frac{12}{28}\) .
2nd Method: =

iii) 1st Method : HCF of 30 and 50 is 10.
\(\frac{30 \div 10}{50 \div 10}=\frac{3}{5}\)
So, \(\frac{3}{5}\) is simplest form of \(\frac{30}{50}\)
2nd method \(\frac{30}{50}=\frac{3}{5}\).

(iv) 1st method: HCF of 40 and 25 is 5
\(\frac{40 \div 5}{25 \div 5}=\frac{8}{5}\)
So \(\frac{8}{5}\) is simplest form of \(\frac{40}{25}\)
2nd method:

(v) 1st method: HCF of 16 and 32 is 16.
\(\frac{16+16}{32+16}=\frac{1}{2}\),
So \(\frac{1}{2}\) is the simplest form of \(\frac{16}{32}\).
2nd method:

vi) 1st method: HCF of 8 and 40 is 8.
\(\frac{8 \div 8}{40 \div 8}=\frac{1}{5}\),
So \(\frac{1}{5}\) is simplest form of \(\frac{8}{48}\).
2nd method:

Question 4.
To get equivalent fractions what should we do a given fraction ?
Answer:
To get equivalent fractions, we multiply / divide both numerator and denominator by the same number.

Question 5.
Write any three equivalent fractions to the given fractions.
(i) \(\frac{5}{8}\)
(ii) \(\frac{32}{64}\)
(iii) \(\frac{3}{7}\)
(iv) \(\frac{125}{255}\)
(v) \(\frac{7}{10}\)
Answer:
(i) Equivalent fractions to \(\frac{5}{8}\) is \(\frac{10}{16}, \frac{15}{24} \text { and } \frac{20}{32}\)
(ii) Equivalent fractions to \(\frac{1}{2}, \frac{2}{4}, \frac{4}{8}, \frac{8}{16}\)
(iii) Equivalent fractions to \(\frac{3}{7}\) is \(\frac{6}{14}, \frac{9}{21}, \frac{12}{28\)
(iv) Equivalent fractions to \(\frac{125}{255}\) is \(\frac{5}{9}, \frac{25}{45} \text { and }\)
(v) Equivalent fractions to \(\frac{7}{10}\) is \(\frac{14}{20}, \frac{21}{30}\) and \(\frac{35}{50}\)

Question 6.
Govindamma distributed her 4 acrse of land to her 3 sons, then write the part of land each got in the form of a fraction.
Answer:
Total land = 4 acres
No. of sons = 3
Part of land each got in the form of fraction = \(\frac{4}{3}\) = 1\(\frac{1}{3}\).

Do these : (TextBook Page No.137)

Question 1.
Observe the example and write the correct fractions in the other circles.

Answer:

Question 2.
Find the sum.
(i) \(\frac{2}{10}+\frac{4}{10}\)
(ii) \(\frac{2}{6}+\frac{3}{6}\)
(iii) \(1 \frac{1}{4}+3 \frac{1}{4}\)
(iv) \(2 \frac{1}{5}+3 \frac{1}{5}\)
Answer:
(i) \(\frac{2}{10}+\frac{4}{10}=\frac{2+4}{10}=\frac{6}{10}\)
(ii) \(\frac{2}{6}+\frac{3}{6}=\frac{2+3}{6}=\frac{5}{6}\)
(iii) \(1 \frac{1}{4}+3 \frac{1}{4}=\frac{5}{4}+\frac{13}{4}=\frac{18}{4}\)
(iv) \(2 \frac{1}{5}+3 \frac{1}{5}=\frac{11}{5}+\frac{16}{5}+\frac{27}{5}\)

Question 3.
\(\frac{1}{2}\) kg of a sugar packet, \(\frac{3}{6}\) kg of jaggery are in a bag. Then what is the
total weight of two items in the bag?
Answer:
Weight of sugar packet = \(\frac{1}{2}\) kg
Weight ofjaggery packet = \(\frac{3}{6}\) kg = \(\frac{1}{2}\) kg
Total weight of two items = \(\frac{1}{2}\) + \(\frac{3}{6}\)
\(\frac{3}{6}+\frac{3}{6}=\frac{3+3}{6}=\frac{6}{6}\) = 1.

Question 4.
Sakru paints \(\frac{1}{5}\) th part of a wall on first day. \(\frac{2}{5}\) th part of the wall on second day Then how much part lie painted in both the days?
Answer:
Painting part of wall on first day = \(\frac{1}{5}\)
Painting part of wall on second day = \(\frac{2}{5}\)
Painting part of wall on both the days = \(\frac{1}{5}+\frac{2}{5}=\frac{1+2}{5}=\frac{3}{5}\)

Question 5.
Polamma had some money. She spent \(\frac{3}{6}\)th part of money on books. \(\frac{1}{6}\) th part of money on pens, pencils and erasers. Then how much part money did she spend ¡n total?
Answer:
Money spent on books = \(\frac{3}{6}\)th part
Money spent on pens = \(\frac{1}{6}\)th part
pencils and erasers.
Money spent in total = \(\frac{3}{6}+\frac{1}{6}=\frac{3+1}{6}\) = \(\frac{4}{6}\)th part

Do these: (TextBook Part No. 139)

Question 1.
Complete this.

Answer:

Question 2.
Find the sum.
(i) \(\frac{1}{5}+\frac{3}{4}\)
(ii) \(\frac{3}{4}+\frac{5}{6}\)
(iii) \(1 \frac{2}{3}+2 \frac{5}{6}\)
(iv) \(3 \frac{1}{8}+2 \frac{5}{6}\)
Answer:

Question 3.
Seetamma read \(\frac{1}{5}\) th part of a book on Monday, \(\frac{4}{10}\) part of the book on Tuesday. Then how much part did she complete on two days ?
Answer:
Book read on Monday = \(\frac{1}{5}\) th part
Book read on Tuesday = \(\frac{4}{10}\) th part
Book read on two days = \(\frac{1}{5}\) + \(\frac{4}{10}\)
LCM of 5 and 10 is 10.
= \(\frac{1}{5} \times \frac{2}{2}+\frac{4}{10} \times \frac{1}{1}\)
= \(\frac{2}{10}+\frac{4}{10}=\frac{2+4}{10}=\frac{6}{10}\)

Question 4.
Polayya painted a wall of \(\frac{3}{4}\) th part on 1st day and \(\frac{3}{6}\) th part of the wall on 2nd day. Then how much part he painted the wall in two days ?
Answer:
Painted part of wal on 1 st day = \(\frac{1}{5}\)
Painted part of wall on 2nd day = \(\frac{3}{6}\)
Painted part in two days = \(\frac{1}{5}\) + \(\frac{3}{6}\)
LCM of 5 and 6 is = 30
= \(\frac{1}{5} \times \frac{6}{6}+\frac{3}{6} \times \frac{5}{5}=\frac{21}{30}=\frac{7^{t h}}{10} \text { Part }\)

Try this: (TextBook Part No.139)

Question 1.
Add \(5 \frac{6}{8}+4 \frac{1}{7}\)
Answer:

Do this: (TextBook Page No.141)

Question 1.
Complete this:

Answer:

Question 2.
Do the following.
(i) \(\frac{6}{10}-\frac{1}{10}\)
(ii) \(\frac{3}{15}-\frac{1}{15}\)
(iii) \(1 \frac{3}{15}-1 \frac{1}{15}\)
(iv) \(2 \frac{4}{7}-1 \frac{2}{7}\)
Answer:
(i) \(\frac{6}{10}-\frac{1}{10}\)
= \(\frac{6-1}{10}=\frac{5}{10}\)

(ii) \(\frac{3}{15}-\frac{1}{15}\)
= \(\frac{3-1}{15}-\frac{2}{15}\)

Question 3.
Eswar painted \(\frac{1}{6}\) th part of a wall on first day. Then how much part will remain to complete?
Answer:
Painted part ofwal on Sunday = \(\frac{1}{6}\)th part
Total part = 1
Remaining part to complete = \(\frac{1}{1}-\frac{1}{6}\)
LCM of 1 and 6 is 6.
= \(\frac{1}{1} \times \frac{6}{6}-\frac{1}{6} \times \frac{1}{1}=\frac{6}{6}-\frac{1}{6}\)
= \(\frac{6-1}{6}-\frac{5}{6}\).

Question 4.
Gown completed \(\frac{1}{4}\) part of her homework on Sunday. \(\frac{5}{12}\) part on Sunday morning. How much part did she complete? How much part of home work is left?
Answer:
Work complete on Saturday = \(\frac{1}{4}\)
Work completed on Sunday = \(\frac{5}{12}\)
Work completed both days = \(\frac{1}{4}\) + \(\frac{5}{12}\)
LCM of 4 and 12 is 12.
= \(\frac{1}{4} \times \frac{3}{3}+\frac{5}{12} \times \frac{1}{1}\)
= \(\frac{3}{12}+\frac{5}{12}=\frac{3+5}{12}=\frac{8}{12}\)
Remainig part left = \(\frac{1}{1}-\frac{8}{12}\)
LCM of 1 and 12 is 12
= \(\frac{1}{1} \times \frac{12}{12}-\frac{8}{12} \times \frac{1}{1}\)
= \(\frac{12}{12}-\frac{8}{12}=\frac{12-8}{12}=\frac{4}{12}=\frac{1}{3} \text { th part }\)

Try these: (TextBook Part No.142)

Question 1.
Complete this:

Answer:

Exercise 2:

Question 1.
Do the following:
a) \(\frac{3}{4}+\frac{7}{4}\)
b) \(1 \frac{1}{2}\)
c) \(\frac{8}{3}+\frac{2}{5}\)
d) \(\frac{6}{3}+\frac{7}{4}\)
e) \(\frac{3}{5}+\frac{9}{11}\)
f) \(\frac{10}{10}+\frac{5}{20}\)
g) \(\frac{9}{10}+\frac{4}{15}\)
h) \(\frac{5}{20}+\frac{13}{30}\)
Answer:

Question 2.
Do the following.
a) \(\frac{3}{7}-\frac{1}{7}\)
b) 6 – \(\frac{1}{3}\)
c) \(\frac{3}{8}-\frac{3}{16}\)
d) \(\frac{8}{7}-\frac{5}{8}\)
e) \(\frac{8}{7}-\frac{5}{8}\)
f) \(\frac{13}{15}-\frac{7}{20}\)
g) \(\frac{63}{40}-\frac{9}{10}\)
h) \(\frac{7}{15}-\frac{3}{10}\)
Answer:
a) \(\frac{3}{7}-\frac{1}{7}\)
= \(\frac{3-1}{7}-\frac{2}{7}\)

Question 3.
Find the difference between 5\(\frac{1}{3}\) and 2\(\frac{4}{7}\)
Answer:
\(\frac{15+1}{3}-\frac{14+4}{7}=\frac{16}{3}-\frac{18}{7}\)
LCM of 3 and 7 = 21
\(\frac{16}{3} \times \frac{7}{7}-\frac{18}{7} \times \frac{3}{3}\)
= \(\frac{112}{21}-\frac{54}{21}=\frac{58}{21}\)

Question 4.
Seetha purchased 1\(\frac{1}{2}\) litre of sunflower oil, \(\frac{3}{4}\) litre of groundnut oil. How much of oil she purchased in total?
Answer:
Purchased quantity of sunflower oil = 1\(\frac{1}{2}\) litr.
Purchased quantity of ground nut oil = \(\frac{3}{4}\) litr.
Purchased quantity of total oil = 1 \(\frac{1}{2}\) + \(\frac{3}{4}\)
= \(\frac{3}{2}\) + \(\frac{3}{4}\)
LCM of 2, 4 is 4
= \(\frac{3}{2} \times \frac{2}{2}+\frac{3}{4} \times \frac{1}{1}\)
= \(\frac{6}{4}+\frac{3}{4}=\frac{6+3}{4}=\frac{9}{4}\)

Question 5.
Vimala purchased 1 \(\frac{3}{4}\) m of cotton cloth for skirt, \(\frac{3}{4}\) m of cloth for blouse. How much cloth is purchased by her?
Answer:
Purchased cloth for skirt = 1 \(\frac{3}{4}\) m
Purchased cloth for blouse = \(\frac{3}{4}\) m
Purchased cloth Total = 1\(\frac{3}{4}\) + \(\frac{3}{4}\)
= \(\frac{7}{4}\) + \(\frac{3}{4}\) = \(\frac{10}{4}\) m.

Question 6.
A water tank is filled with \(\frac{9}{10}\)th part of water, but \(\frac{3}{5}\) th part of water is consumed in a day. Then find the remaining part of water in the tank?
Answer:
Total quantity of water infilled in tank = \(\frac{9}{10}\) th part
Consumed quantity of water is = \(\frac{3}{5}\) th part
Remaining part of water in the tank = \(\frac{9}{10}-\frac{3}{5}=\frac{9-6}{10}=\frac{3}{10} \text { th part }\)

Do these: (TextBook Page No.145)

Question 1.
Read 485.267
Answer:
Four hundred and eighty five (point) two six seven.

Question 2.
Write the place value of all digits in 293.819
Answer:
Given numbers = 293.819
place valu e of 2 = 200
place value of 9 = 90
place value of 5 = 5
place value of 8 = \(\frac{1}{80}\)
place value of 1 = \(\frac{1}{900}\)
place value of 9 = \(\frac{9}{1000}\)

Question 3.
Write any 5 examples for decimal fractions.
Answer:
(i) \(\frac{4756}{100}\) = 47.56
(ii) \(\frac{87685}{1000}\) = 87.685
(iii) \(\frac{763407}{1000}\) = 763.407
(iv) \(\frac{86734}{10000}\) = 8.6734
(v) \(\frac{96302}{10}\)= 9630.2

Exercise 3:

Question 1.
Fill in the blanks:

a) In improper fraction, numerator is ………………… than the denominator.
Answer:
greater

b) \(\frac{6}{6}\) is ………………… fraction (which type?)
Answer:
improper

c) 3\(\frac{1}{2}\) is ………………… fraction (which type?)
Answer:
mixed

d) \(\frac{9}{6}\) ………………… is fraction (which type?)
Answer:
improper

e) \(\frac{2}{5}\) is ………………… fraction (which type?)
Answer:
proper

f) A function having whole number and proper fraction is called ………………… fraction.
Answer:
(mixed)

Question 2.
Convert \(\frac{9}{6}\) into mixed fraction.
Answer:
Mixed fraction of \(\frac{9}{6}=1 \frac{3}{6}\).

Question 3.
Convert 2\(\frac{1}{5}\) into an improper fraction.
Answer:
Improper fraction of \(2 \frac{1}{5}=\frac{2 \times 5+1}{5}=\frac{11}{5}\).

Question 4.
Write any 5 equivalent fractions to \(\frac{2}{3}\).
Answer:
Equivalent fractions of \(\frac{2}{3}\) is \(\frac{4}{6}, \frac{6}{9}, \frac{8}{12}, \frac{10}{15}\) and \(\frac{16}{18}\).

Question 5.
Write simplest form of fraction for \(\frac{25}{75}\).
Answer:
Cancilation method: =
∴ Simplest form of \(\frac{25}{75}\) is \(\frac{1}{3}\).

Question 6.
Write two equivalent fractions to \(\frac{64}{36}\).
Answer:
Equivalent fractions to \(\frac{64}{36}\)
= \(\frac{64 \div 2}{36 \div 2}=\frac{32}{18}\)
= \(\frac{64 \div 4}{36 \div 4}=\frac{16}{9}\).

Question 7.
Classify the following as like and unlike frations.
\(\frac{3}{5}, \frac{2}{7}, \frac{8}{5}, \frac{9}{5}, \frac{8}{4}, \frac{1}{5}\)
Answer:
Like fractions: \(\frac{1}{5}, \frac{3}{5}, \frac{8}{5}, \frac{9}{5}\)
Unlike fractions = \(\frac{2}{7}, \frac{8}{4}\).

Question 8.
Fill in the blanks.
a) \(\frac{15}{20}=\frac{3}{\square}\)
b) \(\frac{2}{5}=\frac{\square}{50}\)
c) \(\frac{3}{5}=\frac{\square}{30}\)
Answer:
a) \(\frac{15}{20}=\frac{3}{4}\)
b) \(\frac{2}{5}=\frac{20}{50}\)
c) \(\frac{3}{5}=\frac{18}{30}\)

Question 9.
Fill in the blanks with = or ≠ (≠ denotes not equal to)
a) \(\frac{1}{2}\) ____ \(\frac{8}{16}\)
b) \(\frac{9}{15}\) ____ \(\frac{27}{30}\)
c) \(\frac{6}{13}\) ____ \(\frac{12}{39}\)
Answer:
a) \(\frac{1}{2}\) = \(\frac{8}{16}\)
b) \(\frac{9}{15}\) ≠ \(\frac{27}{30}\)
c) \(\frac{6}{13}\) ≠ \(\frac{12}{39}\)

Question 10.
Fill in the blanks with equivalent fractions.
a) \(\frac{1}{2}\) —- \(\frac{8}{16}\) ____, ____, ____
Answer:
a) \(\frac{1}{2}\) —- \(\frac{8}{16} \frac{2}{4}, \frac{3}{6}, \frac{5}{10}\).

Question 11.
a) \(\frac{6}{5}+\frac{1}{5}\) = ———-
b) \(\frac{5}{7}+\frac{2}{14}\) = ———-
c) \(\frac{15}{32}+\frac{3}{8}\) = ———-
d) \(\frac{11}{16}+1 \frac{1}{8}\) = ———-
Answer:

Question 12.
Kavitha studied \(\frac{1}{2}\) part of a book on 1st day, \(\frac{1}{3}\) part on 2nd day. then how much part she studied in both the days ?
Answer:
On I st day completed book part = \(\frac{1}{2}\)
On 2nd day completed book part = \(\frac{1}{3}\)
On both days completed book part = \(\frac{1}{2}\) + \(\frac{1}{3}\)
LCM of 2 and 3 is 6.
= \(\frac{1}{2} \times \frac{3}{3}+\frac{1}{3} \times \frac{2}{2}\)
= \(\frac{3}{6}+\frac{2}{6}\)
= \(\frac{3+2}{6}=\frac{5}{6}\).

Question 13.
Koushik went to school km by walk. He went on bicycle with his friend for the remaining distance \(\frac{3}{4}\) km. Then find the distance to school from his house.
Answer:
Distance covered by walk = \(\frac{1}{4}\) km
Distance covered by bicycle = \(\frac{3}{4}\) km
Total distance to school from his house = \(\frac{1}{4}\) + \(\frac{3}{4}\)
= \(\frac{1+3}{4}=\frac{4}{4}\) km

Question 14.
a) \(\frac{8}{10}-\frac{2}{10}\) = ………………
b) \(\frac{1}{3}-\frac{1}{9}\) = ………………
c) \(\frac{15}{32}-\frac{3}{8}\) = ………………
d) \(6 \frac{1}{16}-1 \frac{1}{8}\) = ………………
Answer:

Question 15.
\(\frac{2}{3}\)rd nd part of students in a school is boys. Find the part of girls.
Answer:
Part of boys in a school = \(\frac{2}{3}\)rd
Part of girls inaschool= 1 – \(\frac{2}{3}\).
= \(\frac{3-2}{3}=\frac{1}{3}\)rd.

Question 16.
Subtract \(\frac{21}{4}\) from the total of \(\frac{7}{2}\) and \(\frac{8}{3}\).
Answer:
Total of \(\frac{7}{2}\) and \(\frac{8}{3}\) = \(\frac{7}{2} \times \frac{3}{3}+\frac{8}{3} \times \frac{2}{2}\)
= \(\frac{21}{6}+\frac{16}{6}=\frac{21+16}{6}=\frac{37}{6}\)
Subtract of \(\frac{37}{6}-\frac{21}{4}\) =
LCM of 6 and 4 is 12 = \(\frac{37}{6} \times \frac{2}{2}-\frac{21}{4} \times \frac{3}{3}=\frac{74-63}{12}=\frac{11}{12}\)

Question 17.
Govind studied \(\frac{2}{5}\) th part of a book on 1st day, \(\frac{1}{7}\) th part on 2nd day. Then how much part is yet to be completed?
Answer:
Completed part of book on 1st day = \(\frac{2}{5}\)th
Completed part of book on 2nd day = \(\frac{1}{7}\)th
Completed part of book on two days = \(\frac{2}{5}\) + \(\frac{1}{7}\)
LCM of 5 and 7 is 35.
= \(\frac{2}{5} \times \frac{7}{7}+\frac{1}{7} \times \frac{5}{5}\)
= \(\frac{14}{35}+\frac{5}{35}=\frac{14+5}{35}=\frac{19}{35}\)
remaining part of book yet to be completed = 1 – \(\frac{19}{35}\)
= \(\frac{1 \times 35}{35}-\frac{19}{35} \times \frac{1}{1}\)
= \(\frac{35}{35}-\frac{19}{35}=\frac{35-19}{35}=\frac{14}{35}\)

Question 18.
Write in words 189.257
Answer:
One hundred and eight nine point two five seven.

Question 19.
Write the place value of 6 ¡n 489.167
Answer:
6 is in \(\frac{1}{100}\)th place.

AP State Board Syllabus AP SSC 10th Class Telugu Important Questions Chapter 4 వెన్నెల

AP State Syllabus SSC 10th Class Telugu Important Questions 4th Lesson వెన్నెల

10th Class Telugu 4th Lesson వెన్నెల 2 Marks Important Questions and Answers

ఈ క్రింది ప్రశ్నలకు నాలుగైదు వాక్యాల్లో సమాధానాలు రాయండి.

ప్రశ్న 1.
రాత్రి అనే ఆలోచన రానీక చీకటి అనే పేరు విననీక ‘వెన్నెల’ ను వర్ణించిన కవిని గురించి తెల్పండి. (March 2017)
(లేదా)
చంద్రోదయాన్ని అత్యంత మనోహరంగా వర్ణించిన “వెన్నెల” పాఠ్యభాగ ‘కవి పరిచయం’ వ్రాయండి. (March 2019)
జవాబు:
వెన్నెల పాఠ్యభాగం ఎఱ్ఱన గారు రచించారు. ఆయన తల్లి పేరు పోతమాంబిక, తండ్రి పేరు సూరనార్యుడు. ఆయన 14వ శతాబ్దం ప్రథమార్ధంలో జీవించారు. ఎఱ్ఱన గురువుగారి పేరు శంకరస్వామి. ఎఱ్ఱనకు ప్రబంధ పరమేశ్వరుడు, శంభుదాసుడు అనే బిరుదులు కలవు.

ప్రశ్న 2.
ఎఱ్ఱన రచనల గురించి వ్రాయండి.
జవాబు:
ఎఱ్ఱన ఆంధ్రమహాభారతంలోని అరణ్యపర్వశేషం రచించాడు. హరివంశం, నృసింహపురాణం, రామాయణం మొ||నవి రచించాడు.

ఆంధ్రమహాభారతంలోని అరణ్య పర్వశేషాన్ని రాజరాజ నరేంద్రునికి అంకితమిచ్చాడు. హరివంశం, రామాయణాలను ప్రోలయవేమారెడ్డికి అంకితమిచ్చాడు. నృసింహపురాణమును అహోబిల నృసింహస్వామికి అంకితమిచ్చాడు.

ప్రశ్న 3.
ఎఱ్ఱన రచనా వైశిష్ట్యాన్ని వివరించండి.
జవాబు:
ఎఱ్ఱన రచనలలో వర్ణనలు అధికంగా కనిపిస్తాయి. ఆయన రచించిన నృసింహపురాణంలో ‘అష్టాదశ (18) వర్ణనలు’ ఉన్నాయి. తరువాతి కాలంలో వర్ణనాత్మక కావ్యాలు రావడానికి ఎఱ్ఱన వర్ణనలే కారణం. అందుచేతనే ఎఱ్ఱనను ‘ప్రబంధ పరమేశ్వరుడు’ అని సాహితీలోకం గౌరవించింది.

ఎఱ్ఱన నృసింహపురాణం యొక్క ప్రభావం బమ్మెర పోతన భాగవతంపై ఎక్కువగా కనిపిస్తుంది. పోతన భాగవతంలోని ప్రహ్లాద చరిత్రలో ఎఱ్ఱన ప్రభావం స్పష్టంగా కనిపిస్తుందని పరిశోధకులు నిరూపించారు.

ప్రశ్న 4.
‘కాటుక గ్రుక్కినట్టి కరవటంబున జగదండ ఖండమమెరె’ ఈ మాటలు కవి ఏ సందర్భంలో పేర్కొన్నాడో వివరించండి. ఈ పోలికను విశ్లేషించండి.
జవాబు:
సంధ్యా సమయం తర్వాత విశ్వమంతా చీకటి వ్యాపించిందని చెబుతున్న సందర్భంలో కవి ఈ మాటలను పేర్కొన్నాడు.

చీకటిలో భూమి, ఆకాశం, దిక్కులూ అన్నీ కలిసి పోయాయి. అప్పుడు బ్రహ్మాండ ఖండం కాటుక నింపిన పెద్ద భరిణెలా ఉందన్నాడు. బ్రహ్మాండంలో 14 లోకాలుంటాయి. పై లోకాలు 7. క్రింది లోకాలు 7. మధ్యలోనిది భూలోకం. చీకటి కేవలం భూగోళానికి మాత్రమే ఏర్పడింది. అందుకే పైన, క్రింద సమానమైన మూతలు గల భరిణెతో భూగోళాన్ని పోల్చాడు. దీనిద్వారా ఎఱ్ఱనగారి పరిశీలనా దృష్టి, లోకజ్ఞానం వ్యక్తమౌతున్నాయి.

ప్రశ్న 5.
ఈ పాఠంలో కవి వెన్నెలను వర్ణించడానికి ఏయే అంశాలను ఎన్నుకొన్నాడు?
జవాబు:
ఆకాశమనే చెట్టు దిక్కులనే కొమ్మలతో, వెలుగులీనే నక్షత్రాలనే పూలతో ప్రకాశిస్తోందన్నాడు. అప్పుడే ఉదయిస్తున్న చంద్రుడు తన కిరణములనే చేతులతో ఆ ఆకాశపు చెట్టుకున్న నక్షత్రాలనే పూలను కోయడానికి ప్రయత్నిస్తున్నాడని పేర్కొన్నాడు. ఈ పద్యంలో వెన్నెలను వర్ణించడానికి చెట్టు, కొమ్మలు, పూలు, బాల్యచేష్టలు అనే అంశాలను కవి ఎన్నుకొన్నాడు.

‘వెన్నెల వెల్లి ….’ అనే పద్యంలో వెన్నెల ప్రవాహాన్ని పాలసముద్రంతో పోల్చాడు. చంద్రుడిని ఆదిశేషువుతో పోల్చాడు. చంద్రునిలోని మచ్చను శ్రీమహావిష్ణువుతో పోల్చాడు.

ఈ పద్యంలో తెల్లని వెన్నెలను తెల్లటి పాలసముద్రం ఎన్నుకొని పోల్చాడు. అలాగే తెల్లని చందమామను పోల్చడానికి నల్లటి మచ్చను పోల్చడానికి నల్లని విష్ణువును ఎన్నుకొన్నాడు.

ప్రశ్న 6.
పతిభక్తికి ఆదర్శం పద్మిని అని ఎలా చెప్పగలవు?
జవాబు:
మహా తేజస్సంపన్నుడైన సూర్యుని చూసిన కంటితో అల్పమైన తేజస్సు గల ఇతరులను చూడడానికి మనస్సు అంగీకరించదు. అందుకే కదిలెడి తుమ్మెదలనెడు కనుగ్రుడ్లు గల గొప్పదైన పద్మము అనెడు తన కంటిని వెంటనే మూసుకొన్నది.

ఎంతటి మహా పతివ్రతయైన పరపురుషుని కనీసం చూడనైనా చూస్తుంది. కానీ, పద్మిని మాత్రం కనీసం పరపురుషుని చూడడానికి కూడా అంగీకరించలేదు. అందుకే పద్మిని పతిభక్తికి ఆదర్శం. సూర్యోదయంకాగానే కమలాలు వికసిస్తాయి. సూర్యాస్తమయం కాగానే పద్మాలు ముడుచుకొంటాయి. ఇది లోక సహజం.

ప్రశ్న 7.
బాలచంద్రుడెలా ఉన్నాడు?
జవాబు:
పిల్లలకు ప్రాకడం సహజ లక్షణం. ఎక్కడైనా మెరుస్తున్నవి కనిపిస్తే, అవి లాగడం కూడా సహజమే.

అలాగే ఆకాశంలోని చంద్రుడు కూడా ఉన్నాడు. ఆకాశమే పెద్ద చెట్టులాగ ఉంది. దిక్కులు దానికి కొమ్మలులా ఉన్నాయి. నక్షత్రాలు కొమ్మల చివర పువ్వులులా ఉన్నాయి. అప్పుడే ఉదయించిన బాలచంద్రుడు తన పొడవైన కరములతో ఆ నక్షత్రాలనే పువ్వులు కోయడానికి ప్రయత్నిస్తున్నట్లున్నాడు.

ప్రశ్న 8.
వెన్నెలెలా ఉంది?
జవాబు:
వెన్నెల పాలసముద్రంలాగా తెల్లగా ఉంది. అది విజృంభించి దిక్కులన్నీ ముంచుతోంది. రజనీకరబింబం ఆదిశేషువులాగా తెల్లగా ఉంది. చంద్రునిలోని నల్లని మచ్చ ఆదిశేషునిపై పవ్వళించిన శ్రీమహావిష్ణువులా ఉంది.

పాలసముద్రంలో శ్రీమహావిష్ణువు ఆదిశేషునిపై పవ్వళిస్తాడు. ఈ సందర్భాన్ని వెన్నెలతో సమన్వయం చేస్తూ ఎఱ్ఱనగారు అద్భుతమైన కల్పన చేశారు.

ప్రశ్న 9.
పడమటి దిక్కును ఎఱ్ఱని తెరతో పోల్చడం సరైనదేనా? ఎందుకు?
జవాబు:
సాయం సమయంలో సూర్యుడు అస్తమిస్తాడు. సూర్యాస్తమయం సమయంలో సూర్యుడు పడమటి దిక్కుకు వెడతాడు. అప్పుడు సూర్యబింబం చాలా ఎర్రగా ఉంటుంది. అప్పుడు సూర్యుని కాంతి కిరణాలు కూడా ఎరుపురంగులో ఉంటాయి. ఆ సూర్యకాంతికి పడమటి దిక్కంతా ఎర్రగా అవుతుంది. .

సాధారణంగా తెరలు తెలుపురంగులో ఉంటాయి. ఆయా సన్నివేశాలకు తగినట్లు తెరపైకి కాంతిని ప్రసరింపచేస్తూ, రంగులు మార్చడం నాటకాలలో సహజం. ఇక్కడ కూడా నాట్యం చేసే నిశాసతికి తగినట్లుగా పడమటి దిక్కు అనే తెర ఎరుపురంగులో ఉంది అనడం’ అద్భుతమైన ఊహ.

ప్రశ్న 10.
లోకంలో చీకటి అలుముకున్న సందర్భాన్ని కవి ఎలా వర్ణించాడు?
జవాబు:
కవి ఎట్టా ప్రెగ్గడ ‘వెన్నెల’ పాఠంలో చీకటి యొక్క వ్యాప్తిని ఇలా వర్ణించాడు. సంధ్యాకాలం తరువాత చీకట్లు లోకమంతా వ్యాపించాయి. ఆ చీకటిలో భూమి, ఆకాశం, దిక్కులు అన్నీ ఒకటిగా కలిసిపోయాయి. అప్పుడు బ్రహ్మాండ ఖండము, కాటుక నింపిన పెద్ద బరిణె ఏమో అన్నట్లు ఉంది. అంటే అధికమైన చీకట్ల రాశీ వల్ల, దిక్కులు, భూమి, ఆకాశం ఒకదానిలో ఒకటి కలిసి పోయాయి. విశ్వము బాగా కాటుక నింపిన బరిణె ఏమో అన్నట్లు నల్లగా కనిపించిందని కవి వర్ణించాడు.

10th Class Telugu 4th Lesson వెన్నెల 4 Marks Important Questions and Answers

ఈ క్రింది ప్రశ్నలకు 10 లేక 12 వాక్యాల్లో జవాబులు రాయండి.

ప్రశ్న 1.
రాత్రి అనే ఆలోచన రానీయక, చీకటి అనే పేరును విననీయక ఆనంద తరంగంలా విస్తరించిన వెన్నెల విజృంభణను మీ సొంతమాటల్లో రాయండి.
జవాబు:
చంద్రుడు ఉదయించాడు. వెన్నెల ప్రవాహం పాలసముద్రంలా పొంగి ఆకాశాన్ని ముంచెత్తింది. చంద్రబింబం ఆ పాలసముద్రంలో గుండ్రంగా చుట్టుకొన్న ఆదిశేషుడి శయ్యలా, చంద్రుడిలోని మచ్చ ఆ శయ్య మధ్యన ఉన్న విష్ణువులా కనబడింది.

ఆ వెన్నెలలో కలువల రేకులు విచ్చుకున్నాయి. కలువ పూలలో తేనెలు పొంగి కెరటాలుగా విజృంభించాయి. తుమ్మెదలకు విందు చేస్తూ పూల పరిమళాలు బయలుదేరాయి.

చంద్రకాంత శిలల వానలతో, చకోరాల రెక్కల స్పర్శలతో, స్త్రీల చిరునవ్వుల కాంతులతో అతిశయించి, దిక్కులన్నింటినీ ముంచెత్తి వెన్నెల సముద్రంలా వ్యాపించింది. ఆ వెన్నెల అనే సముద్రపు నీటి నుండి చంద్రుడు ఆవిర్భవించాడు.

ఆ విధంగా అందంగా, గంభీరంగా, నిండుగా చంద్రుని వెన్నెల వ్యాపించింది.

అది రాత్రి అనే ఆలోచనను రానీయలేదు. చీకటి అనే పేరు విననీయలేదు. అవ్యక్తం అనే అనుమానాన్ని కలిగించలేదు. ఆ వెన్నెల కళ్ళకు అమృతపు జల్లులా, శరీరానికి మంచి గంధంలా, మనసుకు ఆనంద తరంగంలా విజృంభించింది.

ప్రశ్న 2.
పద్యభావాలను ఆధారంగా చేసుకొని పాఠ్యభాగ సారాంశాన్ని ఇరవై వాక్యాలకు కుదించి రాయండి.
జవాబు:

1. సాటిలేని కాంతి గల సూర్యుని చూచినట్లుగా అల్ప తేజస్సు కలవానిని చూడలేనని సాయంసంధ్యలో కమలం కళ్లు (రేకులు) మూసుకొన్నది.
2. నక్షత్రాలనే పూలతో అలంకరింపబడిన ఆకాశమనే వేదికపై నాట్యం చేయడానికి రాత్రి అనే స్త్రీకి ఏర్పరచిన ఎఱ్ఱని పరదా వలె పడమటి దిక్కు శోభించింది.
3. ఇంతలో ఆకాశం, భూమి, దిక్కులు ఒక్కటి చేస్తూ చీకటి, ప్రపంచాన్ని కాటుక భరిణిగా మార్చింది.
4. అంతలో తూర్పున చంద్రుడు ఉదయించాడు. అప్పుడు చంద్రబింబం చీకటి అనే పిశాచపు బాధా నివారణకు ముల్లోకాలనెడు స్త్రీ ధరించిన ఎఱ్ఱని బొట్టువలె ఉంది.
5. రాత్రి అనెడు స్త్రీ తూర్పు దిక్కుకు ఇచ్చిన ఎఱ్ఱని గురువింద పూసల దండవలే శశిబింబం ఉంది.
6. ఐరావతం మూపురంపై ఉన్న ఎఱ్ఱని అలంకారంలా, ఇంద్రసభలోని మాణిక్యదీపంలా నెలవంక గోచరించింది.
7. తెల్లకలువ అనురాగ రసంతో నింపిన మాత్రవలె కన్పించింది.
8. స్త్రీలకు ఉత్సాహం పెంచే ఔషధపు ముద్దలా మెరిసింది.
9. అలా అలా చంద్రుడు పెరుగుతూ ఉన్నాడు.
10. ఆకాశమనే వృక్షానికి గల దిక్కులనే కొమ్మలలోని నక్షత్రాలనే పూలు కోయడానికి తన కరములతో ఉత్సాహంగా ఆకాశంలో పాకుతున్నాడు.
11. వెన్నెలనే పాల సముద్రంతో దిశలు, ఆకాశం ముంచుతున్న నెలవంక శేషపాన్పులా కనువిందు చేసింది.
12. చంద్రునిలోని మచ్చ విష్ణువువలె కన్పించింది.
13. ఆ పండు వెన్నెలలో కలువలు బాగా వికసించి, వెన్నెలకు పరవశిస్తూ కొంట్రొత్త సువాసనలు వెదజల్లుతూ నేత్రపర్వం చేస్తున్నాయి.
14. వెన్నెల అనే సముద్రం చంద్రకాంత శిలలను తడుపుతూ, చక్రవాకపక్షుల రెక్కల వేగం పెంచుతూ, విరబూసిన కలువలను ఆకర్షిస్తూ, అనేక విధాల అందగించింది.
15. అందంతో రాత్రి అనే ఆలోచన రానివ్వక వెన్నెల చూచేవారి కళ్లకు అమృతాభిషేకం చేస్తోంది.
16. వెన్నెల తన గంభీరతతో చీకటి అనే పేరు విననివ్వక శరీరానికి గంధపు వర్షం అవుతోంది.
17. ధీరమైన వెన్నెలను చూస్తే పరమాత్మ కూడా గుర్తుకు రాడు.
18. ఎందుకంటే వెన్నెలను చూస్తుంటే అంతరాత్మకు బ్రహ్మానందం కలుగుతుంది.
19. ఈ విధంగా కళ్లకు, శరీరానికి, ఆత్మకు ఆనందకారకమైంది వెన్నెల.
20. ఈ వర్ణన ఎఱ్ఱన కవిత్వంలో కలికితురాయి.

ప్రశ్న 3.
సాయం సంధ్యా సమయంలో ఆకాశం ఎలా ఉందని కవి వర్ణించాడు? అది ఎంతవరకు సమంజసం?
జవాబు:
సూర్యుడు అస్తమించేటపుడు ప్రకృతిని పరిశీలిస్తే మొదట జరిగేది కమలాలు ముడుచుకొంటాయి. అస్తమించే సూర్యునికాంతి వన్నె తగ్గుతుంది. ఆ కాంతిలో వెచ్చదనం తగ్గుతుంది. అందుకే కమలాలు ముడుచుకొంటాయి.

ప్రకృతి సహజమైన దీనిని కవిగారు భార్యాభర్తల సంబంధంతో పోల్చారు. మహా తేజోవంతుడైన సూర్యుని చూసిన కంటితో అల్పతేజస్కులను చూడడానికి అంగీకరించదన్నట్లు పద్మిని తన కమలపు కంటిని మూసుకొన్నది అని చక్కగా వర్ణించారు.

సూర్యుడు పూర్తిగా పడమటి దిక్కుకు చేరి క్రిందకు అస్తమిస్తుంటే పడమటి దిక్కు ఎర్రబారుతుంది. పడమటి దిక్కు ఒక ఎర్రని తెరలా కవిగారు ఊహించారు. ఆ తేర వద్ద నటించడానికి రాత్రి అనే స్త్రీకి ఆకాశమనే వేదిక అలంకరించబడింది. కాలపురుషుడే సూత్రధారి, దిక్పాలకులే ప్రేక్షకులు.

అంటే సూర్యాస్తమయంతో ఒక అంకం ముగిసిపోయింది. క్రొత్త అంకానికి తెరలేచింది. ఇది ఒక అత్యద్భుతమైన నాటక రంగంగా కవి కల్పించాడు. సూర్యుడి అస్తమయంతో ఒక పెద్ద వెలుగు ఆకాశమనే రంగస్థలం నుండి ఖాళీ చేసి వెళ్ళిపోయింది.

రంగస్థలంపై ఎవరైనా కళాకారుడు ఉండాలి తప్ప, వేదిక ఖాళీగా ఉండకూడదు. వెలుగు ఖాళీ చేస్తే ఆ ప్రదేశాన్ని చీకటి ఆక్రమించడం సహజం.

క్రమేణా చీకటి దట్టంగా అలుముకొంది. ఆకాశం, నేల, దిక్కులు అన్నీ చీకటితో నిండిపోయాయి. బ్రహ్మాండంలో , భాగమైన భూగోళం ఒక కాటుక నిండిన భరిణెలా ఉందని ఎఱ్ఱనగారు ఊహించారు.

కవిగారి ఊహ చాలా సమంజసంగా ఉంది.

ప్రశ్న 4.
వెన్నెల వ్యాపించిన విధానాన్ని విశ్లేషించండి.
జవాబు:
వెన్నెల వ్యాపించిన విధానాన్ని ఎఱ్ఱనగారు చక్కగా వర్ణించారు. చంద్రుని ఒక చిన్నబాలుడుగా వర్ణించాడు. చిన్న పిల్లలకు చెట్లెక్కడం సరదా. పూలు, పళ్ళు కోయాలని చేతులు చాపడం సహజం. అది బాల్య చాపల్యం.

ఇక్కడ చంద్రుడు కూడా పుట్టి కొద్దికాలమే అయింది. అందుచే ఆకాశమనే చెట్టు పైకి ప్రాకాడు. దిక్కులనే కొమ్మలకు పూసిన నక్షత్రాలనే చుక్కలను కోయడానికి తన కరములు చాపాడు అని ఎర్రనగారు వర్ణించారు.

చీకటి పడగానే నక్షత్రాలు ఆకాశంలో రావడం సహజం. ఆ తరువాత చంద్రకిరణాలు ఆకాశంలోకి వ్యాపిస్తాయి. చంద్రకిరణాలు వ్యాపించే కొద్దీ వెన్నెల కాంతి పెరుగుతూ ఉంటుంది. వెన్నెల కాంతి పెరుగుతుంటే నక్షత్రాల కాంతి మసకబారుతూ ఉంటుంది. ఇది సహజం. దీనినే ఎర్రనగారు తన ఊహాబలంతో చాలా చక్కగా వర్ణించారు.

తరువాత పద్యంలో వెన్నెలను పాల సముద్రంతో పోల్చారు. చంద్రుడిని ఆదిశేషువుతో పోల్చారు. చంద్రుడిలోని నల్లని మచ్చను శ్రీమహావిష్ణువుతో పోల్చారు.

నిజానికి చంద్రుడి వలన వెన్నెల వ్యాపించింది. అది చంద్రుడిని మించిపోతుంది. చంద్రుడు ఆకాశంలో మాత్రమే కనిపిస్తాడు. వెన్నెల ప్రపపంచమంతా నిండిపోతుంది. దేనికైనా ఆవలి ఒడ్డు, ఈవలి ఒడ్డు ఉంటుంది. కాని అనంతమైనది సముద్రం. సముద్రంలో ఎటుచూసినా నీరే కనిపిస్తుంది. పాలసముద్రంలో ఎటుచూసినా తెల్లదనమే కనిపిస్తుంది. అలాగే ప్రపంచమంతా తెల్లటి వెన్నెలతో నిండిపోయింది.

పాలసముద్రంలో ఆదిశేషుడులా చంద్రుడు కనిపించాడు. ఆదిశేషువు తెలుపు. చంద్రుడు తెలుపే. కాని, చంద్రునిలో నల్లటి మచ్చను కూడా కవి వదలలేదు. అదే మహాకవుల లక్షణం. ఆ మచ్చను శ్రీమహావిష్ణువుతో పోల్చి తన చతురత నిరూపించారు ఎఱ్ఱనగారు.

ప్రశ్న 5.
వెన్నెల పాఠ్యాంశం ఆధారంగా ఎఱ్ఱన రచనా శైలిని విశ్లేషించండి.
జవాబు:
ఎఱ్ఱన వర్ణనలకు గీటురాయి వంటిది వెన్నెల పాఠం. సూర్యాస్తమయం జరిగినపుడు పద్మాలు ముడుచుకోవడం ప్రకృతిలో సాధారణమైన విషయం. కాని, అసమానమైన సూర్యుని చూసిన కంటితో ఇతరులను చూడడానికి అంగీకరించని పద్మిని తన పద్మమనే కంటిని మూసుకొన్నదని చక్కటి వర్ణన చేశాడు. పద్మాన్ని పద్మినీ జాతి స్త్రీతో పోల్చాడు. భార్యాభర్తల సంబంధాన్ని చక్కగా వర్ణించాడు.

‘సురుచిర తారకాకుసుమ శోభి’ అనే పద్యంలో సంధ్యాసమయంలో ఎఱ్ఱబడిన పడమటి దిక్కును పెద్ద తెరతో పోల్చాడు. ఆకాశాన్ని వేదికగా ఊహించాడు. నక్షత్రాలను పుష్పాలుగా ఊహించాడు. కాలమును సూత్రధారిగా భావించాడు. దిక్పాలకులను ప్రేక్షకులను చేశాడు. రాత్రిని నాట్యకత్తెగా మార్చాడు. ఈ పద్యంలో కూడా ఎఱ్ఱన వర్ణనాత్మక శైలిని ప్రదర్శించాడు.

‘దట్టమైన చీకటిని కాటుక భరిణెతో పోల్చి తన పరిశీలనాదృష్టిని ప్రదర్శించాడు.

అప్పుడే ఉదయిస్తున్న చంద్రుడు తన కరములతో ఆకాశమనే వృక్షమునకు గల దిక్కులనే కొమ్మలకు ప్రాకుతూ నక్షత్రాలనే పూలను కోయడానికి ప్రయత్నిస్తున్నాడని వెన్నెల రావడాన్ని చక్కగా వర్ణించాడు.

ఇదే విధంగా ప్రతి పద్యంలోనూ ఎఱ్ఱనగారు తన ప్రత్యేకతను నిరూపించుకొన్నారు.

ప్రశ్న 6.
సూర్యాస్తమయానికి, చంద్రోదయానికి మధ్య కాలాన్ని కవి ఎలా వర్ణించాడు?
జవాబు:
“పద్మిని, తన ప్రియుడయిన గొప్ప తేజస్సు కల సూర్యుడిని చూసిన తన కళ్ళతో, అల్ప కాంతికల పరపురుషులను చూడడం తగదన్నట్లుగా, కదలుతున్న తుమ్మెద అనే నల్లగ్రుడ్డు కల పద్మం వంటి తన కన్నును మూసుకుందని” కవి వర్ణించాడు. అంటే సూర్యాస్తమయం కాగానే పద్మాలు ముడుచుకున్నాయని పద్మాలలో తుమ్మెదలు చిక్కుకున్నాయని భావం. పద్మాలు ముడుచుకోవడం, అవి తమ పతి భక్తిని ప్రదర్శించడానికా అన్నట్లు ఉందని కవి వర్ణించాడు.

1. రాత్రి అనే స్త్రీ చేయబోయే నాట్యానికి, రంగస్థలం మీద కట్టిన ఎఱ్ఱరంగు తోపు తెరవలె, పడమటి దిక్కున ఆకాశంలో సంధ్యారాగం కనబడింది.
2. చుక్కలతో కూడిన ఆకాశం, పువ్వులతో అలంకరించిన నాట్య రంగస్థలం వలె కన్పడింది.
3. రాత్రి అనే స్త్రీ ‘నర్తకి’ వలె ఉంది.
4. కాలము అనేది నాటక సూత్రధారుడు వలె ఉంది. అంటే సంధ్యాకాలమయ్యింది. పడమటి దిక్కున ఆకాశం ఎఱుపురంగులో కనబడుతోంది. ఆకాశంలో నక్షత్రాలు ఉదయించాయి. రాత్రి వస్తోంది. రాత్రి తన చీకటిని అంతా వ్యాపింప చేస్తుందని కవి తెలిపాడు.

ప్రశ్న 7.
చంద్రోదయానికి ముందు ప్రకృతి ఎలా ఉందో వివరించండి.
జవాబు:
1) చంద్రుడు, దిక్కులు అనే కొమ్మలను, తన పొడవైన కిరణాలు అనే చేతులతో పట్టుకొని, పైకి లేచి, ఆకాశము అనే చెట్టు యొక్క నక్షత్రాలు అనే పువ్వులను కోయడం కోసమేమో అన్నట్టు ఆకాశంలోకి పాకాడు. అంటే చంద్రుడి , కాంతి, ఆకాశం అంతా వ్యాపించిందని భావము.

2) వెన్నెల ప్రవాహం పాలసముద్రంలా పొంగి అన్ని దిక్కులనూ, ఆకాశాన్ని ముంచెత్తింది. అప్పుడు చంద్రబింబం, ఆ పాలసముద్రంలో గుండ్రంగా చుట్టుకొన్న ఆదిశేషుడి శయ్యను ఏర్పాటు చేసినట్లు ఉంది. చంద్రుడిలోని నల్లని మచ్చ ఆదిశేషువు శయ్యపై మధ్యలో ఉన్న విష్ణుమూర్తివలె కనబడింది.

అంటే చంద్రుడి వెన్నెల పాలసముద్రంలా కనబడింది. చంద్రబింబం, పాలసముద్రంలో విష్ణుమూర్తి నిద్రించే శేషుని పానుపు వలె కనిపించింది. చంద్రుడిలోని నల్లని మచ్చ, ఆదిశేషువుపై నిద్రించిన నీలవర్ణుడైన విష్ణువువలె కన్పించిందన్న మాట.

ప్రశ్న 8.
కవి వెన్నెలను ఏయే అంశాల ఆధారంగా వర్ణించాడు?
జవాబు:
కవి ఎట్టి ప్రెగ్గడ, వెన్నెలను వర్ణించడానికి క్రింది వస్తువులను ఎన్నుకొన్నాడు.

1. వెన్నెల పాలసముద్రంలా తెల్లగా వ్యాపించిందన్నాడు.
2. నిండు వెన్నెలలో కలువ పువ్వురేకులు విచ్చుకున్నాయని చెప్పాడు.
3. వెన్నెలకు చంద్రకాంత శిలలు కరిగి, జడివానలు కురిశాయన్నాడు.
4. చకోరపక్షులు, తమ రెక్కలతో వెన్నలను స్పృశించాయన్నాడు.
5. వెన్నెల, స్త్రీల అందమైన చిఱునవ్వుల కాంతి వలే ఉందన్నాడు.
6. వెన్నెల, సముద్రజలంలా వ్యాపించిందన్నాడు.
7. చూసే వారి కళ్ళకు వెన్నెల, అమృతపు జల్లులా, శరీరానికి పూసిన మంచిగంధపు పూతలా ఉందన్నాడు.
8. వెన్నెల చూసే వారి మనస్సులకు ఆనంద తరంగము వలె ఉందన్నాడు.

ప్రశ్న 9.
వెన్నెల వ్యాపించినపుడు ప్రకృతిలో కలిగిన మార్పులేమి?
జవాబు:
చంద్రోదయము :

1. చంద్రుడు ఉదయించాడు. వెన్నెల ప్రవాహం, పాలసముద్రంలా పొంగి ఆకాశాన్ని ముంచెత్తింది.
2. చంద్రబింబం, ఆ పాలసముద్రములో గుండ్రంగా చుట్టుకొన్న ఆదిశేషుడి పానుపులా, చంద్రుడిలోని మచ్చ, ఆ శయ్యపై ఉన్న విష్ణువులా కనబడింది.
3. ఆ వెన్నెలలో కలువటేకులు విచ్చుకున్నాయి. కలువపూలలో తేనెలు పొంగి, కెరటాలుగా విజృంభించాయి. తుమ్మెదలకు విందు చేస్తూ పూల పరిమళాలు వ్యాపించాయి.
4. చంద్రకాంత శిలలు కరిగిన జడివానలతో, చకోరాల టెక్కల స్పర్శలతో, స్త్రీల చిఱునవ్వుల కాంతులతో అతిశయించి, దిక్కులన్నింటిలో ముంచెత్తి, వెన్నెల సముద్రంలా వ్యాపించింది. ఈ విధంగా అందంగా, గంభీరంగా, నిండుగా చంద్రుని వెన్నెల వ్యాపించింది.

ప్రశ్న 10.
చంద్రోదయానికి ముందు, తరువాత జరిగిన మార్పులను వివరించండి.
జవాబు:
చంద్రోదయానికి ముందు మార్పులు :

1. సూర్యుడు అస్తమించాడు.
2. పద్మములు ముడచుకున్నాయి.
3. తుమ్మెదలు ఆ పద్మాలలో చిక్కుపడ్డాయి.
4. పడమటి దిక్కున సంధ్యారాగం కనబడింది.
5. ఆకాశంలో నక్షత్రాలు ఉదయించాయి.
6. చీకటి బాగా పెరిగి, దిక్కులూ, భూమ్యాకాశాలు దానిలో కలిసి పోయాయి.
7. విశ్వం, కాటుక నింపిన బరిణెయా అన్నట్లు కనబడింది.

చంద్రోదయము తరువాత జరిగిన మార్పులు :

1. చంద్రుడు ఉదయించగానే, వెన్నెల ప్రవాహం పాలసముద్రం వలె పొంగి ఆకాశాన్ని ముంచెత్తింది.
2. చంద్రబింబం, ఆ పాలసముద్రంలో చుట్టచుట్టుకొన్న ఆదిశేషువు శయ్యవలె కనిపించింది.
3. చంద్రుడిలోని మచ్చ, ఆ శయ్య మధ్యన ఉన్న విష్ణువువలె కనబడింది.
4. వెన్నెలలో కలువరేకులు విచ్చుకున్నాయి. కలువలలో తేనెలు ఉప్పొంగాయి.
5. పువ్వుల పరిమళాలు వ్యాపించాయి. 6) చంద్రకాంత శిలలు కరిగి జడివానలు కురిశాయి.
6. చకోరాల లెక్కల స్పర్శలతో, స్త్రీల చిరునవ్వుల కాంతులతో అతిశయించి, వెన్నెల దిక్కులన్నింటినీ ముంచెత్తి, సముద్రంలా వ్యాపించింది.

10th Class Telugu 4th Lesson వెన్నెల Important Questions and Answers

ప్రశ్న 1.
నీకు నచ్చిన ప్రకృతి దృశ్యాన్ని వర్ణిస్తూ ఒక కవిత రాయండి.
జవాబు:
చంద్రోదయము
1) “చెరువులో నీరు గాలికి తిరుగుచుండె
కలువపుష్పాలు వికసించి కనులు తెరిచె
పూల గంధాల మధుపాలు మురియుచుండె
చంద్రకాంతులు వ్యాపించే జగము నిండ”

2) దివిని చుక్కల పూవులు తేజరిల్లె
రాత్రి సతికట్టె నల్లని రంగు చీర
చంద్రుని రాకకు జగమంత సంతసించె
విశ్వమంతకు వెన్నెల విందు చేసె

ప్రశ్న 2.
మీ పాఠం (వెన్నెల) ఆధారంగా ఎఱ్ఱన రచనా శైలిని వివరిస్తూ మీ తండ్రిగారికి లేఖ రాయండి.
జవాబు:

ప్రశ్న 3.
వెన్నెలను వర్ణిస్తూ కవిత వ్రాయండి.
జవాబు:
“వెన్నెలా ! కన్నెపిల్లల చిన్నారి ముద్దుల చెల్లెలా !
వెన్నెలా ! ప్రేమికుల మనసుల మల్లెచెండులా
వెన్నెలా ! చంద్రుని చిరునవ్వుల పన్నీరులా
వెన్నెలా ! పసిపాపల ముద్దుల బోసి నవ్వులా
వెన్నెలా ! కన్నతల్లుల కమనీయ రాగవెల్లిలా
వెన్నెలా ! మా చిన్నారి పొన్నారి చెల్లెలా
వెన్నెలా ! కలువల చుట్టపు చూపులా
వెన్నెలా ఉన్నావు, చల్లావు చంద్రికలు
నయనారవిందాల నయగార మధురిమలు”

ప్రశ్న 4.
వర్ణన ఎలా ఉంటుందో తెలిసింది కదా! ఉషోదయ కాలం నుండి రాత్రి నిండుపున్నమి వెన్నెల వచ్చే వరకు రోజంతా ఆకాశంలో జరిగే మార్పుల్ని గురించి వర్ణిస్తూ రాయండి.
జవాబు:
ఉషోదయ కాలంలో ఆకాశంలో వేగుచుక్క కనిపిస్తుంది. అక్కడక్కడ చుక్కలు కనబడతాయి. క్రమంగా తూర్పు దిక్కున సూర్యబింబం ఎఱ్ఱని పెద్ద బంతిలా కనబడుతుంది. తెల్లవారేక సూర్యోదయం తూర్పుదిశలో జరుగుతుంది.

ఎండ ఎక్కిన కొద్దీ సూర్యకిరణాలు కరకరలాడుతాయి. మధ్యాహ్నం వేళ సూర్యుడు నడినెత్తిమీదికి వస్తాడు. క్రమంగా సూర్యుడు పడమటి దిశకు జారుతాడు. అక్షాశంలో పడమటి దిశలో పేదరాశి పెద్దమ్మ కుంకుమ ఆరబోసుకున్నట్లు ఆకాశం ఎరుపెక్కుతుంది. సూర్యబింబం ఎరుపు ఎక్కుతుంది.

సాయంత్రం 7 గంటలకు ఆకాశంలోని నక్షత్రాలు ఉదయిస్తాయి. చంద్రబింబం ఎఱుపుగా కనబడుతుంది. చంద్రోదయం కాగానే చల్లదనం ప్రాణాలకు హాయిని ఇస్తుంది. మేఘాల కదలికలో చంద్రబింబం ప్రయాణిస్తున్నట్లు ఉంటుంది.

ప్రశ్న 5.
ఎఱ్ఱన రచించిన వెన్నెల పాఠంలోని వర్ణనలు నీకెందుకు నచ్చిందో వివరిస్తూ నీ మిత్రునికి లేఖ వ్రాయి.
జవాబు:

10th Class Telugu 4th Lesson వెన్నెల 1 Mark Bits

1. దేవతలు దివి నుండి భువికి దిగి వచ్చారు – గీత గీసిన పదానికి అర్థం గుర్తించండి. (June 2017).
A) సముద్రం
B) పాతాళం
C) ఆకాశం
D) చీకటి
జవాబు:
C) ఆకాశం

2. నెమలి ఎగిరి వచ్చి పామును పట్టి చంపింది – గీత గీసిన పదాలకు నానార్థ పదాన్ని గుర్తించండి. (June 2017)
A) వనం
B) ఫలం
C) కులం
D) కుండలి
జవాబు:
D) కుండలి

3. వెన్నెల అన్ని దెసలకు వ్యాపించింది – గీత గీసిన పదానికి ప్రకృతి పదాన్ని గుర్తించండి. (June 2017)
A) దిశ
B) దేశం
C) దివాణం
D) దిక్కు
జవాబు:
A) దిశ

4. ‘UUU’ గుర్తులు గల గణమేది ? (June 2017)
A) య గణం
B) ర గణం
C) మ గణం
D) త గణం
జవాబు:
C) మ గణం

5. రజనీకర బింబమును చూస్తే పిల్లలు ఆనందిస్తారు – (గీత గీసిన పదమునకు అర్థమును గుర్తించుము.) (March 2017)
A) సూర్యబింబము
B) చంద్రబింబము
C) ప్రతిబింబము
D) ఆకాశం
జవాబు:
B) చంద్రబింబము

6. విద్యార్థులు చదువు పట్ల శ్రద్ధాసక్తులు కల్గియుండాలి – (గీత గీసిన పదము ఏ సమాసమో గుర్తించుము. ) (March 2017)
A) కర్మధారయం
B) బహుబ్లిహి
C) తత్పురుషం
D) ద్వంద్వము
జవాబు:
D) ద్వంద్వము

7. ఉత్పలమాల పద్యంలో యతిస్థానం గుర్తించండి. (March 2017)
A) 14వ అక్షరం
B) 10వ అక్షరం
C) 13వ అక్షరం
D) 11వ అక్షరం
జవాబు:
B) 10వ అక్షరం

8. దెసలను కొమ్మ లొయ్య నతిదీర్ఘములైన కరంబులన్ బ్రియం – ఏ పద్యపాదమో గుర్తించండి. (March 2017)
A) ఉత్పలమాల
B) శార్దూలం
C) చంపకమాల
D) మత్తేభం
జవాబు:
C) చంపకమాల

9. భూరుహములు జీవులకు ప్రాణవాయువును ఇస్తున్నాయి – గీత గీసిన పదానికి సరియైన అర్థాన్ని గుర్తించండి. (June 2018)
A) నదులు
B) సముద్రాలు
C) చెట్లు
D) పర్వతాలు
జవాబు:
C) చెట్లు

10. కష్టాలు మిక్కుటమైనందున, రైతులు ప్రభుత్వ సహాయం కొరకు ఎదురుచూస్తున్నారు – గీత గీసిన పదానికి పర్యాయపదాలను గుర్తించండి. (June 2018)
A) అధికం, సమానం
B) క్షామం, మితం
C) ఎక్కువ, అధికం
D) తక్కువ, పరిమితం
జవాబు:
C) ఎక్కువ, అధికం

11. దినకరుడు రాగానే లోకమంతా చైతన్యమౌతుంది. అందుకే సూర్యుడు మనకు ప్రత్యక్ష దైవం – గీత గీసిన పదాలకు పర్యాయపదాన్ని గుర్తించండి. (June 2018)
A) స్నేహితుడు
B) చంద్రుడు
C) ధనికుడు
D) ఇనుడు
జవాబు:
D) ఇనుడు

12. ప్రతి పాదంలో ఒక సూర్యగణం + రెండు ఇంద్రగణములు + రెండు సూర్యగణములు క్రమముగా ఉండే పద్యము పేరు గుర్తించండి. (June 2018)
A) కందం
B) సీసము
C) తేటగీతి
D) ఆటవెలది
జవాబు:
C) తేటగీతి

13. వెన్నెల పిండారబోసినట్లుండటం శరత్కాలపు లక్షణం – గీత గీసిన పదానికి పర్యాయపదాలను గుర్తించండి. (March 2018)
A) కౌముది, జ్యోత్స్న
B) చీకటి, తమస్సు
C) కోరిక, కాంక్ష
D) నిశి, నుసి
జవాబు:
A) కౌముది, జ్యోత్స్న

14. ఎటువంటి పరిస్థితిలోనైన ధర్మం వీడరాదు. (వికృతి గుర్తించండి) (S.N. I – 2018-19)
A) దమ్మం
B) దరమం
C) దరుమం
D) దరువుం
జవాబు:
A) దమ్మం

15. మానవునిచే ప్రకృతి ఒడి పాఠశాలగా చేసుకోబడింది. (కర్తరి వాక్యం గుర్తించండి.) (June 2017)
A) మానవుడు ప్రకృతి ఒడిని పాఠశాలగా చేసుకున్నాడు.
B) మానవుడు ప్రకృతి ఒడిని పాఠశాలగా చేసుకోలేదు.
C) మానవునిచే ప్రకృతి,పాఠశాలగా చేసుకోబడింది.
D) మానవుడు ప్రకృతిలో పాఠశాల నిర్మించాడు.
జవాబు:
A) మానవుడు ప్రకృతి ఒడిని పాఠశాలగా చేసుకున్నాడు.

16. చంద్రుడిలోని మచ్చ విష్ణువులాగ కనిపిస్తూ అలరింపబడుతున్నది – కర్తరి వాక్యం గుర్తించండి. (June 2018)
A) చంద్రుడిలోని మచ్చ విష్ణువులాగా కనిపిస్తూ అలరిస్తున్నది.
B) చంద్రుడిలోని మచ్చ విష్ణువులాగ అలరించడం లేదు.
C) చంద్రుడిలోని మచ్చ శివునిలాగ కనిపిస్తోంది.
D) చంద్రుడిలోని మచ్చ విష్ణువనే భ్రాంతి కలిగిస్తోంది.
జవాబు:
A) చంద్రుడిలోని మచ్చ విష్ణువులాగా కనిపిస్తూ అలరిస్తున్నది.

17. వెన్నెల రాత్రిని పగలుగా మారుస్తున్నది – వ్యతిరేకార్థక వాక్యాన్ని గుర్తించండి. (June 2018)
A) చీకటి రాత్రిని పగలుగా మారుస్తున్నది.
B) వెన్నెల రాత్రిని పగలుగా మార్చడం లేదు.
C) వెన్నెల పగలును రాత్రిగా మారుస్తున్నది.
D) వెన్నెల రాత్రిని రాత్రిగానే ఉంచుతుంది.
జవాబు:
B) వెన్నెల రాత్రిని పగలుగా మార్చడం లేదు.

18. వెన్నెల కాలంలో రాత్రింబవళ్లు వెలుగుంటుంది. (ఇది ఏ రకమైన వాక్యమో గుర్తించండి) (S.N. I – 2018-19)
A) చేదగ్ధక వాక్యం
B) సంక్లిష్ట వాక్యం
C) సంయుక్త వాక్యం
D) నిషేధార్థక వాక్యం
జవాబు:
C) సంయుక్త వాక్యం

19. వానలు వస్తే పంటలు పండుతాయి. (ఇది ఏ రకమైన వాక్యమో గుర్తించండి) (S.I. I – 2018 -19)
A) సంయుక్తం
B) చేదర్థకం
C) సామర్థ్యార్థకం
D) విధ్యర్థకం
జవాబు:
B) చేదర్థకం

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.4

Question 1.
Determine which of the following polynomials has (x + 1) as a factor.
i) x³ – x² – x + 1
Solution:
f(- 1) = (- 1)³ – (- 1)² – (- 1) + 1
= -1 – 1 + 1 + 1 = 0
∴ (x + 1) is a factor.

ii) x⁴ -x³ +x² – x + 1
Solution:
f(- 1) = (- 1)⁴ – (- 1)³ + (- 1)² – (- 1) + 1
= 1 + 1 + 1 + 1 + 1= 5
∴ (x + 1) is not a factor.

iii) x⁴ + 2x³ + 2x² + x + 1
Solution:
f(- 1) = (-1)⁴ + 2 (- 1)³ + 2 (- 1)² + (-1) + 1
= 1 – 2 + 2 – 1 + 1 = 1
∴ (x + 1) is not a factor.

iv) x³ – x² – (3 – √3)x + √3
Solution:
f(- 1) = (- 1)³ – (- 1)² – (3 – √3)(-1) + √3
= – 1 – 1 + 3 – √3 + √3 = 1
∴ (x + 1) is not a factor.

Question 2.
Use the factor theorem to determine whether g(x) is a factor of f(x) in each of the following cases:
i) f(x) = 5x³ + x² – 5x – 1; g(x) = x + 1
[Factor theorem : If f(x) is a polynomial; f(a) = 0 then (x – a) is a factor of f(x); a ∈ R]
Solution:
g(x) = x+ 1 = x- a say
∴ a = – 1
f(a) = f(- 1) = 5 (- 1)³ + (- 1)² – 5 (- 1) – 1
= -5 + 1 + 5 – 1 = 0
∴ x + 1 is a factor of f(x).

ii) f(x) = x³ + 3x² + 3x + 1; g(x) = x + 1
Solution:
g(x) = x + 1 = x – a
∴ a = – 1
f(a) = f(- 1) = (- 1)³ + 3 (- 1)² + 3(-1) + 1
= -1 + 3 – 3 + 1 =0
∴ f(x) is a factor of g(x).

iii) f(x) = x³ – 4x² + x + 6;
g(x) = x – 2
Solution:
g(x) = x- 2 = x- a
∴ a = 2
f(a) = f(2) = 2³ – 4(2)² + 2 + 6
= 8 – 16 + 2 + 6 = 0
∴ g(x) is a factor of f(x).

iv) f(x) = 3x³+ x² – 20x +12; g(x) = 3x – 2
Solution:
g(x) = 3x – 2 = \(x-\frac{2}{3}\) = x – a
∴ a = 2/3

v) f(x) = 4x³+ 20x²+ 33x + 18; g(x) = 2x + 3
Solution:
g(x) = 2x + 3 = x + \(\frac{3}{2}=\) = x – a
∴ a = -3/2

∴ g(x) is a factor of f(x).

Question 3.
Show that (x – 2), (x + 3) and (x – 4) are factors of x³ – 3x² – 10x + 24.
Solution:
Given f(x) = x³ – 3x² – 10x + 24
To check whether (x – 2), (x + 3) and (x – 4) are factors of f(x), let f(2), f(- 3) and f(4)
f(2) = 2³ – 3(2)² – 10(2) + 24
= 8- 12-20 + 24 = 0
∴ (x – 2) is a factor of f(x).

f(- 3) = (- 3)³ – 3(- 3)²– 10(- 3) + 24
= – 27 – 27 + 30 + 24 = 0
∴ (x + 3) is a factor of f(x).

f(4) = (4)³ – 3 (4)² – 10 (4) + 24
= 64 – 48 – 40 + 24
= 88 – 88
= 0
∴ (x – 4) is a factor of f(x).

Question 4.
Show that (x + 4), (x – 3) and (x – 7) are factors of x³ – 6x² – 19x + 84.
Solution:
Let f(x) = x³ – 6x² – 19x + 84
To verify whether (x + 4), (x – 3) and (x – 7) are factors of f(x) we use factor theorem.

Let f(- 4), f(3) and f(7)
f(- 4) = (- 4)³ – 6 (- 4)² – 19 (- 4) + 84
= -64 – 96 + 76 + 84
= 0 .
∴ (x + 4) is a factor of f(x).

f(3) = 3³ – 6(3)² – 19(3) + 84
= 27 – 54 – 57 + 84
= 0
∴ (x – 3) is a factor of f(x).

f(7) = 7³ – 6(7)² – 19(7) + 84
= 343 – 294 – 133 + 84
= 427 – 427
= 0
∴ (x – 7) is a factor of f(x).

Question 5.
If both (x – 2) and \(\left(x-\frac{1}{2}\right)\) of px² + 5x + r, show that p = r.
Solution:
Let f(x) = px²+ 5x + r
As (x – 2) and \(\left(x-\frac{1}{2}\right)\) are factor of f(x), we have f(2) = 0 and f(1/2) = 0
∴ f(2) = p(2)² + 5(2) + r
= 4p + 10 + r = 0
= 4p + r
= – 10 ………………(1)

⇒ p + 10 + 4r = 0
⇒ p + 4r = – 10 ………………. (2)
From (1) and (2);
4p + r = p + 4r
4p – p = 4r – r
3p = 3r
∴ P = r

Question 6.
If (x² – 1) is a factor of ax⁴ + bx³ + cx² + dx + e, show that a + c + e = b + d = 0.
Solution:
Let f(x) = ax⁴ + bx³ + cx² + dx + e
As (x – 1) is a factor of f(x) we have
x² – 1 = (x + 1) (x – 1) hence f(1) = 0 and f(-1) = 0
f(1) = a + b + c + d + e = 0 ……………. (1)
and f(-1) = a- b + c- d + e = 0
⇒ a + c + e = b + d
Substitute this value in equation (1)
a + c + e + b + d=0
b + d + b + d=0
2 (b + d) = 0
⇒ b + d = 0
∴ a + c + e = b + d = 0

Question 7.
Factorise
i) x³ – 2x² – x + 2
Solution:
Let f(x) = x³ – 2x² – x + 2
By trial, we find f(l) = 1³ – 2(1)² – 1 + 2
= 1 – 2 – 1 + 2
= 0 .
∴ (x – 1) is a factor of f(x).
[by factor theorem]
Now dividing f(x) by (x – 1).

f(x) = (x – 1) (x² – x – 2)
= (x – 1) [x² – 2x + x- 2]
= (x – 1) [x (x – 2) + 1 (x – 2)]
= (x – 1) (x – 2) (x + 1)

ii) x³ – 3x² – 9x – 5
Solution:
Let f(x) = x³ – 3x² – 9x – 5By trial,
f(- 1) = (- 1)³ – 3(- 1)² – 9(- 1) – 5
=-1 – 3 + 9 – 5
=0
∴ (x + 1) is a factor of f(x).
[ ∵ by factor theorem]
Now dividing f(x) by (x + 1).

f(x)=(x + 1)(x² – 4x – 5)
But x²– 4x – 5 = x² – 5x + x – 5
= x (x – 5) + 1 (x – 5)
=(x – 5)(x + 1)
∴ f(x)=(x + 1)(x + 1)(x – 5)

iii) x³ + 13x² + 32x + 20
Solution:
Let f(x) = x³ + 13x² + 32x + 20
Let f(- 1)
= (- 1)³ + 13 (- 1)² + 32 (- 1) + 20
= – 1 + 13 – 32 + 20 = 33 – 33 = 0
∴ (x + 1) is a factor of f(x).
[ ∵ by factor theorem] Now dividing f(x) by (x + 1).

iv) y³ + y² – y – 1
Let f(y) = y³ + y² – y – 1
f(1) = 1³+ 1²– 1 – 1 = 0
(y – 1) is a factor of f(y).
Now dividing f(y) by (y – 1).

∴ f(x) = (x + 1)(x² + 12x + 20)
But (x² + 12x + 20) = x²+ 10x + 2x + 20
=x(x + 10)+2(x + 10)
=(x + 10)(x + 2)
∴f(x) = (x + 1)(x + 2)(x + 10)

Question 8.
If ax² + bx + c and bx² + ax + c have a common factor x + 1 then show that c = 0 and a = b.
Solution:
Let f(x) = ax² + bx + c and g(x) = bx² + ax + c given that (x + 1) is a common factor for both f(x) and g(x).
∴ f(-1) = g(- 1)
⇒a(- 1)² + b(- 1) + c
= b(- 1)² + a (- 1) + c
⇒ a – b + c = b – a + c
⇒ a + a = b + b
⇒ 2a = 2b
⇒ a = b
Also f(- 1) = a – b + c = 0
⇒ b – b + c = 0
⇒ c = 0

Question 9.
If x² – x – 6 and x² + 3x – 18 have a common factor x – a then find the value of a.
Solution:
Let f(x) = x² – x – 6 and
g(x) = x² + 3x – 18
Given that (x – a) is a factor of both f(x) and g(x).
f(a) = g(a) = 0
⇒ a² – a – 6 = a² + 3a – 18
⇒ – 4a = – 18 + 6
⇒ – 4a = – 12
∴ a = 3

Question 10.
If (y – 3) is a factor of y³– 2y²– 9y + 18, then find the other two factors.
Solution:
Let f(y) = y³– 2y² – 9y + 18
Given that (y – 3) is a factor of f(y).
Dividing f(y) by (y – 3)

∴ f(y) = (y – 3) (y + y – 6)
But y² + y – 6
= y² + 3y – 2y – 6
= y (y + 3) – 2 (y + 3)
= (y + 3) (y – 2)
∴ f(y) = (y – 2)(y – 3)(y + 3)
The other two factors are (y – 2) and (y + 3).

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.6 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.6

Question 1.
Find the sum of integers which are divisible by 5 from 1 to 100.
Solution:
Numbers which are divisible by 5 from 1 to 100 are 5, 10, 15, …………………95, 100.
∴ Sum of the above numbers = 5+10 + ……………..+ 95 + 100
= 5[1 + 2 + ………………….+ 20]
= 5 [ \(\frac{20 \times(20+1)}{2}\) ]
= \(\frac{5 \times 20 \times 21}{2}\) [∵ Sum of ‘n’ natural numbers = \(\frac{n(n+1)}{2}\) & n = 20 ]
= 1050

Question 2.
Find the sum of integers which are divisible by 2 from 11 to 50.
Solution:
. Numbers which are divisible by 2 from 11 to 50 are 12, 14,48, 50.
Sum of the numbers = 12 + 14 + ……….. + 48 + 50 ‘
= (2 + 4 + ……….. + 50) – (2 + 4 + ……….. + 10)
= 2(1 + 2 +……….. + 25) – 2 (1 + 2 + ……….. + 5)

= 25 × 26 – 5 × 6
= 650 – 30
= 620

Question 3.
Find the sum of integers which are divisible by 2 and 3 from 1 to 50.
Solution:
Numbers which are divisible by 2 and 3 i.-e., which are divisible by 6 from 1 to 50 are 6,12 …………….48.
Sum of the numbers = 6 + 12 + ……..+ 48
= 6(1 + 2 +……… + 8)
= 6 \(\left[\frac{8(8+1)}{2}\right]\)
= 3 × 8 × 9 = 216

Question 4.
(n³ – n) is divisible by 3. Explain the reason.
Solution:

∴ If n = 4, (n³ – n) is divisible by 3.
∴ (n³ – n) is divisible by all the values of n.
Method 2:
n³ – n = n(n² – 1)
= n(n + 1)(n – 1)
∴ (n³ – n) is divisible by ‘3’ for all the values of n.
[∵ (n – 1), n, (n + 1) are three consecutive odd numbers]

Question 5.
Sum of ‘n’ odd number of consecutive numbers is divisible by ‘n’. Explain the reason.
Solution:
Sum of n’ consecutive odd numbers = n²
Since n is a factor of n², It Is divisible by ‘n’.

Question 6.
Is 1¹¹ + 2¹¹ + 3¹¹ + 4¹¹ divisible by 5? Explain.
Solution:
Sum of units digit of number 1¹¹ + 2¹¹ + 3¹¹ + 4¹¹
= 1 + 8 + 7 + 4
= 20 → \(\frac{20}{5}\)(R = 0)
∴ 1¹¹ + 2¹¹ + 3¹¹ + 4¹¹ is divisible by 5.

Question 7.

Find the number of rectangles of the given figure?
Solution:

∴ No.of rectangles in the given figure = 1 + 2 + 3 + 4 + 5 + 6 = 21

Question 8.
Rahul’s father wants to deposit sorne amount of money every year on the day of Rahul’s birthday. On his 1st birth day Rs.100, on his 2nd birth day Rs.300, on his 3 birth day Rs.600, on his 4th birthday Rs. 1000 and so on. What is the amount deposited by his father on Rahul’s 15th birthday.
Solution:

Rahul’s father deposits on every year 200, 300, 400 more than before year.
Then he deposits ₹ 10,500 on 14th birthday.
∴ The amount deposits on 15th birthday
= 10,500 + 1,500
= ₹ 12,000/-

Question 9.
Find the sum of integers from 1 to 100 which are divisible by 2 or 5.
Solution:
Sum of the numbers which are divisible by 2 from 1 to 100
= 2 + 4 + ……….. + 100
= 2(1 + 2 + ………… +50)
= 2 × \(\frac{50 \times(50+1)}{2} \)
= 50 × 51 = 2550
Sum of the numbers which are dMsible by 5froin I to 100
= 5 + 10 + ……….. + 100
= 5(1 + 2 +……….. +20)
= 5 × \(\frac{20 \times(20+1)}{2}\)
=5 × 10 × 21
=1050

Sum of the numbers which are.divisible by both 2 and 5 = 2550 + 1050 =3600
∴ Sum ol the numbers which are divisible by 2 or 5 from 1 to 100
= 10 + 20 + ………..+ 100 ( L.C.M of 2, 5 is 10)
=10(1 + 2 + ………..+ 10)
= 10 × \(\frac{10 \times(10+1)}{2}\)
= 5 × 10 × 11 .
= 550
∴ The sum of required numbers 3600—550 3050

Question 10.
Find the sum of integers from 11 to 1000 which are divisible by 3.
Solution:
Sum ol the numbers which are divisible by 3 from lito 1000
= 12 + 15+ ……….. +099
= 3(4 + 5 + ……….. +333)
= 3(1 + 2 + ……….. + 333) – 3(1 + 2+3)

= 999 × 167 – 9 × 2
= 166833 – 18
= 166815

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.2

Question 1.
Observe the following tables and fmd which pair of variables (x and y) are in inverse proportion

Solution:
i) From the given table if the value of x is decreases then the value of ‘y’ is increases.
∴ x, y are in inverse proportion.
ii) From the given table if the value of x is increases then the value of y is decreases.
∴ x, y are in inverse proportion.
iii) From the given table the value of x is decreases then the value of y is increases.
∴ x, y are in inverse proportion.

Question 2.
A school wants to spend ₹6000 to purchase books. Using this data, fill the following table.

Solution:

Question 3.
Take a squared paper and arrange 48 squares in different number of rows as shown below.

What do you observe? As R increases, C decreases
(i) Is R₁:R₂ = C₂:C₁?
(ii) Is R₃:R₄ = C₄:C₃?
(iii) Is R and C inversely proportional to each other?
(iv) Do this activity with 36 squares.
Solution:
(i) Is R₁:R₂ = C₂:C₁
⇒ 2 : 3 = 16 : 24

(ii) Is R₃:R₄ = C₄:C₃

R₁:R₂ = C₂:C₁

(iii) R₃:R₄ = C₄:C₃
⇒ 4 : 6 = 8 : 12
\(\frac{4}{6}=\frac{8}{12}=\frac{4 \times 2}{6 \times 2}=\frac{4}{6} \Rightarrow \frac{4}{6}=\frac{4}{6}\) =
∴ R₃:R₄ = C₄:C₃

(iv) Do this activity with 36 squares.

From the above table we can conclude that if number of rows are increases then number of columns are decreases.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 10 Direct and Inverse Proportions Ex 10.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 10th Lesson Direct and Inverse Proportions Exercise 10.1

Question 1.
The cost of 5 meters of a particular quality of cloth is ₹ 210. Find the cost of(i) 2 (ii) 4
(iii) 10 (iv) 13 meters of cloth of the same quality.
Solution:
The cost of 5 m of a cloth = ₹ 210
The length of a cloth and its price are in direct proportion.
i) \(\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}=\frac{\mathrm{x}_{2}}{\mathrm{y}_{2}}\)
Here x₁ = 5, y₁ = 210
x₂ = 2, y₂ = ?

Question 2.
Fill the table.

Solution:

Question 3.
48 bags of paddy costs ₹ 16, 800 then find the cost of 36 bags of paddy.
Solution:
Number of bags of paddy and their cost are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 16,800
x₂ = 36 y₂ = ?

= 3 × 4200
y₂ = ₹ 12600
∴ The cost of 36 bags of paddy = ₹ 12600

Question 4.
The monthly average expenditure of a family with 4 members is 2,800. Find the
monthly average expenditure ofa family with only 3 members.
Solution:
Number of family members and their expenditure are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 4
y₁ = 2,800
x₂ = 3 y₂ = ?

= 3 × 700 = 2100
y₂ = ₹ 2100
The expenditure for 3 members = ₹ 2100

Question 5.
In a ship of length 28 m, height of its mast is 12 m. If the height of the mast in its model is
9 cm what is the length of the model ship?
Solution:
The length of ship and the height of its mast are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\) , x₁ = 28
y₁ = 12
x₂ = ? y₂ = 9

x₂ = 7 × 3 = 21
∴ The length of model ship = 21 m

Question 6.
A vertical pole of 5.6 m height casts a shadow 3.2 m long. At the same time find (j) the
length of the shadow cast by another pole 10.5 m high (ii) the height of a pole which casts
a shadow 5m long.
Solution:
length of a vertical pole and length of its shadow are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)

i) x₁ = 5.6
y₁ = 3.2
x₂ = 10.5 y₂ = ?

∴The length of the shadow = 6 cm

ii) x₁ = 5.6 m x₂ = ?
y₁ = 3.2 m y₂ = 5
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)

∴ x₂ = 8.75

Question 7.
A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:
Time and distance are in direct proportion
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 14 km , x₂ = ?
y₁ = 25min = \(\frac{25}{60} \mathrm{hr}=\frac{5}{12} \mathrm{hr}\) = y₂ = 5hrs
⇒ \(x_{2}=\frac{x_{1} \times y_{2}}{y_{1}}=\frac{14 \times 5}{5}=\frac{14 \times \not 5 \times 12}{\not 5}\)
= 168 km
∴ Lorry travelled in 5 hrs = 168km

Question 8.
If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same paper would weigh 16 \(\frac { 2 }{ 3 }\) kilograms?
Solution:
Number of pages and their weight are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 12 km , x₂ = ?
y₁ = 40 gm
y₂ = 16 \(\frac { 2 }{ 3 }\) gm = \(\frac { 50 }{ 3 }\) x 1000 gm
= \(\frac { 50000 }{ 3 }\) gm

From (1)

∴ Number of pages = 5000

Question 9.
A train moves at a constant speed of 75 km/hr.
(i) How far will it travel in 20 minutes?
(ii) Find the time required to cover a distance of 250 km.
Solution:
Speed of the train = 75 km/hr
i) The distance travelled in 20 min.
d = s x t = 75 x 20 min
= 75 x = 25 km
= \(75 \times \frac{20}{60}=\frac{75}{3}\) = 25 km

ii) Time taken to travel 250 km
t = \(\frac{d}{s}=\frac{250}{75}\)
t = \(\frac{10}{3}\) hrs

Question 10.
The design of a microchip has the scale 40:1. The length of the design is 18cm, find the actual length of the micro chip?
Solution:
The scale of the design of a microchip
= 40 : 1
The length of the design = 18 cm
The actual length of microchip = ?
The length of the design and actual length of the microchip are in direct proportion.
⇒ \(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
x₁ = 40 km , x₂ = 18
y₁ = 1
y₂ = ?
⇒ \(\frac{40}{1}=\frac{18}{y_{2}}\)
⇒ \(\frac{18}{40}=\frac{9}{20}\) cm
∴ The original (actual) length of the microchip = [latexs]\frac{9}{20}[/latex]cm

Question 11.
The average age of consisting doctors and lawyers is 40. If the doctors average age is 35 and the lawyers average age is 50, fmd the ratio of the number of doctors to the number of lawyers.
Solution:
Let the number of doctors = x
Number of lawyers = y
The average age of doctors = 35
The total age of doctors = 35 × x
= 35 x years
The average age of lawyers = 50
∴ The total age of lawyers = 50 x y
= 50y
According to the sum
\(\frac{35 x+50 y}{x+y}\) = 40
⇒ 35x + 50y = 40x + 40y
⇒ 40x – 35x = 50y – 40y
⇒ 5x = lOy
⇒ \(\frac{x}{y}=\frac{10}{5}\) (or)
x : y = 2 : 1
∴ The ratio of number of doctors to lawyers = 2:1

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.4

Question 1.
Check whether 25110 is divisible by 45.
Solution:
The given number = 25110
If 25110 is divisible by 45 then it should be divisible by 5 and 9.

∴ The number 25110 is divisible by 45

Question 2.
Check whether 61479 is divisible by 81.
Solution:
If 61479 is divisible by 81 then it is divisible by 9.
If the sum of the digits of a number is dívisible by 9 then the entire number is divisible by 9.
∴ 61479 → 6 + 1 + 4 + 7 + 9 → \(\frac { 27 }{ 9 }\) (R = 0)
∴ 61479 is divisible by 81. [∵ 9 is factor of 81]

Question 3.
Check whether 864 is divisible by 36? Verif,’ whether 864 is divisible by all the factors of 36 ?
Solution:
864 is divisible by 2 and 3.
∴ 864 is divisible by 6.
∴ 864 is divisible by 36 [ ∵ 6 is the factor of 36]
∴ Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18. 36.


∴ 864 is divisible by all the factor of 36.

Question 4.
Check whether 756 is divisible by 42? Verify whether 756 is divisible by all the factors of 42?
Solution:
756 is divisible by 2 and 3.
∴ 756 is divisible by 6.
2a + 3b + c = 2 x 7 + 3 x 5 + 6 = 14 + 15 + 6 → \(\frac { 35 }{ 7 }\) (R = 0)
∴ 756 is divisible by 7.
∴ 756 is divisible by 42. [ ∵ 6, 7 are the factors of 42]
Factors of 42 = 1, 2, 3, 6, 7, 14, 21, 42.


∴ 756 is divisible by all the factor of 42.

Question 5.
Check whether 2156 is divisible by 11 and 7? Verify whether 2156 is divisible by product of 11 and 7?
Solution:

Question 6.
Check whether 1435 is divisible by 5 and 7? Verify if 1435 is divisible by the product of 5 and 7?
Solution:

Question 7.
Check whether 456 and 618 are divisible by 6’? Also check whether 6 divides the sum of 456 and 618 ‘?
Solution:

Question 8.
Check whether 876 and 345 are divisible by 3. Also check whether 3 divides the difference of 876 and 345?
Solution:

Question 9.
Check whether 2² + 2³+2⁴ is divisible by 2 or 4 or by both 2 and 4’?
Solution:

∴ 2² + 2³+2⁴ is divisible by both 2 and 4.

Question 10.
Check whether 32² is divisible by 4 or 8 or by both 4 and 8’?
Solution:

32² is divisible by 4 and 8

Question 11.
If A679B is a 5-dit number is divisible by 72 find ‘A’ and ‘B”?
Solution:
If A679B is divisible by 72 then it should be divisible by 8 and 9.
[ ∵ 8, 9 are the factors of 72]
A679B is divisible by 9 then
A + 6 + 7 + 9 + B = A + B + 22 = 27 (= 9 x 3)
=A + B = 5 ……………. (1)
A679B → \(\frac{79 \mathrm{~B}}{8}\) [From B (2,4,6,8) we take B = 2]
= \(\frac{792}{8}\) (R = 0)
∴ B = 2
From (1) ⇒ A + 2 = 5
∴ A = 3, B = 2

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.3

Question 1.
Check whether the given numbers are divisible by ‘6’ or not?
(a) 273432
(b) 100533
(c) 784076
(d) 24684
Solution:
if a number is divisible by ‘6’, it has to be divisible by 2 and 3.

Question 2.
Check whether the given numbers are divisible by ‘4’ or not?
(a) 3024
(b) 1000
(c) 412
(d) 56240
Solution:

Question 3.
Check whether the given numbers are divisible by ‘8’ or not?
(a) 4808
(b) 1324
(c) 1000
(d) 76728
Solution:

Question 4.
Check whether the given numbers are divisible by ‘7’ or not?
(a) 427
(b) 3514
(e) 861
(d) 4676
Solution:

Question 5.
Check whether the given numbers are divisible by ‘11’ or not?
(a) 786764
(b) 536393
(c) 110011
(d) 1210121
(e) 758043
(f) 8338472
(g) 54678
(h) 13431
(i) 423423
(j) 168861
Solution:

Question 6.
If a number is divisible by ‘8’, then it also divisible by ‘4’. also Explain?
Solution:
If a number is divisible by 8 it ¡s also divisible by 4.
∴ If a number is divisible by 8, then it ¡s also divisible by the factors of 8.
Factors of 8 = 1, 2, 4, 8.
∴ The number which is divisible 8, is also divisible by 4.

Question 7.
A 3-digit number 4A3 is added to another 3-digit number 984 to give four digit number 13B7, which is divisible by 11. Find (A + B).
Solution:
The given 3 – digited numbers are = 4A3, 984
∴ 4A3 + 984 = 13B7. If It is divisible by 11 then,
⇒ 1 3 B 7
(1 + B) – (3 + 7)
⇒ (B+1) – 10 = 0 ⇒ B – 9 = 0
∴ B = 9

⇒ A + 8 = 9 ⇒ A = 9 – 8 = 1
∴ A = 1
A + B= 1+9
∴ A + B = 10

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 15 Playing with Numbers Ex 15.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 15th Lesson Playing with Numbers Exercise 15.2

Question 1.
If 345 A 7 is divisible by 3,supply the missing digit in place of ‘A’.
Solution:
If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.
∴ 345A7 ⇒ 3 + 4 + 5 + A + 7 = 19 + A
19 + A = 3 x 7
⇒ A = 21 – 19 = 2 ⇒ A = 24 – 19 = 5

A + 19 = 3 x 8
⇒ A = 24 – 19 = 5

A + 19 = 3 x 9
⇒ A = 27 – 19 = 8

∴ A = {2,5,8}

Question 2.
If 2791 A,is divisible by 9, supply the missing digit in place of ‘A’.
Solution:
If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.
∴ 2791A = 2 + 7 + 9 + 1 + A = 9 x 3
⇒ 19 + A = 9 x 3 = 27
⇒ A = 27 – 19 = 8
∴ A = 8

Question 3.
Write some numbers which are divisible by 2,3,5,9 and 10 also.
Solution:
90, 180, 270. are divisible by 2, 3, 5, 9 and 10.
[∵ The L.C.M. of 2, 3, 5, 9, 10 is 90]

Question 4.
2A8 is a number divisible by 2, what might be the value of A’?
Solution:
If the units digit of a number be 0, 2, 4, 6, 8 then it is divisible by 2.
∴ 2A8 is divisible by 2 for any value of A.
∴ A = (0, 1, 2 ………………….9)

Question 5.
50B is a number divisible by 5, what might be the value of B?
Solution:
Given number is 50B.
The units digit of a number ¡s either ‘0’ or 5, then it is divisible by 5.
∴ 500 → \(\frac { 0 }{ 5 }\) (R = 0)
505 → \(\frac { 5 }{ 5 }\) (R = 0)
∴ B = {0, 5}

Question 6.
2P is a number which is divisible by 2 and 3, what is the value of P
Solution:
The given number is 2P.
If 2P is divisible by 2, 3 then 2P should be a multiple of 6. [ ∵ L.C.M. of 2, 3 is 6]
∴ 2P = 24, 30 ………….
24 → 2 + 4 → \(\frac { 6 }{ 3 }\) (R = 0)
∴ P = 4

Question 7.
54Z leaves remainder 2 when divided by 5 , and leaves remainder 1 when divided by 3, what is the value of Z’?
Solution:
If 54Z is divisible by 3 then the sum of the digits of the number is divisible by 3.
According to problem 54Z is divisible by 3 and leaves remainder 1’.
∴ 5 + 4 + Z = (3 x 4) + 1
= 9 + Z = 13
∴ Z = 4(or)
9 + Z = (3 x 5) + 1
9 + Z = 16
Z = 7
If 54Z is divisible by 5 then Z should be equal to either ‘0’ or ‘5’.
∴ 54(0 + 2) = 542 (Z = 2)
54(0 + 7) = 547 (Z = 7)
∴ From the above two cases
Z = 7
∵ 547 → \(\frac{7}{5}\)(R = 2)

Question 8.
27Q leaves remainder 3 when divided by 5 and leaves remainder 1 when divided by 2, what is the remainder when it is divided by 3?
Solution:
27Q is divided by 5 gives the remainder 3
Le.,27Q = 27 (0 + 3) = 273(Z = 3)(T)
= 27 (0 + 8) = 278 (Z = 8)
27Q is divided by 2 gives the remainder 1.
i.e., 27Q = 27(0 + 1) = 271 (Z = 1)
27Q = 27 (0 + 3) = 273 (Z = 3) (T)
∴ From above situations Z = 3
∴ 27Q = 273→ 2 + 7 + 3 → \(\frac{12}{3}\)(R = 0)
∴ 273 is divisible by 3 and gives the remainder 0’.