Class 10

AP State Board Syllabus AP SSC 10th Class English Textbook Solutions Chapter 3C What is My Name? Textbook Questions and Answers.

AP State Syllabus SSC 10th Class English Solutions Chapter 3C What is My Name?

10th Class English Chapter 3C What is My Name? Textbook Questions and Answers

Comprehension

Answer the following questions.

Question 1.
What made Mrs. Murthy so restless to know her name?
Answer:
In her scrubbing zeal, Mrs. Murthy had forgotten her name. At once she felt that she had lost her own identity. She had lost her self-respect. That made her so restless to know her name.

Question 2.
How did Mrs. Murthy’s husband look upon her desire to know her name?
Answer:
Mrs. Murthy’s husband laughed at her when she asked him about her name. He did not take it seriously. He wanted her to be called by his name ‘Mrs. Murthy’. He did not give any importance to her feelings.

Question 3.
Do you notice any change in Mrs.Murthy in the first picture and Sarada in the second picture?
Answer:
Yes. I have found a lot of difference in her being Mrs. Murthy and her being Sarada. After she got her name back she got her confidence back. After she recollected her name she felt like a real person because she got her identity and self-respect back. When she was a housewife she was different, and after she has remembered her name she has become flamboyant.

Question 4.
Do you find any similarities between Mrs.Murthy and the women in your family? If yes, list them.
Answer:
Yes. There are so many similarities between Sarada and the members of our family. Not only in our family but almost in all families we can find a woman like Sarada because she is a representative of women of any class of Indian society who are always confined to domestic work like cleaning floors, washing clothes, cooking, and looking after her children, her husband, and other family members.

Question 5.
Why do you think the writer decided to focus on the question of married women’s identity?
Answer:
The writer decided to focus on the question of married women’s identity because she wanted them to live with their own identity and self-respect. She did not want them to confine to their homes. She wanted them to be given equal rights, equal respect. She wanted that they should also assume some responsibility in nation building activity.

Question 6.
Do you really think a woman can forget her name? What do you think is the intention of the author here?
Answer:
No. I don’t think any woman can forget her own name. The intention of the author is that the women should be respected and should not be confined to the domestic work. She feels that women should not lose their own identity.

Question 7.
Which part of the story shows that Mrs. Murthy feels her identity restored?
Answer:
In the last part of the story, when she returned to her husband’s house Mrs. Murthy feels her identity restored. It is clear from her words “…. from now onwards don’t call me yemoi, geemoi. My name is Sarada – call me Sarada, understood ?”

Writing

I. Translation

Read the following news item in Telugu and compare it with its translation in English given after that.

The following is the translated version of the above Telugu news item.
Centre’s Nod to Kasturi Rangan Committee Recommendations on Western Ghats

NEW DELHI :
The Ministry of Environment has accepted the report made by the Kasturi Rangan Committee on the conservation of Western Ghats. The committee, in its recommendations, made it clear that no further development activities be undertaken in the Western Ghats spread across the 60 thousand square kilometers in six states. The committee was appointed by the Union Government and headed by Kasturi Rangan to suggest measures to conserve the rarest ecosystem of the Western Ghat forests. The committee submitted its Report on 15th of April. The Ministry of Environ¬ment , after taking opinions of the six state governments and the people of the states, accepted the recommendations. The Western Ghats extend in Gujarat, Maharashtra, Goa, Karnataka, Kerala and Tamil Nadu states.

Let’s think of the following :

Question 1.
Do you think that translation is just translation of language ? Or does it also include translation of ideas?
Answer:
No. I don’t think that translation is just a translation of language. It includes linguistic, pragmatic and cultural elements. A literary translation must reflect the imaginative intellectual and intuitive writing of the author. Literary translation must reflect all the literary features of the source text such as sound effects, selection of words, figures of speech, etc.

Question 2.
Which translation is better, true translation or free translation?
Answer:
True translation is a dynamic equivalent translation which focuses on creating an equivalent effect in the target languagae.

Free translation is a formal equivalent translation in which the form and content of the originaUmessage is to be preserved. Of these two types of translations, true translation is better. Since true translation yields in the equivalent effect and it conveys the message of the original to the receptor audience and are equivalent to the original text in a dynamic way, true translation is better.

Question 3.
Do you find any change in the order of the sentence? For e.g : We have Subject, Verb,Object in English but the order is Subject, Object, Verb in Telugu.
Answer:
Not only the structure of the sentence but the diction and style and the order of arrangement of phrases also changes from Telugu to English.
Languages have different pragmatic linguistic structures and norms transferring the norms of one language may well lead to pragmatic failure.

Question 4.
Do you think sometimes it creates problems in the choice of vocabulary while attempting to translate a text?
Answer:
Translation should implicate accurate meaning. It may be problematic for translators. Wrong choice of words may cause ambiguity. While choosing the apt words for translating a text the translator should consider the situationality, intentionality and acceptability.

Question 5.
Is it possible to translate a poem from one language to the other?
Answer:
No. Poetry is not possible to translate because no poem means just one thing. It is very difficult to translate a poem into another language because we may not be aware of many of the possible meanings of the poem.

Question 6.
Is it necessary to take cultural aspects into consideration?
Answer:
Yes. It is necessary to take cultural aspects into consideration while translating. Because translation is a kind of activity which inevitably involves at least two languages and two cultural traditions. There may be cultural difference between the source text and the target text.

The lesson, “What Is My Name?” is a translated version in English from Telugu.The following is a part of the Telugu version of the lesson. Read the Telugu version and observe how it was translated into English.

Activities :

Question 1.
Is this a good translation? Yes or no? Give reasons.
Answer:
Yes. This is a good translation because this translation is a formal equivalent to the original text in the form and the content. It is a good translation because it reflected the imaginative intellectual and intuitive writing to the author. But we have found some difficulties in the structure of the sentences, in diction and style.

Question 2.
Now translate the Telugu version on this page into English and list the difficulties you face.
Answer:
A young woman, before she became a housewife had been an educated, cultured, intelligent, capable, and quick-witted with a sense of humour and elegance.

A young man, who liked her beauty and intelligence and was attracted by the dowry offered by her father, tied the three sacred knots around her neck and made her a housewife. After making her his housewife he told her, “Look Ammadu. This house is yours”. On hearing his words the housewife at once pulled the edge of her sari and tucked it in at the waist and swabbed the entire house and decorated the floor with rangoli designs. On seeing this that young man praised her promptly by saying “Ammadu ! You are dexterous in swabbing floors and even more adept in drawing the muggulu. Sabash Keep it up.” He said it in English, giving her a pat on the shoulder in appreciation.

Overjoyed with this, the housewife continued to live with swabbing as the chief mission of her life. Always she used to scrub the floor spotlessly and decorate it with beautiful multi-coloured rangoli designs. Thus her life went on with a sumptuous and ceaseless supply of swabbing cloths and muggu baskets.

But one day while scrubbing the floor, the housewife suddenly asked herself, “What is my name?”. The query shook her up. Leaving the mopping cloth and the muggu basket there itself, she stood near the window scratching her head, lost in thoughts.

“What is my name ? What is my name ?”. The house across the street carried a name-board, Mrs. M. Suhasini, M.A., Ph.D., Principal, ‘X’ College. Yes, she too had a name as her neighbour did. “How could I forget like that? In my scrubbing zeal I have forgotten my name — what shall I do now?” The housewife was perturbed. Her mind became totally restless. Somehow she finished her daubing for the day.

Meanwhile, the maidservant came there. Hoping that at least she would remember her name, the housewife asked her, “Look ammayi, do you know my name ?”

Question 3.
Translate the following extract from the story into Telugu and compare it with the original story in Telugu. (Refer to teacher’s handbook for Telugu version.)
‘Sarada! My dear Sarada!’ she shouted and embraced her. The housewife felt like a person — totally parched and dried up, about to die of thirst — getting a drink of cool water from the new earthen kooja poured into her mouth with a spoon and given thus a new life.The friend did indeed give her a new life — ‘You are Sarada. You came first in our school in the tenth class. You came first in the music competition conducted by the college. You used to paint good pictures too. We were ten friends altogether — I meet all of them some time or other. We write letters to each other. Only you have gone out of our reach! Tell me why you are living incognito?’ her friend confronted her.
Answer:

Project Work

I. Influence of technical gadgets on human relations.
Visit five houses in your neighbourhood and collect the information in the given format related to human relations, i.e. spending quality time with the members of the family and friends, sharing and caring. Analyse the information and write a report by adding your opinion on how the modern gadgets are influencing human relations and present it before the class.

Family-1 (House-1) (Raja Rao’s Family)

Family-2 (House-2) (Rama Rao’s Family)

Family-3 (House-3) (Venkat’s Family)

Family-4 (House-4) (Nageswara Rao’s Family)

Family-5 (House-5) (Bhaskar’s Family)

We all know very well that electronic gadgets such as TVs, Mobile phones, Computers occupy a major place in our day-to-day lives. Though these gadgets have their own advantages, they also have a negative influence on the human relationships. They play a vital role in our lives today. Most of us are addicted to them. There are hundreds of channels which are viewed on TV by us. We start watching TV programmes from a very early time in the morning till midnight. Thus we don’t have any time to talk to our near and dear. In the same way, mobile phones too have advantages as well as disadvantages. Most of the children and youth are spending all their time in using their mobile phones. They talk, watch movies, listen to music, play games on their mobile phones. When they engage in using their mobile phones, how can they find time to spend with their family members? Today, computers have become indispensable to each one of us.

In most of the families, we find a complete different situation in managing human relationships before and after the accessibility of the electronic gadgets. Before the accessibility of these modern gadgets, people share most of their time to spend with their family members. They take care of all the family members. They try to share their feelings with their famiy members. They often visit their friends. Their relationships with their family members and friends are very cordial. They find a lot of time to do all the things leisurely. Thus family relations are hectic. They show love and affection for their family members. They find time to play with their friends. They help their family members in all the matters.

After the accessibility of the modern gadgets, they try to spend all their time in using them only. They spend with them hours together. They don’t find time to spend with their family members. They don’t visit their friends very often. They don’t care for their family members. They give importance to the gadgets only. They become mechanical. They don’t have any affection for their family members. They don’t understand the warmth of relationship. They don’t know how much they are missing the joyous family interaction. They should understand that no artificial media can substitute their family’s warmth and interaction. They should give importance to human relationships.

II. Nowadays, we can easily find children even as young as two years old playing with electronic devices and gadgets anywhere. It is not only the video games that make children stay, it also includes television, mobile phones, computers, tablet computers, PSP (Play Station Portable) games, etc. Parents may find it easier to make their children stay in one place by giving them a gadget to play with.

Work in groups and discuss the following:

Ways of managing children’s electronic devices consumption and preventing
Answer:
Group 1 :
The parents should make their children know the bad effects of spending more time with electronic devices.

Group 2:
The parents should monitor their children’s media consumption – television, mobile phones, computers, tablet computers, PSP games, etc.

Group 3 :
The parents should make their children aware about the advantages of playing games, doing exercises and yoga instead of their spending more time with electronic devices.

Group 4 :
The parents should stop themselves from using the electronic gadgets for longer times. Thus they can set an example for their children.

Group 5 :
The parents should make their children aware about the health problems that would arise with spending longer times with electronic gadgets.

Group 6 :
The parents should share thier feelings with their children. They should discuss with them the ill effects of the games they play. They should try to move closer to their children.

Sum up:
Today most of the children are addicted to the modern gadgets such as television, mobile phones, computers, tablet computers, PSP games, etc. These all become an integral part of children’s lives. Today children are heavily exposed to media. Parents may find it easier to make their children stay in one place by giving them a gadget to play with. The parents should watch carefully what their children are doing with the gadgets and how they are using it. They should prevent their children’s addiction to games. They should find ways to manage their children’s electronic device consumption.

They should make their children know the bad effects of spending more time with electronic devices. They should monitor their children’s media consumption. They should make them play games and do exercise and yoga. The parents should stop themselves from using the electronic gadgets and stand as an example to their children. They should share their feelings with their children. They should try to move closer to their children.

Thus, parents can manage their children’s electronic device consumption and prevent their addiction to games.

What is My Name? Summary in English

Sarada, before she got married, was a well-educated and cultured young woman. She was intelligent, capable, quick-witted and she had a sense of humour and elegance. She used to stand first in her class. She was good at music and dance. She used to paint good pictures.

Falling to her beauty, and intelligence and attracted by the dowry her father offered, a young man married her. Later he showed his house and told her that it was her house. Immediately she began to swab the floors and decorated the floor with rangoli designs. On seeing this, her husband praised that she was dexterous at swabbing the floor. Overjoyed by his applause, Sarada began living with swabbing as a mission of her life. Thus, her life went on scrubbing the house spotlessly and decorating the house with multi-coloured designs. In her scrubbing zeal she had forgotten her name. One day she tried to recollect what her name was. But she could not. She became restless. She asked her maid servant, her neighbours, her husband, and her children about her name. But they all told the name by which they used to call her by using their relation. Her husband laughed and did not take it seriously.

Finally the housewife decided to go to her parents’ house and look for her name in her certificates. But her certificates were kept on the attic. Meanwhile she met her classmate. She called the housewife by her name ‘Sarada’. At once Sarada felt like a person. Because our name gives us our personal identity and self-respect. Our name is our own- unique to us.

Here the author wants to tell that every woman has her own responsibility in nation-building. Women should be given equal rights with men. Women should not be confined to the four walls of the house. She should be let free. She can reach to the heights of sky. She can ascend to the pinnacles of any success and thus she can make any nation greater and stronger.

What is My Name? Glossary

quick-witted (adj) : intelligent; able to think quickly

elegance (n) : a satisfying or admirable neatness; ingenious simplicity or precision in something

dowry (n) : money and property paid by a bride’s family to the bridegroom at the time of marriage

swab (v) : clean

dexterous (adj) : skilful

appreciation (n) : recognizing and enjoying the good qualities of somebody or something

sumptuous (adj) : luxurious, splendid

cease (v) : stop

ceaseless (adj) : continuous

zeal (n) : great energy and enthusiasm

mopping (v) : cleaning/washing

perturb (v) : bother/disturb/trouble

daubing (n) : the act of spreading a substance such as mud thickly

take somebody aback : to shock or surprise somebody very much

immerse (v) : absorb oneself in something

urge (v) : to try hard to persuade

giggling (v) : laughing nervously

anguish (n) : severe pain, unhappiness

frantically (adv) : worriedly/anxiously

chore (n) : a task that you do regularly

attic (n) : a room or space just below the roof of a house often used for storing things ; loft

wail (v) : to make a long loud cry

maternity home (n) : hospital for deliveries

parch (v) : dehydrate

incognito (adv) : having a concealed identity

fish (v) : search

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.4

10th Class Maths 6th Lesson Progressions Ex 6.4 Textbook Questions and Answers

Question 1.
In which of the following situations, does the list of numbers involved in the form a G.P.?
i) Salary of Sharmila, when her salary is Rs. 5,00,000 for the first year and expected to receive yearly increase of 10% .
Answer:
Given: Sharmila’s yearly salary = Rs. 5,00,000.
Rate of annual increment = 10 %.

Here, a = a₁ = 5,00,000
a₂ = 5,00,000 × \(\frac{11}{10}\) = 5,50,000
a₃ = 5,00,000 × \(\frac{11}{10}\) × \(\frac{11}{10}\) = 6,05,000
a₄ = 5,00,000 × \(\frac{11}{10}\) × \(\frac{11}{10}\) × \(\frac{11}{10}\) = 6,65,000

Every term starting from the second can be obtained by multiplying its pre¬ceding term by a fixed number \(\frac{11}{10}\).
∴ r = common ratio = \(\frac{11}{10}\)
Hence the situation forms a G.P.

ii) Number of bricks needed to make each step, if the stair case has total 30 steps. Bottom step needs 100 bricks and each successive step needs 2 bricks less than the previous step.
Answer:
Given: Bricks needed for the bottom step = 100.
Each successive step needs 2 bricks less than the previous step.
∴ Second step from the bottom needs = 100 – 2 = 98 bricks.
Third step from the bottom needs = 98 – 2 = 96 bricks.
Fourth step from the bottom needs = 96 – 2 = 94 bricks.
Here the numbers are 100, 98, 96, 94, ….
Clearly this is an A.P. but not G.P.

iii) Perimeter of the each triangle, when the mid-points of sides of an equilateral triangle whose side is 24 cm are joined to form another triangle, whose mid-points in turn are joined to form still another triangle and the process continues indefinitely.

Answer:
Given: An equilateral triangle whose perimeter = 24 cm.
Side of the equilateral triangle = \(\frac{24}{3}\) = 8 cm.
[∵ All sides of equilateral are equal] ……. (1)
Now each side of the triangle formed by joining the mid-points of the above triangle in step (1) = \(\frac{8}{2}\) = 4 cm
[∵ A line joining the mid-points of any two sides of a triangle is equal to half the third side.]
Similarly, the side of third triangle = \(\frac{4}{2}\) = 2 cm
∴ The sides of the triangles so formed are 8 cm, 4 cm, 2 cm,
a = 8

Thus each term starting from the second; can be obtained by multiplying its preceding term by a fixed number \(\frac{1}{2}\).
∴ The situation forms a G.P.

Question 2.
Write three terms of the G.P. when the first term ‘a’ and the common ratio ‘r’ are given.
i) a = 4 ; r = 3.
Answer:
The terms are a, ar, ar², ar³, ……..
∴ 4, 4 × 3, 4 × 3² , 4 × 3² , ……
⇒ 4, 12, 36, 108, ……

ii) a = √5 ; r = \(\frac{1}{5}\)
Answer:
The terms are a, ar, ar², ar³, ……..

iii) a = 81 ; r = –\(\frac{1}{3}\)
Answer:
The terms of a G.P are:
a, ar, ar², ar³, ……..

⇒ 81, -27, 9,

iv) a = \(\frac{1}{64}\); r = 2.
Answer:
Given: a = \(\frac{1}{64}\); r = 2.

∴ The G.P is \(\frac{1}{64}\), \(\frac{1}{32}\), \(\frac{1}{16}\), …….

Question 3.
Which of the following are G.P. ? If they are G.P, write three more terms,
i) 4, 8, 16, ……
Answer:
Given: 4, 8, 16, ……
where, a₁ = 4; a₂ = 8; a₃ = 16, ……
\(\frac{a_{2}}{a_{1}}=\frac{8}{4}=2\)
\(\frac{a_{3}}{a_{2}}=\frac{16}{8}=2\)
∴ r = \(\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}\) = 2
Hence 4, 8, 16, … is a G.P.
where a = 4 and r = 2
a₄ = a . r³ = 4 × 2³ = 4 × 8 = 32
a₅ = a . r⁴ = 4 × 2⁴ = 4 × 16 = 64
a₆ = a . r⁵ = 4 × 2⁵ = 4 × 32 = 128

ii) \(\frac{1}{3}\), –\(\frac{1}{6}\), \(\frac{1}{12}\), …….
Answer:
Given: t₁ = \(\frac{1}{3}\), t₂ = –\(\frac{1}{6}\), t₃ = \(\frac{1}{12}\), ….
\(\frac{1}{3}\), –\(\frac{1}{6}\), \(\frac{1}{12}\), …….

Hence the ratio is common between any two successive terms.
∴ \(\frac{1}{3}\), –\(\frac{1}{6}\), \(\frac{1}{12}\), ……. is G.P.
where a = \(\frac{1}{3}\) and r = –\(\frac{1}{2}\)

iii) 5, 55, 555, ……..
Answer:
Given: t₁ = 5, t₂ = 55, t₃ = 555, ….

∴ 5, 55, 555, …….. is not a G.P.

iv) -2, -6, -18, ……
Given: t₁ = -2, t₂ = -6, t₃ = -18

∴ -2, -6, -18, is a G.P.
where a = -2 and r = 3
a_n = a . r^n-1 =
a₄ = a . r³ = (-2) × 3³ = -2 × 27 = -54
a₅ = a . r⁴ = (-2) × 3⁴ = -2 × 81 = -162
a₆ = a . r⁵ = (-2) × 3⁵ = -2 × 243 = -486

v) \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\), …….
Answer:

i.e., \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\), ….. is not a G.P.

vi) 3, -3², 3³, ……
Given: t₁ = 3, t₂ = -3², t₃ = 3³, ……


i.e., every term is obtained by multiplying its preceding term by a fixed number -3.
3, -3², 3³, …… forms a G.P,
where a = 3; r = -3
a_n = a . r^n-1
a₄ = a . r³ = 3 × (-3)³ = 3 × (-27) = -81
a₅ = a . r⁴ = 3 × (-3)⁴ = 3 × 81 = 243
a₆ = a . r⁵ = 3 × (-3)⁵ = 3 × (-243) = -729

vii) x, 1, \(\frac{1}{x}\), …….
Answer:
Given: t₁ = x, t₂ = 1, t₃ = \(\frac{1}{x}\), ……

Hence x, 1, \(\frac{1}{x}\), …. forms a G.P.
where a = x; r = \(\frac{1}{x}\)

viii) \(\frac{1}{\sqrt{2}}\), -2, \(\frac{8}{\sqrt{2}}\), …….
Answer:
Given: t₁ = \(\frac{1}{\sqrt{2}}\), t₂ = -2, t₃ = \(\frac{8}{\sqrt{2}}\), ……

ix) 0.4, 0.04, 0.004, ……..
Answer:
Given: t₁ = 0.4, t₂ = 0.04, t₃ = 0.004, ……

∴ 0.4, 0.04, 0.004, …….. forms a G.P.
where a = 0.4; r = \(\frac{1}{10}\)

Question 4.
Find x so that x, x + 2, x + 6 are consecutive terms of a geometric progression.
Answer:
Given x, x + 2 and x + 6 are in G.P. but read it as x, x + 2 and x + 6.
∴ r = \(\frac{\mathrm{t}_{2}}{\mathrm{t}_{1}}\) = \(\frac{\mathrm{t}_{3}}{\mathrm{t}_{2}}\)
⇒ \(\frac{x+2}{x}\) = \(\frac{x+6}{x+2}\)
⇒(x + 2)² = x(x + 6)
⇒ x² + 4x + 4 = x² + 6x
⇒ 4x – 6x = – 4 = -2x = -4
∴ x = 2

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 10 Chemical Bonding

AP State Syllabus SSC 10th Class Chemistry Important Questions 10th Chemical Bonding

10th Class Chemistry 10th Lesson Chemical Bonding 1 Mark Important Questions and Answers

Question 1.
What is a chemical bond? (AP March 2015)
Answer:
An attractive force between two atoms in a molecule is called a chemical bond.

Question 2.
Write the names of any two compounds which have ionic bond. (TS June 2016)
Answer:
1) NaCl
2) MgCl₂

Question 3.
Draw the structural diagram of Ammonia molecule as per the valence-shell electron pair repulsion theory. (TS March 2016)
Answer:

Question 4.
Show the formation of HC/ molecule with Lewis dot structures using the information given below. (TS March 2017)

Answer:

Question 5.
Explain, why bonding angle (HOH) in water is 104° 31′ instead of 109° 28′? (TS June 2018)
Answer:
Due to the repulsion between the lone pair of electrons and bond pair of electrons in water molecule the bond angle will be 104° 31′ rather than 109° 28′.

Question 6.
Imagine and write what type of ion can be formed generally by an atom of element with low ionisation energy, low electron affinity with high atomic size? (AP March 2019)
Answer:
Cation or positivity charged ion can be formed generally by an atom of element with low ionisation energy, low electron affinity with high atomic size.

Question 7.
Draw the structure of Methane molecule and mention bond angle. (TS March 2019)
Answer:
Structure of methane CH₄. Tetrahedral structure with bond angle 109°28′.

Bond Angle is 109°28′

Question 8.
Why do elements form chemical bonds? (TS SCERT: 2019-20)
Answer:

• Elements are unstable if they contain less than eight electrons (octet) in their valency shell.
• Hence, they form chemical bond with other elements to get octet in their valency shell.

Question 9.
How is covalent bond formed?
Answer:
A covalent bond is formed by the sharing of electrons between two atoms.
Ex : Bonding in H₂ molecule, O₂ molecule, N₂ molecule, etc.

Question 10.
What is ‘Octet rule’?
Answer:
Octet rule :
Presence of 8 electrons in the outermost shell of an atom or a molecule is called ‘octet rule’.

Question 11.
What is ‘Bond length’?
Answer:
Bond length :
It is the inner-nucleus distance between the two atoms in a molecule. It is measured in Angstrom, 1 Å = 10^-8 cm.

Question 12.
How is a cation formed?
Answer:
A cation is formed when an atom loses electrons.

Question 13.
How is an anion formed?
Answer:
An anion is formed when an atom gains electrons.

Question 14.
Which type of compounds are more soluble in polar solvents?
Answer:
Ionic compounds are more soluble in polar solvents.

Question 15.
Which compounds exhibit high melting and boiling points?
Answer:
Ionic compounds exhibit high melting and boiling points.

Question 16.
What is electronic configuration?
Answer:
A systematic arrangement of electrons in the atomic orbits is called electronic configuration.

Question 17.
Why are molecules more stable than atoms?
Answer:
Molecules have lower energy than that of the combined atoms. Molecules are more stable than atoms since chemical species with lower energy are more stable.

Question 18.
What is ionic linkage?
Answer:
When two oppositely charged ions are engaged in a bond, it is known as ionic linkage.

Question 19.
On which factors do anions depend?
Answer:

1. Atomic size
2. Ionisation potential
3. Electron affinity
4. Electronegativity

Question 20.
How do you know the valence of a metal?
Answer:
The number of electrons lost from a metal atom is the valence of its element which is equal to its group number.
Ex : Na and Mg have valence 1 and 2 respectively.

Question 21.
How do you know the valence of a non-metal?
Answer:
The number of electrons gained by a non-metal element for its atom is its valency, which is equal to 8 – its group number. Ex : The valency of chlorine is (8 – 7) = 1.

Question 22.
What is Tonic bond’?
Answer:
The electrostatic attractive force that keeps cation and anion together to form a new electrically neutral compounds is called ionic bond.

Question 23.
Why do atoms combine and form molecules?
Answer:
The energy of molecule is less than the total energy of constituent atoms. Therefore atoms combine and go to a stable state of lower energy.

Question 24.
What is ‘orbital concept of bond formation’?
Answer:
Atoms with half-filled or vacant orbitals try to get paired electrons in those orbitals by bond formation, i.e. by losing, gaining or sharing of electrons.

Question 25.
Name the bonds present in the molecules
i) BaCl₂
ii) C₂H₄.
Answer:
i) In BaCl₂ – Ionic.
ii) In C₂H₄ – Covalent (double bond H₂C = CH₂).

Question 26.
Why are ionic compounds good electrolytes?
Answer:

• Electrolytes produce ions in solution, which carry current.
• Ionic compounds in the fused state and aqueous solutions contain ions moving freely. Hence they conduct electricity.

Question 27.
When is ionic bond formed between atoms?
Answer:
Ionic bond is readily formed between atoms of elements with a low ionisation energy and atoms of elements with high electronegativity.

Question 28.
What is crystal lattice’?
Answer:
In a crystal of an ionic compound, each ion is surrounded by oppositely charged ions. The ions arrange themselves at an optimum distance with regular periodicity in a well-defined three-dimensional network called crystal lattice.

Question 29.
What is Lattice energy’?
Answer:
The energy released when gaseous positive and negative ions are brought together from infinity to form one mole ionic crystals is called lattice energy.

Question 30.
How is a σ (sigma) bond formed?
Answer:
By the axial or head on overlap of pure orbitals or hybrid orbitals of two atoms.

Question 31.
How is a π (pi) bond formed?
Answer:
By the lateral or parallel overlap of only pure atomic orbitals after the a bond formation.

Question 32.
Which type of atoms easily enter ionic bonding?
Answer:
An atom with low ionization potential and another with high electron affinity.

Question 33.
What is a polar bond?
Answer:
A covalent bond in a heteroatomic molecule.
Eg : HCl.

Question 34.
What is meant by inter-nuclear axis?
Answer:
The hypothetical line joining the centre of nuclei of two atoms in a molecule.

Question 35.
What are multiple bonds?
Answer:
Double and triple bonds are multiple bonds.

Question 36.
How many a and π bonds are in O₂ molecule?
Answer:
One σ and one π bond.

Question 37.
What are Lewis structures?
Answer:
The symbol of the atom gives the core (or kernel) of the atom, in which valence electrons are shown as dots (•), circles (O) or crosses (x). Thus in Na, Na represents the core of sodium atom and the cross (x) represents the valence electron of sodium.

Question 38.
What are the structures of sodium chloride and calcium chloride crystals?
Answer:
Sodium chloride has face centered cubic structure. Calcium chloride has body centered cubic structure.

Question 39.
NaCl dissolves in water but not in benzene. Explain.
Answer:
NaCl dissolves in water because of hydration. Water being a polar molecule has positive and negative ends which hydrate Na⁺ and Cl^– ions. Benzene being non-polar cannot solvate the ions of NaCl.

Question 40.
What are the bond angles in H₂O and NH₃ molecule?
Answer:
Bond angle in H₂O molecule is 104° 30′.
Bond angle in NH₃ molecule is 107°.

Question 41.
What are the forces present in an ionic bond?
Answer:
Electrostatic forces of attraction are present in ionic bonds.

Question 42.
Which forces are weaker forces and where are they operative?
Answer:

1. Van der Waal’s forces are very weak forces.
2. They are operative between non-polar molecules.

Question 43.
Which compounds exhibit low melting and boiling points?
Answer:
Covalent compounds exhibit low melting and boiling points when compared to ionic compounds.

Question 44.
How are cations formed?
Answer:
The metal atoms lose electrons to form positively charged ions or cations.

Question 45.
Why is a molecule of hydrogen more stable than the uncombined atoms?
Answer:
When a molecule of hydrogen is formed from the atoms, energy is released (104 Kcal/mol). Thus the molecules possessing lower energy are more stable than the atoms.

Question 46.
How many sigma and pi bonds are present in acetylene molecule between carbon atoms?
Answer:
\(\mathrm{HC} \equiv \mathrm{CH}\), one sigma and two pi bonds are present.

Question 49.
In case of ionic substances, a more appropriate term is formula weight, rather than ‘molecular weight’. Why?
Answer:
Molecules are not present in ionic substances. Only ions are present. The formula of an ionic substance represents the simpler ratio of ions in one mole of crystal. Hence a more appropriate term is ‘formula weight’.

Question 50.
Write a short note on bond angles.
Answer:
It is the average angle between two adjacent atoms bonded to the central atom in a molecule. Molecules with larger bond angles are more stable than those with smaller angles.

Question 51.
Sulphur dioxide is a diamagnetic molecule. Explain.
Answer:
Sulphur dioxide (SO₂) is a diamagnetic molecule because it has all electrons paired (no free electrons).

Question 52.
What is an ion?
Answer:
An ion is an electrically charged atom (or group of atoms).

Question 53.
What is a Coordination number?
Answer:
The number of ions of opposite charge that surround a given ion of given charge is known as coordination number of that given ion. –

Question 54.
What factors affect or which factors influence the formation cation or anion?
Answer:
Factors affecting formation of anion or cation are :

1. Atomic size,
2. Ionization potential,
3. Electron affinity,
4. Electronegativity.

Question 55.
Two elements X and Y have the following configurations.
X = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
Y = 1s² 2s² 2p⁶ 3s² 3p⁵
What is the formula of the compound?
Answer:
The electronic configuration of X is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s².
So its valency is 2.
The electronic configuration of Y is 1s² 2s² 2p⁶ 3s² 3p⁵.
So its valency is 1.
∴ The formula of the compound is XY₂.

Question 56.
The electronegativities of two elements are 1 and 3. What type of bond is formed between the two elements? Why?
Answer:
The electronegativity difference of elements = 3 – 1 = 2.
So the electronegativity difference is more than 1.9. Therefore the bond formed between the elements is ionic in nature.

Question 57.
What do you mean by Doublet configuration?
Answer:
If two electrons are present in valence shell, then it is called doublet configuration.
Eg.: Helium exhibits doublet configuration.

Question 58.
Write Lewis symbol of potassium and calcium.
Answer:
Potassium – \(\dot{\mathrm{K}}\) and Calcium – •Ca•

Question 59.
Which of the following is / are true?
1) Ionic compounds exist as solids.
2) Ionic compounds have high melting and boiling points.
3) Ionic compounds conduct electricity in solid state.
4) Ionic compounds have low melting and boiling points.
Answer:
Ionic compounds generally exists as solids and they have very high melting and boiling points and they conduct electricity in aqueous or molten state.
So first and second statements are true.

Question 60.
How many a and n bonds are there in acetylene molecule?
Answer:
The structure of acetylene is given by
\(\mathrm{H}-\mathrm{C} \equiv-\mathrm{H}\)
So they are 3a and 2n bonds.

Question 61.
What is the valency of carbon in CO₂?
Answer:
The valency of oxygen is 2. So the valency of carbon = 2 × 2 = 4.

Question 62.
Which of the following does not get Helium configuration during formation of chemical bonding?
1) Hydrogen,
2) Lithium,
3) Beryllium,
4) Oxygen.
Answer:
Oxygen attains Neon configuration when it undergoes chemical bondings and the rest will attain Helium configuration.

Question 63.
Write the valencies of following elements.
a) Be
b) N and what is the compound formed when these two reacted?
Answer:
a) The electronic configuration of Be is 1s² 2s². So its valency is 2.
b) The electronic configuration of Nitrogen is 1s² 2s² 2p³. So its valency is 3.
∴ Formula of the compound is Be₃N₂.

Question 64.
Which of the following solutions are conductors of electricity?
a) Sugar solution,
b) Alcohol solution,
c) Glucose solution,
d) Salt solution.
Answer:
Salt solution is good conductor of electricity because common salt (NaCl) is an ionic compound and the rest are covalent compounds.

Question 65.
Why are noble gases (inert gases) stable?
Answer:

1. Noble gases are stable because their outermost orbit contains 8 or 2 electrons.
2. Noble gases exist an individual atoms.

10th Class Chemistry 10th Lesson Chemical Bonding 2 Marks Important Questions and Answers

Question 1.
Represent each of the following molecules using Lewis notation. (TS March 2015)
(i) Calcium and Chlorine to form Calcium chloride.
(ii) Formation of Oxygen molecule from Oxygen atoms.
Answer:

Question 2.
Between a neutral atom and its cations which has bigger size? Why? (TS June 2016)
Answer:

• A neutral atom has bigger in size than its cation.
• A cation has more protons and nucleus of the cation attracts the electrons in the outermost orbital more.
• Hence, the radius of the anion decreases. It means size of the anion decreases.
• So, a neutral atom is bigger than its cation.

Question 3.
Draw the diagram to show the formation of Oxygen molecule by Valence bond theory. (TS March 2017)
Answer:

Question 4.
Explain Ionic bond with suitable example. (TS June 2018)
Answer:
1) Sodium (Na) looses one electron and forms Sodium ion (Na⁺).
Na → Na⁺ + 1e^–

2) Chlorine (Cl) gains one electron and forms chloride ion (Cl^–).
Cl + 1e^– → Cl^–

3) Positive sodium ion (Na+) and negative chloride (Ct) ion come together due to electrostatic forces, participates in ionic bond and form NaCl.
Na⁺ + Cl^– → NaCl

Question 5.
Give two examples of each to ionic and covalent compounds. (AP SCERT: 2019-20)
Answer:
1) Ionic compounds eg : NaCl, MgCl₂
2) Covalent compounds eg : Cl₂, HCl

Question 6.
Distinguish between a sigma and a pi bond.
Answer:

Sigma bond	Pi bond
1. It is formed by the end-end on overlap of orbitals.	1. It is formed by the lateral overlap of orbitals.
2. It has independent existence.	2. It has no independent existence.
3. It is a strong bond. Because axial overlap is more.	3. It is a weak bond. Because lateral overlap is less.
4. There can be only one a bond.	4. There can be one or two 7t bonds between two atoms.
5. All orbitals form ‘o’ bond.	5. Only p, d orbitals form a bond.

Question 7.
Write about ‘Hydrogen bond’
Answer:

• Hydrogen bond is formed between molecules in which hydrogen atom is attached to an atom of an element with large electronegativity and very small size (F, O, N). Because in hydrogen bond the molecules associate themselves and hence possess higher B.P’s and M.P’s.
• The hydrogen bond formed between two molecules is called inter-molecular hydrogen bond.
• The hydrogen bond formed between different groups of the same molecule is called intra-molecular hydrogen bond.

Question 8.
Bring out the difference between ionic and covalent bonds.
Answer:

Ionic bond	Covalent bond
1. It is formed by transference of electrons from one atom to the other.	1. It is formed by the sharing of electron pairs by two atoms.
2. Electrostatic.	2. Not electrostatic, but rigid.
3. Ionic substances are formed by ionic bonds.	3. Molecules are formed by covalent bonds.
4. Non-directional.	4. Directional.

Question 9.
What is hybridisation?
Answer:
In the formation of molecules, the atomic orbitals of the atoms may hybridise.

1. It is the process of mixing up of atomic orbitals of an atom to form identical hybrid orbitals.
2. This takes place only during the formation of bond.
3. There should not be much difference in the energies of the orbitals that hybridise.
4. The number of hybrid orbitals formed is equal to the number of hybridising atomic orbitals.
5. Hybrid orbitals form a bonds only not n bonds.

Question 10.
What is ionisation? Give one example.
Answer:
1) The process of removal of electron (s) from an atom or molecule is termed as ionisation.
Eg : Na → Na⁺ + e^–

2) Dissociation of an ionic solid into constituent ions upon its dissolution in a suitable solvent, is also called ionisation.
Eg : NaCl_(aq) → Na⁺_(aq) + Cl^–_(aq)

Question 11.
Write a short note on octet rule with example.
Answer:
Atoms try to get 8 electrons in their outermost shell (inert gas structure) by combining with other atoms. In order to get 8 outer electrons, atoms may lose to, gain from or share electrons with other atoms.

Question 12.
What are the important characteristic features of hybridisation?
Answer:

• Orbitals on a single atom only would undergo hybridization.
• The orbitals involved in hybridisation should not differ largely in their energies.
• The number of hybrid orbitals formed is equal to the number of hybridising orbitals.
• The hybrid orbitals form stronger directional bonds than the pure s, p, d atomic orbitals.
• It is the orbitals that undergo hybridisation and not the electrons.
• Concept of hybridisation is useful in explaining the shape of molecules.

Question 13.
Why is a large amount of energy needed to remove an electron from a neutral gaseous neon atom than the energy needed to remove an electron from gaseous sodium atom?
Answer:
Na (g) +118.4 K.cal → Na⁺ + e^–
Ne (g) + 497.0 K.cal → Ne⁺ + e^–

Orbitals which are fully-filled are very stable, so need large amount of energy to remove an electron from them.

Question 14.
What are the drawbacks of electronic theory of valence?
Answer:
1) When covalent bond formed between any two atoms, irrespective of the nature of the atoms, the bond lengths and bond energies’ are expected to be same. But practically it was observed that bond lengths and bond energies are not same when the atoms that form the bond are different.

2) This theory fails to explain the shapes of molecules.

Question 15.
What is the structure of NaCl and write its coordination number of its constituent ions?
Answer:
The structure of NaCl is face centred cubic lattice.
Its coordination number is 6 for Na⁺ and Cl_–.

Question 16.
Why is there absorption of energy in certain chemical reactions and release of energy in other reactions?
Answer:
If bond dissociation energy of reactants is more than bond energy of products, then energy is absorbed in the chemical reaction.

If bond dissociation energy of reactants is less than bond energy of products, then energy is released in the chemical reaction.

Question 17.
Why do ionic compounds dissolve in polar solution and covalent compounds dissolve in non-polar solution?
Answer:
Ionic compounds are polar in nature so they are soluble in polar solvents whereas covalent compounds are non-polar in nature so they are soluble in non-polar solvents.

Question 18.
Identify the following whether they are either oxidation or reduction reactions.

Answer:
i) Oxygen atom is gaining electrons. So it is reduction reaction.
ii) Potassium atom is losing electron. So it is oxidation reaction.
iii) Ferric ion is gaining electron. So it is reduction reaction.

Question 19.
Elements A, B, and C have atomic numbers 9, 20, and 10 respectively.
a) State which one is (i) a metal, (ii) a non-metal, (iii) chemically inert.
b) Write down the formula of the compound formed by two of the above elements.
Answer:
a) i) Element with atomic number 20 is a metal because its electronic configuration is 2, 8, 8, 2. It belongs to 2^nd group and the element is calcium.
ii) Element with atomic number 9 is a non-metal because its electronic configuration is 2, 7. So it belongs to 17^th group and it is Fluorine.
iii) Element with atomic number 10 is inert gas because its electronic configuration is 2, 8 and the element is Neon.

b) The formula of compound formed between elements Calcium and Fluorine is CaF₂.

Question 20.
In the formation of the compound XY₂, an atom X gives one electron to each Y atom. What is the nature of bond in XY₂? Give two properties of XY₂.
Answer:
Two electrons are transferred from X to Y. X forms positive ion and Y forms a negative ion. So the bond formed is ionic in nature.
Properties of ionic compounds :
a) They are hard crystals.
b) They have high boiling point and melting point.
c) They are soluble in water.

Question 21.
Element X is a metal with valency 2. Element Y is a non-metal with valency 3.
i) Write equations to show X and Y form ions.
Answer:
X is a metal with valency 2.

ii) If Y is a di-atomic gas, write the equation for the direct combination of X and Y to form a compound.
Answer:
4X + 3Y₂ → 2X₂Y₃

Question 22.
What are the differences between sodium atom and sodium ion?
Answer:

• Sodium atom is neutral whereas sodium ion has unipositive charge.
• The size of sodium ion is smaller than sodium atom.
• The properties of sodium ion is different when compared with sodium.

Question 23.
What is the percent of p-character in sp, sp² and sp³ hybrid orbitals?
Answer:
In sp the p-character is 50%.
In sp² the p-character is 66.66%.
In sp³ the p-character is 75%.

Question 24.
What is the percent of s-character in sp, sp² and sp³ hybrid orbitals?
Answer:
In sp hybrid orbitals s-character is 50%.
In sp² hybrid orbitals s-character is 33.33%.
In sp³ hybrid orbitals s-character is 25%.

Question 25.
A chemical compound has following Lewis notation.

a) How many valence electrons does element Y have?
b) What is the valency of element Y?
c) What is the valency of element X?
d) How many covalent bonds are there in the molecule?
e) Suggest a name for the element X and Y.
Answer:
a) The valence electrons in Y are 5.
b) The valency of Y is 3.
c) The valency of X is 1.
d) There are three covalent bonds.
e) The element Y is Nitrogen and X is Hydrogen.

Question 26.
The electronic configurations of three elements.
X is 2, 6; Y is 2, 8, 7; Z is 2, 8,1.
In each ease given below state whether the bonding is ionic or covalent and give the formula of molecules of the compound formed,
a) Between two atoms of X.
Answer:
As we move from left to right in a period non-metallic character increases. So Z is a metal and remaining two are non-metals.
Therefore between two atoms of X, there would be covalent bond. The formula of compound is X₂.

b) Between the atom X and atom Z.
Answer:
X is a non-metal and Z is a metal. So between these two atoms, there would be an ionic bond.
The valency of X is 2 and the valency of Z is 1.
So the formula of compound is Z₂X.

c) Between the atom Y and atom Z.
Answer:
Y is also non-metal and we know Z is metal.
The valency of Y is 1.
So the compound formed is ZY.

Question 27.
Electronic configuration of X is 2, 8, 1 and electronic configuration of Y is 2, 8, 7. Explain what type of bond is formed between them.
Answer:
The element with electronic configuration 2, 8,1 is sodium which is a metal. Similarly the element with electronic configuration 2, 8, 7 is chlorine which is a non-metal. The electronegativity difference between these two elements is more than 1.9 so they form ionic bond.

Question 28.
Complete the following statements.
i) The bond which is formed by transfer of an electron from one atom A to another atom B is called ……………
Answer:
Ionic bond

ii) The atoms A and B become ions. Thus A – 2e^– = …………….. and B + e^– = ………….
Answer:

iii) The formula of the compound is
Answer:
The valency of A is 2 and valency of B is 1. So the compound formed is AB₂.

Question 29.
Answer the following.
i) What is the valency of Nitrogen in Ammonia?
Answer:
The formula of Ammonia is NH₃. We know valency of hydrogen is 1. So valency of Nitrogen is 3.

ii) What is the valency of Oxygen in Water?
Answer:
The formula of water is H₂O. The valency of hydrogen is 1. So the valency of oxygen is 2.

iii) What is the valency of Carbon in Carbon tetra chloride?
Answer:
The formula of carbon tetra chloride is CCl₄. The valency of chlorine is 1. So the valency of carbon is 4.

10th Class Chemistry 10th Lesson Chemical Bonding 4 Marks Important Questions and Answers

Question 1.
What is Hybridisation? Explain the formation of BeF, molecule using hybridisation. (AP June 2017)
Answer:
Hybridisation :
Hybridisation is a phenomenon of intermixing of atomic orbitals of almost equal energy which are present in the outer shells of the atom and their reshuffling or redistribution into the same number of orbitals but with equal properties like energy and shape.

Formation of BeF₂ molecule :

1. ₄Be has electronic configuration 1s² 2s².
2. In excited state electronic configuration of ₄Be is 1s² 2s¹ 2p_x¹
3. ₄Be allows its 2s orbitals and 2p_x orbitals which contain unpaired electrons to intermix and redistribute to two identical orbitals.
4. The hybridisation is called ‘sp’ hybridisation and form two ‘sp’ hybrid orbitals. Each orbital contains one electron.
5. ₉F has electronic configuration 1s² 2s² 2p⁵ (or) 1s² 2s² 2px² 2py² 2pz¹
6. F contains one unpaired electron.
7. Two Fluorine atoms come closure to Be and form two covalent bonds.
   σsp – p and σsp – p
8. Now the BeF₂ has (F \(\widehat{\mathrm{Be}}\) F) angle of 180°.
   

Question 2.
The arrangement of electrons in different shells of atoms of 18th Group elements is given in the table.

Answer the following : (AP March 2017)
(i) What is the general electronic configuration of the above elements except He?
(ii) What is the valency of Argon?
(iii) Write Lewis dot structure of Neon.
(iv) Why the above elements do not take part in bond formation?
Answer:
(i) ns² np⁶
(ii) Zero
(iii)

(iv) They are stable as they have 8 electrons (except Helium) in their outer most orbit.

Question 3.
“Nitrogen and Hydrogen react to form a molecule of Ammonia (NH₃). Carbon and Hydrogen react to form a molecule of Methane (CH₄).”
For each reaction :
a) What is the valency of each of the atom involved in the reaction?
b) Draw the dot structure of the products that are formed. (AP June 2018)
Answer:
a) In ammonia : Nitrogen valency – 3, Hydrogen valency – 1
In methane : Carbon valency – 4, Hydrogen valency – 1

b)

Question 4.
Explain the formation of Boron tri-fluoride molecule by Hybridization. (TS June 2018)
Answer:
1) Electron configuration of Boran at ground state.

2) Electron configuration of Boran at excited state is

3) As it forms three identical B-F bonds in BF₃.
4) In excited state of Boran, one ‘s’ orbital and two ‘p’ orbitals undergoes hybridization and forms three sp² orbitals.
1s + 2p → 3sp²
5) Three sp² orbitals make end – on – end overlap with three ‘p’ orbitals of fluorine atoms.
6) They form three sp² – p sigma bonds.
7) In this way Boran tri-fluoride molecule is formed.
8)

Question 5.
Explain the formation of BF₃ molecule with the help of Valence Bond theory? (TS March 2018)
Answer:
Formation of BF₃ molecule :

1. ₅B has electronic configuration 1s² 2s² 2p_x.
2. The excited electronic configuration of ₅B is 1s² 2s¹ 2p_x¹ 2p_y¹.
3. As it forms three identical B-F bonds in BF₃
4. It is suggested that excited ‘B’ atom undergoes hybridisation.
5. There is an intermixing of 2s, 2p_x, 2p_y orbitals and their redistribution into three identical orbitals called sp² hybrid orbitals.
6. For three sp² orbitals to get separated to have minimum repulsion the angle between any two orbitals is 120° at the central atom and each sp² orbital gets one electron.
7. Now three fluorine atoms overlap their 2p_z orbitals containing unpaired electrons (F₉ 1s² 2s² 2p_x² 2p_y² 2p_z²) the three sp² orbitals of ‘B’ that contain unpaired electrons to form three ssp²-p bonds.

Question 6.
Explain the formation of N₂ molecule using Valence Bond theory. (TS June 2019)
Answer:
Formation of N, molecule (Valence Bond theory) :

1. ₇N has electronic configuration 1s² 2s² 2p_x¹ 2p_y¹ 2p_z.
2. Suppose that p_x orbital of one ‘N’ atom overlaps the ‘p_x‘ orbital of the other ‘N’ atom giving σ p_x – p_x bond along the inter – nuclear axis.
3. The p_y and p_z orbitals of one ‘N’ atom overlap the p_y and p_z orbital of other ‘N’ atom laterally, respectively perpendicular to inter – nuclear axis giving π p_y – p_y and π p_z-p_z bonds.
4. Therefore, N₂ molecule has a triple bond between two nitrogen atoms.

Question 7.
If the electronic configurations of atoms A and B are 1s², 2s², 2p₆, 3s², 3p¹ and 1s², 2s², 2p₄ respectively, then
a) which atom forms negative ion?
Answer:
Given electronic configuration of atom A is 1s² 2s² 2p⁶ 3s² 3p¹, i.e. Aluminium and B is 1s² 2s² 2p⁴, i.e. Oxygen.

The atom ‘B’ tends to form negative ion by gaining two electrons in order to get nearest inert gas. Neon’s configuration is 1s² 2s² 2p⁶. So its valency is 2.

b) which atom forms positive ion?
Answer:
The atom ‘A’ tends to form positive ion by losing three electrons in order to get nearest inert gas. Neon’s configuration is 1s² 2s² 2p⁶.

c) what is the valency of atom A?
Answer:
Valency of atom ‘A’ is ‘3’.

d) what is the molecular formula of the compound formed by atoms A and B ?
Answer:
According to Criss-Cross method, the molecular formula of the compound formed by atoms both A and B is A₂B₃, i.e. Al₂O₃.

Question 8.
Writs the salient features of VSEPRT.
(OR)
Explain VSEPR theory.
Answer:
The full form of VSEPRT is Valence Shell Electron Pair Repulsion Theory.

1. VSEPRT considers electrons in the valence shells which are in covalent bonds and in lone pairs as charge clouds that repel one another and stay as far apart as possible which will give specific shapes to molecules.
2. The presence of lone pair in the central atom causes slight distortion of bond angles from expected regular shape.
3. If the angle between lone pair and bond pair increases at the central atom due to more repulsion, the actual bond angles between atoms must be decreased.
4. If two bond pairs are present without any lone pair, the shape of the molecule is linear with bond angle 180°.
5. If three bond pairs are present without any lone pair, the shape of the molecule is trigonal planar with bond angle 120°.
6. If there are four bond pairs in the valency shell of central atom without lone pair, the shape is tetrahedron and the bond angle is 109°28′.
7. If there are three bond pairs and one lone pair, the shape of the molecule is pyramidal.
8. If there are two bond pairs and two lone pairs, the shape is V.

Question 9.
Do you think that a pair of Na⁺Cl^– as units would be present in the solid crystal? Explain.
Answer:

• No, the electrostatic forces are non-directional.
• Therefore, it is not possible for one Na⁺ to be attracted by one Cl^– and vice-versa.
• Depending upon the size and charge of particular ion, number of oppositely charged ions get attracted by it, but, in a definite number.
• In sodium chloride crystal each Na⁺ is surrounded by 6 Cl^– and each Cl^– by six Na⁺ ions.
• Ionic compounds in the crystalline state consists of orderly arranged cations and anions held together by electrostatic forces of attractions in three dimensions.

Question 10.
Explain the valence bond theory.
Answer:
Valence bond theory :

1. A covalent bond between two atoms is formed when the two atoms approach each other closely and one atom overlaps its valence orbital containing unpaired electron with the valence orbital of other atom that contains the unpaired electron of opposite spin.
2. The greater the overlapping of the orbitals, the stronger will be bond.
3. Valence bond theory gives a directional character to the bond when other than ‘s’ orbitals are involved.
4. Electron pair in the overlapping orbitals is shared bv both the atoms involved in the overlapping.
5. If orbitals overlap along inter-nucleus axis, they form a strong bond called sigma (σ) bond.
6. If the orbitals overlap laterally, they form a weak bond called pi (π) bond.

Question 11.
Answer the following.
a) Name the charged particles which attract one another to form ionic or electro-valent compounds.
Answer:
The charged particles are ions which are formed due to transfer of electrons from atom of one element to atom of other element.

b) In the formation of ionic compounds, electrons are transferred from one element to another. How are electrons involved in the formation of covalent compounds?
Answer:
In the formation of covalent bond electrons are mutually shared between atoms of elements.

c) The electronic configuration of Nitrogen is 2, 5. How many electrons in the outer- shell of a nitrogen atom are not involved in the formation of Nitrogen molecule?
Answer:
In the formation of Nitrogen molecules three pairs of electrons are mutually shared by the two nitrogen atoms. So each Nitrogen has two electrons which do not take part in bonding. Those two electrons are called lone pair.

d) In the formation of magnesium chloride, name the substance that is oxidized and the substance that is reduced.
Answer:
Two electrons are transferred from Magnesium atom to Chlorine atom in the formation of Magnesium chloride.

So Magnesium is oxidised (loss of electrons is oxidation) and Chlorine is reduced (gain of electrons is reduction).

Question 12.
The symbols of two elements with their atomic numbers are given below. ₁₆A and ₂₀B
i) Which element will form a cation?
Answer:
The atomic number of B is 20. So its electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² or 2, 8, 8, 2. So it easily loses two electrons to form cation thereby attain inert gas stability.

ii) Which element will form an anion?
Answer:
The atomic number of A is 16. So its electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁴ or 2, 8, 6. So it easily gains two electrons to form anion. Therefore it attains inert gas stability.

iii) Show how each element forms an ion.
Answer:

iv) Which element will be oxidised in forming the ion?
Answer:
B is losing electrons. So it is oxidised. .

v) When the elements react, what will be the nature of the bond formed?
Answer:
When the elements react they form ionic bond because B is metal and A is non-metal.

Question 13.
Give one example for each of the following.
i) A polar covalent bond formed from two dissimilar atoms one of which is oxygen.
Answer:
In the formation water molecule, polar covalent bond is formed between two dissimilar atoms hydrogen and oxygen.

ii) A non-polar covalent compound formed from two similar atoms.

Answer:
The examples for a non-polar covalent compound formed from two similar atoms are Fluorine (F₂), Chlorine (Cl₂).

iii) A solid non-polar covalent compound.
Answer:
The example for solid non-polar covalent compound is Iodine (I₂).

iv) A polar covalent compound containing three shared pairs of electrons.

Answer:
The polar covalent compound containing three shared pairs of electrons is Ammonia.

Question 14.
The list of some substances is given below.
HCl, H₂O, Cl₂, NaBr, CH₄, NH₃, N₂, O₂, CaO, HF, I₂ and Br₂.
From the list above choose the substance / substances.

i) Which have only ionic bond in the molecules?
Answer:
NaBr, CaO have only ionic bond in the molecules,

ii) Which has a triple covalent bond in its molecule?
Answer:
N₂ has a triple covalent bond in its molecule (\(\mathbf{N} \equiv \mathbf{N}\)).

iii) Which are solids?
Answer:
Generally ionic compounds are solids. So NaBr, CaO are solids. I₂ being a covalent compound still it is solid.

iv) Which has a double covalent bond in its molecule?
Answer:
O₂ has a double covalent bond in its molecule (O = O).

v) Which is non-polar covalent liquid?
Answer:
The non-polar covalent liquid is Br₂.

Question 15.
There are elements with atomic numbers 4, 14, 8, 15 and 19. From this information answer the following questions.
a) A solid non-metal of valency 3.
Answer:
The element with atomic number 15 is phosphorus. Its electronic configuration is 2, 8, 5. So its valency is 3 and it is a solid non-metal.

b) A gas of valency 2.
Answer:
The element with atomic number 8 is oxygen. Its electronic configuration is 2, 6. So its valency is 2 and it is a gas.

c) A metal of valency 1.
Answer:
The element with atomic number 19 is potassium. Its electronic configuration is 2, 8, 8, 1. So its valency is 1 and it is a metal.

d) A non metal of valency 4.
Answer:
The element with atomic number 14 is silicon. Its electronic configuration is 2, 8, 4. Its valency is 4 and it is a non-metal.

Question 16.
Explain the following.
a) Ionic compounds conduct electricity.
Answer:
Ionic compounds can be electrolysed to give their constituent ions. The ions obtained by dissociation move freely in solution and hence conduct electricity. When ionic compound is dissolved in water or melted, it becomes a good conductor of electricity.

b) Ionic compounds have high melting and boiling points while covalent compounds have low melting and boiling points.
Answer:
The forces existing between ionic compounds are electrostatic in nature. They are strong forces. So in order to break these forces lot of energy is required. Therefore ionic compounds have higher boiling and melting points.

c) Ionic compounds dissolve in water whereas covalent compounds do not.
Answer:
Ionic compounds are polar in nature. So they are soluble in polar solvent like water whereas covalent compounds are non-polar in nature. So they are insoluble in water.

d) Ionic compounds are usually hard crystals.
Answer:
Due to strong attractive forces they are usually hard crystals.

Question 17.
If A, B, C and D are elements given that B is an inert gas (not Helium).

Element	Atomic Number
A	Z- 1
B	Z
C	Z + 1
D	Z + 4

What type of bonding would take place between (i) A and C and (ii) D and A and write their formulae?
Answer:
1) Given that B is inert gas.
So the valency of A is 1 and it is a non-metal and valency of C is 1 and it is metal and valency of D is 4 and it is a non-metal.

2) The bond formed between C and A is ionic nature and its formula is CA.
The bond formed between D and A is covalent nature and its formula is DA₄.

Question 18.
State whether the following statements are true or false. If statement is false, correct it.
i) In polar compounds the electropositive atom attracts the electron pair towards it.
Answer:
This statement is wrong. In polar compounds the more electronegative atom attracts the electron pair towards it which is known as electronegativity.

ii) Hydrogen chloride gas is a di-polar molecule.
Answer:
Chlorine is more electronegative than Hydrogen. So the electron pair shifts more towards chlorine atom. Thus partial positive charge is formed on Hydrogen and partial negative charge is formed on Chlorine. So it is dipolar molecule. So the statement is true.

iii) Covalent compounds are generally gases due to presence of weak electrostatic forces of attraction.
Answer:
This statement is false. The forces are not electrostatic and also they are weak.

iv) Atoms achieve stable electronic configuration only by transfer of electrons from one atom to another.
Answer:
This statement is false because they can acquire by not only electron transfer but also mutually sharing of electrons.

v) Ionic compounds are soft, solids or liquids at ordinary temperature.
Answer:
This statement is false because ionic compounds are generally hard solids.

Question 19.
Five atoms are labelled from A to E.

Atoms	Atomic Number	Mass Number
A	20	40
B	9	19
C	3	7
D	8	16
E	7	14

a) Which one of these atoms (i) contain 7 protons, (ii) has an electronic configuration 2, 7?
Answer:’
i) The atom E contains 7 protons because its atomic number is 7.
ii) The atom with electronic configuration 2, 7 is B because its atomic number is 9.

b) Write down the formula of the compound formed between C and D.
Answer:
The electronic configuration of C is 2, 1 since its atomic number is 3. The electronic configuration of D is 2, 6 since its atomic number is 8.
So the valency of C is 1. –
The valency of D is 2 (8 – 6 = 2).
∴ The compound formed between C and D is C₂ D.

c) Predict which are (i) metals, (ii) non-metals?
Answer:
Metals are A, C.
Non-metals are B, D, E.

Question 20.
Can you suggest an experiment to prove that ionic compounds possess strong bonds when compared to that of covalent bonds? Explain the procedure.
Answer:

• Take a small amount of sodium chloride (NaCl) on a metal spatula (having an insulated handle).
• Heat it directly over the flame of a burner.
• We will see that sodium chloride (NaCl) does not melt easily.
• Sodium chloride melts (and becomes a liquid) only on strong heating.
• This shows that sodium chloride which is an ionic compound possesses strong bonds.
• So it has a high melting point.
• Whereas covalent compound like naphthalene and carbon tetra chloride has low boiling points such as 80° C and 77° C respectively.
• The force of attraction between the molecules of a covalent compound is very weak.
• Only a small amount of heat energy is required to break these weak molecular forces, due to which covalent compound has low melting points and low boiling points.

Question 21.
The electronic configurations of following elements are given below.

From these values complete the table.

Compound	Type of bonding
Lithium chloride
Lithium hydride
Hydrogen chloride
Carbon tetrachloride

Answer:
a) Lithium is metal and chlorine is non-metal. So the bond is ionic in nature.
b) Lithium is metal and hydrogen is non-metal. So the bond is ionic in nature.
c) Hydrogen and chlorine are two dissimilar non-metals. So the bond formed is polar covalent bond.
d) Carbon and chlorine are dissimilar non-metals but electronegativity difference is less. So they form non-polar covalent bond.

Compound	Type of bonding
Lithium chloride	Ionic bond
Lithium hydride	Ionic bond
Hydrogen chloride	Polar covalent bond
Carbon tetrachloride	Non-polar covalent bond

Question 22.
Draw the structure of O₂ by using valence bond theory.
Answer:
O₂ Formation :

Question 23.
Draw the structure of N₂.
Answer:

Question 24.
Draw the structure of molecules by Lewis method.
Answer:
Formation of F₂:

Formation of O₂ :

Formation of N₂ :

Formation of CH₄ :

Formation of NH₃ :

Formation of H₂O :

Formation of BeCl₂ :

Formation of BF₃ :

Question 24.
Explain the formation of the following molecules using Lewis theory.
a) N₂
b) O₂
(OR)
Write the formation of double bond and triple bond according to Lewis theory.
Answer:
a) Formation of N₂ molecule by Lewis theory :

1. The electronic configuration of ’N’ atom is 2, 5 and to have octet in the valence shell it requires three more electrons.
2. When two nitrogen atoms approach each other, each atom contributes 3 electrons for bonding.
3. There are six electrons shared between two nitrogen atoms in the form of three pairs.
4. Therefore, there is a triple bond between two nitrogen atoms in N₂ molecule.

b) Formation of O₂ molecule by Lewis theory :

1. The electronic configuration of ₈O is 2, 6.
2. Oxygen atom has six electrons in its valence shell.
3. It requires two more electrons to get octet in its valence shell.
4. Therefore oxygen atoms come close and each oxygen atom contributes two electrons for bonding.
5. Thus, there exist two covalent bonds between two oxygen atoms in O₂ molecule as there are two pairs of electrons shared between them.
6. Two pairs of electrons are distributed between two oxygen atoms.
7. So, we can say that a double bond is formed between two oxygen atoms in O₂ molecule.
8. By viewing the following diagram, both the oxygen atoms have octet in the valence shell.

AP State Board Syllabus AP SSC 10th Class English Textbook Solutions Chapter 4B Maya Bazaar Textbook Questions and Answers.

AP State Syllabus SSC 10th Class English Solutions Chapter 4B Maya Bazaar

10th Class English Chapter 4B Maya Bazaar Textbook Questions and Answers

Comprehension

Answer the following questions.

Question 1.
You have read the review of the film ‘Maya Bazaar’. List the things that the review focuses on.
Answer:
The things that the review focuses on are :
a) its director and producers
b) the actors acted in the film
c) other artists who made it a great success
d) its contribution to Telugu culture, language, and customs
e) its dialogues and songs
f) its contribution to the Telugu dictionary
g) its theme
h) the director’s greatness in making it the “Greatest Indian Film”
i) how it is helpful to an NRI.

Question 2.
Do you think this review of the film is positive or negative? Substantiate your view.
Answer:
I think this review of the film is a positive one. In the beginning lines itself, it is told that the film ’Maya Bazaar’ has been voted as the Greatest Indian Film’. The expressions such as unleashing’, ‘landmark movie’, sterling performances’, ‘tribute to Telugu culture, language, and customs’, immortalized dialogues’, ‘reverberation of songs’, ‘repertoire added to Telugu dictionary’, audience laughed heartily’, ‘womenfolk recalling their tribulations’, ‘feast for the eyes and soul’, introducing Telugu culture to an NRI’, etc., show us that this review is positive. Each and every part of this review clearly tells us it is positive.

Question 3.
What made ‘Maya Bazaar’ a landmark film?
Answer:
The film ‘Maya Bazaar’ is considered one of the enduring classics of Indian cinema and was christened as a landmark achievement in Indian film’s cinematography, art direction, and VFX with the available technology during that time. The great director K.V. Reddy, producers Nagireddy and Chakrapani, sterling performances of the star-ensemble, all time great artists, the dialogues of Pingali Nagendra Rao, the songs and its contribution to Telugu culture, language, and customs made ‘Maya Bazaar’ a landmark film. With all these great qualities, Maya Bazaar’ hit the screen in 1957 to become a landmark movie.

Question 4.
Why is ‘Maya Bazaar’ watched repeatedly?
Answer:
The viewers identify every character of the film with someone they knew in their immediate vicinity. Hence, ‘Maya Bazar’ is watched repeatedly. The audience still do the same now.

Question 5.
What is the central theme of ‘Maya Bazaar’? How have the Telugu speakers settled abroad looked upon ‘Maya Bazaar’?
Answer:
The central theme of ‘Maya Bazaar’ is to make the audience aware of their culture, language and customs by showing the striking similarities. The story itself is woven around the love of Sasirekha-Abhimanyu. With Krishna and Balarama having difference of opinion over it, their wives too take sides. Telugu speakers settled abroad have looked upon ‘Maya Bazaar’ as a masterpiece of their language. They try to introduce the learners (or the children) to ‘Maya Bazaar’ to learn Telugu culture, language and customs.

Question 6.
The purpose of the review is …
a) …to give an account of the stars of the film.
b) …to establish the element of Telugu culture in the film.
c) …to help NRIs understand Telugu culture.
d) …to praise the producer of the film.
e) …to advertise the film.
(Put a tick (✓) on the options that indicate the purpose of the review.)
Answer:
a) …….to give an account of the stars of the film. (✓)
b) …….to establish the element of Telugu culture in the film. (✓)
c) …….to help NRIs understand Telugu culture. (✓)

Vocabulary

I. In the review of ‘Maya Bazaar’ the expression, ‘language and custom’, has two words linked with the conjunction ‘and’. We also use expressions like ‘cup and saucer’, ‘near and dear’ etc.

These pairs of words joined by a conjunction ‘and’ are called ‘binomials’. In these expressions, the word order never changes. For example, we say ‘near and dear’ and not ‘dear and near’.

Here are some sentences with ‘binomials’ underlined.

1. Sports is a part and parcel of education.
2. The new theatre is rough and ready.
3. Music is not only Rahman’s bread and butter but also his passion.
4. The film industry is expanding in leaps and bounds.
5. The customer can pick and choose anything he likes. It is the customer to decide.
6. ‘Give and take’ policy is always helpful.
7. Ray gave his heart and soul to reading books.
8. He stood by me through thick and thin.
9. The main goods were shifted first. Then the odds and ends were taken later.

Match the following binomials with their meanings and use them in your own sentences.

Column – A	Column – B
1. part and parcel	a) unimportant things
2. rough and ready	b) livelihood
3. leaps and bounds	c) in difficult times
4. bread and butter	d) dedicated
5. pick and choose	e) help one another
6. give and take	f) a large choice
7. heart and soul	g) big leaps
8. thick and thin	h) an integral part
9. odds and ends	i) almost finished

Answer:

Column – A	Column – B
1. part and parcel	h) an integral part
2. rough and ready	i) almost finished
3. leaps and bounds	c) in difficult times
4. bread and butter	b) livelihood
5. pick and choose	f) a large choice
6. give and take	e) help one another
7. heart and soul	d) dedicated
8. thick and thin	g) big leaps
9. odds and ends	a) unimportant things

Own Sentences:

• Teaching students how to behave and how to deal with others is part and parcel of education.
• The corporate hospital is rough and ready.
• Mr. Rao’s career is progressing in leaps and bounds.
• Teaching is my bread and butter. I don’t want to go away from it.
• The parent can pick and choose any school he/she likes.
• They always follow give and take policy.
• Narayana Murthy gave his heart and soul to studying engineering course at IIT.
• Both Ramu and Somu have been together through thick and thin: they don’t want to desert each other.
• He is carrying a suitcase which is full of odds and ends.

Some other binomials with their meanings :

1) neat and tidy : clean
2) sick and tired : annoyed/frustrated
3) wine and dine : entertain someone with a good quality meal
4) up and down : moving between the same two points repeatedly
5) skin and bone : to be very thin
6) loud and clear : very clear and very easy to understand
7) by and large : on the whole
8) back and forth : moving first in one direction then in another
9) alive and kicking : well and healthy
10) bits and pieces : small things of different types
11) by and by : after a while
12) body and soul : believe something completely
13) far and wide : across a large area
14) fair and square : honest
15) hard and fast : inevitable, fixed
16) free and easy : unconventional
17) hustle and bustle : confusion; busy activity
18) day and night : all the time
19) pros and cons : advantages and disadvantages
20) spick and span : neat, trim, impeccable
21) tooth and nail : with all one’s resources or energy
22) wear and tear : loss, damage
23) far and wide : a large number of places, across a large graphical area
24) high and mighty : behaving as though one is more important than others
25) short and sweet : when something is very quick and to the point; of minimum length and no longer than it needs.

II. Read the following conversation that took place in the classroom of a film institute.
Professor Paul : As a part of our ‘Diploma in film-making’ we have discussed certain aspects of film technology, right? I hope by now all of you are familiar with those concepts. If you have some more queries, we can discuss them. Now the class is open for questions.

Rishi : I think what is included and excluded in an individual shot is called ‘Frame’. Am I right?

P P : Yes, you are. What is brought to the viewer is called a frame.

Prudhvi : Which shot is used to show a happy reunion, Sir?

P P : It is called Arc’. The camera moves in an arc around the subject (artistes). Any more questions…?

P P : Here is a list of camera shots that helps you to have a clear idea about different shots in film-making.

Shots and Angles :

1. Establishing shot : It is usually from a greater distance to establish setting, (familiarises the scene or setting)
2. Close-up : It refers to the image occupying at least 80 percent of the frame.
3. Two shot : Two people in the frame are showed equally.
4. Dutch angle : It is neither vertical nor horizontal. It‘s oblique.
5. Bird’s eye shot : It is shot directly and vertically down at the subject.
6. High angle : Camera is above the subject. It creates an impression that the subject is weak or powerless.
7. Low angle : The camera is placed below the subject. The subject appears larger than normal.
8. Rack focus : Here the focus is shifted from one subject to the other where the subject that is not in focus is blurred.
9. Footage : It is the total exposed film.
10. Montage : Different images are assembled to build an impression.

Here are some visuals. Identify their features and label them. You may choose the labels given above.

Answer:

1. Low angle shot
2. Dutch angle shot
3. Establishing shot
4. Bird’s eye view shot

Grammar

I. Modals

The following is a conversation between two artists.

Shankar : Shanti, yesterday the ABC Movies called me up seeking my opinion.
Shanti : What is that?
Shankar : They’ve a role for a heroine in their film that is going to sets very soon. They wanted to offer that role to you. Will you accept it?
Shanti : Did they really mean it? I’ve two doubts about it. First, can I do it? The second, should I do it?
Shankar : Yes, you can certainly do it. Regarding your second doubt -I think it’s not that you should do it, you must do it.
Shanti : Why?
Shankar : We may like some roles and others we may not. But we accept them for the sake of our career.
Shanti :Mmm… .
Shankar : Certain roles we should accept for the sake of satisfaction. That’s why, I said you must do it.
Shanti : Then 1 take your word as final to me. I shall do it. Tell the ABC Movies that I have accepted the offer.

The words given in bold letters in the above conversation perform different functions.They are called modal auxiliaries or modals. Their functions are given below.

Modal	Function
shall	obligation, offer, order, suggestion,
should	obligation, advice
can	ability, possibility, permission
could	request, suggestion, permission, possibility, ability
will	certainty, intention, futurity, purpose
would	offer, preference, past habit, future of the past
may	permission, possibility, wish
might	possibility
must	compulsion, inference

Some important points about modals :

1. Modals take any subject before them. There is no Subject-Verb agreement problem.

2. No two modals co-occur in a sentence.
e.g.: The film will be released next week.

3. The main verb will always be the V₁ form after the modals.
e.g.: He cannot move the big stone.

4. Past form of some modals stands for ‘politeness’.
e.g.: i) Would you like to have some tea?
ii) Could you lend me your book?

Read the following sentences and identify the functions performed by the modals.

a) He can sing for eight hours at a stretch.
b) You could take an umbrella, it’s raining outside.
c) We should complete the shooting by tomorrow evening.
d) Shall I wait till you come?
e) The time was up but the makeup person would not turn up.
f) The car hasn’t arrived yet. It must’ve got a flat tyre.
g) There may be heavy rains tomorrow. So, why couldn’t we go for indoor shooting?
h) Will you join us for tea?
i) The agreement between the two parties shall be in force for two years.
j) It might be an idea to postpone the release of the film.
Answer:
a) ability
b) suggestion
c) obligation
d) offer
e) future of the past
f) inference
g) suggestion
h) request/invitation
i) obligation
j) possibility

Writing

I. Now read the review once again. Write a review of a film/a TV Programme/an episode of a TV serial you have seen.

I have recently seen the Telugu film ‘Swathimuthyam’ which is directed by K.Viswanath. ‘Swathimuthyam’ is a wonderfully crafted masterpiece. K.Viswanath has shown all his artistic brilliance in weaving the story. Each frame of the film has roused curiosity to know what would be the next scene. Kamal Hassan’s action is simply superb. He gets into the skin of the character very much and brings the liveliness not only to his character but also to the film itself. His heroine, Radhika too acted brilliantly. The music of Ilayaraja and the prowess of K.Viswanath are completely overshadowed by the excellence of Kamal.

The music composed by Ilayaraja is unforgettable. He tries to give his best in this film. The songs such as “Suwi Suwi”, “Vatapatra Sai ki…” reverberate in each and every viewer’s mind. The theme of the movie is based on how a naive, child-like person, emerges unscathed in this selfish world, managing to uplift the lives of some troubled souls (Radhika, who is a widow and her son). There is an underlying reflection of goodwill still left in the society – it is skilfully depicted through the characters of the dhobi, Nirmalamma, Radhika and so on. One last thing about the music which is composed to “Rama Kanavemira” takes the potential of Ilayaraja to mix various traditions of music. This movie is produced by Edida Nageswara Rao.

II. Read the following letter.

Kondamudusu Palem, Kandukuru. Nov. 27, 2013. To The Editor, The Hindu, Vijayawada. Sir/Madam, I am a regular reader of your esteemed newspaper. I am very much interested in reading the news in ‘Friday Review’ in your newspaper every week. I am writing to bring to your notice certain feelings I had when I watched the colour version of the movie ‘Maya Bazaar’. Recently ‘Mayabazaar’ was released again but this time it was a different ‘Maya Bazaar’, Maya Bazaar-in colour! Thanks to the Goldstone Technologies, I was carried away to a different world of colourful visuals. Of course the audio part remained the same. But here and there the effects of digital track sound have made some difference. Such an experiment began in Hindi with ‘Mughal-E-Azam’ and received applause from the audience. I hope many such experiments are welcome to the film lovers. Thanking you, Yours faithfully, N. Sarathchandra, Z.P.HIGH SCHOOL, KANDUKURU, PRAKASAM (Dist.)

III. Write a reply to the letter that appeared in the Hindu in response to the review of ‘Maya Bazaar’. In your letter write whether the opinion expressed in above letter is a sufficient appreciation of ‘Maya Bazaar’.

Uppalapadu, Guntur. Dec. 2, 2013. To The Editor, The Hindu, Vijayawada. Sir/Madam, This is a rejoinder to the letter in the Hindu in response to the review of ‘Maya Bazaar’. First of all I would like to thank Mr. N. Sarathchandra for his appreciation of the movie ‘Maya Bazaar’ that was released in colour. I totally agree with his view when he wrote to say that experiments are welcome to the film lovers. In this fast developing world, experimentation is necessary in each and every field to get better results. But I am not happy as the film’s naturality has been destroyed by touching some colours to it. The characteristics such as picturization, dialogues, direction, audio-effects, etc., are all the important components of a film’s success. If we try to experiment with any one of them, it will lose its originality. Hence, I opine that we should welcome the technological development but at the same time we shouldn’t try to make a difference which takes away over Telugu culture, language and customs. I hope all will agree with me. Thanking you, Yours faithfully, XXXX Z.P. High School, Uppalapadu, Pedakakani Mandal, Guntur District.

IV. Let’s read the following skit.

Suhas : Amma, Amma……. .
Mother : What, Suhas?
Suhas : My friends are planning to watch a movie. I too want to join them.
Mother : Movies, movies, movies. No, I can’t let you go to movies too often.
Suhas : My friends are making fun of me. They call me ‘a bird in a cage’. (Goes out disgusted taking his school bag.)
Mother : Money, money, money. He always needs money. This is not a kid. It’s an apparition.
Father : (coming) Why are you shouting, Sarala? What did you say?
Mother : Didn’t you hear? An apparition.
Father : Who is the apparition?
Mother : Who else could I mean? It’s your son.
Father : Not your son? He needs money?
Mother : Yes, he needs money. He watches movies. His life is movies. He doesn’t care to study. All the children in the street are doing very well. But this devil (Starts banging her head against the wall. No wounds.No blood)
Father : No, Sarala. No, don’t do it.(Mother falls down and father brings a glass of water.)

Father : Sarala…, Sarala…., have some water. (A pause)
Father : Sarala, we have to understand certain things.
Mother : (Having drunk water) As if you have understood You don’t worry about anything. He doesn’t study at all. He is after movies.
Father : Wait! I’ll find out.
Father : Suhas, you didn’t take your bicycle today. You didn’t take your lunch box. You look tired ……… .
Suhas : Amma doesn’t understand me. She always makes fuss of me. She always wants me to do one thing; study…. study….. study. I want to go to a film. That too once a week. All my friends are going. I too want to go. Is it wrong on my part?
Father : Why don’t you try to understand your mother’s concern? She is worried a lot about your future.
Suhas : But I am all right in my studies. I think, watching a film once a week, that too, watching a good film is not a sin.
Father : Which film did you want to watch?
Suhas : “Life of Pi”. It’s an interesting movie with high technical values.
Father : All right, my boy.

Father : Sarala, the boy didn’t want to do anything bad.
Mother : But I have my own fears.
Father : I do understand. Your fears may have reasons. But what Suhas wanted was just 100 rupees. And that too for watching an interesting film.
Mother : Maybe I was wrong. We’ll ask Suhas to go to the film. He is such a nice boy. He never troubles us.
Father : Come on. Let’s have supper. Come, Suhas. Now let’s ponder over these points :

Keeping in mind the features of a skit given below, prepare a skit on any subject you like most.
A skit has all the characteristics of a play. Usually, it does have a message to convey. It has a plot (a beginning, a crisis and a resolution). However, there are certain differences with respect to time, number of characters, setting, etc. The skit is more of an informal nature. The number of characters is less. The setting does not have much importance.
Answer:
(Rakesh, Prabhas and Suhas are the students of tenth class, studying at ZPH School, Uppalapadu.)
Rakesh : (with sorrowful face) Oh, no! What can I do now?
Prabhas : What happened, Rakesh? Why are you so sad?
Rakesh : (searching his pockets) I have lost my money. How can I get home ?
Suhas : Hi friends ! Don’t you come home? It’s already late.
Prabhas : Rakesh has lost his money.
Suhas : What? When did it happen?
Rakesh : I have just known about it.
Prabhas : Calm down. It will be OK How much money have you lost?
Rakesh : I have lost my fifty-rupee note !
Prabhas : Don’t worry; I found a fifty-rupee note in the conference hall in the lunch break. I wanted to hand over the money to our H.M. I think that it is yours only! Here it is your note.
Rakesh : Yes, it is the note I have lost. I even wrote my name on it. See my name!
Suhas : Yes, your name is on the note. Certainly it is yours only.
Rakesh : Thank you very much, Prabhas. You are really a good boy. You could have bought something about the money you got but you didn’t do that.
Prabhas : But, I wouldn’t have felt good about it. I think that it is not a good thing to do. Wanting others’ money is nothing but stealing. 1 know very well that it is someone else’s and he/she would be upset. I don’t want to make sorrowful anyone. I too had the same bitter experience. I wouldn’t forget how I was disappointed when I lost my money.
Rakesh : I won’t forget your honesty. I am very glad and proud to have a friend like you.
Prabhas : OK, friends. Come, our bus is ready to depart.

Maya Bazaar Summary in English

“Maya Bazaar” is a review published in “The Hindu” on 21st April, 2006 on the^ occasion of its Golden Jubilee. K.V.Reddy’s Maya Bazaar’ has been voted as the ‘Greatest Indian Film” in an online poll conducted by a television news channel. Its director K.V.Reddy and producers Nagireddy and Chakrapani made it both in Telugu and Tamil.

The film became a super hit because of the performances of the star-actors S.V.Ranga Rao, Savitri, NTR, ANR and Gummadi and the efforts of K.V. Reddy. The great persons like Marcus Batley, Ghantasala, M.L. Vasantha Kumari, Leela, Suseela, Madhavapeddi, Gokhale, Pasumarthy and Pitambaram played their roles together in making up of such a landmark movie in Telugu film industry. It shows us Telugu culture, language and customs in every frame. When people saw it, they identified every character of the film with someone they knew in their area. The dialogues written by Pingali Nagendra Rao became immortalized. The songs such as ‘Aha naa pelli anta,” “Vivaha bhojanambu”, etc., have a strong effect on people for a long time. This film made people familiarize with the words such as “Talpam”, “Gilpam”, “Asamadiyulu”, “Tasamadiyulu”, etc. We can’t forget the expressions such as “veyandira veediko veeratadu”, “hai hai sodara” and “hai hai naayaka”.

Its story deals with the love of Sasirekha-Abhimanyu. The director uses a magic box to introduce the theme. It creates a wonderful effect on viewers. The scenes shown on its screen made the audience laugh heartily. The dialogues of Balarama made the women viewers recall their own problems in the hands of such brothers. The director greatly displays all the follies of human beings through all the characters except Ghatothkacha and Krishna. This film is a feast to us with its simile, imagery, adage, sarcasm and wit. Though it is the story of Pandavas and Kauravas, Pandavas are never seen throughout the movie. Even an NRI will know about Telugu culture if he/she happens to watch this Imovie.

Maya Bazaar Glossary

enduring (adj) : continuing for a very long time

christened (v) : gave something or someone a name

cinematography (n) : the skill or study of making films

VFX : visual Effects shortened to Visual FX and then to VFX

unleashing (v) : suddenly letting a strong force, feeling, etc. have its full effect

sterling (adj) : very good

ensemble (n) : a small group of musicians, actors or dancers who perform together regularly

facet (n) : one of several parts of someone’s character, a situation, etc.

the vicinity (n) : the area around a particular place

reverberate (v) : have a strong effect on people for a long time

repertoire (n) : all that a performer can do

advent (n) : arrival

array (n) : an impressive collection of things

indignation (n) : a feeling of anger

sarcasm (n) : a way of using words that are the opposite of what one means to make fun of somebody

laps up (phr. v.) : to enjoy something without worrying about whether it is good, true, etc.

curtly (adv) : abruptly

confronted (v) : dealt with something very difficult or unpleasant in a brave and determined way

tribulation (n) : formal serious trouble or a serious problem

folly (n) : a very stupid thing to do, especially one that is likely to have serious results

simile (n) : an expression that describes something by comparing it with something else

imagery (n) : the use of words or pictures to describe ideas or actions in poems, books, films, etc.

adage (n) : a well-known phrase that says something wise about human experience

wit (n) : the ability to say things that are clever and amusing

AP State Board Syllabus AP SSC 10th Class Telugu Solutions 10th Class Telugu Grammar Samasalu సమాసాలు Notes, Questions and Answers.

AP State Syllabus SSC 10th Class Telugu Grammar Samasalu సమాసాలు

సమాసాలు

సమాసం :
వేరు వేరు అర్థాలు కల రెండు పదాలు కలసి, ఏకపదంగా ఏర్పడితే దాన్ని ‘సమాసం’ అంటారు.

గమనిక :
అర్థవంతమైన రెండు పదాలు కలిసి, కొత్త పదం ఏర్పడడాన్ని సమాసం అంటారు. సమాసంలో మొదటి పదాన్ని పూర్వ పదం అంటారు. రెండవ పదాన్ని ఉత్తర పదం అంటారు.
ఉదా :
‘రామ బాణము’ అనే సమాసంలో, ‘రామ’ అనేది పూర్వ పదం. ‘బాణము’ అనేది ఉత్తర పదం.

1. ద్వంద్వ సమాసం :
రెండు కాని, అంతకంటే ఎక్కువ కాని నామవాచకాల మధ్య ఏర్పడే సమాసాన్ని, “ద్వంద్వ సమాసం” అంటారు.

ఈ కింది వాక్యాల్లోని ద్వంద్వ సమాస పదాలను గుర్తించి రాయండి.

1) ఈ అన్నదమ్ములు ఎంతో మంచివాళ్ళు,
జవాబు:
అన్నదమ్ములు

2) నేను మార్కెట్ కు వెళ్ళి కూరగాయలు తెచ్చాను.
జవాబు:
కూరగాయలు

3) ప్రమాదంలో నా కాలుసేతులకు గాయాలయ్యాయి.
జవాబు:
కాలుసేతులు

అభ్యాసం:

I. ఈ కింది ద్వంద్వ సమాసాలకు విగ్రహవాక్యాలు రాయండి.

సమాసపదం	విగ్రహవాక్యం
1) ఎండవానలు	ఎండయు, వానయు	ద్వంద్వ సమాసాలు
2) తల్లిదండ్రులు	తల్లియు, తండ్రియు
3) గంగాయమునలు	గంగయు, యమునయు

II. ఈ కింది విగ్రహవాక్యాలను సమాసపదాలుగా మార్చండి.

సమాసపదం	విగ్రహవాక్యం
1) కుజనుడూ, సజ్జనుడూ	కుజనసజ్జనులు
2) మంచి, చెడూ	మంచిచెడులు
3) కష్టమూ, సుఖమూ	కష్టసుఖములు

2. ద్విగు సమాసం :
సమాసంలో మొదటి (పూర్వ) పదంలో సంఖ్య గల సమాసాన్ని ద్విగు సమాసం అంటారు.

అభ్యాసం :
కింది సమాస పదాలను ఉదాహరణలో చూపిన విధంగా వివరించండి.
ఉదా : నవరసాలు – నవ (9) సంఖ్య గల రసాలు
1) రెండు జడలు – రెండు (2) సంఖ్య గల జడలు
2) దశావతారాలు – దశ (10) సంఖ్య గల అవతారాలు
3) ఏడు రోజులు – ఏడు (7) సంఖ్య గల రోజులు
4) నాలుగువేదాలు – నాలుగు (4) సంఖ్య గల వేదాలు

గమనిక :
పైన పేర్కొన్న సమాసాలలో సంఖ్యావాచకం పూర్వ పదంగా ఉండటాన్ని గమనించండి. ఇలా మొదటి పదంలో సంఖ్య గల సమాసాలు ద్విగు సమాసాలు.

3. తత్పురుష సమాసం :
విభక్తి ప్రత్యయాలు విగ్రహవాక్యంలో ఉపయోగించే సమాసాలు “తత్పురుష సమాసాలు.” అభ్యాసము : కింది పదాలను చదివి, విగ్రహవాక్యాలు రాయండి.

సమాసపదం	విగ్రహవాక్యం
1) రాజభటుడు	రాజు యొక్క భటుడు
2) తిండి గింజలు	తిండి కొఱకు గింజలు
3) పాపభీతి	పాపము వల్ల భీతి

గమనిక :
‘రాజభటుడు’ అనే సమాసంలో ‘రాజు’ పూర్వ పదం. ‘భటుడు’ అనే పదం ఉత్తర పదం. ‘రాజభటుడు’ కు విగ్రహవాక్యం రాస్తే ‘రాజు యొక్క భటుడు’ అవుతుంది. దీంట్లో యొక్క అనేది షష్ఠీ విభక్తి ప్రత్యయం. భటుడు రాజుకు చెందినవాడు అని చెప్పడానికి షష్ఠీ విభక్తి ప్రత్యయాన్ని వాడారు. ఈ విధంగా ప్రత్యయాలు విగ్రహవాక్యంలో ఉపయోగించే సమాసాలు తత్పురుష సమాసాలు.

గమనిక :
పూర్వ పదం చివర ఉండే విభక్తిని బట్టి తత్పురుష సమాసాలు వస్తాయి.

తత్పురుష సమాసం రకాలు	విభక్తులు	ఉదాహరణ, విగ్రహవాక్యం
1) ప్రథమా తత్పురుష సమాసం	డు, ము, వు, లు	మధ్యాహ్నము – అహ్నము యొక్క మధ్య
2) ద్వితీయా తత్పురుష సమాసం	ని, ను, ల, కూర్చి, గురించి	జలధరం – జలమును ధరించినది
3) తృతీయా తత్పురుష సమాసం	చేత, చే, తోడ, తో	బుద్ధిహీనుడు – బుద్ధిచేత హీనుడు
4) చతుర్డీ తత్పురుష సమాసం	కొఱకు, కై	వంట కట్టెలు – వంట కోఱకు కట్టెలు
5) పంచమీ తత్పురుష సమాసం	వలన (వల్ల), కంటె, పట్టి	దొంగ భయం – దొంగ వల్ల భయం
6) షష్ఠీ తత్పురుష సమాసం	కి, కు, యొక్క, లో, లోపల	రామబాణం – రాముని యొక్క బాణం
7) సప్తమీ తత్పురుష సమాసం	అందు, న	దేశభక్తి – దేశము నందు భక్తి
8) నఞ్ తత్పురుష సమాసం	నఞ్ అంటే వ్యతిరేకము	అసత్యం – సత్యం కానిది

అభ్యాసం :
కింది సమాసాలు చదివి, విగ్రహవాక్యాలు రాయండి. అవి ఏ తత్పురుష సమాసాలో తెలపండి.

సమాస పదం	విగ్రహవాక్యం	సమాసం పేరు
1) రాజ పూజితుడు	రాజుచే పూజితుడు	తృతీయా తత్పురుషం
2) ధనాశ	ధనము నందు ఆశ	సప్తమీ తత్పురుషం
3) పురజనులు	పురమందు జనులు	సప్తమీ తత్పురుషం
4) జటాధారి	జడలను ధరించినవాడు	ద్వితీయా తత్పురుషం
5) భుజబలం	భుజముల యొక్క బలం	షష్ఠీ తత్పురుషం
6) అగ్నిభయం	అగ్ని వల్ల భయం	పంచమీ తత్పురుషం
7) అన్యాయం	న్యాయం కానిది	తత్పురుష సమాసం

తత్పురుష సమాసాలు :
విభక్తులు ఆధారంగా ఏర్పడే తత్పురుష సమాసాలను గూర్చి తెలిసికొన్నారు. కింది వాటిని కూడా పరిశీలించండి.
1) మధ్యాహ్నము – అహ్నము యొక్క మధ్యము (మధ్య భాగం)
2) పూర్వకాలము – కాలము యొక్క పూర్వము (పూర్వ భాగం)

గమనిక :
పై వాటిలో మొదటి పదాలైన మధ్య, పూర్వ అనే పదాలకు ‘ము’ అనే ప్రథమా విభక్తి ప్రత్యయం చేరడం వల్ల
‘మధ్యము’, ‘పూర్వము’గా మారతాయి. ఇలా పూర్వపదానికి ప్రథమా విభక్తి ప్రత్యయం రావడాన్ని ‘ప్రథమా తత్పురుష సమాసం’ అంటాము. కింది వాటిని పరిశీలించండి.
1) నఞ్ + సత్యం = అసత్యం – సత్యం కానిది
2) నఞ్ + భయం = అభయం – భయం కానిది
3) నఞ్ + అంతము = అనంతము – అంతము కానిది
4) నఞ్ + ఉచితం = అనుచితం – ఉచితము కానిది

గమనిక :
సంస్కృతంలో ‘నఞ్’ అనే అవ్యయం, వ్యతిరేకార్థక బోధకం. దీనికి బదులు తెలుగులో అ, అన్, అనే ప్రత్యయాలు వాడతారు. పై ఉదాహరణల్లో వాడిన ‘నం’ అనే అవ్యయాన్ని బట్టి, దీన్ని “నః తత్పురుష సమాసం” అంటారు.

అభ్యాసము :
కింది పదాలకు విగ్రహవాక్యాలు రాసి, సమాస నామము పేర్కొనండి.

సమాస పదం	విగ్రహవాక్యం	సమాసం పేరు
1) అర్థరాత్రి	రాత్రి యొక్క అర్ధము	ప్రథమా తత్పురుషం
2) అనూహ్యము	ఊహ్యము కానిది	నఞ్ తత్పురుషం
3) అక్రమం	క్రమము కానిది	నఞ్ తత్పురుషం
4) అవినయం	వీనయం కానిది	నఞ్ తత్పురుషం

4. కర్మధారయ సమాసం :
‘నల్లకలువ’ అనే సమాస పదంలో ‘నల్ల’, ‘కలువ’ అనే రెండు పదాలున్నాయి. మొదటి పదం ‘నల్ల’ అనేది, విశేషణం. రెండో పదం ‘కలువ’ అనేది నామవాచకం. ఇలా విశేషణానికీ, సామవాచకానికీ (విశేష్యానికీ) సమాసం జరిగితే, దాన్ని కర్మధారయ సమాసం అంటారు.

4. అ) విశేషణ పూర్వపద కర్మధారయ సమాసం :
విశేషణం పూర్వపదంగా (మొదటి పదంగా) ఉంటే, ఆ సమాసాన్ని “విశేషణ పూర్వపద కర్మధారయ సమాసం’ అంటారు.
ఉదా :
1) తెల్ల గుర్రం – తెల్లదైన గుర్రం.
తెలుపు (విశేషణం) (పూర్వపదం) – (మొదటి పదం) గుర్రం – (నామవాచకం) (ఉత్తరపదం)- (రెండవ పదం)

ఆ) విశేషణ ఉత్తరపద కర్మధారయ సమాసం :
మామిడి గున్న’ అనే సమాసంలో మామిడి, గున్న అనే రెండు పదాలున్నాయి. మొదటి పదం (పూర్వపదం) ‘మామిడి’ సోమవాచకం, రెండో పదం (ఉత్తరపదం) గున్న అనేది విశేషణం. ఇందులో విశేషణమైన ‘గున్న’ అనే పదం ఉత్తరపదంగా – అంటే రెండో పదంగా ఉండడం వల్ల, దీన్ని ‘విశేషణ ఉత్తరపద కర్మధారయ సమాసం’ అంటారు. అభ్యాసము : కింది పదాలను చదివి, విగ్రహవాక్యాలు రాసి, ఏ సమాసమో రాయండి.

సమాస పదం	విగ్రహవాక్యం	సమాసం పేరు
1) పుణ్యభూమి	పుణ్యమైన భూమి	విశేషణ పూర్వపద కర్మధారయ సమాసం
2) మంచిరాజు	మంచి వాడైన రాజు	విశేషణ పూర్వపద కర్మధారయ సమాసం
3) కొతపుస్తకం	కొత్తదైన పుస్తకం	విశేషణ పూర్వపద కర్మధారయ సమాసం
4) పురుషోత్తముడు	ఉత్తముడైన పురుషుడు	విశేషణ పూర్వపద కర్మధారయ సమాసం

ఇ) సంభావనా పూర్వపద కర్మధారయ సమాసం :
‘తమ్మివిరులు’ అనే సమాసంలో, మొదటి పదమైన ‘తమ్మి’, ఏ రకం విరులో తెలియజేస్తుంది. ఇలా పూర్వపదం, నదులు, వృక్షాలు, ప్రాంతాలు మొదలైన వాటి పేర్లను సూచిస్తే దాన్ని సంభావనా పూర్వపద కర్మధారయ సమాసం అంటారు.
ఉదా : మట్టి చెట్టు – మట్టి అనే పేరు గల చెట్టు – సంభావనా పూర్వపద కర్మధారయ సమాసం
గంగానది – గంగ యనే పేరు గల నది – సంభావనా పూర్వపద కర్మధారయ సమాసం
భారతదేశం – ‘భారతం’ అనే పేరు గల దేశం – సంభావనా పూర్వపద కర్మధారయ సమాసం

ఈ) ఉపమాన పూర్వపద కర్మధారయ సమాసం :
‘కలువ కనులు’ అనే సమాసంలో కలువ, కనులు అనే రెండు పదాలున్నాయి. దీనికి ‘కలువల వంటి కన్నులు’ అని అర్థం. అంటే కన్నులను కలువలతో పోల్చడం జరిగింది. సమాసంలోని మొదటి పదం (పూర్వపదం) ఇక్కడ ‘ఉపమానం’ కాబట్టి దీన్ని ఉపమాన పూర్వపద కర్మధారయ సమాసం అంటారు.

ఉ) ఉపమాన ఉత్తరపద కర్మధారయ సమాసం :
‘పదాబ్దము’ అనే సమాసంలో పద (పాదం) మరియు, అబ్జము (పద్మం) అనే రెండు పదాలున్నాయి. వీటి అర్థం పద్మము వంటి పాదము అని. ఇక్కడ పాదాన్ని పద్మం (తామరపూవు)తో పోల్చడం జరిగింది. కాబట్టి పాదం ఉపమేయం. పద్మం ఉపమానం. ఉపమానమైన అబ్జము అనే పదం, ఉత్తరపదంగా (రెండవపదం) ఉండడం వల్ల దీన్ని ఉపమాన ఉత్తరపద కర్మధారయ సమాసం అంటారు.

అభ్యాసం :
కింది సమాసాలకు విగ్రహవాక్యాలు రాసి, సమాస నామాలు పేర్కొనండి.

సమాస పదం	విగ్రహవాక్యం	సమాసం పేరు
1) తేనెమాట	తేనె వంటి మాట తేనె – ఉపమానం, మాట – ఉపమేయం	ఉపమాన పూర్వపద కర్మధారయం
2) తనూలత	లత వంటి తనువు తనువు – ఉపమేయం, లత – ఉపమానం	ఉపమాన ఉత్తరపద కర్మధారయం
3) చిగురుకేలు	చిగురు వంటి కేలు చిగురు – ఉపమానం, కేలు – ఉపమేయం	ఉపమాన పూర్వపద కర్మధారయం
4) కరకమలములు	కమలముల వంటి కరములు కరములు – ఉపమేయం కమలములు – ఉపమానం	ఉపమాన పూర్వపద కర్మధారయం

5. రూపక సమాసం :
‘విద్యాధనం’ – అనే సమాసంలో విద్య, ధనం అనే రెండు పదాలున్నాయి. పూర్వపదమైన విద్య, ధనంతో పోల్చబడింది. కాని ‘విద్య అనెడి ధనం’ అని దీని అర్థం కనుక, ఉపమాన, ఉపమేయాలకు భేదం లేనంత గొప్పగా చెప్పబడింది. ఈ విధంగా ఉపమాన, ఉపమేయాలకు భేదం లేనట్లు చెబితే అది ‘రూపక సమాసం’.

సమాస పదం	విగ్రహవాక్యం
ఉదా : 1) హృదయ సారసం	హృదయం అనెడి సారసం
2) సంసార సాగరం	సంసారం అనెడి సాగరం
3) జ్ఞాన జ్యోతి	జ్ఞానము అనెడి జ్యోతి
4) అజ్ఞాన తిమిరం	అజ్ఞానము అనెడి తిమిరం
5) సాహితీ జగత్తు	సాహిత్యమనెడి జగతు – రూపక సమాసం

6. బహుప్రీహి సమాసం :
అన్య పదార్థ ప్రాధాన్యం కలది. కింది ఉదాహరణను గమనించండి. ”
చక్రపాణి – చక్రం పాణియందు (చేతిలో) కలవాడు. ‘విష్ణువు’ అని దీని అర్థం. దీంట్లో సమాసంలోని రెండు పదాలకు అనగా “చక్రానికి” కాని “పాణికి” కాని ప్రాధాన్యం లేకుండా, ఆ రెండూ మరో అర్థం ద్వారా “విష్ణువును” సూచిస్తున్నాయి. ఇలా సమాసంలో ఉన్న పదాల అర్థానికి ప్రాధాన్యం లేకుండా, అన్యపదముల అర్థాన్ని స్ఫురింపజేసే దాన్ని బహుప్రీహి సమాసం అంటారు. అన్య పదార్థ ప్రాధాన్యం కలది. ‘బహుబ్లి హి సమాసం’. అభ్యాసం : కింది పదాలకు విగ్రహవాక్యాలు రాసి, సమాసం పేరు రాయండి.

సమాస పదం	విగ్రహవాక్యం	సమాసం పేరు
1) నీలవేణి	నల్లని వేణి కలది	బహుబీహి సమాసం
2) నీలాంబరి	నల్లని అంబరము కలది	బహుబీహి సమాసం
3) ముక్కంటి	మూడు కన్నులు గలవాడు	బహుథీహి సమాసం
4) గరుడవాహనుడు	గరుత్మంతుడు వాహనంగా కలవాడు	బహుబీహి సమాసం
5) దయాంతరంగుడు	దయతో కూడిన అంతరంగము కలవాడు	బహుప్రీహి సమాసం
6) చతుర్ముఖుడు	నాలుగు ముఖములు కలవాడు	బహుప్రీహి సమాసం

7. అవ్యయీభావ సమాసం :
అవ్యయం పూర్వపదముగా ఉన్న సమాసాలను, “అవ్యయీభావ సమాసాలు” అంటారు.

అవ్యయం :
అవ్యయాలు అనగా లింగ, వచన, విభక్తి లేని పదాలు. ఈ విధమైన భావంతో ఉన్న సమాసాలను అవ్యయీభావ సమాసాలు అంటారు. ఈ క్రింది సమాస పదాలను, వాటికి రాయబడిన విగ్రహవాక్యాలను పరిశీలించండి.

సమాస పదం	విగ్రహవాక్యం	సమాసం పేరు
అ) ప్రతిదినము	దినము దినము	అవ్యయీభావ సమాసం
ఆ) యథాశక్తి	శక్తి ఎంతో అంత (శక్తిననుసరించి)	అవ్యయీభావ సమాసం
ఇ) ఆబాలగోపాలం	బాలుడి నుండి గోపాలుడి వరకు	అవ్యయీభావ సమాసం
ఈ) మధ్యాహ్నం	అహ్నం మధ్యభాగం	అవ్యయీభావ సమాసం
ఉ) అనువర్షం	వర్షముననుసరించి	అవ్యయీభావ సమాసం

గమనిక : ‘మధ్యాహ్నము” అనే సమాస పదానికి, విగ్రహం ‘మధ్యము – అహ్నము’ అని చెప్పాలి. ఇది ‘ప్రథమా తత్పురుష సమాసం’ అవుతుంది. అవ్యయీభావ సమాసం కాదు.

పాఠ్యపుస్తకంలోని ముఖ్య సమాసాలు – విగ్రహవాక్యాలు :

సమాస పదం	విగ్రహవాక్యం	సమాసం పేరు
1)  నలుదెసలు	నాలుగైన దెసలు	ద్విగు సమాసం
2) సూర్యచంద్రులు	సూర్యుడును,చంద్రుడును	ద్వంద్వ సమాసం
3) చంపకవతి పట్టణం	‘చంపకవతి’ అనే పేరు గల పట్టణం	సంభావనా పూర్వపద కర్మధారయ సమాసం
4) మహాభాగ్యం	గొప్పదైన భాగ్యం	విశేషణ పూర్వపద కర్మధారయం సమాసం
5) సేవావృత్తి	సేవ అనెడి వృత్తి	రూపక సమాసం
6) పదాబ్దములు	పద్మముల వంటి పదములు	ఉపమాన ఉత్తరపద కర్మధారయం
7) కలువ కన్నులు	కలువల వంటి కన్నులు	ఉపమాన పూర్వపద కర్మధారయం
8) మామిడి గున్న	గున్నయైన మామిడి	విశేషణ ఉత్తరపద కర్మధారయం
9) మృదుమధురము	మృదువును, మధురమును	విశేషణ ఉభయపద కర్మధారయం
10) సత్యదూరము	సత్యమునకు దూరము	షష్ఠీ తత్పురుష సమాసం
11) అమెరికా రాయబారి	అమెరికా యొక్క రాయబారి	షష్ఠీ తత్పురుష సమాసం
12) వాదనాపటిమ	వాదన యందు పటిమ	సప్తమీ తత్పురుష సమాసం
13) అసాధ్యం	సాధ్యము కానిది	నఞ్ తత్పురుష సమాసం
14) నెలతాల్పు	నెలను తాల్చువాడు	ద్వితీయా తత్పురుష సమాసం
15) గురుదక్షిణ	గురువు కొఱకు దక్షిణ	చతుర్డీ తత్పురుష సమాసం
16) వయోవృద్ధుడు	వయస్సు చేత వృద్ధుడు	తృతీయా తత్పురుష
17) దొంగభయము	దొంగ వలన భయము	పంచమీ తత్పురుష సమాసం
18) ధూపదీపములు	ధూపమును, దీపమును	ద్వంద్వ సమాసం
19) శివభక్తి	శివుని యందు భక్తి	సప్తమీ తత్పురుష సమాసం
20) రుద్రాక్షభూషలు	‘రుద్రాక్షలు’ అనే పేరు గల భూషలు	సంభావనా పూర్వపద కర్మధారయ సమాసం
21) వితంతు వివాహం	వితంతువు యొక్క వివాహం	షష్ఠీ తత్పురుష సమాసం
22) విద్యాధికులు	విద్యచేత అధికులు	తృతీయా తత్పురుష సమాసం
23) భారతదేశం	భారతం అనే పేరు గల దేశం	సంభావనా పూర్వపద కర్మధారయ సమాసం
24) ముప్పయి సంవత్సరాలు	ముప్ఫై సంఖ్య గల సంవత్సరాలు	ద్విగు సమాసం
25) స్త్రీ పురుషులు	స్త్రీయును, పురుషుడును	ద్వంద్వ సమాసం
26) ప్రముఖదినం	ప్రముఖమైన దినం	విశేషణ పూర్వపద కర్మధారయం
27) నాలుగు గీతలు	నాలుగు సంఖ్య గల గీతలు	ద్విగు సమాసం
28) అసాధారణం	సాధారణం కానిది	నఞ్ తత్పురుష సమాసం
29) మానవచరిత	మానవుల యొక్క చరిత	షష్ఠీ తత్పురుష సమాసం

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 11 Electric Current.

AP State Syllabus SSC 10th Class Physics Important Questions 11th Lesson Electric Current

10th Class Physics 11th Lesson Electric Current 1 Mark Important Questions and Answers

Question 1.

Find the quantity of current in the above circuit. (AP March 2017)
Answer:
R = 3 + 5 + 2 = 10 Ω
I = \(\frac{1.5}{10}\) = 0.15 A.

Question 2.

Three resistors A, B and C are connected as shown in the figure. Each of them dissipates energy to a maximum of 18 W. Find the maximum current that can flow through the three resistors. (TS March 2015)
Answer:

Question 3.
What happens if we use a fuse made up of same wire which is used to make the electric circuit? (TS March 2017)
Answer:
It doesn’t work as a fuse. If high voltage occurs fuse do not melt and circuit will not be opened / breaked. So home appliances will be damaged.

Question 4.
Write any two differences between ohmic and non- ohmic conductors. (TS June 2018)
Answer:

Ohmic conductors	Non-ohmic conductors
Ohmic conductors follow the Ohms law.	Non-ohmic conductors do not follow the Ohms law.
Ohmic conductors are electric conductors.	Non-ohmic conductors are semicon­ductors.
V-I graph of ohmic conductors is a straight line.	V-I graph of non-ohmic conductors is a curve.

Question 5.
Draw the electric circuit with the help of a Battery, Voltmeter, Ammeter, Resistance and connecting wires. (TS March 2018)
Answer:

Question 6.
What happens, if the household electric appliances are connected in series? (TS March 2019)
Answer:
If all household appliances are connected in series, then if one appliances stop working due to failure then all the appliances stops working due incomplete circuit.

Question 7.
Define lightning.
Answer:
Lightning is an electric discharge between two clouds or between cloud and earth.

Question 8.
Define drift speed or drift velocity.
Answer:
The electrons in the conductor move with a constant average speed called drift speed or drift velocity.

Question 9.
Define conductors.
Answer:
The materials which can conduct electricity are called conductors.
Eg : Copper, Silver, Aluminium.

Question 10.
Define insulators.
Answer:
The materials which can’t conduct electricity are called insulators or non – conductors.
Eg: Wood, Rubber.

Question 11.
Define semi-conductors.
Answer:
The materials whose resistivity is 10⁵ to 10¹⁰ times more than that of metals and 10¹⁵ to 10¹⁶ times less than that of insulators.
Eg: Silicon, Germanium,

Question 12.
How does a battery work?
Answer:
In a circuit, the battery stores chemical energy and this energy converts into electric energy. Thus a battery works.

Question 13.
Define lattice.
Answer:
Conductors like metals contain a large number of free electrons while the positive ions are fixed in their locations. The arrangement of the positive ions is called lattice.

Question 14.
Define potential difference.
Answer:
Work done by the electric force on unit positive charge to move it through a distance is called potential difference.

Question 15.
Define electromotive force.
Answer:
Electromotive force (emf) is defined as the work done by the chemical force to move unit positive charge from negative terminal to positive terminal of the battery.

Question 16.
Define resistance of a conductor.
Answer:
The obstruction to the motion of the electrons in a conductor is called resistance of a conductor.

Question 17.
Define resistivity (ρ).
Answer:
Resistivity is a constant.

Question 18.
Write Ohm’s formula.
Answer:
V = IR, where V is the potential difference (voltage), I is the electric current and R is the resistance.

Question 19.
Define resistor.
Answer:
The material which offers resistance to the motion of electrons is called a resistor.

Question 20.
What is a multimeter?
Answer:
It is an electronic measuring instrument that combines several measurement functions in one unit.

Question 21.
Define electric power.
Answer:
The product of voltage and electric current is called electric power.

Question 22.
Define electric energy.
Answer:
The product of electric power and time is called electric energy.

Question 23.
What is lattice?
Answer:
The arrangement of positive ions in a conductor is known as lattice.

Question 24.
Name two special characteristics of fuse wire.
Answer:
High resistivity and low melting point.

Question 25.
Name two special characteristics of heating coil.
Answer:
High resistivity and high melting point.

Question 26.
How does resistivity vary with material of conductor?
Answer:
The resistivity is less for a good conductor and is large for a bad conductor.

Question 27.
If length of a particular conductor increased by two times and its area of crosssection decreased by four times, then what happens to its resistivity?
Answer:
The resistivity of the conductor does not change because resistivity does not depend on dimensions of conductor.

Question 28.
What does the slope of V -1 graph for an Ohmic conductor represent?
Answer:
For an Ohmic conductor the slope of V -1 graph represents the resistance.

Question 29.
Express the units of ohm in terms of volt and ampere.
Answer:
1) The SI unit of resistance is ‘Ohm’.
2) Ohm = \(\frac{\text { Volt }}{\text { Ampere }}\)

Question 30.
What is the resultant resistance of this combination?

Answer:
R, R, R Ω resistances are in parallel.
⇒ Resultant resistance = \(\frac{R}{3}\)

Question 31.
What are the quantities are conserved in Kirchhoff’s and Is’ laws?
Answer:
According to Ist law, charge, and 2^nd law, energy are conserved.

Question 32.
If the length and radius of a conductor are both halved. What happend resistance of wire?
Answer:
Length and radius are halved.

Resistance is doubled.

Question 33.
If work done is W and the charge that flows through is Q, then what is the equation 1 of potential difference?
Answer:

Question 34.
How many electrons constitutes a current of one ampere?
Answer:
6.25 × 10¹⁸ electrons in one second.

Question 35.
What are the maximum and minimum resistances are prepared by 30Ω, 30Ω, 30Ω?
Answer:
Maximum resistance = 30 + 30 + 30 = 90Ω
Minimum resistance = \(\frac{30}{3}\) = 10 Ω

Question 36.
Electric current I = nqA,v_d. Write the representation of letters.
Answer:
n = Electron density
A = Area of cross – section
v_d = drift velocity
q = charge of electron .

Question 37.
What is the resistance of bulb marked 60W and 120V?
Answer:

Question 38.
From figure, if V_A = 10V, then V_B = ?

Answer:
V_A – IR – E = V_B
10 – 1 × 5 – 3 = V_B
10 – 5 – 3 = V_B
2 = V_B
∴ V_B = 2 volts

Question 39.
Write the examples of non-ohmic conductors.
Answer:
Non-ohmic conductors are electrolytes, semi conductors, vacuum tubes.

Question 40.
What is meant by electric shock?
Answer:
When the current flows through the body the functioning of organs inside the body gets disturbed. This disturbance inside the body is felt as electric shock.

Question 41.
There is no electric shock on bird, when stand on the Electric wires. Why?
Answer:
When the bird stands on a high voltage wire, there is no potential differences between the legs of the bird, so no current passes through the bird.

Question 42.
Which factors are influence the resistivity of wire?
Answer:
a) Temperature,
b) Nature of material.

Question 43.
The length of wire is doubled and area of cross-section also doubled. What is the change in resistivity.
Answer:
Resistivity is independent of length and Area of cross-section.
Resistivity is not change.

Question 44.
A battery of 6V is applied across a resistance of 15Ω. Find the current flowing through the circuit.
Answer:

Question 45.
The formula V = I R is applicable for what substances?
Answer:
V = IR is applicable for
a) Ohmic conductors,
b) Non-ohmic conductor.

Question 46.
Resistance of a conductor of length 75 cm is 3.250. Calculate the length of a similar conductor, whose resistance is 16.25Ω.
Answer:

Question 47.
If four wires of each resistance R are joined to form a square, then what is the resistance between its opposite vertices?
Answer:

Resistance of ABC = 2R
Resistance of ADC = 2R
2R & 2R are parallel ⇒ Resultant = ‘R’

Question 48.
How much power consumption in a day of 100W television utilised 10 hours?
Answer:
Power consumption = \(\frac{100 \mathrm{~W} \times 10 \mathrm{~h}}{1000}\)
1 KWH or 1 unit.

Question 49.
How should we connect a voltmeter to measure voltage?
Answer:
The voltmeter must be connected in parallel to the electric device to measure the potential difference across the ends of the electric device.

Question 50.
How are ammeter and voltmeter connected in a circuit?
Answer:
Ammeter is always connected in series and voltmeter is always connected in parallel in a circuit.

Question 51.
Is the voltmeter connected in series or parallel in circuit? Why?
Answer:
Voltmeter should be connected parallel in the circuit to measure the potential difference between two points of conductor.

Question 52.
State whether the home appliances like Fridge, TV, Computer are connected in series or parallel Why?
Answer:
They are connected in parallel because if any one device is damaged, the rest will work as usual because the circuit does not break.

Question 53.
Why are copper wires used as connecting wires?
Answer:
Copper is a good conductor of electricity so copper wires are used as connecting wires.

Question 54.
Why is the fuse wire fitted in a porcelain casing?
Answer:
Porcelain is an insulator of electricity. So fuse wire is fitted in a porcelain casing.

Question 55.
Draw the diagram of potential – current of a copper conductor.
Answer:

Temperature T = 27°C
It is a straight line passing through origin.

Question 56.
Draw the shape of V – I graph for a silicon.
Answer:

Silicon is a semi conductor.
It is not obey the Ohm’s law.

Question 57.
Is there any application of Ohm’s Law in daily life?
Answer:

• Electrical device like electric bulb, iron box and regulators are some applications of Ohm’s Law.
• Fuse in household circuits is also another application of Ohm’s Law.

Question 58.
Two wires of the same material and same length have radii r₁ and r₂ respectively. Compare (i) their resistance, (ii) their resistivities.
Answer:

Resistivity for a same material is same. So their ratio =1:1.

Question 59.
A wire of resistance is doubled on itself, then what is its new resistance?
Answer:
Suppose length is l, area of cross-section is A and resistance is R.

Question 60.
Calculate effective resistance between points A and B.

Answer:
R₁ = 1 Ω, R₂ = 2 Ω are in series. So R’ = 1 + 2 = 3 Ω.
Now R₃ and R’ are in parallel.

Question 61.
A fuse is rated 8A. Can it be used with an electrical appliance of rating 5 KW and 200 V?
Answer:
Given P = 5 KW = 5 × 1000 = 5000 W; V = 200 V.

So a fuse of rate 8 A is not suitable because it uses current of 25 A.

Question 62.
A current of 2A is passed through a coil of resistance 75 Ω for 2 minutes. How much heat energy is produced?
Answer:
Given i = 2A, R = 75 Ω and t = 2 min. = 2 × 60 sec. = 120 sec.
Heat energy produced H = i² Rt = 2² × 75 × 120 = 36000 J = 36 KJ.

Question 63.
What is ,the resistance under normal working conditions of a 240 V electric lamp rated at 60 W?
Answer:
P = 60 W; V = 240V.

Question 64.
State the use of Ammeter. How should the Ammeter be connected in electric circuit?
Answer:
Ammeter should be connected in series in a circuit.

Question 65.
What is current value of x?

Answer:
2 + 3 = 5 of ‘A’
5 – 1 = 4 of ‘B’
2x = 4
x = 2 A at ‘C’

Question 66.
What is the value of V_A, when V_B = 8V?

Answer:
V_A – 6 × 1 – 3 = V_B
V_A – 6 – 3 = 8
V_A = 8 + 9
V_A = 17 volts

Question 67.
In the figure, how much current passing through 6Ω resistor?

Answer:
Answer:
R₁ : R₂ = 6 : 4
R₁ : R₂ = 3 : 2
i₁ : i₂ = R₂ : R₁ = 2 : 3
∴ i₁= 2A
i₂ = 3A

10th Class Physics 11th Lesson Electric Current 2 Marks Important Questions and Answers

Question 1.
Observe the graph of potential difference (V) drawn between two ends of a conductor and current (I) passing through it. Answer the following questions :

a) Which law is used to explain the graph? State it.
b) What is the resistance of the conductor? (AP June 2017)
Answer:
a) Ohm’s law is used to explain the given graph.
Ohm’s Law :
The potential difference between the ends of a conductor is directly proportional to the electric current passing through it at constant temperature.

Question 2.
Draw the experimental set-up to verify that V/I is constant for a conductor. (TS March 2016)
Answer:

Question 3.
A house has 3 tube-lights of 20 watts each. On the average, all the tube-lights are kept on for five hours. Find the energy consumed in 30 days. (AP SCERT : 2019-20)
Answer:
Number of tube lights = 3
Wattage = 20 watts each
Consumed hours = 5 hrs

Consumed energy for 30 days = 0.3 × 30 = 9 KWH (or) 9 units.

Question 4.
How do the resistors 6 Ω, 10 Ω to be connected in a circuit to get minimum resistance? Find the resultant resistant of the circuit. (TS June 2019)
Answer:

1. Resistors 6Ω, 10Ω should be connected in parallel connection in a circuit to get minimum resistance.
2. Resultant resistance
   

Question 5.
Write two examples for ohmiq and non-ohmic materials each. (TS June 2019)
Answer:
Ohmic materials : Copper, Aluminium
Non – ohmic materials : Germanium, Silicon

Question 6.
Give reasons for using lead in making fuses.
Answer:

• Lead is used in making fuses because it has low melting point EK resistivity.
• If the current in the lead wire exceeds certain value, the wire will heat up and melt, so the circuit in the households is opened and all the electric devices are saved.

Question 7.
How can we decide the direction of electric current in a conductor?
Answer:
We know I = nqAv_d. In this n and A are positive. Hence the direction of current is determined by the signs of the charge ‘q’ and drift speed v_d.

1. For electrons, q is negative and v_d is positive. Then the product of q and v_d is negative. So the direction of electric current is opposite to the flow of negative charge.
2. For positive charge, the product of q and v_d is positive. Hence, the direction of electric current can be taken as the direction of flow of positive charges.

Question 8.
What are the devices used in a circuit?
Answer:
1) Ammeter :
It is used to measure current.

2) Volt meter:
It is used to measure potential difference across the ends of conductor.

3) Rheostat:
It varies current in a circuit.

4) Switch :
It is useful to make a circuit or break a circuit.

5) Cell:
It is source of electric energy in the circuit.

6) Multimeter :
It is useful to measure current, voltage and resistance in the circuit.

Question 9.
A student says “Potential difference and Emf are same.” Justify your answer.
Answer:
Both are different because potential difference is the work done by the electric force on unit positive charge to move it through a distance between two points whereas emf is the chemical force to move unit positive charge from negative terminal to positive terminal of the battery.

Question 10.
Define Ohmic and non-ohmic conductors and give two examples each of them.
Answer:
Ohmic conductors :
The conductors which obey Ohm’s law are called ohmic conductors, e.g, : Copper, Iron.

Non-ohmic conductors :
The conductors which do not obey Ohm’s law are called non-ohmic conductors, e.g. : Semiconductors, Electrolytes.

Question 11.
Explain the Junction law of Kirchhoff.
Answer:

1. At any junction point the sum of the currents into the junction must be equal to the sum of currents leaving the junction.
2. There is no accumulation of electric charges at any junction in a circuit.
   I₁ + I₄ + I₆ = I₅ + I₂ + I₃

Question 12.
Write differences between overloading and short circuiting.
Answer:
Current chooses a path which has least resistance. So sometimes electrical appliances get damaged by passage of excess of current due to short circuit.

When so many electrical appliances are connected to the same electrical main point, maximum current can be drawn from the mains which causes overheating and may cause a fire which is called overloading.

Question 13.
The V-I graph for a series combination and for parallel combination of two resistors is shown in figure. Which of the two A or B, represent parallel combination? Give reason for your answer.

Answer:
i) For same change in I, change in V is less for the straight line A than for the straight line B (i.e., straight line A is less steeper than B).
ii) So the straight line A represents small resistance, while straight line B represents more resistance.
iii) In parallel combination, the resistance decreases, while in series combination, the resistance increases, so A represents a parallel combination.

Question 14.
Two resistors are joined with a battery such that
a) same current flows in each resistor.
b) potential difference is small across each resistor.
c) equivalent resistance is less than either of the two resistors.
d) equivalent resistance is more than either of the resistors.
State how are the resistors connected in each of the above case.
Answer:
a) Since same current is flowing in each resistor, they are connected in series.

b) Potential difference is same across same resistor that shows they are connected in parallel.

c) Equivalent resistance is less than the either of the two resistances which shows they are connected in parallel because in parallel connection resistance decreases.

d) Equivalent resistance is more than the individual resistances which indicate that they are connected in series. Since in series connection resistance increases.

Question 15.
Which of the cables, one rated 5A and other 15 A will be thicker wire? Give reason for your answer.
Answer:
i) To carry larger current, the resistance of wire should be low, so its area of cross-section should be large.

ii) Because the second cable carrying 15 A ampere current which indicates it has low resistance that is more surface area which implies it is thicker wire.

Question 16.
When a potential difference 30V is applied across a resistor, it draws a current of 3A. If 20V is applied across the same resistor, what will be the current?
Answer:
Situation : 1

Situation : 2

Question 17.
Is Ohm’s Law universally applicable for all conducting elements? If not, give example of elements which do not obey the Ohm’s Law.
Answer:

• Ohm’s Law is not applicable for all conducting elements.
• For example, some semi-conductors like silicon, germanium do not obey the Ohm’s Law.
• Those which do not obey Ohm’s Law are called non-ohmic materials.
• LED’s are non-ohmic materials.

Question 18.
Alloys are used in electrical heating devices rather than the pure metal. Why?
Answer:
Alloys are homogeneous mixtures of two or more metals. Generally alloys have more resistivity than the metals from which they have been prepared. As the resistivity increases heating effect of conductor also increases. So alloys are preferred in heating devices.

Question 19.
A switch should not be touched with wet hands. Why?
Answer:

• A switch should not be touched with wet hands.
• If water reaches the live wire, it forms a conducting layer between the hand and the live wire of the switch through which the current passes to the hand, and the person may get a total shock.

Question 20.
Which material is used for power transmission? Why?
Answer:

• The wires which are used for connections and for power transmission are made of material such as copper or aluminium.
• The reason is their resistivity is very small and they are made thick so that their resistance can be considered to be negligible.
• Further, the loss of electrical energy due to heating is also negligible in them.

Question 21.
Which material is preferred for heating element? Why?
Answer:

• The heating elements or resistance wires (or standard resistors) are made of material such as nichrome, manganin, constantan, etc. for which the resistivity is quite large and the effect of change in temperature on their resistance is negligible.
• So electrical energy is converted into heat energy when current passes through the wire.

Question 22.
Draw the symbols of the following.
i) Battery
ii) Resistance
iii) Ammeter
iv) Voltmeter
v) Key
vi) Rheostat
Answer:

Question 23.
Draw V-I graphs of Ohmic and non-ohmic conductors.
Answer:

Question 24.
Draw a circuit diagram with a cell, an electric bulb, an ammeter and plug key.
Answer:

Question 25.
Draw a circuit diagram to verify the Ohm’s Law.
Answer:

Question 26.
If 60 C of charge passes through a conductor for 1 minute, find the current through the conductor.
Answer:

Question 27.
The work done in moving 6 C of charge through a circuit is 12 J. Find the potential difference in the circuit.
Answer:

Question 28.
Resistance of two resistors are 6 Ω, 12 Ω respectively. Find the resultant resistance if the resistors are connected (1) in series (2) in parallel.
Answer:

Question 29.
10 equal resistors of resistance 20 are connected in a circuit. Find the resultant resistance if they are connected in series or in parallel ?
n = 10; R = 20 Ω
In series connection resultant resistance R’ = nR = 10 × 20 = 200 Ω

Question 30.
From the figure find the current through 6 Ω, 12 Ω resistors and find the resultant resistance in the circuit and also find current in the circuit.

Answer:
R₁ =6 Ω; R₂ = 12 Ω; V = 6 V

Question 31.
Find which has greater resistance. 1 KW heater or a 100 W tungsten bulb, both marked for 230 V.
Answer:

Question 32.
Two wires (one is copper and other is aluminium) have equal area of cross-section and have the same resistance. Find which one is longer.
Answer:
Suppose the resistance of copper and aluminium wires are R₁ and R₂. Suppose their area of cross-section is A.
The resistivity of copper (ρ₁) = 1.68 × 10^-8
The resistivity of aluminium (ρ₂) = 2.82 × 10^-8

Question 33.
Three equal resistances are connected In series, then in parallel. What will be the ratio of their resultant resistances?
Answer:
Suppose the resistance of equal resistors is ‘R’. Suppose they are connected in series. Then their equivalent resistance R’ = R + R + R = 3R
If they are connected in parallel their equivalent resistance R” = \(\frac{R}{3}\)
∴ Ratio of resultant resistances = R’: R” = 3R : \(\frac{R}{3}\) = 9 : 1

Question 34.
How is Ideal earthing helpful during short circuiting?
Answer:

• During short circuiting an excessive current flows through the live wires.
• It will pass to earth through the earth wire if there is local earthing otherwise it may cause a fire due to overheating of the live wires.

Question 35.
You have three resistor values 2Ω, 3Ω and 5Ω. How will you join them Sd that the total resistance Is less than 2Ω? Find resultant resistance.
Answer:
Given R₁ = 2Ω, R₂ = 3Ω, R₃ = 5Ω.
The resistors should be joined in parallel.

Question 36.
An electric kettle Is rated 3 KW, 250 V. Give reason whether this kettle can be used in a circuit which contains a 13 A fuse?
Answer:
V = 250 V, P = 3KW = 3000 W

The fuse is suitable because safe limit of current for kettle is 12 A.

Question 37.
Two resistors of 4 Ω and 6 Ω are connected parallel. The combination is connected across 6 V battery of negligible resistance. Find, i) the power supplied by the battery and ii) the power dissipated in each resistance.
Answer:
i) R₁ = 4 Ω, R₂ = 6 Ω, V = 6V.

ii) In parallel connection the resultant resistance

Question 38.
In the circuit shown below calculate the value of x if the equivalent resistance between A and B is 4 Ω.

Answer:
Given R₁ = 4 Ω, R₂ = 8Ω, R₃ = x Ω and R₄ = 5 Ω.
And resultant resistance R = 4 Ω.
R₁ and R₂ are in series.
Resultant resistance R’ = R₁ + R₂ = 4Ω + 8Ω = 12 Ω.
R₃ and R₄ are also in series.
Resultant resistance R” = R₃ + R₄ = x + 5
R’ and R” are in parallel.

Question 39.
A wire of 9 Ω resistance having 30 cm length is tripled on itself. What is its new resistance?
Answer:
Given R = 9 Ω, l = 30 cm and suppose area of cross-section A.

10th Class Physics 11th Lesson Electric Current 4 Marks Important Questions and Answers

Question 1.
A house has four tube-lights, three fans and a television. Each tube-light draws 40 W. The fan draws 80 W and the television draws 100 W. On an average, all the tube-lights are kept on for 5 hours, all fans for 12 hours and the television for 6 hours everyday. Find the cost of electric energy used in 30 days at the rate of Rs. 3.00 per KWH. (AP March 2015)
Answer:
The power used by
1) Four tube-lights of 40 W for 5 hours in 30 days
= 4 × 40 × 5 × 30 = 2400 WH

2) Three fans of 80 W for 12 hours in 30 days
= 3 × 80 × 12 × 30 = 86400 WH

3) One television of 100 W for 6 hours in 30 days
= 1 × 100 × 6 × 30 = 18000 WH

4) Total electric energy used
= 24000 + 86400 + 18000 = 128400 WH

W.H. converted into K.W.H. = 128.4 K.W.H. \(\left[\because \frac{128400}{1000}\right]\)
Cost of 1 unit = Rs. 3.00
Amount to be paid for 128.4 units = 128.4 × 3 = Rs. 385.20

Question 2.
Observe the given circuit. (AP June 2017)
R₁ and R₂ are two resistors and R₁ = R₂ = 4Ω. Emf of the battery E is 10 V.

Answer the following questions.
a) How are the resistors R₁ and R₂ connected in the circuit ?
b) What is the potential difference across R₁?
c) What is the effective resistance of the circuit?
d) What is the total current drawn from the battery?
Answer:
a) Resistors R₁ and R₂ are connected in parallel in the given circuit.
b) The potential difference across R₁ is ‘E’ volts = 10 V.

Question 3.
State Kirchhoff’s Loop law and explain. (AP June 2018)
Answer:
Loop law :
The algebraic sum of the increases and decreases in potential difference across various components of the circuit in a closed circuit loop must be zero.

Explanation :
Let us imagine in a circuit loop the potential difference between the two points at the beginning of the loop has a certain value. As we move around the circuit loop and measures the potential difference across each component in the loop, the potential difference may decrease or increase depending upon the nature of the element like a resistor or battery. But when we have completely traversed the circuit loop and arrive back at starting point. The net change in the potential difference must be zero. Thus the algebraic sum of changes in potential difference is to be zero.
(OR)
Let us apply loop law to a circuit as below.

for the loop ACDBA
– V₂ + I₂R2₂ – I₁R₁ + V₁ = 0

for the loop EFDCE
-(I₁ + I₂)R₃ – I₂R₂ + V₂ = 0

for the loop EFBAE
-(I₁ + I₂)R₃ – I₁R₁ + V₁ = 0

Question 4.

Observe the above diagram and answer the following. (AP March 2018)
a) Are all the resistors connected in parallel or series?
b) What is the equivalent resistance of the combination of three resistors?
c) In this system, which physical quantity is constant?
d) If R₁ = 2Ω, R₂ = 3Ω, R₃ = 4Ω. find equivalent resistance.
Answer:
a) Connected in series.
b) R_eq = R₁ + R₂ + R₃
c) Current (I)
d) R_eq = R₁ + R₂ + R₃
=2 + 3 + 4
= 9Ω

Question 5.
How do you verify that resistance of a conductor of uniform cross-section area is proportional to the length of the conductor at constant temperature? (TS March 2015)
Answer:
Aim :
To verify the relation between re.M+iance and lenath of the conductor.

1) Required material :
Wires or spokes of different lengths with same cross-section area of the same metal.
2) Battery
3) Ammeter
4) Key
5) Connecting wires.

Procedure :
1) Construct a circuit with Battery, Ammeter, Switch (key) and connecting wires, keeping some space at the both ends.
2) Connect the selected wires or spokes at the ends to complete the circuit.

3) Connect the wires or spokes individually and record the current using ammeter.

Conclusion :
If the current flowing in the circuit decreases with an increase in the length of the wire or spokes (Resistance increases), we can say that the resistance of the conductor is proportional to the length of the conductor.

Question 6.
What are the factors affecting the resistance of an electric conductor? Explain any two factors. (TS June 2015)
Answer:
The factors affecting the resistance of an electrical conductor are

1. Nature of material
2. Temperature
3. Length of the conductor
4. Area of cross-section of conductor

Explanation :

1. As the temperature increases the resistance increases and vice versa.
2. As the material changes resistance changes.
3. Resistance is directly proportional to length of the conductor (if T and A are kept constant). R ∝ l
4. Resistance is inversely proportional to area of cross-section (if 1 and T are kept constant). R ∝ \(\frac{1}{A}\)

Question 7.
What is the relationship between length of a conductor and its resistance? Write the experimental procedure to verify that relationship. (TS Junc 2017)
Answer:

• The resistance of a conductor is directly proportional to its lenght for a constant potential difference.
• Take iron spokes of different lengths with same cross-sectional arreas.
• Make a circuit by connecting an iron spoke with battery, ammeter and switch in series.
• Put the switch on and allow the current to pass in the circuit. Measure the ammeter reading.
• Repeat this for other lengths of the iron spokes. Note the corresponding values of currents.
• The resistance of each spoke increases with increase in the length of the spoke.

Question 8.
Find the resultant resistance for the following given arrangement. Find the current when this arrangement is connected with 9 V battery. (TS March 2017)

Answer:
The diagram is not clear so award 4 marks in the public examinations.

Question 9.
12 V battery is conncected in a circuit and to this 4Ω, 12Ω resistors are connected in parallel, 3Ω resistor is connected in series to this arrangement. Draw the electric circuit from this information and find the current in the circuit. (TS June 2018)
Answer:

Question 10.
In a circuit,60V battery, three resistances R₁ = 10 Ω, R₂ = 20 Ω and R₃ = x Ω are connected in series. If 1 ampere current flows in the circuit, find the resistance in R₃ by using Kirchhoffs loop law. (TS March 2018)
Answer:

Given
I = 1 amp
R₁ = 10Ω
R₂ = 20Ω
R₃ = xΩ
V = 60V
In ADCBA loop
– IR₃ – IR₂ – IR₁ + V = 0
– lx – 1 × 20- 1 × 10 + 60 = 0
– x – 30 + 60 = 0
– x + 30 = 0
x = 30Ω

Question 11.
List out the material required for the experiment “The effect of increasing of cross-section of a conductor upon its resistance” and write the experimental procedure. (TS June 2019)
Answer:
Aim :
To show that the effect of increasing of cross-section of a conductor upon its resistance.

Required Material :
Battery eliminator, Key, Ammeter, Manganin wires of equal lengths but different cross sectional areas, Copper connecting wires.

Procedure :

• Collect manganin wires of equal lengths but different cross sectional areas.
• Make a circuit as shown in the figure.

• Connect one of the wires between points ‘P’ and ‘Q’.
• Note the value of the current using the ammeter connected to the circuit and note it.
• Repeat this with other wires.
• Note the corresponding values of currents in each case and note them.

Conclusion :
We can notice that the current flowing through the wire increases with increasing their cross sectional area.

Question 12.
Derive an expression to find drift velocity of electrons.
Answer:

• Consider a conductor with cross sectional area A. Assume that the two ends of the conductor are connected to a battery to make the current flow through it.
• Let ‘v_d‘ be the drift speed of the charges and ‘n’ be the number of charges present in the conductor in a unit volume.
• The distance covered by each charge in one second is ‘v_d‘

• Then the volume of the conductor for this distance = Av_d
• ∴ The number of charges contained in that volume = n.Av_d
• Let q be the charge of each carrier.
• Then the total charge crossing the cross sectional area at position D in one second is ‘n q A_d‘.
  This is equal to electric current.
  ∴ Electric current I = n q Av_d.
  ∴ v_d = \(\frac{I}{nqA}\)

Question 13.
Derive an expression to measure emf of a battery.
Answer:
Electromotive force (emf) is defined as the work done by the chemical force to move unit positive charge from negative terminal to positive terminal of the battery.

1. Let this chemical force be F_c.

2. This chemical force does some work to move a negative charge ‘q’ from positive terminal to negative terminal against the electric force Fe. Let this work be ‘W’.

3. ∴ The work done by the chemical force to move 1 coloumb of charge from the

This S.I unit of emf is ‘volt’ and is measured using voltmeter.

Question 14.
What are the factors on which the resistance of conductor depends? Give the corresponding relation.
Answer:

• The value of resistance of a conductor depends on temperature for constant potential difference.
• Resistance of a conductor depends on the material of the conductor.
• Resistance of a conductor is directly proportional to its length, i.e., R ∝ l.
• Resistance of a conductor is inversely proportional to the area of cross-section of the material, i.e., R ∝ \(\frac{1}{A}\)

Question 15.
What do you mean by (i) short circuit (ii) overloading? What are the safety precautions taken to avoid these problems in domestic electric circuits?
Answer:
Short circuit:
Sometimes current chooses a path which has least resistance which is called short circuit.

Overloading :
The over heating due to drawing excess of current from a single main is called overloading.
Precautions to avoid damage due to short circuit and overloading.

1. Using fuse
2. Connecting electrical appliances in various mains.
3. Earthing

Question 16.
A circuit is set up as shown in the figure. Calculate the current and potential difference across R₁, R₂ and R₃, when
a) keys K₁ and K₂ are both closed,
b) key K₁ is closed and K₂ is open,
c) key K₁ is open and K₂ is closed.

Answer:
a) When both the keys K₁ and K₂ are closed :
The resistors R₁ and R₃ are parallel.
So resultant resistance R_p = \(\frac{R_{1} R_{3}}{R_{1}+R_{3}}=\frac{6 \times 4}{6+4}\) = 2.4 Ω.
The resistor R₁ and R_p are in series as shown in figure.

Total resistance of circuit R_s= R₁ + R_p = 6 + 2.4 = 8.4 Ω
Current I = \(\frac{V}{R_{S}}=\frac{4.2}{8.4}\) = 0.5 =A.
Potential difference across R₁ is V₁ = IR₁ = 0.5 × 6 = 3V.
Potential difference across the combination of R₂ and R₃ is
V’ = V – V₁ = 4.2 – 3 = 1.2 V.
Now since R₂ and R₃ are in parallel,
Potential difference across R₂ = Potential difference atross R₃ = V’ = 1.2 V

b) When the key K₁ is closed and key K₂ is open : The resistor R₃ will not be in circuit.
The resistors R₁ and R₂ are in series.
Total, resistance R_s = R₁ + R₂ = 6 + 6=12 Ω.
Current I = \(\frac{V}{R_{s}}=\frac{4.2}{12}\) = 0.35 A
The same current will flow through each resistor R₁ and R₂.
Potential difference across R₁ is V₁ = IR₁ = 0.35 × 6 = 2.1 V.
Potential difference across R₂ is V₂ = IR₂ = 0.35 × 6 = 2.1 V.
The current and potential difference across R₃ will be zero.

c) When key K₁ is open and key K₂ is closed :
No current flows through R₁ R₂ and R₃ since the circuit is incomplete. Hence potential difference across R₁, R₂ and R₃ is zero.

Question 17.
For the combination of resistance shown in figure, find the equivalent resistance between (a) C and D, (b) A and B.

Answer:
a) Between C and D :
The resistors R₂, R₃ and R₄ are in series. They can be replaced by an equivalent resistance R_s where
R_s = R₂ + R₃ + R₄ = 3 + 3 + 3 = 9 Ω
The resistance R₅ and R_s are in parallel between the points C and D.
The equivalent resistance between C and D then

Thus the equivalent resistance between C and D is 2.25 Ω.

b) Between A and B :
Now the resistors R₁, R_p and R₆ are in series between the points a and B.

The equivalent resistance between A and B is
R_AB = R₁ + R_p + R₆ = 3 +2.25 + 3 = 8.25 Ω.

Question 18.
How does resistance and resistivity vary with temperature?
Answer:

• For a metallic conductor, the resistance increases w ith the increase in temperature. The resistance of filament of bulb is more w hen it is glowing than when it is not glowing. The specific resistance or resistivity also increases with increase in temperature.
• For alloys (such as constantan and manganin), the resistance and the resistivity remains practically unchanged with the increase in temperature.
• For semi conductors (such as silicon, germanium, etc.), the resistance decreases with the increase in temperature.
  eg.: the resistance of carbon also decreases with the increase in its temperature.

Question 19.
Why is a copper wire unsuitable for making fuse wire?
Answer:

• A fuse is a short piece of wire made up of a material of high resistivity and of low melting point so that it may easily melt due to overheating when current in excess to the prescribed limit passes through it.
• The thickness of wire is different in different fuses depending on the amount of current which is permitted to flow through them.
• Generally an alloy of lead and tin is used as the material of the fuse wire because it has a high resistivity and low’ melting point.
• A copper wire is unsuitable for using as fuse wire because copper has low resistivity and high melting point.
• Therefore the use of an ordinary’ wire as a fuse must be avoided.

Question 20.
Why is current rating of a fuse required?
Answer:
1) The electric wiring for light and fan circuit uses a thin fuse wire of low current carrying capacity because the line wire has a current carrying capacity of 5A.

2) Thicker fuse wires of higher current carry ing capacity (15 A) are used for large current consuming appliances such as air conditioner, geyser, washing machine, etc. because the line wire for such dev ices have current carrying capacity of 15A.
The current rating of a fuse in a circuit can be obtained by the following relation.

Question 21.
Describe the activity with the help of diagram to establish the relationship between Current (I) flowing in a conductor and potential difference (V) maintained across its ends.
Answer:
Aim :
To establish the relationship between Current (I) flowing in a conductor and potential difference (V) maintained across its ends.

Material required :
5 dry cells of 1.5 V each, conducting wires, an ammeter, a volt meter, thin iron spoke of length 10 cm, LED and key.
Diagram :

Procedure :

• Connect a circuit as shown in the figure.
• Solder the conducting wires to the ends of the iron spoke.
• Close the key.
• Note the readings- of current from ammeter and potential difference from volt meter in the given table,

• Now connect two cells (in series) instead of one cell in the circuit.
• Note the respective readings of the ammeter and voltmeter and record the values in the table.
• Repeat the same for three cells, four cells and five cells respectively.
• Record the values of V and I corresponding to each case in the table.
•  Find \(\frac{V}{I}\) for each set of values.
  Conclusion : The ratio of \(\frac{V}{I}\) is constant.
• From this activity we can conclude that the potential difference between the ends of the iron spoke (conductor) is directly proportional to the current passing through it. This means V ∝ I.

Question 22.
In an experiment to verify Ohm’s Law the following values are given below. Draw a graph of ‘I’ versus ‘V’. Show that the graph conforms Ohm’s Law and find the resistance of the resistor.

Answer:
1) Graph between ‘V’ and ‘I’.

2) From the above graph,
Straight line §hows that the relation between potential difference (V) and current (I) as \(\frac{V}{I}\) is constant.

3) V ∝ I.

4) The potential difference between the ends of a conductor is directly prbportional to the electric current passing through it at constant temperature.

5) This is the Ohm’s Law and the graph conforms it.
Resistance of the resistor

Question 23.
Identify the defects 1ft the circuit. Redraw it.

Answer:
Defects in the circuit:

1. Ammeter was connected in parallel in the circuit.
2. Volt meter was connected in series in the circuit.
3. Positions of resistor and battery were reversed.

Correct diagram :

Question 24.
What is the advantage of MCB over fuse?
Answer:

• These days instead of fuses, Miniature Circuit Breakers (MCB) are used for each lighting circuit.
• They switch off the circuit in a very short time duration in case of short-circuiting or some fault in the line.
• After repairing the fault in the circuit, the MCB is again switched on.
• Thus, the use of MCB is better than a fuse. It avoids the inconvenience of connecting a new fuse wire and it is much safer due to its quick response.

Question 25.
A circuit is shown in the picture.

The current passing through A is I.
a) What is the potential difference between A and B?
b) What is the equivalent resistance between A and B?
c) What amount of current is flowed through C and D?
Answer:
a) According to KirchhofPs loop law the algebraic sum of increase and decrease in potential difference across various components of the circuit in a closed circuit loop must be zero.
So the potential difference across CD is zero because it is a closed loop.

b) Here 20 Ω, 5 Ω are parallel to each other and resultants are in series to each other. Resultant resistance of 20 Ω and 5 Ω.

Question 26.

Observe the picture. The potential values at A, B, C are 70 V, 0 V, 10 V
a) What is the potential at D?
b) Find the ratio of the flow of current in AD, DB, DC.
Answer:

a) By following Ohm’s law potential difference is (V) = IR
In the given circuit we are applying junction laws.
‘D’ works as junction so, I = I₁ + I₂
Let p.d at D is V₀.

b)

Question 27.
Observe the circuit R₁ = R₂ = R₃ = 200 Ω. If reading of voltmeter is 100 V, resistance of voltmeter is 1000 Ω.
Find the Emf of the battery.

Answer:
Given values are R₁ = R₂ = R₃ = 200 Ω.
and Voltmeter reading = V = 100 V.
Resistance of Voltmeter = R_v = 1000 Ω.
In the given circuit R₁ and R₂ are in series R = R₁ + R₂ = 200 + 200 = 400 Ω
Resultant resistance (400 Ω) and voltmeter (1000 Ω) are always in parallel.

Question 28.
A circuit is made with a copper wire as shown in the diagram. We know that conductor’s resistance is directly proportional to its length. Calculate the equivalent resistance between points 1 and 2.

Answer:

Let the resistance of the wire be ‘R’ and length of the wire be ‘l’.
The shape of the circuit be square length of the side (l) = R
In a square, diagonal is \(\sqrt{2}\) times its length = \(\sqrt{2}\)l
Resistance towards diagonal is \(\sqrt{2}\)R
The circuit diagrams for the given arrangement are along PTR and QTS is ineffective as no current flows through it.
PQ and PS are in series so effective resistance is R₁ + R₂ = R + R = 2R.
QR and SR are in series so effective resistance is R₁ + R₂ = R + R = 2R.
Redrawn of the circuit again as resultant resistance between the points 1 and 2 is

Question 29.
From the adjacent figure,

i) Find the potential at D.
ii) Find the current that passes through AD, DB and DC.
Answer:
Suppose from A and C current is flowing through the circuit and at B current is flowing away from the circuit. Suppose potential at D is V.
∴ I₁ + I₂ = I₃ (where I₁ I₂ are current into the junction. I₃ is the current away from the junction)

Question 30.
Find the electric current drawn from the battery of emf 8V from the given circuit.
Answer:

According Kirchhoffs loop law
6I₁ + 3I₁ – 8 = 0
9I₁ = 8
I₁ \(\frac{8}{9}\) = 0.89 A
∴ Cuttent in 8 V is 0.89 A

Question 31.
A household uses the following electric appliances.
i) Refrigerator of rating 400 W each for ten hours each day.
ii) Two electric fans of rating 80 W each for 12 hours each day.
iii) Six electric tubes of rating 18 W each for 6 hours each day. Calculate the electric bill of the household in a month if the cost per unit electric energy is Rs. 3.00.
Answer:
i) Electrical energy consumed by Refrigerator in a month (in KWH)

ii) Electrical energy consumed by two electrical fans in a month (in KWH)

iii) Electrical energy consumed by electric tubes in a month (in KWH)

Total electrical energy consumed by all the electrical appliances
= 120 + 57.6 + 19.44 = 197.04 units
Cost of 1 unit = ₹ 3
Cost of 197.04 units = 197.04 × 3 = ₹ 591.12

Question 32.
What is the reason for connecting the fuse in the live wire?
Answer:

• The fuse is always connected in the live wire of the circuit.
• If the fuse is put in the neutral wire, due to excessive flow of current the fuse burns, current stops flowing in the circuit.
• But the appliance remains connected to the high potential point of the supply through the live wire.
• Now if a person touches the appliance, he may get a shock as the person will come in contact with the live wire through the appliance.

Question 33.
In the diagram given below, two resistors R₁ and R₂ of 3Ω and 6Ω respectively are connected in parallel across a battery of potential difference 12 V. Calculate the electrical energy consumed in 1 minute in each resistance.

Answer:
Given R₁ = 3 Ω, R₂ = 6 Ω, V= 12 V, t = 1 min 60 sec.
The resistors R₁ and R₂ are connected in parallel, so the voltage V across each resistor is equal to 12 V.

Question 34.
Calculate the electrical energy consumed in a month, in a house using 2 bulbs of 100 W each and 2 fans of 60 W each, if the bulbs and fans are used for an average of 10 hours each day.
If the cost per unit is ₹ 3, calculate the amount of electrical bill to be paid per month.
Answer:
Given power of each bulb = 100 W
Power of each fan = 60 W
Time t =10 hours each day
Power of 2 bulbs = 2 × 100 = 200 W
Power of 2 fans = 2 × 60 = 120 W
Total power = 200 + 120 = 320 W = \(\frac{320}{1000}\) = 0.32 KW
Time duration of consumption 10 hours per day per month
= 10 hours × 30 = 300 hours
∴ Total energy consumed = Total power x Time duration
= 0.32 KW × 300 h = 96 KWh
∴ Total cost = 96 x 3 = ₹ 288.

Question 35.
What resistance must be connected to a 15 Ω resistance to provide an effective resistance of 6 Ω?
Answer:
Given R₁ = 15 Ω, R₂ = ?, Effective resistance R_p = 6 Ω.
Since the effective resistance has decreased, R2 must be connected in parallel with R₁

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 7 Human Eye and Colourful World.

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

10th Class Physics 7th Lesson Human Eye and Colourful World 1 Mark Important Questions and Answers

Question 1.
Define the power of lens. (AP June 2016)
Answer:
The reciprocal of focal length is called “power of lens”.

Question 2.
What physical quantity can be found in an experiment done with prism? (AP June 2017)
Answer:

1. Angle of deviation.
2. Refractive index of a prism.

Question 3.
What is the relation between Power and Focal length of the lens? (AP June 2018)
Answer:
\(P=\frac{1}{f(\text { in metres })}\) (OR) \(P=\frac{100}{f(\text { in cms })}\)

Question 4.
Draw the diagram of a lens which will be recommended by an eye doctor to a long sighted patient. (TS June 2015)
Answer:

The lens is convex lens.

Question 5.
What is the cause of Presbyopia? (TS March 2015)
Answer:
Presbyopia is vision defect when the ability of accommodation of the eye decreases with ageing.

Question 6.
Draw a ray diagram to show the angle of deviation when a ray of light passes through a glass prism. (TS March 2015)
Answer:

d = angle of deviation

Question 7.
Suggest reasons for the phenomenon associated with the following. The sky appearing blue. (TS March 2015)
Answer:
The reason for blue sky is due to the scattering of light by the molecules of N₂ and O₂ whose size is comparable to the wavelength of blue light.

Question 8.
+50 cm focal length bi-convex lens is recommended to correct the defect of vision of a mart. Find the power of the lens. (TS June 2016)
Answer:
f = +50 cm
Power (P) = \(\frac{100}{\mathrm{f}}\) D (in cm); P = \(\frac{100}{50}\) = 2 D
Power of the bi-convex lens is 2D.

Question 9.
What happens if the eye lens of a person cannot accommodate its focal length more than 2.4 cm? (TS March 2017)
Answer:
The person can able to see certain distance only he cannot see distance objects. For correction, he should use concave lens.

Question 10.
Write the reason for Sun appears red during the Sun-rise and Sun-set. (TS June 2018)
Answer:
Due to the high velocity (wave length) of red right, it reaches our eye without under go scattering. So, sun appears red during sunrise and sunset.

Question 11.
A person is unable to see distant objects. Show the defect of vision of the person with the help of ray diagram. (TS March 2018)
Answer:
1) His vision defect is myopia.
2) Ray diagram

Question 12.
Which molecules of atmosphere act as scattering centres are responsible for the blue sky? (TS June 2019)
Answer:
Oxygen and Nitrogen molecules.

Question 13.
Write any one application of a prism. (AP SA-I; 2019-20)
Answer:

1. Prism is used in scopes like binoculars, telescopes and light houses.
2. Prism is used to create artificial rainbow.

Question 14.
Mention the function of retina in a human eye.
Answer:
It acts as a screen, (which the image is formed) for image formed.

Question 15.
State the role of ciliary muscles in accommodation.
Answer:
It can adjust the focal length of the eye lens.

Question 16.
Why is normal eye not able to see clearly the objects kept closer than 25 cm?
Answer:
The maximum accommodation of a normal human eye is reached when the object is at a distance of 25 cm from the eyes. The focal length of the eye lens cannot be decreased this minimum limit.

Question 17.
What is “Power of accommodation of the eye” (or) “Least distance of clear vision”?
Answer:
The ability of the eye lens to adjust its focal length to see nearby and distant objects clearly.

Question 18.
What is “Iris” and “Pupil”?
Iris :
The muscular diaphragm between the aqueous humour and the lens is called ‘iris’. Iris is the coloured part that we see in an eye.

Pupil:
The small hole in iris is called ‘pupil’.

Question 19.
What are three common defects of vision?
Answer:
The common defects of vision are i) Myopia ii) Hypermetropia iii) Presbyopia.

Question 20.
What is “Far point”?
Answer:
The point of maximum distance at which the eye lens can form an image on retina is called ‘far point’.

Question 21.
What is power of lens?
Answer:
The reciprocal of focal length is called power of lens. Power of lens focal length = \(\frac{1}{\text { focal length }}\)

Question 22.
What is the function of pupil in human eye?
Answer:
It allows the light falling on iris.

Question 23.
What is the purpose of human eye?
Answer:
The purpose of eye is to see and perceive the objects around us.

Question 24.
What is the principle of the working of human eye?
Answer:
It acts as camera having a lens system forming an invented real image on the light sensitive screen, retina.

Question 25.
What is the nature of the image formed on the retina?
Answer:
Real, inverted and same sized.

Question 26.
What is meant by dispersion of light?
The splitting of white light into its component colours when it passes through the prism is called dispersion of light.

Question 27.
On which factors does the colour of the scattered white light depend?
Answer:

1. Angle of scattering
2. Distance travelled by light
3. Size of the molecules.

Question 28.
Why is normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
The focal length of eye lens cannot decrease below 25 cm.

Question 29.
A person is advised to wear spectacles with concave lenses. What type of defect of vision is he suffering from?
Answer:
Myopia (or) short sightedness.

Question 30.
A person suffering from an eye-defect uses lenses of power – 1D. Name the defect he is suffering from and the nature of lens is used.
Answer:
Defect:
Myopia, Nature of lens, Concave/divergence.

Question 31.
Name the two phenomena involved in the formation of rainbow.
Answer:
Dispersion of light and internal reflection.

Question 32.
Identify the following part of the human eye.
1. Where is image of an object formed?
2. Which controls size of pupil?
Answer:

1. Image of an object is formed on retina.
2. Iris controls the size of pupil.

Question 33.
What is the relation between power of lens and focal length (f)?
Answer:
Power of lens (concave/convex)
\(P=\frac{1}{f(\text { in metres })}\) (OR) \(P=\frac{100}{f(\text { in cms })}\)

Question 34.
Explain about “Bending of light”.
Answer:

1. When a light travels from rarer to denser medium, it bends towards the normal.
2. And from denser to rarer medium, it moves away from the normal.

Question 35.
What is sclerotic?
Answer:
It is the outermost covering of the eye. It protects the vital internal parts of the eye.

Question 36.
What is “Retina”?
Answer:
Retina is the internal part and the light sensitive surface of the eye. It is equivalent to the photographic film in a camera.

Question 37.
What is “Atmospheric refraction”?
Answer:
When the light rays pass through the atmosphere having layers of different densities and refractive indices, refraction of light takes place. This refraction of light by the earth’s atmosphere is called atmospheric refraction.

Question 38.
What is a “Telescope”?
Answer:
The instrument which is used to see the distant objects such as a star, planet (moon, sun) or distant tree is called telescope.

Question 39.
Have you seen a rainbow in the sky after rain? How is it formed?
Answer:

1. A rainbow is a natural spectrum of sunlight in the form of bows appearing in the sky when the sun shines on raindrops after the rain.
2. It is formed due to reflection, refraction and dispersion of sunlight by tiny water droplets present in the atmosphere.

Question 40.
“To look at the twinkling of stars is a wonderful experience.” How does it happen?
Answer:
The continuous changing atmosphere (due to varying atmospheric temperature and density) refracts the light from the stars by varying amounts and in different directions from one moment to the next.

Question 41.
Some things appear blue on a misty day. Give two examples.
Answer:

1. The long distance hills covered with thick growth of trees appear blue.
2. The smoke coming from a cigarette or an incense stick (agarbatti) appears blue on a misty day.

Question 42.
Which coloured suits do rescue workers wear?
Answer:
Rescue workers wear orange coloured suits during any rescue operations.

Question 43.
Which colour is best for school buses?
Answer:
Orange or yellow colour is best for school busses.

Question 44.
What is persistence of vision?
Answer:
The time for which the sensation of vision (of an object) continues in the eye is called persistence of vision. It is about 1/16^th part of a second.

Question 45.
Why can’t some people identify some colours?
Answer:
Rods identify the colours in the retina. If some rods are absent, the distinction of colours is not possible. In such cases, persons can’t identify some colours.

Question 46.
Write the reasons for colour blindness.
Answer:

1. Absence of colour responding rod cells in the retina.
2. Due to genetic disorder.

Question 47.
Why does it take some time to see objects in a dim room when we enter the room from bright sunlight outside?
Answer:
In bright light, the size of the pupil is small to control the amount of light entering the eye. When we enter a dim room, it takes some time so that the pupil expands and allows more light to enter and helps to see things clearly.

Question 48.
What is tyndall effect?
Answer:
The phenomenon of scattering of white light by colloidal particles is known as ‘Tyndall effect”.

Question 49.
Give two examples illustrating “Tyndall effect”.
Answer:

• A fine beam of sunlight entering a smoke filled room through a hole. Smoke particles scatter the white light and hence the path of light beam becomes visible.
• Sunlight passing through the trees in forest.
• Tiny water droplets through the trees in forest.

Question 50.
An eye camp was organised by the doctors in a village. What were the benefits to organise such camps in rural areas?
Answer:

1. To make people aware of eye diseases
2. To take proper and balanced diet.

Question 51.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
In the eye, the image distance (distance between eye lens and retina) is fixed and it cannot be changed. So when we increase the distance of an object, there is no change in the image distance.

Question 52.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
The maximum accommodation of a normal eye is at a distance of 25 cm from the eye.
The focal length of the eye lens cannot be decreased below this. Thus an object placed closer than 25 cm cannot be seen clearly by a normal eye.

Question 53.
Write the relation between intensity of scattered light (I) and wavelength (λ).
Answer:
Light of short wavelength is scattered more than the light of long wavelength.
i.e., Intensity of scattered light (I) ∝ \(\frac{1}{\text { wavelength }(\lambda)}\)

Question 54.
Why would the sky look dark if the earth had no atmosphere?
Answer:
If the earth has no atmosphere, no particles present either. Thus no scattering of light. Then, the sky appears dark.

Question 55.
Why do different coloured rays deviate differently in the prism?
Answer:
Because the angle of refraction of different colours is different while passing through the glass prism.

Question 56.
What prevents rainbow from being seen as complete circles?
Answer:
The earth comes in the way of the rainbow and prevents it to form a complete circle.

Question 57.
When a monochromatic light passes through a prism will it show dispersion?
Answer:
No, it will not show any dispersion but show deviation.

Question 58.
Will a star appear to twinkle if seen from free space (say moon)?
Answer:
No, because there is no atmosphere in free space for refraction to take place.

Question 59.
A short-sighted person may read a book without spectacles. Comment.
Answer:
The statement is true, because a short-sighted person has difficulty in observing far off objects.

Question 60.
What is angle of vision?
Answer:
The maximum angle, at which we can see the whole object is called angle of vision.

Question 61.
What is cornea?
Answer:
The front curved portion of eye, which is covered by a transparent protective membrane is called the cornea.

Question 62.
Which lens do you use to correct the eye defect, Myopia?
Answer:
Bi-concave lens are used to correct the eye defect, Myopia.

Question 63.
What happens to power of lens if (i) focal length is increased, (ii) focal length is decreased?
Answer:

1. If focal length is increased, then power of lens decreases.
2. If focal length is decreased, then power of lens increases.

Question 64.
What type of image is formed by magnifying glass?
Answer:
It forms virtual, erect and magnified image.

Question 65.
How is power of lens related to its focal length?
Answer:
Power of lens is inversely proportional to its focal length.

Question 66.
What is “Scattering of light”?
Answer:
The process of re-emission of light in all directions with different intensity is called scattering of light. The re-emitted light is called scattered light.

Question 67.
Ramu is unable to see letters on the blackboard sitting at the last bench in the class¬room. What is the defect from which Ramu is suffering?
Answer:
Ramu is suffering from Myopia.

Question 68.
Vinay is able to read letters in a book beyond certain distance from least distance of distinct vision. What is the eye defect of Vinay?
Answer:
Vinay is suffering from Hypermetropia.

Question 69.
How can an eye lens accommodate its focal length?
Answer:
To see an object comfortably and distinctly, one must hold at a distance about 25 cm from his/her eyes. This distance is called least distance of distinct vision.

Question 70.
Write the uses of “rods” and “cones”.
Answer:
Retina contains about 125 million receptors called rods and cones. Rods identify the colour and cones identify the intensity of light.

Question 71.
Frame some questions, that you are going to ask your friend who is suffering from eye sight.
Answer:

1. Are you not able to see near objects or far objects?
2. Are you not able to see both near and far objects?

Question 72.
Write some more examples to find dispersion of light as VIBGYOR.
Answer:

1. Formation of rainbow.
2. When white light passes through water drop.

Question 73.
Why do stars twinkle?
Answer:
Due to change in atmospheric conditions, density changes so position keeps on changing.

Question 74.
“Sky appears dark to passengers flying at very high altitudes.” Why?
Answer:
At very high altitude, there is no atmosphere. So there is no scattering of light at such heights. So sky appears dark to passengers.

Question 75.
Why are danger signals red? (OR) Why are danger signals shown in red colour?
Answer:
Among the colours of visible light, red has more wavelength and least scattered. Thus, red colour can easily go through fog or mist or smoke without getting scattered. It can be seen from long distance. So red colour is used in universal danger signal.

Question 76.
How can you identify regarding the type of defect a person is suffering from by physically touching his spectacles?
Answer:
By touching the spectacles we can find out whether the lens is concave or convex lens and hence the defect from which he suffers.

Question 77.
A person cannot see objects beyond 1-2 m distinctly. What should be the nature of the defect and what type of lens should be used to correct the defect?
Answer:
The person is suffering from Myopia. It can be corrected by using concave lens.

Question 78.
How do you appreciate the working of “Retina”?
Answer:

1. It is the innermost delicate membrane having a large number of receptors called ‘rods’ and ‘cones’.
2. The rods identify the colour and the cones identify the intensity of light.
3. The retina is a part on which the image of an object is formed.

Question 79.
How do you appreciate the working of “Optic nerve”?
Answer:

1. Optic nerve consists of a large number of fibres.
2. These optic nerve fibres are connected to the rods and cones.
3. Optic nerve fibers transmit the light signals to the brain.

Question 80.
We see advertisements for eye donation on television or in newspaper. Write the importance of such advertisements.
Answer:
Eye donation advertisements are important as :

1. The people aw;are about donation of organs after their death.
2. Sympathetic nature towards others.

Question 81.
An eye donation camp is being organised by social workers in your locality. How and why would you help in this cause?
Answer:

1. We can intimate other people to participate in the camp.
2. As a human being, we should also register our eyes for donation after death.

Question 82.
A short sighted person cannot see clearly beyond 2m. Calculate the power of lens required to correct his vision.
Answer:

Question 83.
How does the power of a lens change if its focal length is doubled?
Answer:
The power get halved.

Question 84.
A person cannot see distinctly objects kept beyond 2m. Which among these lens is useful to correct the defect
a) – 0.2D
b) – 0.5 D
c) + 0.2D
d) +0.5D?
Answer:
The person is suffering from Myopia.
The lens is concave and its focal length f = – 2m.
\(P=\frac{1}{5}=\frac{1}{-2}=-0.5 D\)
So to correct the defect concave lens of – 0.5 D power should be used (b).

Question 85.
Express the power of concave lens of focal length 20 cm with its sign.
Answer:

10th Class Physics 7th Lesson Human Eye and Colourful World 2 Marks Important Questions and Answers

Question 1.
How do you appreciate the working of iris in the eye? (AP June 2018)
Answer:
Iris helps in controlling the amount of light entering the eye through pupil.

Question 2.
What is the reason for the blue colour of the sky? (AP March 2018)
How do you appreciate the role of molecules in the atmosphere in this regard?
Answer:
Blue colour of sky :

1. Atmosphere contains more O₂ and N₂ molacules and they are caused to blue colour of the sky.
2. The size of molecules of O₂ and N₂ are comparable with the blue colour and scatter blue colour only.

Role of molecules in the atomosphere in scattering :

1. Light of certain frequency falls on that atom or molecule.
2. This molecule responds to the light whenever the size of the molecule is comparable to the wavelength of light and vibrates.
3. Due to these vibration, molecule reemits a certain fraction of absorbed energy in all directions. The emitted light is called scattered light.
4. The atoms or molecules are called scattering centres.
5. I appreciate the role of the molecules in scattering.

Question 3.
In which conditions does a rainbow form? Why? (TS June 2015)
Answer:
1) Rainbow forms and appears when,
i) tiny water droplets present in the atmosphere (after rain shower),
ii) sunlight falls on the droplets,
iii) observer watches the rainbow in a specific direction.
2) Rainbow forms due to dispersion of sunlight by tiny droplets, present in the atmosphere, which acts as small prisms.

Question 4.
Draw the ray diagram, showing the correction of defect of vision hyper metropia by using a convex lens. (TS June 2016)
Answer:

Question 5.
Least distance of distinct vision of a person is observed as 35 cm. What lens is useful for him to see his surroundings clearly? Why? (TS March 2016)
Answer:
The least distance of distinct vision is 35 cm. This is more than the least distance of distinct vision of an ordinary person. Hence the person is suffering from Hypermetropia. He has to use double convex lens to see his surroundings clearly.

Question 6.
What happens, if Ciliary muscles do not perform contraction and expansion? Guess and write. (TS June 2018)
Answer:

1. If Ciliary muscles do not perform contraction and expansion, foal length of eye lens do not change.
2. Human eye can see the objects at specific distance only, eye cannot see the object either nearer or far distance.

Question 7.
Write any two situations to observe dispersion of light in your daily life. (TS June 2018)
Answer:
We can observe the dispersion of light in the following situations in our daily life.

1. In the formation of rainbow.
2. When observing sun light through the triangular transparent material like prism, scale edge.
3. At the time of curing of walls of new houses with water.
4. Dispersion of light by inclained plane mirror which is in water.

Question 8.
Write the material that you use to find out the value of refractive index of a prism. What is the necessity of the graph in this experiment? (AP March 2019)
Answer:
The material used to find out the value of refractive index of a prism:
Prism, Piece of white chart, Pencil, Pins, Scale and Protractor.

Necessity of the graph :
To find the angle of minimum deviation graph is required.

Question 9.
Draw a ray diagram showing the correction of myopia eye defect. (TS March 2019)
Answer:
Diagram of Myopia correction :
Note : Draw the diagram using Bi Concave Lens and show the far point (M). Image should form on Retina.

Question 10.
What happen if dispersion and scattering of light do not occur? (TS March 2019)
Answer:
If dispersion does not occur in nature, then there is no rainbow formation and splitting of light into seven different colors. If there is no scattering then the oceans and sky appears to be black. The sun appears white all the time (Including Sunrise and Sunset).

Question 11.
When Mohan viewed white light through a transparent scale, he observed some colours. Predict and write the phenomenon involved in his observation. (AP SCERT: 2019-20)
Answer:

1. The phenomenon involved in his observation is dispersion of light.
2. Splitting of white light into different colours (VIBGYOR) is called dispersion.

Question 12.
A boy who is suffering from eye defect has been given a prescription as -2D. Based on the information given, answer the following questions.
a) Identify the eye defect he is suffering.
b) Write the nature and focal length of the lens. (AP SCERT: 2019-20)
Answer:
a) The boy is suffering from myopia.
b) Nature of the lens :
The lens is biconcave lens.
It is thin at the middle and thicker at the edges.
Focal length of the lens :
Given that power is – 2D

Question 13.
How does eye lens change its focal length? (AP SA-I:2019-20)
Answer:

• Eye lens changes its focal length by the ciliary muscle attached to it.
• By relaxing ciliary muslces, the focal length of the eye lens is reached its maximum value.
• By straining ciliary muscles, the focal length of the eye lens is reached its minimum value.

Question 14.
Kishore wore spectacles. When you saw through his spects the size of his eyes seemed bigger than their original size.
a) Which lens did he use?
b) Explain that defect of vision.
Answer:
a) When we saw through Kishore’s spects the size of his eyes seemed bigger than their original size. This is possible with convex lens only because magnification of the lens is greater than T.

b) The defect he suffers is hypermetropia. This is also called farsightedness.

A person who suffers with this type of defect, he can’t see the objects clearly which are placed near distance because the image is formed beyond the retina. So by using convex lens the rays can be converged on retina.

Question 15.
“God has given the gift for us to see the sunrise and sunset.” Explain the feeling of it.
Answer:

• The sun is visible two minutes before the actual sunrise and remains visible two minutes after the actual sunset.
• The actual sunrise takes place when the sun is just above the horizon.
• The actual sunset takes place when the sun is just below the horizon.

Question 16.
“Smoke coming out of coal fired chimney appears blue on a misty day.” Why?
Answer:

• On a misty day, the air has large amount of tiny particles of water droplets, dust and smoke.
• These tiny particles present in the air scatter blue colour of the white light passing through it.
• When this scattered blue light reaches our eyes the smoke appears blue.

Question 17.
“Motorists use orange light on a foggy day rather than normal white light.” Why?
Answer:

• On a foggy day, the air has large amount of water droplets.
• If a motorist uses white light, the water droplets present in the air scatter large amount of the blue light.
• This on reaching our eyes decreases visibility and hence driving becomes extremely difficult.
• Whereas orange light has longer wavelength and hence it is least scattered.

Question 18.
A rainbow viewed from an airplane may form a complete circle. Where will the shadow of the airplane appear? Explain.
Answer:

• A rainbow viewed from an airplane form a complete circle because the earth does not come along the way of the airplane and rainbow.
• A rainbow is a three dimensional cone of dispersed light it appear as a complete circle.
• The shadow of the airplane appears within the circle of the rainbow.

Question 19.
How do we see colours?
Answer:

• The retina of human eye has a large number of receptors.
• These receptors are of two types i.e., rods and cones.
• The rod cells recognise the colour of light rays, while the cones identify the intensity of light.
• It is these cone cells, which make it possible for a men to see different colours and distinguish between them.

Question 20.
Why do we use lenses in spectacles to correct defects of vision?
Answer:
The process of adjusting focal length is called “accommodation”. This process has to be done by eye itself. Sometimes the eye may gradually lose its power of accommodation. In such condition, the person cannot see the object clearly and comfortably. In this situation, we have to use lenses in spectacles to correct defects of vision.

Question 21.
What is a) far point of the eye and b) near point of the eye?
Answer:
a) The farthest point up to which the eye can see objects clearly without strain (in the eye) is called the far point of the eye. For a normal eye, the far point is at infinity.

b) The minimum distance at which objects can be seen most clearly without strain (in the eye) is called the least distance of distinct vision or the near point of the eye.

Question22.
Write the difference between “Myopia” and “Hypermetropia”.
(OR)
Distinguish between Myopia and Hypermetopia.
Answer:
The eye defect in which people cannot see at long distances but can see nearby objects clearly is called Myopia. The eye defect in which people cannot see near distant objects but can see distant objects is called hypermetropia.

Question 23.
Define the following words.
a) Prism
b) Dispersion of light
c) Scattering of light
Answer:
a) Prism :
A prism is a transparent medium separated from the surrounding medium by at least two plane surfaces which are inclined at a certain angle.

b) Dispersion of light :
The splitting of white light into different colours is called dispersion of light.

c) Scattering of light:
The process of reemission of absorbed light in all directions with different intensities by atoms or molecules is called scattering of light.

Question 24.
Define the words associated with prism with the help of figure.

1) Angle of incidence :
The angle between incident ray and normal is called angle of incidence.

2) Angle of emergence :
The angle between normal and emergent ray is called angle of emergence.

3) Normal:
Perpendicular drawn to the surface of prism.

4) Angle of deviation:
The angle between extended incident ray and emergent ray is called angle of deviation.

Question 25.
State the cause of dispersion, when white light enters a glass prism. Explain with a diagram.
Answer:

• Light is made up of different colours. Each colour travels at its own speed inside a prism.
• Due to this different colours of light bends through different angles with respect to the incident ray, as it passes through a prism.
• The red light bends the least while the violet most.
• Thus, the rays of each colour emerge along different paths and become distinct.
• It is the bond of distinct colours that we see in a spectrum.

Question 26.
What happens to the lens and the ciliary muscles when you are looking at distant objects and near objects?
Answer:
a) The ciliary muscles become relaxed and the lens becomes thin, i.e. its radius of curvature increases. So focal length of eye lens increases for distant object.

b) The ciliary muscles contract and the lens becomes thick, i.e. its radius of curvature decreases. So focal length of eye lens decreases for near objects.

Question 27.
Why does it take some time to see objects in cinema hall when we just enter the hall from bright sunlight? Explain in brief.
Answer:

• The pupil regulates and controls the amount of light entering eye.
• In bright sunlight, the size of pupil is small and when we enter the cinema hall it takes some time for the pupil to expand in size due to dim light.

Question 28.
What are the factors which influence the total angle of deviation?
Answer:

• The angle of incidence at the first surface (i).
• The angle of prism (A).
• Refractive index of the material.

Question 29.
How does eye change its focal length take place in the eyeball?
Answer:

• Eye lens changes its focal length by the ciliary muscle attached to it.
• By relaxing ciliary muslces, the focal length of the eye lens is reached its maximum value.
• By straining ciliary muscles, the focal length of the eye lens is reached its minimum value.

Question 30.
Stars twinkle while planets do not. Why?
Answer:
1) Continuously changing atmosphere refracts light from the stars by different amounts from one moment to the other, when atmosphere refracts more starlight towards us and the stars appear to be bright and when the atmosphere refracts less star-light then the stars appear to be dim.

2) However the planets are nearer to us than the stars, they appear to be comparatively bigger to us so they cannot be considered as a point source, hence no twinkling is seen.

Question 31.
How do earth and stars appear for a person who is on the moon?
Answer:

• For the person who is on the moon, the earth appears blue due to blue colour of sunlight scattered by the earth’s atmosphere reaching him.
• Stars and other heavenly bodies are seen as usual, but without twinkling.

Question 32.
Why does the sky appear dark and black to an astronaut instead of blue?
(OR)
Why does the sky appear dark to the passenger flying at high altitudes?
Answer:

• This is because there is no atmosphere containing air in the outer space to scatter light.
• Since there is no scattered light which can reach our eyes in outer space, the sky looks dark and black there.
• This is why the astronauts who go to outer space find the sky to be dark and black instead of blue.

Question 33.
A person is able to see objects clearly only when these are lying at distance between 60 cm and 250 cm from his eye. What kind of defect of vision is he suffering from?
Answer:
For a normal eye, the near point is at 25 cm and the far point is at infinity. The given person cannot see object clearly either close to the eye or far away from the eye. So he is suffering from presbyopia.

Question 34.
Write the material required in finding the refractive index of a prism.
Answer:
Materials required:
Prism, piece of white chart of size 20 x 20 cm, pencil, pins, scale and protractor.

Question 35.
Draw the graph between angle of incidence and angle of deviation.
Answer:

Question 35.
Draw the diagram of scattering of sunlight.
Answer:

Question 35.
How do you appreciate the “Eye Donor”?
Answer:
The human eye is one of the most important sense organs. Without eye we are unable to see the beautiful world. So we have to appreciate “eye donor” for his kindness to give sight to blind people.

Question 36.
How can we appreciate the working of “Iris”?
Answer:

• Iris consists of muscles which help in controlling the amount of light entering the eye through pupil.
• In case of low light the iris makes the pupil to expand and allow more light to enter the eye.
• In case of bright or excess light the iris makes the pupil to contract in order to decrease the amount of light entering the eye. The iris consists of muscles that expand and contract the pupil.

Question 37.
The power of lens is 2.0 D. Find its focal length and state what kind of lens it is.
Answer:
P = \(\frac{1}{f}\) (f in metres)
Given P = 2D
∴ f= \(\frac{1}{P}\) = \(\frac{1}{2}\) = 0. 5 m = 50 cm.
The focal length positive indicates it is a convex lens.

Question 38.
Two convex lenses of powers 1D and 2D are combined together to form a new lens. Then what is the resultant power and focal length of lens?
Answer:
P₁ = 1 D ; P₂ = 2D
Resultant power P = P₁ + P₂ = 1 + 2 = 3D
P = \(\frac{1}{f}\) (f in metres)
∴ f= \(\frac{1}{P}\) = \(\frac{1}{3}\) = 0.3333 m = 33.33 cm.

Question 40.
Figure shows the refraction of light through an equilateral prism. Incident at an angle of 30°. The ray suffers a deviation of 37°. What are the angles marked at A, e and f respectively?
Answer:

Since the prism is an equilateral prism,
A = 60°, also D = 37° and i = 30° (i.e., i₁), e = i₂ = ?
We know that i₁ + i₂ = A + D
30° + e = 60 + 37
e = 97 – 30 = 67°
Also A + f = 180°
f = 180°-60° = 120°

10th Class Physics 7th Lesson Human Eye and Colourful World 4 Marks Important Questions and Answers

Question 1.
Kavya can see distant objects clearly but cannot see objects at near distance. With what eye defect is she suffering? Draw the diagrams showing the defected eye and its correction. (AP June 2016)
Answer:
Kavya is suffering from hypermetropia.
The following diagram shows the defective eye and its correction.

Question 2.
Revathi is a front bench student. She is unable to draw the picture drawn on the blackboard. She got permission from the teacher and sat in the back row. What could be the defect that Revathi is suffering from? Draw the diagram, which shows the correction of the above defect? (AP Mareh 2017)
Answer:
Hypermetropia

Question 3.
An eye specialist suggested a + 2D lens to the person with defect in vision. Which kind of defect in vision does he have? Draw the diagrams to show the defect of vision and its correction with a suitable lens. (TS June 2017)
Answer:
Eye specialist suggested +2D lens, that is convex lens is used to correct Hypermetropia. So the person has Hypermetropia.
Deffect of Vision:

Correction :

Question 4.
How will you calculate the focal length of a biconvex lens that is used to correct the defect of Hypermetropia? Explain it mathematically. (TS March 2017)
Answer:
The person who has hypermetropia cannot see near objects. He can see the objects those are beyond near point (H). For correction of this eye defect the image of the object placed at “least distance of distinct vision (L)” should be at near point (H).
u = -25 cm; v = -d cm

here d > 25, so ‘f gets positive value.
Hence, convex lens should be used.

Question 5.
Mention the required material and chemicals for the experiment of “scattering of light.” Write the experiment procedure. (TS March 2018)
(OR)
Write the required apparatus and chemicals to show the scttering of light experimentally and write the experimental process.
(OR)
How can you demonstrate scatteing of light by an experiment?
Answer:
Aim :
To show the scattering of light.

Material required :
Beaker, sodium thiosulphate, sulphuric acid.

Procedure:

1. Take a solution of sodium-thio-sulphate (hypo) and sulphuric acid in a glass beaker.
2. Place the beaker in an open place where abundant sun light is available.
3. Watch the formation of grains of sulphur and observe the changes in the beaker.
4. We will notice that sulphur participates as the reaction is in progress. At the beginning, the grains of sulphur are smaller in size and as the reaction progress, their size increases due to precipitation.
5. Sulphur grains appear blue in colour at the beginning and slowly their colour becomes white as their size increases.
6. The reason for this is scattering of light.
7. At the beginning, the size of grains is small and almost comparable to the wavelength of blue light. Hence they appear blue in colour.
8. As the size of grains increases, their size becomes comparable to wavelength of other colours.
9. As a result, they acts as scattering centres of all colours.

Question 6.
A boy has been playing games in mobile phone and is suffering from eye defect. The doctor prescribed him to use spectacles of power – 5D. What eye defect is he suffering from? (AP SA-I:2018-19)
Draw a neat diagram which shows the correction of above eye defect.
Answer:
Doctor suggested the boy – 5D lens, that is concave lens. Concave lens is used to correct myopia. So the boy has myopia.
Defect of Vision :

Correction:

Question 7.

Answer the following questions from the above information. (TS June 2019)
i) What is the defect of vision in ‘D’ suffering from? Why does it happen?
ii) Whose defect of vision can be corrected by using Biconcave lens?
iii) Who is suffering with similar defect of vision as of ‘B’?
iv) Who among the above do not have any defect of vision?
Answer:
i) 1) The vision defect of person ‘D’ is presbyopia.
2) Presbyopia happens due to gradual weakening of ciliary muscles and diminishing flexibility of the eye lens. This effect can be seen in aged people.
ii) Person A’ defect of vision can be corrected by using Biconcave lens.
iii) Person ’C’ is suffering with similar defect of vision as of ’B’.
iv) Person ‘E’ has no defect of vision.

Question 8.
A prism causes dispersion of white light while a rectangular glass block does not. Explain.
Answer:

• In a prism the refraction of light occurs at two plane surfaces.
• The dispersion of white light occurs at the first surface of prism where its constituent colours are deviated through different angles.
• At the second surface, these split colours suffer only refraction and they get further separated.
• But in a rectangular glass block, the refraction of light takes place at the two parallel surfaces.
• At the first surface, although the white light splits into its constituent colours on refractions, but they split colours on suffering refraction at the second surface emerge out in the form of a parallel beam, which gives an impression of white light.

Question 9.
A convex lens of power 4D is placed at a distance of 40 cm from a wall. At what distance from the lens should a candle be placed so that its image is formed on the wall?
Answer:

So candle should be placed 66.66 cm from the lens.

Question 10.
Explain briefly the reason for the blue of the sky.
Answer:

• Our atmosphere contains different types of molecules and atoms.
• The reason for blue sky is the molecules N₂ and O₂.
• The sizes of these molecules are comparable to the wavelength of blue light.
• These molecules act as scattering centres for scattering of blue light.

Question 11.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:

• For a normal eye, image distance in the eye is fixed.
• This is equal to the distance of retina from the eye lens.
• When we increase the distance of the object from the eye, focal length of eye lens is changed on account of power of accommodation of the eye so as to keep the image distance constant.

Question 12.
A person is able to see objects clearly only when they are lying at distance between 60 cm and 250 cm from his eye. What kind of lenses will be required to increase his range of vision from 25 cm to infinity? Explain.
Answer:
A bi-focal lens consists of concave and convex lens of suitable focal lengths will be required to correct the defect and to increase his range of vision from 25 cm to infinity. In bi-focal lens, the upper half of the lens is concave lens which corrects distant vision and the lower half is convex which corrects near vision.

Question 13.
Discuss why sun is visible before actual sunrise and after actual sunset?
Answer:

• The sun is visible to us about 2 minutes before the actual sunrise and 2 minutes after the actual sunset because of atmospheric refraction.
• By actual sunrise we mean the actual crossing of the horizon by the sun.

• Figure shows the actual and apparent positions of the sun with respect to the horizon.
• The time difference between actual sunset and the apparent sunset is about 2 minutes.
• The apparent flattering of the sun’s disc at sunrise and sunset is also due to the same phenomenon.

Question 14.
When does the colour of sky appear black for an observer?
Answer:

• In the absence of atmosphere, there will not be any scattering of light and so light will reach our eye, i.e. the sky will appear black instead of blue at night in the absence of light.
• On the moon, since there is no atmosphere, there is no scattering of sunlight reaching the moon surface. Hence to an observer on the surface of moon, no light reaches except the light directly from sun. Thus sky will have no colour and will appear black to an observer on the moon surface. This is applicable for any planet which does not have atmosphere.
• When an astronaut goes above the atmosphere of the earth in rocket he sees the sky black.

Question 15.
Why does the colour of clouds appear black?
Answer:

• The clouds are nearer the earth surface and they contain dust particles and aggregates of water molecules of size bigger than the wavelength of visible light.
• Therefore, the dust particles and water molecules present in clouds scatter all colours of incident white light from sun to the same extent and hence when the scattered light does not reach our eye, the clouds seem black.

Question 16.
Give daily life examples of scattering of light by earth’s atmosphere.
Answer:
Some daily life effects of scattering of sunlight by earth’s atmosphere are

1. Red colour of sun at sunrise and sunset.
2. White colour of sky at noon.
3. Blue colour of sky is due to molecules N₂ and O₂.
4. Black colour of sky is due to the absence of atmosphere.
5. Use of red light for the danger signal because it is least scattered by particles due to its greater wavelength.
6. White colour of clouds is due to rise in temperature.

Question 17.
What is the relation between scattering and wavelength of light? Explain.
Answer:

• Scattering is the process of absorption and then re-emission of light energy.
• The air molecules of size smaller than the wavelength of incident light absorb the energy of incident light and then re-emit it without change in its wavelength.
• The intensity of scattered light is found to be inversely proportional to fourth power of wavelength of light.
• The wavelength of violet is least and red light is most, therefore from the incident white light, violet light is scattered the most and red light is scattered the least.

Question 18.
How can we get this (in human eye) same image distance for various positions of objects?
Answer:

• The ciliary muscle attached with eye lens helps to change the focal length of eye lens.
• When the eye is focussed on a distant object, these are relaxed. So the focal length of eye lens increases to its maximum value.
• F_(max) – 2.5 cm,u = oo, v = 2.5 cm. The parallel rays coming from a distant object are focussed on the retina with 2.5 cm image distance.
• When the eye is focussed on a nearer object the muscles are strained. So the focal length of the eye lens decreases its minimum value.
• F_(min)= 2.27 cm, u = 25 cm, v = 2.5 cm. The rays from an object (u = 25 cm) at L (point of least distance of distinct vision) are focussed on the retina with 2.5 cm image distance.

Question 19.
Does eye lens form a real image or virtual image? Explain it.
Answer:

• Eye lens forms a real image.
• The light that enters the eye forms an image on the retina.

• The image is obtained on a screen (retina) and it is an inverted image.
• So, we can say eye lens forms a real image.

Question 20.
How does the image formed on retina help us to perceive the object without change its shape, size and colour?
Answer:

• The eye-lens forms a real and inverted image of an object on the retina.
• The retina is a delicate membrane which contains about 125 million receptors called ‘rods’ and ‘cones’ which receive the light signals.
• Rods identify the colour.
• Cones identify the intensity of light.
• These signals are transmitted to the brain through about 1 million optic nerve fibres.
• The brain interprets these signals and finally processes the information so that we perceive the object in terms of its shape, size and colour.

Question 21.
How do you prove that a prism does not produce colours itself?
Answer:

• A white light from a slit ‘S’ is made to pass through prism P which forms spectrum on a white screen AB.
• A narrow slit H is made on the screen AB, parallel to slit S to allow the light of particular colour to pass through it.
• The light of a particular colour is made to fall on the second prism Q placed with its base in opposite direction to that of the prism P.
• The light after passing through the second prism Q is received on another white screen M.
• It is observed that the colour of light obtained on the screen M is same as that of the light incident on the second prism Q through the slit H.

Question 22.
Draw the structure of human eye and explain its parts.
Answer:

Question 23.
What is the part indicated by an arrow mark? What is its working function?
Answer:

• The part indicated by the arrow mark is ciliary muscle.
• Ciliary muscles help to change the focal length of the eye lens.
• When it relaxes the focal length of the eye lens is maximum.
• When it strains, the focal length of the eye lens is minimum.
• In this way the ciliary muscles to which eye lens is attached help to give us clear vision.

Question 24.
Two observers standing apart from each other do not see the “same” rainbow. Explain.
Answer:

• All the rain drops that disburse the light to form rainbow lie within a cone of semi vertical angle 40° to 42°.
• If two observers are standing at a distance apart, they will observe rainbow at different parts on the surface of the cone.
• So the portion of the rainbow observed by an observer depends on the position of the observer.
• Two different observers will form two different cones with the observer standing at the vertex of the cone, therefore rainbow seen by them will be different.

Question 25.
A prism with an angle A = 60° produces an angle of minimum deviation of 30°. Find the refractive index of material of the prism.
Answer:
Given, A = 60° and D = 30°

Question 26.
A person cannot see the objects distinctly, when placed at a distance less than 50 cm. Calculate the power and nature of the lens he should be using to see clearly the object placed at a distance of 25 cm from his eyes.
Answer:

Question 27.
A person cannot see the objects distinctly, when placed beyond 2 m.
Calculate the power and nature of the lens he should be using to see the distant objects clearly.
Answer:
For myopia the focal length = – far point distance = – 2m
Power = \(\frac{1}{\mathrm{f}}=\frac{1}{-2}\) = 0.5 D.

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 4 Acids, Bases and Salts.

AP State Syllabus SSC 10th Class Chemistry Important Questions 4th Acids, Bases and Salts

10th Class Chemistry 4th Lesson Acids, Bases and Salts 1 Mark Important Questions and Answers

Question 1.
Take some water in a test tube and add concentrated H₂SO₄ to it. Shake the test tube well. If you touch the bottom of the test tube, you feel it as hot. Now, instead of H₂SO₄, if you add NaOH pellets to water in another test tube and touch the bottom, what do you observe? (TS June 2015)
Answer:
The bottom of test tube is also hot because reactions of acids, bases with water are exothermic reactions.

Question 2.
What happens if the copper sulphate crystals taken into dry test tube are heated? (TS June 2016)
Answer:

• When copper sulphate crystals are heated, water present in crystals is evaporated and the salt turns white.
• Evaporated water appears as droplets on the walls of the test tube.
• Blue coloured copper sulphate (CuSO₄ 5H₂O) is turned into white colour because 5H20 molecules are evaporated from crystals.

Question 3.
Why does the soil of agricultural lands get tested for pH?
Answer:
Plants require a specific pH range for their healthy growth. So, finding pH of a soil suggested the farmers to treat the fields with acidic or basic substances to maintain the required pH range.

Question 4.
Write the molecular formulae of common salt and baking soda which are widely used at home. (TS June 2017)
Answer:
Common Salt: NaCl; Backing soda : NaHCO₃

Question 5.
Mention the precautions to take while conducting an experiment to prove acids produce ions only in aqueous solutions. (TS June 2018)
Answer:

• Testing of the evolved gas by using dry litmus paper first. Then with wet litmus paper.
• Use gaurd tube containing calcium chloride.

Question 6.
What are antacids?
Antacids are mild alkalies. These are used for getting relief from acidity and indigestion and sometimes even headache. When taken orally, it reacts with hydrochloric acid present in the stomach and reduces its strength by consuming some of it.
Ex: Milk of magnesia.

Question 7.
Tap water conducts electricity whereas distilled water does not. Why?
Answer:
Tap water contains some impurities in the form of salts. Due to presence of salts, it conducts electricity. Distilled water is free from all kinds of salts and hence does not conduct electricity.

Question 8.
What do you mean by dilution of an acid or base? Why is it done?
Answer:
Dilution of an acid or base means mixing an acid or base with water. This is done to decrease the concentration of ions per unit volume. In this way the acid or the base is said to be diluted.

Question 9.
What is a universal indicator?
Answer:
An indicator which passes through a series of colour changes over a wide range of H₃O⁺ ions concentration is called universal indicator.

Question 10
What is tooth decay?
Answer:
Tooth enamel is chemically calcium phosphate Ca₃(PO₄)₂. It starts corroding when pH falls below 5.5. Food particles left in the mouth degrade to produce acid which lower the pH of the mouth. This is called tooth decay.

Question 11.
Define Alkalis and give some examples.
Answer:
Alkalis : An alkali is a base that dissolves in water.

Examples :
i) Sodium Hydroxide (NaOH),
ii) Potassium Hydroxide (KOH),
iii) Magnesium Hydroxide (Mg(OH)₂)

Question 12.
Why should we not taste or touch alkalis?
Answer:
We should not taste or touch alkali. Because they are corrosive.

Question 13.
Salts conduct electricity. Why?
Answer:
Salts contains ions. So they conduct electricity.

Question 14.
Why are calcium sulphate hemihydrates called Plaster of Paris?
Answer:
Calcium sulphate hemihydrates are used as plaster for supporting fractured bones in the right position. So, it is called Plaster of Paris.

Question 15.
Why does an aqueous solution of acid conduct electricity?
Answer:
An aqueous solution of acid liberates H⁺ ions. This makes the aqueous solution of acid to conduct electricity.

Question 16.
How is the concentration of hydronium ions (H₃O⁺) affected when a solution of an acid is diluted?
Answer:
When a solution of an acid is diluted, concentration of H₃O⁺ ions decreases.

Question 17.
What is pH?
A. pH is a scale for measuring hydrogen ion concentration in a solution. It is the negative logarithm of H⁺ concentration.
pH = – log [H⁺].

Question 18.
How is pH of a solution related to the [H₃O⁺] of that solution?
Answer:
The presence of H₃O⁺ ions indicate us whether it is a strong acid or weak acid.

Question 19.
There are two solutions of pH values 6 and 8. Which solution has more hydrogen ion concentration? Which of this is acidic and which one is basic?
Answer:

• The solution whose pH value 6 is acid and has more hydrogen ion concentration.
• The solution of pH value 8 is basic and has less hydrogen ion concentration.

Question 20.
Can you give example for use of olfactory indicators in daily life?
Answer:
Examples of olfactory indicators : Onion, vanilla extract.

Question 21.
How do acids neutralize bases?
(OR)
How do acids and bases react with each other?
Answer:
According to Arrhenius theory acids produce H⁺ ions and bases produce OH^– ions in aqueous media.
The combination of H⁺ and OH^– ions is called ‘neutralization’.
Thus acids neutralize bases.

Question 22.
How strong are acids and base solutions?
Answer:
The acids of pH value as much less as possible have more concentration [pH < 7], Basic nature increases as pH value increases.

Question 23.
What do you say about salts of both weak acid and weak base?
Answer:
The pH of aqueous solutions of salt obtained from both weak acid and weak base is nearly 7.

Question 24.
Which base is used for removing permanent hardness of water?
Answer:
Sodium carbonate is used for removing permanent hardness of water.

Question 25.
Name two antacids used to get rid of our indigestion problem.
Answer:
Magnesium hydroxide and a mild base (baking soda).

Question 26.
Under what soil conditions would a farmer would treat the soil of his fields with quicklime (calcium hydroxide) or calcium carbonate?
Answer:
When the field has acidic nature, the farmer uses quicklime or calcium carbonate to neutralize it.

Question 27.
Write the formulas of Gypsum and Plaster of Paris.
Answer:
The formulae of Gypsum is CaSO₄ . 2H₂O and Plaster of Paris is CaSO₄ . ½H₂O.

Question 28.
Write any two Acid Base indicators.
Answer:

1. Methyl orange
2. Phenolphthalein.

Question 29.
Which salt is used in the manufacture of borax?
Answer:
Washing soda (Na₂CO₃.10H₂O)

Question 30.
What is family of salts? Give examples.
Answer:
Salt having the same positive or negative radical is called family of salts.
Eg : Family of sodium salts : NaCl, Na₂SO₄
Family of chloride salts : NaCl, KCl.

Question 31.
‘A’ is a substance which is acidic and it is added in solution to preserve pickles. What is A and what is the name given to its dilute solution?
Answer:
‘A’ is acetic acid and its dilute solution is called vinegar.

Question 32.
What are the chemical names of the following?
1) Baking soda
2) Gypsum
Answer:

• The chemical name of baking soda is sodium hydrogen carbonate (NaHCO₃).
• The chemical name of gypsum is calcium sulphate dihydrate (CaSO₄ • 2 H₂O).

Question 33.
Write the water of crystallisation of following compound.
a) Hydrous copper sulphate
b) Washing soda
c) Gypsum
d) Plaster of Paris
Answer:

Compound	Formula	Water of crystallisation
1) Hydrous copper sulphate	CuSO4 . 5H2O	5
2) Washing soda	Na2CO3 . 10 H2O	10
3) Gypsum	CaSO4 . 2 H2O	2
4) Plaster of Paris	CaSO4 . ½H2O	½

Question 34.
Why don’t we use a strong base like NaOH as antacid?
Answer:
Strong bases like potassium hydroxide (KOH), sodium hydroxide (NaOH) are corrosive in nature. So they can harm the internal organs. Therefore we should not use them as antacid.

Question 35.
What do you mean by HsO⁺ ion?
Answer:
Hydrogen ions cannot exist as base ions. They associate with water molecules and exist as hydrated ions with each H⁺ attached by 4 to 6 water molecules. For this we represent H⁺ as hydronium ion, H³O⁺.
H⁺ + H₂O → H₃O⁺

Question 36.
Which indicator is useful at all pH? Why?
Answer:
Universal indicator is useful to test solutions of all pH because it gives different colours at different pH range.

Question 37.
Which substance is useful in removing permanent hardness of water?
Answer:
The permanent hardness of water is due to chloride and sulphate salts of magnesium and calcium, which can be removed by adding washing soda.

Question 38.
Given two examples for strongest bases.
Answer:
Sodium hydroxide – NaOH
Potassium hydroxide – KOH

Question 39.
What is the confirmation test for hydrous and anhydrous salt?
Answer:

• On heating hydrous salt in a test tube it will form water droplets on the sides of test tube.
• On heating anhydrous salt in a test tube it will not form water droplets on the sides of test tube.

Question 40.
P.O.P, cement, calcium chloride should be stored in moisture proof containers. Why?
Answer:

• P.O.P, cement and calcium chloride react with moisture (H₂O) in the atmosphere and set into hard solid masses.
• To avoid availability of moisture they should be stored in moisture proof containers.

Question 41.
Give some examples for hydrous and anhydrous salts.
Eg : For hydrous salts :

1. CuSO₄ . 5H₂O
2. Na₂CO₃ . 10H₂O
3. CaSO₄ . 2H₂O

Eg : for anhydrous salts :

1. NaCl
2. MgCl₂
3. Na₂CO₃

Question 42.
How are bitter and sour taste substances tested without testing?
Answer:

• Sour taste substances turn blue litmus to red.
• Bitter taste substances turn red litmus to blue. By these tests we can test them as acids and bases.

Question 43.
Do the metallic oxides react with acids?
Answer:
Yes, metallic oxides are basic in nature. They react with acids and form salt and water.

Question 44.
Does non-metallic oxide react with base?
Answer:
Yes, non-metallic oxide is acidic in nature. It reacts with base and forms salt and water.

Question 45.
Why does dry HCl gas not change the colour of the dry litmus paper?
(OR)
Prove that dry HCl gas is not an acid but HCl aqueous solution is an acid using an activity.
Answer:

• Dry hydrogen chloride gas is not an acid. Hence it can’t turn blue litmus into red.
• Hydrochloric acid is an aqueous solution. Hence it can turn blue litmus into red.

Question 46.
Do you know that the atmosphere of Venus is made up of thick white and yellowish clouds of sulphuric acid? Do you think life can exist on this planet?
Answer:

1. No, it is not possible.
2. When pH value decreases, the survival of living organisms becomes difficult.
3. Hence there is not any possibility of life on Venus.

Question 47.
Why do acids not show acidic behaviour in the absence of water?
Answer:
Acids don’t show acidic behaviour in the absence of water as H⁺ ions are absent in them.

Question 48.
How is the concentration of hydroxide ions (OH^–) affected when excess base is dissolved in a solution of sodium hydroxide?
Answer:
When a base like NaOH (Sodium hydroxide) is dissolved in water, it liberates (OH^–) ions.
Equation :

OH^– ion concentration increases.

Question 49.
How does the nature of the solution change with change in concentration of H⁺_(aq) ions?
Answer:
The concentration of H⁺ ions is responsible for the acidic nature of a substance.
If [H⁺] > 1 × 10^-7 mol/lit the solution is acidic.
If [H⁺] < 1.0 × 10^-7 mol/lit the solution is basic.

Question 50.
Do basic solutions also have H⁺_(aq) ions? If yes, then why are these basic?
Answer:
Yes. Basic solutions have OH^–_(aq) ions and bases have less number of H₃O⁺ ions. H⁺_(aq) ions are less in base. In basic solution [OH^–] > [H⁺].

Question 51.
What do acids have in common?
Answer:
(Acids have sour taste and conduct electricity. They release H₂ (Hydrogen) gas on reacting with metals). All acids have H⁺_(aq) ions.

Question 52.
What do bases have in common?
Answer:
Bases are slippery to touch and bitter to taste. All bases have OH^–_(aq) ions.

Question 53.
Why does pure acetic acid not turn blue litmus to red?
Answer:
Pure acetic acid is a weak acid so it does not have sufficient H⁺_(aq) ion to change the colour of blue litmus to red.

Question 54.
What will happen if the pH value in your body increases?
Answer:
It affects our digestion system.

Question 55.
A student checked pH of a salt solution and found that its pH is more than 7. How is that type of salt formed?
Answer:

• When a strong base reacts with weak acid then the solution is basic in nature. So its pH is more than 7.
• For example when acetic acid reacts with sodium hydroxide the salt formed has basic nature.

Question 56.
Explain the procedure that you follow to reduce water from a given salt.
Answer:
Procedure to reduce water from a given salt :

1. Take a boiling test tube.
2. Drop given salt in the test tube.
3. Heat the test tube gently.
4. Water from salt evaporates.
5. In this way we can reduce the water from salt.

Question 57.
Write the observations, when the hydrated salt or unhydrated salt is heated.
Answer:

• When hydrated salt is heated water droplets form inside the walls of test tube and sometimes blue or green colour salt turns into white colour.
• When unhydrated salt is heated it does not form water droplets inside the test tube walls and colour also does not change.

Question 58.
On heating the hydrated salt it loses water molecules present in it. To show this what are the equipment required?
Answer:
1) On heating the hydrated salt it loses water molecules present in it.
2) To show this the given equipment are required

1. Boiling tube
2. Test tube holder
3. Burner

Question 59.
Try to collect the information to reasons for calling calcium sulphate hemihydrates as Plaster of Paris (POP).
Answer:

• Gypsum plaster (or) Plaster of Paris (POP) is produced by heating gypsum to about 300°F.
• A large gypsum deposit is found at Montmartre in Paris (France).
• This gave the name Plaster of Paris to calcium sulphate hemihydrates.
• The term plaster can refer to gypsum.

Question 60.
Is the substance present in antacid tablet acidic or basis?
Answer:
The substance present antacid is weakly basic.

Question 61.
Give pH of neutral, acid and base.
Answer:

Nature of substance	pH range
Neutral	7
Acid	0 – 7
Base	7 – 14

Question 62.
Which nature of Plaster of Paris makes its importance? Appreciate it.
Answer:
Plaster of Paris is a white powder. It is very soft and can be used to make toys, materials for decoration and to make surfaces smooth.

But on mixing with water, it changes to a hard solid mass (Gypsum). This is the important character of Plaster of Paris (POP).

Question 63.
What is acid rain? How does it affect our aquatic life?
Answer:
When the pH of rain water is less than 5.6 it is called acid rain. When acid rain flows into the rivers, it lowers the pH of the river water. Since our body works within a narrow pH range close to 7 the survival of aquatic life in river water mixed with rain water becomes difficult.

Question 64.
Why are pickles and sour substances not kept in brass and copper vessels?
Answer:
Pickles and sour substances contain acidic nature which may react with brass and copper vessels to produce toxic substances.
So, we don’t keep them in brass arid copper vessels.

Question 65.
Can you suggest some examples of use of pH in everyday life?
Answer:
Uses of pH in everyday life :

1. pH value helps us to identify acids, bases and neutrals.
2. If pH value is less in our mouth, it leads to tooth decay. We can find it as the reason for our tooth decay.
3. pH value helps us to know about acid rain.

Question 66.
Write any two uses of Bleaching powder.
Answer:

• It is used for disinfecting drinking water to make it free of germs.
• It is used as a reagent in the preparation of chloroform.

10th Class Chemistry 4th Lesson Acids, Bases and Salts 2 Marks Important Questions and Answers

Question 1.
What value of pH in the mouth leads to tooth decay? Why? (TS June 2015)
Answer:

• Tooth decay starts when the pH of the mouth is lower than 5.5.
• Tooth enamel, made of calcium phosphate is the hardest substance in the body.
• It does not dissolve in water, but is corroded when the pH in the mouth is below 5.5.
• Bacteria present in the mouth produce acids by degradation of sugar and food particles remaining in the mouth.

Preventions :

1. Clean the mouth after eating food.
2. Using tooth pastes, which are generally basic neutralize the excess acid and pre¬vent tooth decay.

Question 2.
Equal lengths of Magnesium ribbons are taken in two test-tubes X and Y. Hydro¬chloric acid is added to test-tube X and Acetic acid is added to test-tube Y. In which test-tube, the reaction will be more vigorous? Why? (TS March 2015)
Answer:
The speed of the reactions is higher in X test tube than Y test tube.

Reason :
Due to strong acidic nature, Hydrochloric acid reacts very fast with magnesium ribbon.

Question 3.
Name the four chemicals that are obtained from common salt and write their molecular formulae. (TS March 2015)
Answer:
Chemicals that can be obtained from common salt are

1. Sodium Hydroxide – NaOH
2. Baking soda / Cooking soda / Caustic soda / Sodium bicarbonate / Sodium Hydrogen carbonate. – NaHCOv
3. Washing soda / Sodium carbonate – Na₂CO₃ 10H₂O
4. Bleaching powder / Calcium Oxy Chloride – CaOCl₂

Question 4.
Observe the information given in the table and answer the questions given below the table. (TS March 2017)

Substance (in aqueous solution)	Colour change with Blue Litmus	Colour change with Red Litmus
A	Red	No change
B	No change	Blue
C	No change	No change

i) Which one of them may be the neutral salt among A, B, C?
ii) What may happen when some drops of phenolphthalein is added to the substance B?
Answer:
i) C
ii) Pink Colour

Question 5.
Why do we use antacids? Write it’s nature. (TS March 2018)
Answer:
Pain and irritation will be caused in stomach during the acidity problem/indigestion problem. Antacids used to neutralize the excess acid in the stomach and gives relief from acidity Antacids are basic in nature.

Question 6.
Which product will form when CaO is dissolved in water? How do you find the nature of product? (TS March 2018)
Answer:
CaO reacts with water and gives calcium hydroxide [Ca(OH₂)]. The nature of the calcium hydroxide will be tested with red litmus paper or pH paper.

Calcium Hydroxide turns red litmus into blue. Thus we can say that ca(OH)₂ is basic in nature.

Ca(OH)₂ shows pH value more than 7. Thus we can say that Ca(OH)₂ is basic in nature.

Question 7.
How do you test the nature of the solution formed by dissolving CaO in water? What is the nature of the solution? (TS June 2019)
Answer:

• The solution formed by dissolving CaO in water is tested with red litmus paper, it turns into blue colour, (or) It is tested with methyl orange, it turns into yellow in colour.
• The solution of CaO and water is basic in nature.

Question 8.
Write the experimental procedure to test carbon dioxide gas. (AP SCERT: 2019-20)
Answer:

1. Pass the CO₂ gas through lime water [Ca(OH)₂].
2. The lime water appears as milky white.
3. The reaction is Ca(OH)₂ + CO₂ → CaCO₃ ↓ + H₂O.
4. The milky white is caused by CaCO₃.

Question 9.
Write two reactions of acids with carbonates and metal hydrogen carbonates. (AP SA-I : 2019-20)
Answer:
Reactions :

Question 10.
What do acids have in common?
Answer:
Common characteristics of acids :

1. Similar chemical properties.
2. Acids generate hydrogen gas on reacting with metals.
3. Hydrogen is common to acids.
4. Acids are sour in taste and turn blue to red when react with bases form salt and water.

Question 11.
What do bases have in common?
Answer:
Common characteristics of bases :

1. Bitter in taste.
2. Soapy in nature.
3. Turn red to blue colour.
4. On heating decompose into metal oxides and water.
5. React with acids to form salt and water.
6. Produce OH^– ions in aqueous solution.

Question 12.
How is bleaching powder produced?
Production of bleaching powder :
Bleaching powder is produced by the action of chlorine on dry slaked lime (Ca(OH)₂).
Equation :

Question 13.
What can you conclude about the ideal soil pH for the growth of plants in your region?
Answer:
1) Soil is considered the ‘skin of the earth’. The soil pH plays a vital role in the growth of a plant and it influences plant nutrition.
2) Soil pH strongly affects the nutrients required for the plant growth.
3) The nutrients may be stored on soil colloids, and live or dead organic matter, but may not be accessible to plants due to extremes of pH.

Conclusion :
For optimum plant growth, the generalized content of soil components by volume should be roughly 50% solids (45% mineral & 5% organic matter), and 50% voids of which half is occupied by water and half by gas.

Question 14.
How can you prepare turmeric indicator? What is the use of it?
Answer:
i) Turmeric indicator is prepared from turmeric.
ii) It has red colour in basic solution.

Question 15.
Name two salts and write their formulae which possess water of crystallization.
Answer:

1. Hydrous copper sulphate. Its formula is CuSO₄ . 5H₂O
2. Gypsum. Its formula is CaSO₄ . 2H₂O

Question 16.
What is neutralization reaction? Give two examples.
Answer:
When an acid reacts with base it forms salt and water. This reaction is called neutralisation reaction.
e.g.: NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)
H₂SO_4(aq) + 2Na0H_(aq) → Na₂SO_4(aq) + 2H₂O_(l)

Question 17.
All alkalis are bases but all bases are not alkalis. Do you agree with the statement? If yes, why?
Answer:
Yes, I agree with the statement. Because alkalis are those bases which are soluble in water. So all alkalis are bases but all bases are not alkalis.

Question 18.
Why are solutions of acids, bases and salts good conductors of electricity?
Answer:
For passage of electricity through a material or substance charged particles are required. In metals charged particles are electrons whereas in solutions ions are charged particles which carry electrical energy. Solutions of acids, bases and salts undergo ionisation and produce ions. So they are good conductors of electricity.

Question 19.
What is strength of acid ? What are the factors that influence strength of acid?
Answer:
The extent which an acid undergoes ionisation is called strength of acid.
Factors influence strength of acid :

1. Degree of ionisation.
2. Concentration of hydronium ions produced by acid.

Question 20.
Why are organic acids weak acids when compared with mineral acids?
Answer:

• Strength of acid depends on extent of ionisation.
• Organic acids do not undergo 100% ionisation. Their ionisation is less than 30%. There is equilibrium between ionised and unionised molecules whereas mineral acids undergo complete ionisation.
• So mineral acids behave like strong acids when compared with organic acids.

Question 21.
“Acids do not contain OH^– ions”. Do you agree with this statement? If not, why?
Answer:
No. I do not agree with the statement because all acids also contain OH^– ions but in acid solutions, H⁺ ions are more than OH^– whereas in bases OH^– ions are more than H⁺.

Question 22.

If B is calcium chloride, what are A, C, D and E?
Answer:
A is calcium carbonate or calcium hydrogen carbonate.
C is water and D is carbon dioxide.
E is calcium carbonate and F is water.

Question 23.
Some salts are given below. Classify them into hydrous and anhydrous salts. Sodium carbonate, Sodium chloride, Sodium hydrogen carbonate, Copper sulphate, Hypo, Magnesium Sulphate (epsum salt)
Answer:
Hydrous salts :

1. Hypo (Na₂S₂O₃ • 2H₂O)
2. Epsum (MgSO₄ • 7H₂O)
3. Copper sulphate (CuSO₄ • 5H₂O)

Anhydrous salts :

1. Sodium chloride (NaCl)
2. Sodium carbonate (Na₂CO₃)
3. Sodium hydrogen Carbonate (NaHCO₃)

Question 24.
If someone in the family is suffering from a problem of acidity, which of the following would you suggest as a remedy : lemon juice, vinegar or baking soda solution? Which property do you think of while suggesting the remedy?
Answer:

• I suggest baking soda solution. As acidity can be neutralized by baking soda solution, we can use it.
• Neutralizing property of baking soda solution.

Question 25.
Why are curd and sour substances not kept in copper vessels?
Answer:
Curd and sour substances contain acids which react with copper vessels and form poisonous substances. So curd and sour substances should not be kept in copper vessels.

Question 26.
Which gas is liberated when acids react with metals? Give one example.
Answer:
When acids react with active metals they release hydrogen gas.
Zn_(s) + 2HCl_(aq) → Zncl_2(aq) + H₂ ↑_(g)

Question 27.
Why should pickles not be stored in metallic containers?
Answer:
Pickles contain acids which react with metallic containers and form poisonous substance. So they are kept in plastic containers.

Question 28.
Solution x turned blue litmus red and Solution y turned red litmus blue.
a) What products could be formed when x and y are mixed?
b) Which gas is released when we put magnesium pieces in solution x?
c) Will any chemical reaction take place when zinc pieces are put in solution y?
d) Which of the above solutions contain more hydrogen ions?
Answer:
Given solution Y turned blue litmus into red so, Y is an acid.
Given solution ‘y’ turned red litmus into blue so, ‘y’ is a base.
a) The reaction of an acid (x) with a base to give a salt and water.
b) When we put magnesium pieces in solution releases hydrogen gas.
c) When zinc pieces are put in solution y, a chemical reaction will take place there.
d) Acids contain more H+ ions in the given solutions, Y has more H+ ions because it is an acid.

Question 29.
Acid should be added to water but not water to the acid. Why?
Answer:

• The dissolving of an acid or base in water is an highly exothermic process. Care must be taken while mixing concentrated HNO₃ or concentrated H₂SO₄ with water.
• The acid must always be added slowly to water with constant stirring.
• If water is added to a concentrated acid, the heat generated may cause the mixture to splash out and cause bums.
• The glass container may also break due to excessive local heating.

Question 30.
Explain the procedure to confirm the given salt is a hydrous or anhydrous.
Answer:

1. Take given salt in a test tube
2. Observe the colour of salt
3. Heat the test tube gently
4. Observe the colour of salt and also moisture (droplets) inside of the test tube walls.
5. If its colour changes or forms water droplets, it is hydrous salt.
6. Otherwise, it is anhydrous salt.

Question 31.
Categorize the following as acids, bases, and salts :
Lemon juice, salt water, soap water, tamarind juice, surf water, lime water.
Answer:
Acids :

1. Lemon juice
2. Tamarind juice

Bases :

1. Soap water
2. Surf water
3. Lime water

Salts :

1. Salt water

Question 32.
Classify the following salts as family of salts having same cation or anion and prepare a table.
Potassium sulphate, Sodium sulphate, Calcium sulphate, Magnesium sulphate, Copper sulphate, Sodium chloride, Sodium nitrate, Sodium carbonate and Ammonium chloride.
Answer:

Question 33.
Observe the table and answer the following questions.

Solutions	pH value
Blood	7.3
Pure water	7.0
Gastric fluid	1.2
Sodium hydroxide	13

1) Which of the solutions among these is a strongest base?
2) Which body fluid has slightly basic nature?
3) What is the nature of pure water?
4) Which body fluid is strongest acid?
Answer:

1. Sodium hydroxide because its pH is 13.
2. Blood because its pH is 7.3.
3. Pure water is neutral in nature because its pH is 7.
4. Gastric juice because its pH is 1.2.

Question 34.
The diagram given below shows the removal of water crystallisation. Find error in the diagram.

Answer:
Error in the diagram

1. Test tube is placed towards observer. It causes bums on his hands.
2. So, test tube should be placed away from the observer.

Question 35.
What are the uses of Plaster of Paris?
Answer:

• The substance which doctors use as plaster for supporting fractured bones in the right position.
• Plaster of Paris is used for making toys, materials for decoration and for making surfaces smooth.

Question 36.
What are the applications of pH in daily life?
Answer:
1) In medical science :
The pH values of urine and blood are taken for diagnosis of various diseases.

2) In dairies :
Milk has pH of 6.6. A change in the pH of milk indicates that milk has turned sour.

3) In agriculture :
For better growth of crops the pH of the soil is regularly tested.
For examples :

1. Citrus fruits require slightly alkaline soil.
2. Rice requires acidic medium.
3. Sugarcane requires neutral soil.

4) In technology:
Organic and biochemical reactions are carried out under control pH.

10th Class Chemistry 4th Lesson Acids, Bases and Salts 4 Marks Important Questions and Answers

Question 1.
Draw a neat diagram showing a base solution in water conducts electricity. Why the solution of sugar/glucose in water do not conduct electricity? (AP March 2017)
Answer:

The solution of sugar/glucose in water do not conduct electricity because there is no H⁺ ions in the solution.

Question 2.
Explain an activity to show the water of crystallisation in CuSO₄ • 5H₂O. (AP June 2018)
Answer:

• Take a few crystals of copper sulphate in a dry test tube and heat the test tube.
• We observe water droplets on the walls of the test tube and salt turns white.
• Add 2 – 3 drops of water on the sample of copper sulphate obtained after heating.
• We observe the blue colour of copper sulphate crystals is restored.

Question 3.
Read the information given in the table and answer the following questions. (TS March 2016)

a) List out the acids in the above table.
Answer:
The acids are HCl and lemon juice.

b) What is the nature of the solution which gives pink colour with Phenolphthalene solution?
Answer:
The nature of the solution which turns pink colour with phenolphthalene solution is basic.

c) List out the neutral solutions in the above table.
Answer:
The neutral solutions are distilled water and NaCl.

d) Name the strongest acid and the strongest base among the given solutions.
Answer:
The strongest acid is HCl and the strongest base is NaOH.

Question 4.
Observe the following table and answer the questions given below. (TS June 2o17)
The table contains the aqueous solutions of different substances with the same concentrations and their respective pH values.

i) Which one of the above acid solutions is the weakest acid? Give a reason.
Answer:
Weakest acid is ‘C’. Because its pH value is less than 7 and it is nearer to 7.

ii) Which one of the above solutions is the strongest base? Give a reason.
Answer:
Strongest base is ‘D’. Because it’s pH value is near to 14.

iii) Which of the above two produce maximum heat when they react ? What does that heat energy called?
Answer:
B, D produce maximum heat when they react. This heat energy is known as neutralization energy.

iv) Which one of the above solutions has the pH equal to that of the distilled water? What is the name given to solutions of that pH value?
Answer:
‘G’ has the pH equal to that of the distilled water. These type of solutions are known as neutral solution.

Question 5.
List out the materials required to test whether the solutions of given acids and bases contain ions or not. Explain the procedure of the experiment. (TS March 2017)
Answer:
Required Materials :
Beaker, Bulb, Graphite rods, connecting wires, 230 V AC current, water, different acids, bases.

Experimental Procedure :

1. Connect the two connecting wires to the graphite rods.
2. Keep the graphite rods into the beaker, take care that two graphite rods do not touch each other.
3. Arrange a bulb in the circuit.
4. Pour dilute acid into the beaker.
5. Connect the ends of the connectors to 230 V AC.
6. In this way, change the acid / base and do the experiment.

The bulb glows in the experiment when the beaker contains acid or base. Hence, when the bulb glows we can say that acid or base contain ions.

Question 6.
List out the material for the experiment “when Hydrochloric acid reacts with NaHCO₃ and evolves CO₂“. Write the experimental procedure. (TS March 2018)
Answer:
Required material : Stand, test tubes, delivery tube, thistle funnel, two hole rubber corks, Ca(OH)₂, NaHCO₃, HCl.

Experimental procedure :

1. Take NaHCO₃ in a test tube and fix two hole cork to the test tube.
2. Fix thistle funnel in one hole of cork and insert delivery tube in the second hole of the cork. Insert the second end of the delivery tube in the other test tube which is containing Ca(OH)₂/lime water.
3. Pour dil. HCl into the test tube using thistle funnel.
4. Due to chemical reaction, gas is evolved and pass into the Ca(OH)₂ through delivery tube. It turns in to milky. We can conclude that it is CO₂ gas.

Question 7.
Prepare a table based on the colour responses of acid, base and salt with indicators such as indicators. (AP SA-I:2018-19)
Answer:

Question 8.
Draw universal pH value indicator and identify different substances. (AP SA-I : 2019-20)
(OR)
Draw a neat diagram showing variation of pH with the chage in concentration of H⁺_(aq) ion and OH^–_(aq) ions.
Answer:

Variation of pH with the change in concentration of H⁺_(aq) ions and OH^–_(aq) ions.

Question 9.

Answer the following questions by using above information. (TS June 2019)
1) Which of the above is neutral solution?
2) Which of the above is used to neutralize the acidity in stomach?
3) Which is the strong acid among the above solutions?
4) What is the colour of Phenolphthalein indicator in NaOH solution?
Answer:

1. Distilled water
2. Milk of Magnesia
3. Gastric juice
4. Pink

Question 10.
If the pH values of solutions X, Y and Z are 13, 6 and 2 respectively, then
a) Which solution is a strong acid? Why?
b) Which solution contains ions along with molecules of solution?
c) Which solution is a strong base? Why?
d) Does the pH value of a solution increase or decrease when a base is added to it? Why?
Answer:
The strength of an acid (or) an alkali can be tested by using pH value of a solution. If the value of a pH of a solution is less, then that solution exhibits acidic nature.

If the value of a pH of a solution is more, then that solution exhibits basic nature.
pH value of a solution “X” is 13
pH value of a solution “Y” is 6
pH value of a solution “Z” is 2

a) Solution ‘Z’ is strong acid because its pH is 2.
b) Among given solutions, solution X is weakest acid. Weak solution contains ions along with molecules of solution. So X exhibits like this character.
c) Solution X is strong base. Because its pH is 13.
d) If base is added to solution ‘Z’, then its pH will increase.

Question 11.
Distinguish between acids and bases.
Answer:

Acids	Bases
1) They are sour to taste.	1) They are better to taste and soapy to touch.
2) When non-metallic oxides dissolved in water they form acids.	2) When metallic oxides dissolved in water they form bases.
3) They react with bases to form salt and water.	3) They react with acids to form salt and water.
4) They produce aqueous H+ ions.	4) They produce aqueous OH- ions.
5) They turn blue litmus into red.	5) They turn red litmus into blue.
6) They turn methyl orange indicator to red.	6) They turn methyl orange indicator to yellow.
7) The turn phenolphthalein indicator to colourless.	7) They turn phenolphthalien indicator to pink.

Question 12.
Explain chlor-alkali process.
Answer:
When electricity is passed through an aqueous solution of sodium chloride, it decomposes to form sodium hydroxide. The process is called chloralkali process because of the products formed chlor for chlorine and alkali for sodium hydroxide.
2 NaCl (aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g)

Chlorine gas is given off at the anode and hydrogen gas at the cathode and sodium hydroxide is formed near the cathode.

Question 13.
Define the following. Give one example for each.
a) Strong acid
b) Strong base
c) Weak acid
d) Weak base.
Answer:
a) Strong acid :
The acid which undergoes 100% ionisation is called strong acid.
e.g.: HCl, H₂SO₄

b) Strong base :
The base which undergoes 100% ionisation is called strong base.
e.g.: NaOH, KOH

c) Weak acid:
The acid which undergoes less than 100% ionisation is called weak acid.
e.g.: CH₃COOH, H₂CO₃

d) Weak base:
The base which undergoes less than 100% ionisation is called weak base.
e.g.: NH₄OH, Mg(OH)₂

Question 14.
Write any four chemical properties of acids.
Answer:
Chemical properties of acids :
1) Active metals react with acids and liberate hydrogen gas.
Zn + HCl → ZnCl₂ + H₂ ↑

2) Acids react with bases to form salt and water.
HCl_(aq) + NaOH_(aq) → NaCl_(aq) + H₂O_(l)

3) Acids react with metallic oxides to form salt and water.
MgO_(s)+ 2HCl_(aq) → MgCl_2(aq) + H₂O_(l)

4) Acids react with carbonates and hydrogen carbonates and release carbon dioxide gas.
CaCO_3(s) + 2HCl_(aq) → CaCl_2(aq)+ H₂O_(l) + CO_2(g) ↑
Ca(HCO₃)_2(l) + 2HCl_(aq) → CaCl_2(aq) + 2H20((| + 2CO_2(g) ↑

Question 15.
Write the formulae of the following salts.
a) Sodium sulphate
b) Ammonium chloride
Identify the acids and bases for which the above salts are obtained. Also write chemical equations for the reactions between such acids and bases. Which type of chemical reactions are they?
Answer:
a) Formula of sodium sulphate is Na2S04. When sulphuric acid reacts with sodium hydroxide it forms sodium sulphate.
H₂SO_4(aq)+ 2NaOH_(aq) → Na₂SO_4(aq) + 2H₂O_(l)

b) Formula of Ammonium chloride is NH4C/. When Ammonium hydroxide reacts with hydrochloric acid it forms Ammonium chloride.
NH₄OH_(aq) + HCl_(aq) → NH₄Cl_(aq) + H₂O_(l)

Question 16.
Write balanced equations to satisfy each statement,
a) Acid + Active metal → Salt + Hydrogen
Answer:
Zn_(s) + 2HCl_(aq) → ZnCl_2(aq) + H_2(g)

b) Acid + Base → Salt + Water
Answer:
NaOH_(aq) + HCl_(aq) → NaCl_(aq) + H₂O_(l)

c) Acid + Carbonate / Hydrogen carbonate → Salt + Water + Carbon dioxide
Answer:
CaCO_3(s) + 2 HCl_(aq) → CaCl_2(aq) + H₂O_(l) + CO_2(g) ↑
NaHCO_3(s) + HCl_(aq) → NaCl_(aq) + H₂O_(l) + CO_2(g) ↑

d) Metal oxide + Acid → Salt + Water
Answer:
CaO_(s) + 2 HCl_(aq) → CaCl_2(aq) + H₍₂₎O_(l)

e) Non metal oxide + base → Salt + Water
Answer:
CO_2(g) + 2 NaOH_(aq) → Na₂CO_3(aq) + H₂O_(l)

Question 17.
Give important products obtained from chloralkali process.

Answer:
Result:
Five water molecules are present in one formula unit of copper sulphate. Water of crystallization proves that the crystals contain a fixed quantity of water in them.

Question 18.
Give the equations for the preparation of each of the following.
i) Copper sulphate from copper (II) oxide.
Answer:
CuO + H₂SO₄ → CuSO₄ + H₂O

ii) Potassium sulphate from potassium hydroxide solution.
Answer:
2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O

iii) Lead chloride from lead carbonate.
Answer:
PbCO₃ + 2 HCl → PbCl₂ + H₂O + CO₂

Question 19.
How are the following salts prepared?
1) Calcium sulphate from calcium carbonate
Answer:
When calcium carbonate is treated with sulphuric acid it forms calcium sulphate.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

2) Lead carbonate from lead nitrate
Answer:
When lead nitrate is treated with carbonic acid we will get lead carbonate.
Pb(NO₃)₂ + H2CO₃ → PbCO₃ + 2 HNO₃

3) Sodium nitrate from sodium hydroxide
Answer:
When sodium hydroxide is reacted with nitric acid it will form sodium nitrate.
NaOH + HNO₃ → NaNO₃ + H₂O

4) Magnesium carbonate from magnesium chloride
Answer:
When magnesium carbonate is reacted with hydrochloric acid it forms magnesium chloride.
MgCO₃ + 2 HCl → MgCl₂ + H₂O + CO₂

Question 20.
Which of the following reactions are considered as neutralization reactions? Why?
1) NaOH + HCl → NaCl + H₂O
2) CaO + 2 HCl → CaCl₂ + H₂O
3) CO₂ + 2 NaOH → Na₂CO₃ + H₂O
4) SiO₂ + CaO → CaSiO₃
Answer:
All of them are considered as neutralization reactions.

1. An acid (HCl) reacts with base (NaOH) and forms salt and water. So it is a neutralization reaction.
2. Here metallic oxide which is basic in nature reacts with acid and forms salt and water. So it is also a neutralization reaction.
3. In third case non-metallic oxide (acidic oxide) reacts with base (NaOH) and forms salt and water. So it is also a neutralization reaction.
4. In fourth case a metallic oxide (CaO) reacts with non-metallic oxide (SiO₂) and forms salt. So it is also a neutralization reaction in the absence 6f water.

Question 21.
Which metals produce hydrogen gas when they are reacted with bases like NaOH and KOH? Write the chemical equations for the reactions.
Answer:
Zinc, aluminium and lead react with bases like NaOH and KOH and produce hydrogen gas.

Question 22.
i) A solution has a pH of 7. How would you increase its pH and decrease its pH? Explain.
Answer;
We can increase the pH of a solution by adding base because we know that bases have pH > 7. We can decrease the pH of a solution by adding an acid because acidic solution have pH < 7.

ii) If a solution changes the colour of litmus from red to blue, then what can you say about its pH?
Answer:
Bases can change red litmus into blue. So the pH of the solution is greater than 7.

iii) What can you say about pH of a solution that liberates carbon dioxide from sodium carbonate?
Answer:
Acids react with carbonates and liberate hydrogen gas. So the pH of the solution is less than 7.

Question 23.
Write the pH values of some solutions.
Answer:

pH value	Solutions
0	Battery Acid
1	Con. H2S04
2	Lemon juice
3	Orange juice
4	Tomato juice
5	Black coffee, Bananas
6	Milk, urine
7	Pure water
8	Sea water, eggs
9	Baking soda
10	Milk of magnesia
11	Ammonia solution
12	Soapy water
13	Bleach oven cleaner
14	Liquid drain cleaner

Question 24.
Fill the following table of results of reactions between some substances (acids, bases, neutral substances) and indicators.

Answer:

Question 25.
The pH values of six solutions A, B, C, D, E, F are given as 5,2,1,3,7 and 9 respectively. Which solution is
a) Neutral
b) Strongly alkaline
c) Strongly acidic
d) Weakly acidic?
Arrange the pH in increasing order of Hydrogen ion concentration.
Answer:
a) Solution E is neutral.
b) Solution F is Alkaline.
c) Solution C is strongly acidic.
d) Solution A is weakly acidic.
e) Solution B is strongly acidic.
f) Solution D is strongly acidic.
g) Ascending order of increase of Hydrogen ion concentration is F, E, A, D, B, C.

Question 26.
Collect information about various organic acids different occurring naturally and prepare a table.
Answer:

1. Acetic acid	Vinegar (obtained from fruits after fermentation).
2. Citric acid	Citrus fruits like orange and lemons.
3. Butyric acid	Butter gone bad or rancid
4. Lactic acid	Curd
5. Malic acid	Apples
6. Oleic acid	Olive oil
7. Tartaric acid	Fruits such as grapes, apples and tamarind
8. Stearic acid	From fats
9. Succinic acid	From vegetables like lettuce and unripe fruits
10. Uric acid	From urine

Question 27.
Complete the table.

Answer:

Question 28.
Fill the table.

Answer:

Question 29.
Draw a diagram to show the reaction of acids with metals.
Answer:

Reaction of Zinc granules with dil. HCl and testing hydrogen gas by a burning candle

Question 30.
Draw a diagram to show that all metal carbonates and react hydrogen carbonates
Answer:

Question 31.
What are the uses of Bleaching powder?
Answer:

• It is used for bleaching cotton and linen in the textile industry for bleaching wood pulp in paper industry and for bleaching washed clothes in laundry.
• Used as an oxidizing agent in many chemical industries.
• Used for disinfecting drinking water to make it free of germs.
• Used as a reagent in the preparation of chloroform.

Question 32.
What are the uses of Baking soda?
Answer:
1) Baking powder is a mixture of baking soda and a mild edible acid such as tartaric acid. When baking powder is heated or mixed in water, the following reaction takes place.

Carbon dioxide produced during the reaction causes bread or cake to rise making them soft and spongy.

2) Sodium hydrogen carbonate is also an ingredient in antacids. Being alkaline, it neutralizes excess acid in the stomach and provides relief.

3) It is also used in soda-acid, fire extinguishers.

4) It acts as mild antiseptic.

Question 33.
What are the uses of Washing soda?
Answer:

• Sodium carbonate (washing soda) is used in glass, soaps and paper industries.
• It is used in the manufacture of sodium compounds such as borax.
• Sodium carbonate can be used” as a cleaning agent for domestic purposes.
• It is used for removing permanent hardness of water.

Question 34.
Write the chemical formulae of the following :
i) Bleaching powder
ii) Sodium Chloride
iii) Slaked lime
iv) Baking Soda
v) Washing Soda
vi) Gypsum
vii) Plaster of Paris
viii) Acetic acid
ix) Sodium Hydroxide
x) Limestone
Answer:
i) Bleaching powder = CaOCl₂
ii) Sodium Chloride = NaCl (Common Salt)
iii) Slaked lime (or) lime water = Ca(OH)₂
iv) Baking Soda = NaHCO₃
v) Washing Soda = Na₂CO₃, 10H₂O
vi) Gypsum = CaSO₄ . 2H₂O
vii) Plaster of Paris = CaSO₄ . ½H₂O
viii) Acetic acid = CH₃COOH
ix) Sodium Hydroxide = NaOH
x) Limestone = CaCO₃

Question 35.
What are the various applications of neutralization?
Answer:

• The acidity of soil is reduced by adding slaked lime.
• The sting of yellow wasps contains alkalis. If acetic acid is rubbed on affected area, they are neutralized.
• Ants and bees have formic acid in their stings which can be neutralised by applying soap and some other alkali.
• Antacids tablets contain magnesium hydroxide, persons suffering from acidity are administered these tablets.
• The affect of nettle plant leaves is neutralized by leaves of dock plant.

Question 36.
Mention two situations where you use hydrated and unhydrated salts in your daily life.
Answer:

• NaCl is unhydrated salt. It flows freely when filled in a container.
• NaHCl (Baking soda) is unhydrated salt. It flows freely when filled in a container.
• NaCO₃ . 10H₂O (washing soda) is hydrated salt. It leaves wetness inside the container.
• CaSO₄ . 2H₂O (Gypsum) is also hydrated salt. On careful heating of gypsum it loses water molecules partially to become (CaSO₄ . ½H₂O) P.O.P. It is used in hospitals as plaster for supporting fractured bones in right position.

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 12 Electromagnetism.

AP State Syllabus SSC 10th Class Physics Important Questions 12th Electromagnetism

10th Class Physics 12th Lesson Electromagnetism 1 Mark Important Questions and Answers

Question 1.
Draw the diagram showing the magnetic field lines of bar mannet. (TS June 2016)
Answer:

Question 2.
Correct the diagram according to Lenz law and draw it again. (TS March 2017)
Answer:

Question 3.
What is the use of slip – ring in AC motor? (TS June 2018)
Answer:
Uses of slip rings :
Slip rings are used to change the direction of current in the coil continuously.

Question 4.
Draw the direction of magnetic lines force, assuming that the current is flowing into the page. (AP SCERT: 2019-20 )
Answer:

Question 5.
What happens when a current carrying coil is placed in a uniform magnetic field? (TS June2019)
Answer:
The rectangular coil comes into rotation in clock – wise direction because of equal and opposite pair of forces acting on the two sides of the coil.

1. If the direction of the current in the coil unchanged it rotates half clock – wise and comes to half and rotates in anticlock – wise direction.
2. If the direction of the current in the coil changed after the first half rotation, the coil continuously rotates in a same direction.

Question 6.
Write the name of the device that converts mechanical energy into electrical energy. (AP June 2019)
Answer:
Generator.

Question 7.
Name some sources of direct current.
Answer:
Dry cell, lead-acid battery.

Question 8.
Which sources produce alternating current?
Answer:
A.C generator, thermal power station, hydroelectric stations.

Question 9.
What is the role of split ring in an electric motor?
Answer:
The split rings are used to change the direction of current flowing through the coil.

Question 10.
Write one method of inducing current in the coil.
ANswer:
By pushing or pulling a bar magnet into or away from the coil we can induce current. Name two safety measures commonly used in electric circuit, i) Fuse ii) Earthing

Question 11.
On what factors does the magnetic induction at the centre of the coil depend?
Answer:
It depends on current, number of turns and radius of the coil.

Question 12.
Which is more dangerous AC or DC?
Answer:
AC is more dangerous.

Question 13.
State two serious hazards of electricity.
Answer:

• If a person touches the live wire, he gets severe shock which may prove fatal.
• Short-circuiting can cause a spark which may lead to fire in a building.

Question 14.
Why is earthing of electrical appliances recommended?
Answer:
To protect the user from any accidental electrical shock caused due to leakage of current.

Question 15.
Why is a spark produced at the place of short circuit? Why is the spark in white colour?
The resistance of circuit decreases, and a sudden flow of large current heats up the live wire and vapourises the metal. This causes spark. The metal of wire becomes very hot and naturally emits white light.

Question 16.
What is electromagnetic induction?
Answer:
Mechanical energy can be converted into electrical energy by moving a magnet inside a coil.

Question 17.
What is Maxwell’s right hand screw rule?
Answer:
The direction of current is the direction in which the tip of the screw advances and direction of ration of the screw gives the direction of magnetic lines of force.

Question 18.
What type of energy transformation takes place in electric generator?
Answer:
Electrical energy from mechanical energy.

Question 19.
Where are the electromagnets used?
Answer:
In electric generators and televisions.

Question 20.
What is electromagnet?
Answer:
When current carrying conductor is wound over a magnetic material like soft iron it gets magnetized.

Question 21.
What are the different types of power stations?
Answer:
Electrical energy is produced in different power stations from mechanical energy of water, meat energy, and nuclear energy.

Question 22.
If the current in the coil is in anti-clockwise, then what would be the face of the coil?
Answer:
It behayes as north pole.

Question 23.
If the current in the ceil is in clockwise, then what would be the face of the coil?
Answer:
It behaves as south pole.

Question 24.
What is the frequency of the A.C. supplied in your house?
Answer:
It is approximately 50 Hz.

Question 25.
What type of current is generated in electric power station?
Answer:
Alternating current.

Question 26.
What is the shape of magnetic lines due to straight current carrying conductor?
Answer:
They are concentric circles.

Question 27.
What is a transformer?
Answer:
It is a device which increases or decreases the voltage.

Question 28.
State two ways by which speed or rotation of electric motor can be increased.
Answer:

1. By increasing strength of the current.
2. By increasing number of turns in the coil.

Question 29.
What happens if an iron piece is dropped between two poles of strong magnet?
Answer:
Eddy current is produced in it. These eddy currents oppose the motion of the piece of iron. So it falls as it is moving through a viscous liquid.

Question 30.
If a copper rod carries a direct current, then where will be the magnetic field in the conductor?
Answer:
It will be both inside and outside the rod.

Question 31.
In what form is the energy in a current carrying coil stored?
Answer:
It is stored in the form of magnetic field.

Question 32.
What is solenoid?
Answer:
A solenoid is a long wire wound in a close packed helix.

Question 33.
What is the pattern of field lines inside a solenoid around when current carrying solenoid?
Answer:
Parallel to each other.

Question 34.
List any two properties of magnetic field lines.
Answer:

• Inside the magnet they start from south pole and end at north pole whereas outside the magnet they start at north pole and end at south pole.
• Two magnetic lines of force never intersect each other.

Question 35.
Why does the picture appear distorted when the bar magnet is brought close to the screen of a television?
Answer:
Picture on a television screen is due to motion of the electrons reaching the screen. These electrons are affected by magnetic field of bar magnet.

Question 36.
What is meant by electromagnetic induction?
Answer:
Whenever there is continuous change of magnetic flux linked with a closed coil, current generated in the coil is called electromagnetic induction.

Question 37.
State the Lenz’s principle.
Answer:
The induced current will appear in such a direction that it opposes the changes in the flux in the coil.

Question 38.
What is induced emf?
Answer:
Change in magnetic flux produces emf in the circuit called induced emf.

Question 39.
What do you mean by magnetic effect of current?
Answer:
Current carrying conductor produces a magnetic field around it. This is called magnetic effect of current.

Question 40.
What is the direction of magnetic field at the centre of the coil carrying current in (i) clockwise, (ii) anti-clockwise direction?
Answer:
i) Along the axis of coil inwards.
ii) Along the axis of coil outwards.

Question 41.
Why does a current carrying freely suspended solenoid rest along a particular direction?
Answer:
A current carrying solenoid behaves like a bar magnet.

Question 42.
What effect will there be on a magnetic compass when it is brought near a current carrying solenoid?
Answer:
The needle of the compass will rest in the direction of magnetic field due to solenoid at that point.

Question 43.
How is magnetic field due to solenoid carrying current affected, if a soft iron bar is introduced inside the solenoid?
Answer:
The magnetic field increases when iron bar is introduced inside the solenoid.

Question 44.
What happens to magnetic field if we reverse the current direction?
Answer:
The magnetic field also gets reversed.

Question 45.
How do magnetic field lines inside a current carrying solenoid appear?
Answer:
They form along the axis and parallel to each other.

Question 46.
In which of the following cases does the electromagnetic induction occur?
i) A current is started in a wire held near a loop of wire.
ii) The current is stopped in a wire held near a loop of wire.
iii) A magnet is moved through a loop of wire.
iv) A loop of wire is held near a magnet.
Answer:
In first three cases there is a change in magnetic flux. So electromagnetic induction occurs in first three cases.

Question 47.
Why must an induced current flow in such a direction so as to oppose the change producing it?
Answer:
So that the mechanical energy spent in producing the change, is transformed into the electrical energy in form of induced current.

Question 48.
What is the maximum force acting on the current (i) carrying conductor of length (l) in the presence of magnetic field (B)?
Answer:
F = Bil

Question 49.
A charged particle q is moving with a speed v perpendicular to the magnetic field of induction B?
Find the equation of time period of the particle.
Answer:
\(\mathrm{T}=\frac{2 \pi \mathrm{m}}{\mathrm{Bq}}\)

Question 50.
Name two safety measures commonly used in electric circuit.
Answer:
a) Fuse
b) Earthing

Question 51.
On what factors the magnetic induction at the centre of coil depends?
Answer:
a) Number of turns
b) Radius of the coil.

Question 52.
Write the formula for magnetic flux passing through an Area A with an angle θ.
Answer:
Flux ΦV= BA cos θ

Question 53.
Write the Lenz’s law.
Answer:
The induced current will appear in such a direction that it opposes the changes in the flux in the coil.

Question 54.
What is the difference between AC and DC generator?
Answer:
In AC generator, ends of coil are connected to two slip rings.
In DC generator ends of coil are connected to two half split rings.

Question 55.
What are the uses of electromagnet?
Answer:
It is used in electric bells, electric motors, telephone diaphragms, etc.

Question 56.
What is the principle of Electric motor?
Answer:
When a rectangular coil is placed in magnetic field and current is passed through it, two equal and opposite forces act on the coil which rotates it continuously.

Question 57.
What factors are influence the speed of rotation of the motor?
Answer:

1. Strength of current
2. Number of turns
3. Area of the coil
4. Strength of magnetic field

Question 58.
Which two physical quantities are interrelated in Oerstead experiment?
Answer:
Electricity and magnetism are interrelated.

Question 59.
Which property of proton can change while it moves freely in a magnetic field?
Answer:
When a proton (the charge) moves in a magnetic field, then magnetic force is acting on proton. So its momentum changes.

Question 60.
Which type of conductors are producing magnetic field?
Answer:

1. Long straight current carrying conductor.
2. Circular loop.
3. Solenoid.

Question 61.
How much force acting on a neutron particle is moving with velocity V in a mag¬netic field with induction B?
Answer:
Zero, because neutron is charge less particle.

Question 62.
What are the instruments used in A.C Generator?
Answer:
Rectangular coil, brushes, slip rings, and magnetic poles.

Question 63.
What are the ways to produce the induced current in a coil?
Answer:
When a magnet is moved towards or away from coil or there is a relative motion between coil and magnet a current is induced in the coil.

Question 64.
At the time of short circuit, what happens to the current?
Answer:
At the time of short circuit, the current in the circuit increases heavily becomes the resistance of the conductor becomes almost zero.

Question 65.
A wire with green insulation is usually the line wire of an electric supply. Is it true?
Answer:
It is false, the wire with green insulation is the earth wire, not the line wire.

Question 66.
Two circular coils A and B are placed close to each other. If the current in the coil A is changed, will some current be induced in the coil B? Give reason.
Answer:
Yes. current will be induced in coil ‘B’, because flux linked the coil ‘B’ changes with respect to time.

Question 67.
Name two safety measures commonly used in electric circuits and appliances.
Answer:
Electric fuse and Miniature Circuit Breaker (MCB).

Question 68.
An attemating current has frequency of 50Hz. How many times does it change its direction in one second?
Answer:
Frequency of AC = 50 Hz ⇒ 50 cycles in one sec. So it reverses its direction 100 times in one second.

Question 69.
Under what orientation, the induced current produced in moving conductor in a magnetic field can be maximum?
Answer:
The current induced in a conductor is maximum when direction of motion of conductor is at right angle to the magnetic field.

Question 70.
How could you make the coil rotate continuously in motor?
Answer:
The direction of current through the coil is reversed every half rotation, the coil will rotate continuously and the same direction.

Question 71.
What is the formula for induced cmf when change the magnetic flux?
Answer:
Induced EMF = \(-\frac{\Delta \phi}{\Delta t}\)

Question 72.
The magnetic flux is varying with time. Which cases E.M.F is induced?

Answer:
In OA and BC cases, E.M.F is induced.

Question 73.
In above problem, how much EMF is induced in BC curve?
Answer:

Question 74.
What is the equation of motional E.M.F?
Answer:
Motional E.M.F = Blv;
where B = magnetic induction,
l = length of rod,
v = velocity of rod.

Question 75.
When a magnet and a coil are moving same direction with same speed. Then induced E.M.F in coil is zero. Why?
Answer:
E.M.F = \(\frac{-\Delta \phi}{\Delta \mathrm{t}}\), but both moving same direction, so change in flux linked with coil is zero i.e., ∆Φ = 0

Question 76.
What is shape of conductor is drawn when current is passing through conductor?
Answer:

Question 77.
Draw the symbols of North and South pole when depends on current.
Answer:

Question 78.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:

Question 79.
Draw the diagram of AC and DC current varying with time.
Answer:

Question 80.
Where are electric motors used?
Answer:
Electric fans, water-pumps, coolers.

Question 81.
Mention two uses of solenoid.
Answer:
It is used in electric bells, fans, and motors.

Question 82.
Mention applications of electromagnetic induction.
Answer:
It is used in devices which convert mechanical energy into electrical energy.

Question 83.
What are the advantages of an electromagnet over permanent magnet?
Answer:

1. An electromagnet can produce a strong magnetic field.
2. The strength of electromagnet can be changed.
3. The polarity of electromagnet can be changed.

10th Class Physics 12th Lesson Electromagnetism 2 Marks Important Questions and Answers

Question 1.
Anand appreciated the law behind the making of ‘generator’. Name the law and state it. (AP June 2017)
Answer:
1) The law behind the making of ‘generator’ is Faraday’s law.

2) Faraday’s Law :
“Whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil”.

Question 2.
Explain Oersted experiment to show that Electricity and Magnetism were related phenomena. (AP June 2018)
Answer:

• Place a compass needle underneath a wire and then turn on electric current.
• Immediately the needle of compass shows the deflection. By this we can conclude that electricity and magnetism are related phenomena.

Question 3.
With the help of the given figure, the teacher explained that magnetic field lines are closed lines and not open lines. Write the questions which you will ask to rest whether the given statement is right or wrong. (TS June 2015)

Answer:

• Are there any magnetic field lines inside the magnet?
• If magnetic field lines are there inside the magnet, what is the direction of field lines inside the magnet?
• What is the direction of field lines outside the magnet?

Question 4.
State Right-hand rule with a labelled diagram. (TS March 2015)
Answer:

(OR)

• Right hand rule indicates the direction of magnetic force acting on a moving charge.
• It is used when velocity and field are perpendicular to each other. If the fore finger points towards the direction of velocity of charge or current (I), middle finger points to the direction of field (B), then thumb gives the direction of force (F).

Question 5.
A coil of insulated Copper wire is connected to a Galvanometer. (TS March 2015)
What happens, if a bar magnet is ………….
1) pushed into the coil?
2) withdrawn from inside the coil?
3) held stationary inside the coil?
Answer:

1. If a bar magnet is pushed into the coil, then the needle in Galvanometer gets deflected. Because current is generated in the coil.
2. If a bar magnet is withdrawn from inside the coil, then the needle in Galvanometer gets deflected. Because current is generated in the coil.
3. If a bar magnet is held stationary inside the coil, then the needle in Galvanometer does not get deflected. Because current is not generated in the coil.

Question 6.
Compare the magnetic field lines of force formed around a current carrying solenoid with the magnetic field lines of force of a bar magnet.
Answer:

Magnetic field lines of a bar magnet	Magnetic field lines of a solenoid
2) The direction of the field lines of the outside the magnet is from north to south pole.	2) The direction of the field lines of the outside the solenoid is from north to south pole.
3) The direction of the field lines of the inside of the magnet field looks like south to north pole.	3) The direction of the field lines of the inside of the solenoid is from south to north pole.
4) These lines are closed loops.	4) These lines are also closed loops.
5) We cannot find the field lines inside the bar magnet.	5) We can find the field lines inside the solenoid.
6) These field lines are same as field lines formed by a solenoid.	6) These field lines are also same as field lines formed by a bar magnet.
7) More field lines are found at poles.	7) More field lines are found at poles.

Question 7.
Which energy we get from an electric motor? Write two daily life applications of the electric motor. (TS June 2017)
Answer:

• We get mechanical energy from electric motor.
• In our daily life we use motor in
  i) Mixies
  ii) Grinders
  iii) Water Pumps
  iv) Fans / Coolers, etc.

Question 8.
List out the material required for Oersted experiment and mention the precautions to be taken in the experiment. (TS March 2019)
Answer:
Materials required for Oersted experiment are :

1. Thermocol sheet.
2. Wooden sticks.
3. Copper wire of 24 gauge.
4. Battery.
5. Key.
6. Magnetic compass.

Precautions to be taken are :

1. Copper Wire is made through the slits of the wooden sticks tightly.
2. Arrange/complete the circuit correctly.

Question 9.
What happens when magnetic flux passing through a coil changes continuously? Where does this process is used? (TS June 2019)
Answer:
1) When there is a continuous change of magnetic flux passing through a coil a current is generated in the coil.

2) This process is used in

1. Induction stoves
2. Security checking entrance/exit doors.
3. ATM cards scanners.

Question 10.
Distinguish between AC and DC.
Answer:

AC	DC
1) AC means alternate current.	1) DC means direct current.
2) The current direction changes	2) The current direction does not
always.	change. It is always a single direction.
3) The magnitude of current changes from minimum to maximum.	3) Always it has maximum value.

Question 11.
Distinguish between AC motor and DC motor.
Answer:

AC motor	DC motor
1) It works with alternate current.	1) It works with direct current.
2) It does not require change in current direction.	2) It requires change in current direction which is provided by split rings which act as commutator.

Question 12.
Distinguish between AC generator and DC generator.
Answer:

AC generator	DC generator
1) It generates alternate current.	1) It generates direct current.
2) It has two slip rings.	2) It has two half slip rings.

Question 13.
What is Faraday’s law of electromagnetic induction? Write its expression.
(OR)
State the Faraday’s law of electromagnetic induction. Write the equation of this law.
Answer:
The induced EMF generated in a closed loop is equal to the rate of change of magnetic flux passing through it.
Induced EMF = Change in flux / time
ε = ∆Φ / ∆t

Let Φ₀ be flux linked with single turn. If there are N turns of the coil, the flux linked with the coil is NΦ₀.
∴ ε = NAΦ₀ / ∆t

Question 14.
State the right-hand thumb rule. How does the rule help us?
Answer:
When you curl your right hand fingers in the direction of current thumb gives the direction of magnetic field.

It is useful to find the magnetic field direction as well as current direction.

Question 15.
A flat rectangular coil is rotated between the pole pieces of a horseshoe magnet. In which position of coil with respect to magnetic field, will the emf (i) be maximum, (ii) be zero, (iii) change its direction.
Answer:
i) The emf is maximum when the plane of coil is parallel to the magnetic field.
ii) The emf is zero when the plane of coil is normal to the magnetic field.
iii) The emf will change its direction when the plane of coil passes from the position normal to the magnetic field.

Question 16.
State two factors on which the magnitude of induced emf depend?
Answer:
The magnitude of induced emf depends on the following two factors.

1. The change in the magnetic flux.
2. The time in which the magnetic flux changes i.e., the rate of change of magnetic flux.

More the change in magnetic flux, more is the emf induced. Further if more rapid the magnetic flux changes, more is the emf induced.

Question 17.
How do you increase the magnetic field of solenoid?
Answer:
The magnetic field of solenoid can be increased by the following two ways.

1. by increasing the number of turns of winding in the solenoid.
2. by increasing the current through the solenoid.

Question 18.
State the function of split ring in a DC motor.
Answer:

• The split ring acts as a commutator in a DC motor.
• With the split ring, the direction of current through the coil is reversed after every half rotation of coil.
• Thus the direction of couple rotating the coil remains unchanged and the coil continues to rotate in the same direction.

Question 19.
A DC motor is rotating in a clockwise direction. How can the direction of rotation be reversed?
Answer:
The direction of rotation of motor can be reversed by interchanging the connections at the terminals of the battery joined to the brushes of the motor.

Question 20.
State the effect of inserting a soft iron core within the coil of DC motor.
Answer:

• On inserting a soft iron core within the coil of a DC motor, the speed of rotation of coil increases.
• The reason is that the strength of magnetic field between the pole pieces of magnet increases due to which the deflecting couple on coil increases.

Question 21.
State condition when magnitude of force on a current-carrying conductor placed in a magnetic field is (a) zero, (b) maximum.
Answer:
a) When current in conductor is in direction of magnetic field,
b) When current in conductor is normal to the magnetic field.

Question 22.
A flat coil ABCD Is freely suspended between the pole pieces of a U – shaped permanent magnet with the plane of coil parallel to the magnetic field.
a) What happens when current is passed in the coil?
b) When will the coil come to rest?
c) Name the instrument which makes use of the principle stated above.
Answer:
a) The coil will experience a torque due to which it will rotate.
b) The coil will come to rest when its plane becomes normal to the magnetic field.
c) DC motor.

Question 23.
Why is it more difficult to move a magnet towards a coil which has a large number of turns?
Answer:

1. Emf induced in the coil becomes more when the number of turns in the coil are made large.
2. So it is difficult to move a magnet towards a coil which has a large number of turns.

Question 24.
A coil ABCD mounted on an axle is placed between the poles N and S of permanent magnet as shown in figure.

a) In which direction will the coil begin to rotate when the current is passed through the coil in direction ABCD by connecting a battery at the ends A and D of the coil.
b) Why is commutator necessary for continuous rotation of the coil?
Answer:
a) Anti-clockwise direction.
b) Commutator is needed to change the direction of current in the coil after each half rotation of coil.

Question 25.
Draw the magnetic field lines in uniform magnetic field.
Answer:

Question 26.
Draw a labelled diagram to make an electromagnet from soft iron bar AB. Mark the polarity at its ends. What precaution would you observe?
Answer:

The labelled diagram is shown in figure. The polarity at the end A where the current is clockwise, is south (S), while at the end B where the current is anti-clockwise is north (N).
Precaution :
The source of current must be the DC source.

Question 27.
State the principle of an electric motor. Name some appliances in which the Electric motor is used.
Answer:
Current carrying coil rotates when it is kept in a uniform magnetic field. It is the working principle of electric motor.
Appliances containing electric motor are :

1. Fans,
2. Mixies,
3. Grinders,
4. Machines, etc.

10th Class Physics 12th Lesson Electromagnetism 4 Marks Important Questions and Answers

Question 1.
How can you verify with experiment “The magnetic field lines are closed loops”? (AP March 2017)
Answer:

• Place a retort stand on the plank as shown in the figure.
• Pass a copper wire through hole of the plank and rubber knob of the retort stand in such a way that the wire be arranged in a vertical position and not touch the stand.
• Connect the two ends of the wire to a battery via switch. Allow the current flows through wire.
• By sprinkling the iron fillings around the wire, we can observe the magnetic field lines are in circular shape.

Conclusion :
Hence it is proved that “Magnetic field lines are closed loops”.

Question 2.
Name the device that converts electrical energy into mechanical energy. Draw its diagram and label the parts. (AP March 2018)
Answer:
1) The device that converts electrical energy into mechanical energy is motor.
2) Diagram of motor

Question 3.
List out the materials required for the Oersted experiment of electromagnetism. Write the procedure of the experiment. What do you understand by this experiment? (TS March 2016)
Answer:
List of material required for Oersted experiment:
A thermocol sheet, two small sticks, insulated copper wires, 9 V battery, switch, magnetic compass.

Procedure :

1. Take a thermocol sheet and fix two thin wooden sticks of height 1 cm which have small slit at the top of their ends.
2. Arrange a copper wire so that it passes through these slits and makes a circuit.
3. The circuit consists of a 9 V batten’, key, and copper wires connected in series.
4. Now keep a magnetic compass below the wire.
5. Now switch on the circuit and observe the compass needle.
6. Change the directions of current and observe the compass needle.

Observation :

1. When current is passed through circuit, we observe deflection of compass needle in one direction.
2. When the direction of current is changed, the compass needle deflects in another direction.
3. This shows that a current carrying conductor possesses magnetic field around it.

Question 4.
Write the experimental procedure and observations of the experiment that is to be performed to observe the magnetic field formed due to solenoid. (TS June 2017)
Answer:

• Fix a white paper on a wooden plank.
• Make two holes to the plank at appropriate distance.
• Make some more holes parallel to these two holes.
• Insert a copper wire through these holes. It looks as a coil.
• Connect the two ends of the coil to a battery and a switch in series.
• When swich is on current flows through the wire.
• Sprinkle some iron filings around the coil and tap the plank.

Observation :
An orderly pattern of iron filing is seen on the paper. These are magnetic field lines. The magnetic field lines set up by solenoid resemble those that of a bar magnet.

Question 5.
Why the current-carrying straight wire which is kept in a uniform magnetic field, perpendicularly to the direction of the field bends aside? Explain this process with a diagram showing the direction of forces acting on the wire. (TS March 2017)
Answer:

Uniform magnetic field (due to horse shoe magnet)

Magnetic field due to current carrying straight wire

Net field formed due to the above two fields

Explanation :
The net field in upper part is strong and in lower part it is weak. Hence a non-uniform field is created around wire. Therefore the wire tries to move to the weaker field region.

Question 6.
List out the apparatus and experimental procedure for the experiment to observe a current-carrying wire experiences a magnetic force when it is kept in uniform magnetic field. (TS June 2018)
Answer:
Required apparatus :
i) Horseshoe magnet,
ii) Conducting wire,
iii) Battery, switch

Experimental procedure :
1) Arrange the circuit :
Take a wooden plank and arrange two wooden sticks with slits and arrange a conductor through the sticks and make a circuits with switch, battery.

2) Put horse shoe magnet:
Arrange the horse shoe magnet on the conductor in such a way that the conductor should be in between the two poles of the magnet.

3) Deflections in conductor :
Allow the current to pass through the circuit. We can find that conductor deflects up wards or down wards.

4)

Question 7.

Answer the following questions by observing above diagram. (TS March 2018)
1. Which device function of working does the above figure gives?
Answer:
Motor

2. What is the angle made by AB and CD with magnetic field?
Answer:
90°

3. What are the directions of magnetic forces on sides AB and CD?
Answer:
By applying right hand rule to get the directions of magnetic force. At ‘AB’, the magnetic force acts inward perpendicular to field of the magnet and on ‘CD’, it acts outwards.

4. What is the net force acting on the rectangular coil?
Answer:
The net force on ‘AB’ is equal and opposite to the force on ‘CD’ due to external magnetic field because they carry equal currents in the opposite direction. Sum of these forces is zero. Similarly, the sum of the forces on sides ‘BC’ and ‘DA’ is also zero.

So, net force on the rectangular coil is zero.

Question 8.
Explain the working process of induction stove. (TS March 2019)
Answer:

• An induction stove works on the principle of electromagnetic induction.
• A metal coil is kept just beneath the cooling surface.
• It carries alternating current (AC) so that AC produce an alternating magnetic field.
• When you keep a metal pan with water on it, the varying magnetic field beneath it crosses the bottom surface of the pan, and EMF is induced in it.
• Induced EMF produces induced current in the metal pan.
• The pan has a finite resistance.
• The flow of induced current produces heat in it.
• That heat is conducted to the water. In this way, induction stove works and water will be heated.

Question 9.
Which device is used to convert mechanical energy into electrical energy? Draw a neat diagram and label the parts of this device. (TS March 2019)
Answer:
1) Dynamo is used to convert mechanical energy into electrical energy. They are also called Generators.

2) AC Generator (or) DC Generator

AC Generator

DC Generator

Question 10.
A coil is hung as shown in the figure. A bar magnet with north pole facing the coil is moved perpendicularly

a) How does the magnetic flux passing through the coil change?
b) State the direction of the flow of the current induced in the coil, keeping the direction of bar magnet in view.
c) Draw the diagrams showing the magnetic field formed due to bar magnet at the surface of the coil and the magnetic field formed due to induced current.
d) Explain the reason for induced current.
Answer:
a) A bar magnet with north pole facing the coil is moved perpendicularly, the magnetic flex increases when passing through the coil.
b) The direction of the flow of the. current induced in the coil, keeping the direction of bar magnet is anti-clockwise due to north pole. KT
c) Φ = 0

Plane of coil is parallel to ‘B’.
d) Electromagnetic induction is the reason for induced current.

Question 11.
Conductor of length T moves perpendicular to its length with the speed V. Length of the conductor is perpendicular to the magnetic field of the conductor. Let us assume that electrons could move freely in the conductor and the charge of an electron is ‘e’.

a) What is the magnetic force acting on electron in the conductor?
b) In which direction does the above force act?
c) What effect does this force have on motion of electrons?
Answer:
a) Magnetic field acting on the electron inside the conductor is = F_m = e (\(\bar{V} \times \bar{B}\)) = BeV
This field acts from P to Q.

b) Consider in the field P and Q are ends of a conductor. ‘Q’ will act as negative end and ‘P’ will act as positive end then flow passes from P to Q means downwards.

c) The force on electrons shows an effect creates a potential difference at the ends of the rods.
∴ BeV = eE ⇒ E = BY

Question 12.
A charged particle q is moving with a speed V perpendicular to the magnetic field induction B. Find the radius of the path and time period of the particle.
Answer:

1. Let us assume that the field is directed into the page.
2. Then the force experienced by the particle F = qvB.
3. We know that this force is always directed perpendicular to velocity.
4. Hence the particle moves along a circular path and the magnetic force on a charged particle acts like a centripetal force.
5. Let r be the radius of the circular path.

Question 13.
Explain different ways to induced current in a coil.
Answer:

• Moving a north pole of a magnet into a coil.
• Withdrawing north pole from a coil.
• Moving a south pole of magnet into a coil.
• Withdrawing a south pole of a magnet from a coil.
• Moving a coil towards a magnet.

Question 14.
What are the similarities between a current-carrying solenoid and a bar magnet?
Answer:
1) The magnetic field lines due to current carrying solenoid are identical to those of bar magnet. Thus a current-carrying solenoid behaves just like a bar magnet with fixed polarities at the ends. The end at which the direction of current is clockwise behaves like a south pole and the end at which current is anti-clockwise behaves like a north pole.

2) A current-carrying solenoid when suspended freely, will set itself in the north south direction exactly in the same manner as a bar magnet does.

3) A current-carrying solenoid also acquires the- attractive property of magnet. If iron filings are brought near the solenoid, it attracts them when current flows through the solenoid.

Question 15.
What are the dissimilarities between a current-carrying solenoid and a bar magnet?
Answer:

• The magnetic field strength due to solenoid can be altered by altering current in it, while the magnetic field strength of a bar magnet cannot be changed.
• The direction of magnetic field due to solenoid can be reversed by reversing the direction of current in it, but the direction of magnetic field of the bar magnet cannot be changed.

Question 16.
Compare electromagnet with a permanent magnet.
Answer:

Electromagnet	Permanent Magnet
1) It is made of soft iron.	1) It is made of steel.
2) It produces the magnetic field so long as current flows in its coils.	2) It produces permanent magnetic field.
3) The magnetic field strength can be changed.	3) The magnetic field strength cannot be changed.
4) The electromagnet can be made as strong as needed.	4) The permanent magnets are not so strong.
5) The polarity of an electromagnet can be reversed.	5) The polarity of permanent magnet cannot be reversed.
6) It can be easily de-magnetised by switching off the current.	6) It cannot be easily de-magnetised.

Question 17.
What are the characteristics of magnetic field lines due to current in a loop (or circular coil)?
Answer:

• The magnetic lines are nearly circular in the vicinity of coil.
• Within the space enclosed by the wire the magnetic field lines are in the same direction.
• Near the centre of loop, the magnetic field lines are nearly parallel and the magnetic field may be assumed to be uniform in a small space near the centre.
• At the centre, the magnetic lines are along the axis of the loop and at right angles to the plane of the loop.
• The magnetic field lines become denser if the strength of current in the loop is increased and there are more number of turns in the loop.

Question 18.
A straight conductor passes vertically through a cardboard having some iron filings sprinkled on it.
a) Show the setting of iron filings when current is passed in the downward direction and then the cardboard is gently tapped. Draw arrows to represent the direction of magnetic field lines.
b) What changes occur if
i) current is increased ?
ii) the single conductor is replaced by several parallel conductors each carrying same current flowing in the same direction?
c) Name the law used by you to find the direction of magnetic field lines.
Answer:

a) The figure shoy/s, the pattern in which iron filings will set themselves. The arrows show the direction of magnetic field lines.
b) i) The arrangement of iron filings remains unchanged, but they become denser and Cardboar get arranged up to a larger distance from the conductor when the strength of current is increased and it is effective up to a larger distance from the conductor.
ii) The magnetic field at a point due to each conductor will be in same direction, so they will be added up. Thus the magnetic field strength is increased and it is effective up to a large distance so the magnetic field lines come closer and iron filings get arranged up to a larger distance.
c) Right hand thumb rule.

Question 19.
In figure A and B represent two straight wires carrying equal currents in a direction normal to the plane of paper inwards.
a) Sketch separately the magnetic field lines produced by each current.
b) Give a reason why the magnetic field at K (mid point of the line joining A and B) will be zero.
c) What will be the effect on the magnetic field at the point K if the current in wire B is reversed?

Answer:
a) Figure shows the sketch of magnetic field A and B.

b) The point K is equidistant from the wires A and B, and the wires A and B carry equal currents. So the magnetic fields at K due to wires A and B are equal in magnitude, but opposite in direction. Due to the wire A, it is downwards in the plane of paper, while due to the wire B, it is upwards in the plane of paper. So the net magnetic field at the point K is zero as the two fields cancel each other.
c) On reversing the direction of current in the wire B, the direction of magnetic field due to current is reversed at the point K, i.e. it becomes downwards in the plane of paper.

Question 20.
The diagram given below shows two coils X and Y. The coil X is connected to a battery S and a key K. The coil Y is connected to a galvanometer G.

When the key K is closed state the polarity.
i) At the end B of the coil X.
ii) At the end C of the coil Y.
iii) At the end C of the coil Y if the coil Y is (a) moved towards the coil X, (b) moved away from the coil X.
Answer:
i) On closing the key K, the current at the end B of the coil X is anti-clockwise, therefore at this end there is a north pole.

ii) While closing the key, polarity at the end C of the coil Y will be north. But there will be no polarity at the end C of the coil Y when the current becomes steady in the coil X

iii) a) With the key K closed, while the coil Y is moved towards the coil X, the polarity at the end C of the coil Y is north.
b) With the key K closed, while the coil Y is moved away from the coil X, the polarity at the end C of the coil Y is south.

Question 21.
How do you increase the speed of rotation of coil in a DC motor?
Answer:
The speed of rotation of coil can be increased by

1. Increasing the strength of current.
2. Increasing the number of turns in coil.
3. Increasing the strength of magnetic field. To increase the strength of magnetic field a soft iron core can be inserted within the coil.

Question 22.
Explain Lenz’s law with an activity.
Answer:
Lenz law :
The induced current will appear in such a direction that it opposes the change in the flux in the coil.

Explanation :

1. We know that when a bar magnet is pushed towards a coil with its north pole facing the coil an induced current is set up in the coil.
2. Let the direction be clockwise then current carrying loop behaves like a magnet with its south pole facing the north pole of bar magnet.
3. In such a case, the bar magnet attracts the coil. Then it gains kinetic energy. This is contradictory to conservation of energy.
4. Hence our assumption is wrong. So correct induced current direction is anti-clockwise.
5. Let us see a case where the bar magnet is pulled away from the coil with the north pole facing the coil. In such case, the coil opposes the motion of bar magnet to balance the conversion of mechanical energy into electric energy.
6. It happens only when the north pole of the magnet faces the south pole of the coil.
7. So, the direction of induced current in the coil must be in anti-clockwise direction.
8. In simple terms, when flux increases through coil, the coil opposes the increase in the flux and when flux decreases through coil, it opposes the decrease in the flux. This is Lenz law.

Question 23.
What are the uses of electromagnet?
Answer:
Electromagnets are mainly used for the following purposes.

1. For lifting and transporting the large masses of scrap, girders, plates, etc. especially to the places where it is not convenient to take the help of human labour.
2. For loading furnaces with iron.
3. For separating the magnetic substances such as iron from other debris.
4. For removing pieces of iron from wounds.
5. In several electrical devices such as electric bell, telegraph, electric thumb, electric motor, relay, microphone, loudspeaker, etc.
6. In scientific research, to study the magnetic properties of a substance in a magnetic field.

Question 24.
Write the differences between AC generator and DC motor.
Answer:

AC Generator	DC Motor
1) A generator is a device which converts the mechanical energy into the electrical energy.	1) A DC motor is a device which converts electrical energy into the mechanical energy.
2) A generator works on the principle of electromagnetic induction.	2) A DC motor works on the principle of force acting on a current carrying conductor placed in a magnetic field.
3) In a generator, the mechanical energy is used in rotating the armature coil in a magnetic field so as to produce electricity.	3) In a DC motor electrical energy is provided by the DC source to flow current in the armature coil placed in a magnetic field due to which coil rotates.
4) A generator makes use of two separate coaxial slip rings.	4) A DC motor makes use of two parts of a slip ring (i.e., split rings) which acts as commutator.

 

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.3

10th Class Maths 8th Lesson Similar Triangles Ex 8.3 Textbook Questions and Answers

Question 1.
Equilateral triangles are drawn on the three sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.
Answer:

Let △PQR is a right angled triangle, ∠Q = 90°
Let PQ = a, QR b and
PR = hypotenuse = c
Then from Pythagoras theorem we can
say a² + b² = c² ……… (1)
△PSR is an equilateral triangle drawn on hypotenuse
∴ PR = PS = RS = c,
Then area of triangle on hypotenuse
= \(\frac{\sqrt{3}}{4}\)c² ……… (2)
△QRU is an equilateral triangle drawn on the side ‘QR’ = b
∴ QR = RU = QU = b
Then area of equilateral triangle drawn on the side = \(\frac{\sqrt{3}}{4}\)b² …….. (3)
△PQT is an equilateral triangle drawn on another side ‘PQ’ = a
∴ PQ = PT = QT = a
Area of an equilateral triangle drawn an another side ‘PQ’ = \(\frac{\sqrt{3}}{4}\)a² …….. (4)
Now sum of areas of equilateral triangles on the other two sides

= Area of equilateral triangle on the hypotenuse.
Hence Proved.

Question 2.
Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangles described on its diagonal.
Answer:

Let PQRS is square whose side is ‘a’ units then PQ = QR = RS = SP = ‘a’ units.
Then the diagonal
\(\overline{\mathrm{PR}}\) = \(\sqrt{a^{2}+a^{2}}\) = a√2 units.
Let △PRT is an equilateral triangle, then PR = RT = PT = a√2
∴ Area of equilateral triangle constructed on diagonal

Let △QRZ is another equilateral triangle whose sides are
\(\overline{\mathrm{QR}}\) = \(\overline{\mathrm{RZ}}\) = \(\overline{\mathrm{QZ}}\) = ‘a’ units
Then the area of equilateral triangle constructed on one side of square = \(\frac{\sqrt{3}}{4}\)a² ……. (2)
∴ \(\frac{1}{2}\) of area of equilateral triangle on diagonal = \(\frac{1}{2}\left(\frac{\sqrt{3}}{2} a^{2}\right)\) = \(\frac{\sqrt{3}}{4}\)a² = area of equilateral triangle on the side of square.
Hence Proved.

Question 3.
D, E, F are midpoints of sides BC, CA, AB of △ABC. Find the ratio of areas of △DEF and △ABC.
Answer:

Given in △ABC D, E, F are the midpoints of the sides BC, CA and AB.
In △ABC, EF is the line join of mid-points of two sides AB and AC of △ABC.
Thus FE || BC [∵ \(\frac{AF}{FB}\) = \(\frac{AE}{EC}\) Converse of B.P.T.]
Similarly DE divides AC and BC in the same ratio, i.e., DE || AB.
Now in □ BDEF, both pairs of opposite sides (BD || EF and DE || BF) are parallel.
Hence □ BDEF is a parallelogram where DF is a diagonal.
∴ △BDF ≃ △DEF ……… (1)
Similarly we can prove that
△DEF ≃ △CDE ……… (2)
[∵ CDFE is a parallelogram]
Also, △DEF ≃ △AEF …….. (3)
[∵ □ AEDF is a parallelogram]
From (1), (2) and (3)
△AEF ≃ △DEF ≃ BDF ≃ △CDE
Also, △ABC = △AEF + △DEF + △BDF + △CDE = 4 . △DEF
Hence, △ABC : △DEF = 4 : 1.

Question 4.
In △ABC, XY || AC and XY divides the triangle into two parts of equal area. Find the ratio of \(\frac{AX}{XB}\).
Answer:

Given: In △ABC, XY || AC.
XY divides △ABC into two points of equal area.
In △ABC, △XBY
∠B = ∠B
∠A = ∠X
∠C = ∠Y
[∵ XY || AC; (∠A, ∠X) and ∠C, ∠Y are the pairs of corresponding angles]
Thus △ABC ~ △XBY by A.A.A similarity condition.
Hence \(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{XBY}}=\frac{\mathrm{AB}^{2}}{\mathrm{XB}^{2}}\)
[∵ The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides]

⇒ \(\frac{AX}{XB}\) + 1 = √2
⇒ \(\frac{AX}{XB}\) = √2 – 1
Hence the ratio \(\frac{AX}{XB}\) = \(\frac{√2 – 1}{1}\).

Question 5.
Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Answer:

Given: △ABC ~ △XYZ
R.T.P: \(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{XYZ}}=\frac{\mathrm{AD}^{2}}{\mathrm{XW}^{2}}\)
Proof : We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Hence the ratio of areas of two similar triangles is equal to the squares of ratio of their corresponding medians.

Question 6.
△ABC ~ △DEF. BC = 3 cm, EF = 4 cm and area of △ABC = 54 cm². Determine the area of △DEF.
Answer:

Given: △ABC ~ △DEF
BC = 3 cm
EF = 4 cm
△ABC = 54 cm²
∴ △ABC ~ △DEF, we have
\(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{DEF}}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}\)
[∵ The ratio of two similar triangles is equal to the ratio of the squares of the corresponding sides].
\(\frac{54}{\Delta \mathrm{DEF}}\) = \(\frac{3^{2}}{4^{2}}\)
∴ △DEF = \(\frac{54 \times 16}{9}\) = 96 cm²

Question 7.
ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm and BP = 3 cm, AQ =1.5 cm, CQ = 4.5 cm. Prove that area of △APQ = \(\frac{1}{16}\) (area of △ABC).
Answer:

Given: △ABC and \(\overline{\mathrm{PQ}}\) – a line segment meeting AB in P and AC in Q.
AP = 1 cm; AQ =1.5 cm;
BP = 3 cm; CQ = 4.5 cm.
\(\frac{AP}{PQ}\) = \(\frac{1}{3}\) ……… (1)
\(\frac{AQ}{QC}\) = \(\frac{1.5}{4.5}\) = \(\frac{1}{3}\) ……..(2)
From (1) and (2),
\(\frac{AP}{BP}\) = \(\frac{AQ}{CQ}\)
[i.e., PQ divides AB and AC in the same ratio – By converse of Basic pro-portionality theorem]
Hence, PQ || BC.
Now in △APQ and △ABC
∠A = ∠A (Common)
∠P = ∠B [∵ Corresponding angles for the parallel lines PQ and BC]
∠Q = ∠C
∴ △APQ ~ △ABC [∵ A.A.A similarity condition]
Now, \(\frac{\Delta \mathrm{APQ}}{\Delta \mathrm{ABC}}=\frac{\mathrm{AP}^{2}}{\mathrm{AB}^{2}}\)
[∵ Ratio of two similar triangles is equal to the ratio of the squares of their corresponding sides].
= \(\frac{1^{2}}{(3+1)^{2}}\) = \(\frac{1}{16}\) [∵ AB = AP + BP = 1 + 3 = 4 cm]
∴ △APQ = \(\frac{1}{16}\) (area of △ABC) [Q.E.D]

Question 8.
The areas of two similar triangles are 81 cm² and 49 cm² respectively. If the altitude of the bigger triangle is 4.5 cm. Find the corresponding altitude of the smaller triangle.

Answer:
Given: △ABC ~ △DEF
△ABC = 81 cm²
△DEF = 49 cm²
AX = 4.5 cm
To find: DY
We know that,
\(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{DEF}}=\frac{\mathrm{AX}^{2}}{\mathrm{DY}^{2}}\)
[∵ Ratio of areas of two similar triangles is equal to ratio of the squares of their corresponding altitudes]

∴ DY = 3.5 cm.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.3

10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.3 Textbook Questions and Answers

Question 1.
A chord of a circle of radius 10 cm. subtends a right angle at the centre. Find the area of the corresponding: (use π = 3.14)
i) Minor segment ii) Major segment
Answer:

Angle subtended by the chord = 90° Radius of the circle = 10 cm
Area of the minor segment = Area of the sector POQ – Area of △POQ
Area of the sector = \(\frac{x}{360}\) × πr²
\(\frac{90}{360}\) × 3.14 × 10 × 10 = 78.5
Area of the triangle = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 10 × 10 = 50
∴ Area of the minor segment = 78.5 – 50 = 28.5 cm²
Area of the major segment = Area of the circle – Area of the minor segment
= 3.14 × 10 × 10 – 28.5
= 314 – 28.5 cm²
= 285.5 cm²

Question 2.
A chord of a circle of radius 12 cm. subtends an angle of 120° at the centre. Find the area of the corresponding minor segment of the circle.
(use π = 3.14 and √3 = 1.732)
Answer:
Radius of the circle r = 12 cm.
Area of the sector = \(\frac{x}{360}\) × πr²
Here, x = 120°

\(\frac{120}{360}\) × 3.14 × 12 × 12 = 150.72
Drop a perpendicular from ‘O’ to the chord PQ.
△OPM = △OQM [∵ OP = OQ ∠P = ∠Q; angles opp. to equal sides OP & OQ; ∠OMP = ∠OMQ by A.A.S]
∴ △OPQ = △OPM + △OQM = 2 . △OPM
Area of △OPM = \(\frac{1}{2}\) × PM × OM

= 18 × 1.732 = 31.176 cm
∴ △OPQ = 2 × 31.176 = 62.352 cm²
∴ Area of the minor segment
= (Area of the sector) – (Area of the △OPQ)
= 150.72 – 62.352 = 88.368 cm²

Question 3.
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm. sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. (use π = \(\frac{22}{7}\))
Answer:
Angle made by the each blade = 115°
Total area swept by two blades
= Area of the sector with radius 25 cm and angle 115°+ 115° = 230°
= Area of the sector = \(\frac{x}{360}\) × πr²
= \(\frac{230}{360}\) × \(\frac{22}{7}\) × 25 × 25
= 1254.96
≃  1255 cm²

Question 4.
Find the area of the shaded region in figure, where ABCD is a square of side 10 cm. and semicircles are drawn with each side of the square as diameter (use π = 3.14).

Answer:
Let us mark the four unshaded regions as I, II, III and IV.

Area of I + Area of II
= Area of ABCD – Areas of two semicircles with radius 5 cm
= 10 × 10 – 2 × \(\frac{1}{2}\) × π × 5²
= 100 – 3.14 × 25
= 100 – 78.5 = 21.5 cm²
Similarly, Area of II + Area of IV = 21.5 cm²
So, area of the shaded region = Area of ABCD – Area of unshaded region
= 100 – 2 × 21.5 = 100 – 43 = 57 cm²

Question 5.
Find the area of the shaded region in figure, if ABCD is a square of side 7 cm. and APD and BPC are semicircles. (use π = \(\frac{22}{7}\))

Answer:
Given,
ABCD is a square of side 7 cm.
Area of the shaded region = Area of ABCD – Area of two semicircles with radius \(\frac{7}{2}\) = 3.5 cm
APD and BPC are semicircles.
= 7 × 7 – 2 × \(\frac{1}{2}\) × \(\frac{22}{7}\) × 3.5 × 3.5
= 49 – 38.5
= 10.5 cm²
∴ Area of shaded region = 10.5 cm

Question 6.
In figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm., find the area of the shaded region, (use π = \(\frac{22}{7}\)).
Answer:
Given, OACB is a quadrant of a Circle.
Radius = 3.5 cm; OD = 2 cm.
Area of the shaded region = Area of the sector – Area of △BOD

= 9.625 – 3.5 = 6.125 cm²
∴ Area of shaded region = 6.125 cm².

Question 7.
AB and CD are respectively arcs of two concentric circles of radii 21 cm. and 7 cm. with centre O (See figure). If ∠AOB = 30°, find the area of the shaded region. (use π = \(\frac{22}{7}\)).

Answer:
Given, AB and CD are the arcs of two concentric circles.
Radii of circles = 21 cm and 7 cm and ∠AOB = 30°
We know that,
Area of the sector = \(\frac{x}{360}\) × πr²
Area of the shaded region = Area of the OAB – Area of OCD

∴ Area of shaded region = 102.66 cm²

Question 8.
Calculate the area of the designed region in figure, common between the two quadrants of the circles of radius 10 cm each, {use π = 3.14)
Answer:
Mark two points P, Q on the either arcs.
Let BD be a diagonal of ABCD
Now the area of the segment

= 28.5 + 28.5 = 57 cm²

Side of the square = 10 cm
Area of the square = side × side
= 10 × 10 = 100 cm²
Area of two sectors with centres A and C and radius 10 cm.
= 2 × \(\frac{\pi r^{2}}{360}\) × x = 2 × \(\frac{x}{360}\) × \(\frac{22}{7}\) × 10 × 10
= \(\frac{1100}{7}\)
= 157.14 cm²
∴ Designed area is common to both the sectors,
∴ Area of design = Area of both sectors – Area of square
= 157 – 100 = 57 cm²
(or)
\(\frac{1100}{7}\) – 100 = \(\frac{1100-700}{7}\)
= \(\frac{400}{7}\)
= 57.1 cm²