Bhagya

Punjab State Board PSEB 8th Class English Book Solutions English Grammar Adjective and Degree Exercise Questions and Answers, Notes.

PSEB 8th Class English Grammar Adjective and Degree

An Adjective is a word which qualifies or adds something to the meaning of a Noun or a Pronoun.

जो शब्द किसी Noun अथवा Pronoun की विशेषता प्रकट करता है उसे Adjective (विशेषण) कहते हैं। Adjective प्रायः Noun के साथ प्रयोग होता है। यह या तो किसी व्यक्ति, पशु अथवा स्थान का वर्णन करता है या किसी संख्या अथवा मात्रा का बोध कराता है; जैसे-

1. Many years ago there was a miser. (संख्या)
2. He never ate good food. (गुण)
3. He spent little money. (मात्रा)

इन वाक्यों में Many, good तथा little शब्द क्रमशः संख्या, गुण एवं मात्रा का बोध कराते हैं। अतः ये शब्द Adjectives हैं।
Adjective मुख्यतः 9 प्रकार के होते हैं-
1. Adjectives of Quality-जो विशेषण किसी व्यक्ति अथवा वस्तु के गुण या प्रकार का बोध कराते हैं, उन्हें Adjectives of Quality (गुणवाचक विशेषण) कहते हैं ; जैसे-good, fresh; bad आदि।

2. Proper Adjectives_Proper Nouns से बनने वाले विशेषण Proper Adjectives (व्यक्तिवाचक विशेषण) कहलाते हैं; जैसे-Indian, Chinese.

3. Adjectives of Quantity-जो विशेषण किसी वस्तु की मात्रा का बोध कराते हैं उन्हें Adjectives of Quantity (परिमाणवाचक विशेषण) कहते हैं। इस प्रकार के Adjective प्रायः कितने (How much) शब्द का उत्तर देते हैं; जैसे-some, no, great.

4. Adjectives of Number-जो विशेषण संख्या अथवा क्रम का बोध कराते हैं, उन्हें Adjectives of Number (संख्यावाचक विशेषण) कहते हैं; जैसे-two, all आदि।

5. Distributive Adjectives-जो विशेषण किसी विशेष वर्ग के व्यक्ति अथवा वस्तु को अलग-अलग प्रकट करें, उन्हें Distributive Adjectives (विभाजन विशेषण) कहते हैं; जैसे-each, either, every आदि।

6. Demonstrative Adjectives-जो विशेषण सम्बन्धित व्यक्ति अथवा वस्तु की ओर संकेत करते हैं, उन्हें Demonstrative Adjectives (संकेतवाचक विशेषण) कहते हैं; जैसे-this, that आदि।

7. Interrogative Adjectives -जो विशेषण प्रश्न पूछने के लिए प्रयोग किया जाता है, उसे Interrogative Adjective (प्रश्नवाचक विशेषण) कहते हैं; जैसे-which, what, whose आदि।

8. Possessive Adjectives- किसी वस्तु या व्यक्ति पर अधिकार व्यक्त करने वाले विशेषण को Possessive Adjectives (सम्बन्धवाचक विशेषण) कहा जाता है; जैसे, My book, Your brother, आदि।

9. Emphasizing Adjectives- जो शब्द Possessive Adjectives पर बल देने के लिए प्रयोग किए जाते __ . हैं, वे Emphasizing Adjectives (निजवाचक विशेषण) कहलाते हैं। इसके लिए प्रायः own शब्द प्रयोग किया जाता है; जैसे,
I became my own servant.

A- Degrees of Adjectives

There are three degrees of comparison:

1. Positive Degree.
2. Comparative Degree.
3. Superlative Degree.

B-Degrees के प्रयोग के नियम

1. Adjective (विशेषण) की पहली Degree (Positive) का प्रयोग उस समय होता है जब कोई तुलना न करनी हो या तुलना की दोनों वस्तुएं (Noun या Pronoun) एक जैसी हों। इसमें as’ का प्रयोग होता है, जैसे-

• Ram is a good boy.
• He is as strong as I.
• This book is as good as Ram’s.

2. दूसरी Degree (Comparative) का प्रयोग उस समय होता है जब दो वस्तुओं या दो व्यक्तियों की तुलना करनी हो। इसमें than’ का प्रयोग होता है; जैसे-

• He is more intelligent than his brother.
• Ram is more intelligent of the two brothers.
• This is better of the two pens.

3. तीसरी Degree (Superlative) का प्रयोग उस समय होता है जब कोई व्यक्ति या वस्तु अन्य सभी व्यक्तियों या वस्तुओं की तुलना में अच्छी हो। इसका प्रयोग एक वस्तु की दो से अधिक वस्तुओं के साथ तुलना करने में होता है; जैसे,-

• She is the most beautiful of all the sisters.
• Ram is the eldest of the four brothers.

C–Use of Some Adjectives

(Note : Adjectives are given in italics.)
1. Your elder brother will captain our team.
2. Our school is older than yours.
3. He is older than I.
4. Who is the eldest member in the family ?
5. It is the oldest temple of the area.
6. My bat is superior to yours.
7. This wood is inferior to that.
8. Is John junior to Gopal ?
9. Ishwar is senior to me.
10. There is little milk in the jug.
11. I have a little money in my pocket.
12. The little money I saved is spent.
13. He makes few mistakes.
14. He makes a few mistakes.
15. The few boys who were present in the morning have run away.
16. He has some books on American literature.
17. I have some friends here in this town.
18. Do you have any books on grammar ?
19. Have you any money?
20. I have not received any help from him.
21. I reached the school later than Shyam.
22. What is the latest score ?
23. The latest issue of this magazine is awaited.
24. Mohan narrated two tales but I found the latter more interesting.
25. The last train leaves at 11-50 at night.
26. The last batsman played extremely well.
27. We have much amount to spend on clothing.
28. Many students in our class have won prizes.

Exercises (Solved)

I. Write out the story, putting one of the adjectives from the list in each blank:
Long delightful high sour ripe warm hungry

It was a ………….. day. A fox had just entered a garden. Among the many……….. things in the garden was a vine laden with bunches of grapes. One of them was a ……..3…….. bunch which hung invitingly over the fox’s head. The very sight of it made the fox feel ……..4……… He leapt up at the grapes several times. But they were too ……..5……. for him to reach. After a ……..6…… time he decided to give up trying. He went away declaring that the grapes were ……..7…….. This is the origin of the expression ‘sour grapes’.
Hints:
1. warm
2. delightful
3. ripe
4. hungry
5. high
6. long
7. sour.

II. Fill in the blanks with the comparative or superlative degree of the adjectives given in the brackets:

1. The Taj is ……………… than any other building. (beautiful)
2. This is the ……………… book I have ever read. (good)
3. The climate of Shimla is ………………. than that of Delhi. (good)
4. Mohan is the …………….. boy in his class. (tall)
5. Hardwar is one of the ……………… places for the Hindus. (holy)
6. Ram is …………… than his sister. (intelligent)
7. Mumbai is ……………… than any other city in Maharashtra. (large)
8. This dress is ………………. than that. (costly)
9. Radha has …………….. sense than her sister. (much)
10. My table is the ……………… of all. (big)
Hints:
1. more beautiful
2. best
3. better
4. tallest
5. holiest
6. more intelligent
7. larger
8. costlier
9. more
10. biggest.

III. Put the correct word in the blank spaces:

(a) 1. He came …………….. than Sham. (later, latter)
2. The ……………… train leaves the station at 11 p.m. (last, later)
3. What is the …………….. news ? (latest, last)
4. I met Shila and Rita yesterday. The former promised to come to the party but the ………………… gave a flat refusal. (later, latter)
5. Ram is ………………… than Hari. (elder, older)
6. He is the ……………. member of the family. (eldest, oldest)
7. The ……………… brother controls the entire business. (older, elder)
8. Delhi is ………………… from Ludhiana than Ambala. (further, farther)
9. You are not going to get any ……………… help from us. (further, farther)
10. History is ……………….. difficult than Mathematics. (less, fewer)
11. ……………… boys attended the meeting this time. (less, fewer)
12. These chairs cost sixty rupees ……….. (every, each)
13. ……………….. man loves him. (every, each)
14. His days are numbered; he has ……………… hope to survive. (little, a little)
15. …………….. effort on your part will help me a lot. (little, a little)
Hints :
1. later
2. last
3. latest
4. latter
5. older
6. eldest
7. elder
8. farther
9. further
10. less
11. Fewer
12. each
13. Every
14. little
15. a little.

IV. Correct the following sentences:

1. The poors always deserve mercy.
2. The three friends helped each other.
3. Shakespeare is greater than any dramatist.
4. There were no less than fifty persons present.
5. Every of the two friends is ill.
6. Iron is more useful than any metal.
7. My hat is more superior than yours.
8. The three brothers understand each other well.
9. He is a best boy in the class.
10. The higher you go the coolest it is.
Hints:
1. The poor always deserve mercy.
2. The three friends helped one another.
3. Shakespeare is greater than any other dramatist.
4. There were no fewer than fifty persons present.
5. Each of the two friends is ill.
6. Iron is more useful than any. other metal.
7. My hat is superior to yours.
8. The three brothers understand one another well.
9. He is the best boy in the class.
10. The higher you go, the cooler it is.

Punjab State Board PSEB 8th Class English Book Solutions English Grammar The Verb Exercise Questions and Answers, Notes.

PSEB 8th Class English Grammar The Verb

A word that describes action or state of a person, place, thing, or an event is called verb.
वह शब्द जो शब्द किसी व्यक्ति, स्थान, वस्तु अथवा घटना की क्रिया अथवा स्थिति के बारे में कुछ बतलाए उसे Verb (क्रिया) कहते हैं।

Verbs मुख्यतः तीन प्रकार के होते हैं:

1. Transitive
   Ram took tea.
2. Intransitive
   Mohan laughs.
3. Auxiliary

Primary Auxiliaries

Be
1. Be का प्रयोग Continuous Tense बनाने के लिए किया जाता है:-
He is playing. I was playing.

2. Passive Voice बनाने के लिए
The horse was sold.

3. Infinitive सहित be का प्रयोग किसी आयोजन, प्रबन्ध अथवा समझौते के लिए किया जाता है, जैसे-
I am to go to Delhi tomorrow.

4. आदेश के लिए
You are to see me in the evening.

5. Be का प्रयोग भूतकाल में Perfect Infinitive के रूप में भी किया जाता है। इस प्रयोग में किसी ऐसी व्यवस्था का प्रदर्शन किया जाता है जिसको व्यवहार में नहीं लाया जा सका।
He was to have finished his work last evening but had to leave it on the arrival of the guests.

Have
1. Have का प्रयोग Perfect Tenses के लिए किया जाता है।
He has done his.work.
They have been doing their work.

Do
1. Do का प्रयोग Simple Present तथा Simple Past के नकारात्मक तथा प्रश्नवाचक वाक्य बनाने के लिए किया जाता है।
I do not play.
Do I play?
He does not play.
Does he play?
He did not come.
Did he come?

2. Do का प्रयोग किसी शब्द की दोहराई को रोकने के लिए किया जाता है।
Does she come daily? Yes, she does.
यहां does का अर्थ come शब्द का बोध कराता है अर्थात् Yes, she does. का अर्थ है-हां, वह आती है।

3. Do का प्रयोग किसी बात पर जोर देने के लिए भी किया जाता है।
Please do come.
You do look handsome.

Modal Auxiliaries

1. May and Might

May का प्रयोग होता है-
(a) अनुमति (Permission) देने या लेने के लिए; जैसे,
May I go now?
You may go now.
He may go now.

(b) सम्भावना (Probability) (ऐसी सम्भावना के लिए जिसमें कुछ घटित हो भी सकता है या नहीं भी हो सकता); जैसे,
It may rain today.
It may not rain today.

(c) इच्छा (Wish) को व्यक्त करने के लिए; जैसे,
May you succeed!

(d) May के Past Tense (भूतकाल) के रूप में
He said he might come the next day. (Probability)
He asked if he might borrow my bicycle. (Permission)

(e) अनुमति के लिए (यदि अनुमति वर्तमान में मांगी जा रही हो और इसके मिलने की सम्भावना बहुत ही कमहो); जैसे,
Might I ask a question ?

(f) सम्भावना के लिए (यदि सम्भावना वर्तमान में की जा रही हो और सम्भावना नाममात्र हो); जैसे,
It might not rain tonight.

2. Can and Could

Can का प्रयोग होता है-
(a) अनुमति के लिए (यदि यह अनुमति साधारण बातचीत के दौरान मांगी जाए उस समय can प्राय: may का स्थान लेता है); जैसे,
Can I smoke here?
Yes, you can.
Or
No, you can’t. (यदि अनुमति न देनी हो या यदि ऐसा करने की अनुमति न हो)
(b) सम्भावना के लिए
Pets can be troublesome.
Electricity can be dangerous.

(c) क्षमता या योग्यता के लिए
Can you see the bird in the tree?
I can do this for you.

Could का प्रयोग किया जाता है:
(a) Can के भूतकाल (Past Tense) के रूप में:
He could drive a car at the age of ten.
अतीत में अस्थायी अथवा निजी सफलता के लिए हम could के स्थान पर was or was not able का प्रयोग करते हैं:
He was able to climb the hill without any difficulty.
परन्तु हम could या couldnt का प्रयोग करते हैं:
He couldn’t make much progress in his studies while at school.

(b) विनयपूर्वक प्रार्थना के लिए (यदि यह प्रार्थना वर्तमान में प्रश्नवाचक वाक्य के रूप में की गई हो); जैसे,-
Could I ask you a question ?
Note- भूतकाल की सम्भावना (जो पूर्ण न हुई हो) प्रकट करने के लिए Could का भूतकाल अर्थात् Could have का प्रयोग होता है; जैसे,
You could have caught the train, if you hurried.”

3. Would

Would का प्रयोग होता है-
(a) Will तथा Shall के भूतकाल (Past Tense) के रूप में; जैसे,
He said he would be moving into his new house next month.
(He said, “I shall be moving……….”)

(b) विनयपूर्वक प्रार्थना (a polite request) के लिए; जैसे,
Would you please close the door?

(c) किसी की इच्छा जानने के लिए (an enquiry about someone’s wish; जैसे,-
Would you like to have a cup of tea ?

(d) ऐसी घटनाएं (Happenings) या कार्य (Actions) जो भूतकाल में बार-बार या कभी-कभी घटित होती या होते थे, जैसे-
Sometimes he would come home drunk and shout at his children.
Note-भूतकाल में सम्भावना व्यक्त करने के लिए would का past tense, would have का प्रयोग किया जाता है; जैसे-,
Had he run fast, he would have caught the train.

4. Should

Should का प्रयोग होता है-
(a) कर्त्तव्य (Duty) अथवा नैतिक उत्तरदायित्व (Moral obligation) प्रकट करने के लिए; जैसे-
We should stand united.

(b) परामर्श (Advice) अथवा निर्देश (Instructions) अथवा दृष्टिकोण (Opinion) व्यक्त करने के लिए; जैसे-
You should see a doctor at once.

5. Must

Must का प्रयोग होता है-
(a) आवश्यकता (Necessity) प्रकट करने के लिए; जैसे,
You must save money for old age.

(b) उत्तरदायित्व (Obligation) प्रकट करने के लिए; जैसे,
You must pay your fees before Tuesday.

(c) तर्कसंगत निष्कर्ष अथवा निश्चितता प्रकट करने के लिए; जैसे,
The fan in the headmaster’s room is on. He must be in.
Note:
1. निषेधवाचक वाक्यों के लिए must not का प्रयोग करना चाहिए; जैसे,
You must not come here again.
2. भूतकाल में निश्चितता अथवा निष्कर्ष प्रकट करने के लिए must का स्थान must have ले लेता है; जैसे,
Mohan must have passed. He had been working quite hard.
3. भूतकाल में आवश्यकता व्यक्त करने के लिए must का Past Tense, had to का प्रयोग करना पड़ता है; जैसे,
The candidates had to appear for an interview after the written test.

6. Need

Need का प्रयोग दो प्रकार का होता है-
(i) नियमित verb के रूप में
We need a lot of money for the wedding.

(ii) Modal के रूप में (ऐसा केवल प्रश्नवाचक तथा नकारात्मक वाक्यों में होता है); जैसे,
Need I come again?
No, you needn’t.
Note- भूतकाल में Need का स्थान need का Past Tense, need have ले लेता है; जैसे,
Need he have taken all that trouble ?

7. Ought

Ought का प्रयोग प्रायः Should जैसी भावना व्यक्त करने के लिए होता है:
(a) परामर्श अथवा सिफारिश व्यक्त करने के लिए-जैसे,
You ought to read this book. It is very exciting.

(b) नैतिक अथवा सामाजिक कर्तव्य के लिए-जैसे,
People ought not to take or offer bribes.

(c) निष्कर्ष के लिए-जैसे,
The train ought to be at the platform. Passengers are coming out of the station with their luggage.

Note-
1. Ought का Past Tense, Ought to have होता है; जैसे- The driver ought to have been more careful. (This has the meaning that the driver
was not careful.)
2. Indirect Speech में Past Tense की क्रिया के बाद ought में कोई परिवर्तन नहीं किया जाता; जैसे, The captain of the school team stated that they ought to win the trophy.

8. Shall.

Shall का प्रयोग (प्रथम पुरुष-I Person) के साथ होता है:

1. भविष्य के लिए-I shall go to Delhi tomorrow.
2. प्रार्थना के लिए-Shall we use this pen?
3. इच्छा जानने के लिए-Shall I accompany you to your house?
4. इच्छा व्यक्त करने के लिए-I shall gladly do it.
5. चेतावनी या धमकी देने के लिए-You shall be fined for coming late. (इस अर्थ में shall का प्रयोग II या III person के साथ किया जाता है।)
6. आदेश देने के लिए-You shall not come here again.
7. वचन के लिए-He shall be rewarded for his good work. (इस अर्थ में भी shall का प्रयोग II या III person में होता है।)

9. Will

Will का प्रयोग (II या III Person के साथ) होता है-

1. भविष्य के लिए-You will get good marks.
2. प्रार्थना के लिए-Will you help me?
3. निमन्त्रण के लिए-Will you dine with us tonight?
4. आदेश के लिए- “You will reach here at 8,” said the Principal.
5. आदत के लिए-A dog will always remain faithful to his master.
6. भविष्यवाणी के लिए- You will fail if you do not work hard.

(I Person के साथ Will का प्रयोग)

1. वाचन देने के लिए-I will see you tomorrow without fail.
2. संकस्थ के लिए- I will not bow before them.
3. इरादा व्यक्त करने के लिए- From here we will go to the circus ground direct.

Forms of the Verb

Verb के निम्नलिखित चार रूप होते हैं:

1. Present form (First form)
2. Past form (Second form)
3. Past Participle form (Third form)
4. Present Participle form (ing form)
   आगे कुछ महत्त्वपूर्ण verbs की First, Second एवं Third forms दी गई हैं। इन्हें ध्यान से पढ़ें एवं AIG off :

Orms (Conjugation of the Verb

Agreement of The Verb With The Subject

अंग्रेजी भाषा को शुद्ध लिखने के लिए वाक्य में Subject (कर्ता) का Verb (क्रिया) के साथ ठीक ताल-मेल होना अनिवार्य है। कहने का अभिप्राय यह है कि Verb वाक्य के Subject के वचन (Number) तथा पुरुष (Person) के अनुसार होना चाहिए; जैसे,
Ram goes.
Ram and Sham go.
They go.
She goes.
The boys are going
A boy is going.
They were going.
He/She was going.
Subject तथा Verb के उपयुक्त तालमेल के लिए अग्रलिखित नियमों का अध्ययन करें:

(1) यदि दो या दो से अधिक एकवचन Subjects को and से जोड़ा जाए तो प्रायः उनके साथ बहुवचन Verb का प्रयोग होता है; जैसे,
Mohan and Ram are friends.
Mohan, Sham and Ram play together.
He and his brother have done the work.
Sham and Gopal were there in the meeting.
He and his friend do not work together.

(2) यदि दो एकवचन संज्ञाएं (Nouns) किसी एक व्यक्ति या किसी एक चीज़ का बोध कराती हों तो ऐसी एकवचन संज्ञाओं (Nouns) के साथ एकवचन Verb का प्रयोग होता है; जैसे,-
My brother and helper has arrived.
The leader and speaker is sitting on the stage.
A great Gandhian and leader is dead.

नोट-
(i) यदि एक ही व्यक्ति का बोध कराना हो तो Article का प्रयोग दोनों संज्ञाओं अथवा विशेषणों में से पहली संज्ञा या विशेषण के साथ ही किया जाता है। ऊपर के वाक्य (2) में केवल leader से पहले
The और वाक्य (3) में Gandhian से पहले A का प्रयोग किया गया है।

(ii) यदि वाक्य दो संज्ञाओं से दो अलग व्यक्तियों का बोध कराता हो तो दोनों संज्ञाओं के साथ उचित Article का प्रयोग किया जाता है। ऐसे वाक्य में बहुवचन क्रिया का प्रयोग होता है।
The leader and the speaker are sitting on the stage.

(3) यदि दो Subjects मिलकर किसी एक ही विचार (one idea) को व्यक्त करते हों तो एकवचन क्रिया (Verb in Singular) का प्रयोग होता है; जैसे,-
Fish and meat is the food of the Bengalis.
Slow and steady wins the race.
Twenty kilometres is not a long distance.
The horse and the carriage has arrived.
Early to bed and early to rise makes a man healthy, wealthy and wise.

(4) यदि एकवचन कर्ता (Subject) से पूर्व each अथवा every का प्रयोग किया जाए तो एकवचन क्रिया (Verb in Singular) का प्रयोग होता है; जैसे,-
Every boy and girl was well dressed.
Every man, woman and child was present.
Each day and each hour takes its account.

(5) यदि दो या दो से अधिक एकवचन Subjects को or, nor, either…….. or, neither………..nor से जोड़ा जाए तो एकवचन क्रिया का प्रयोग होता है; जैसे,
Work or play has no difference for me.
Either Mohan or Sohan has broken the slate.
Neither he nor I was present.
Neither food nor water was available there.

(6) जब or तथा nor से जोड़े गए Subjects का वचन भिन्न-भिन्न हो तो बहुवचन क्रिया का प्रयोग किया जाता है तथा बहुवचन Subject को क्रिया के साथ रखा जाता है; जैसे,
Sham or his brothers have broken the slate.
Neither the monitor nor the students were present.
Neither the speaker nor the listeners were serious.
Either the man or his sons have gone wrong.

(7) जब विभिन्न कर्ता (Subjects) or अथवा nor से जुड़े हुए हों, तो क्रिया (Verb) अपने निकटतम् Subject के Person (पुरुष) से मेल खाती है; जैसे,
Either he or I have to go to Delhi.
Neither you nor he is to blame.
Either your father or you are paying for it.

(8) जब Subjects के वचन अथवा पुरुष (Person) में अन्तर हो और वे and से जुड़े हों तो क्रिया (Verb) बहुवचन में प्रयोग होती है; जैसे,
He and I are good friends.
My father and I have jointly done this.
You and he are always together.
You and I are always in time.

(9) समूहवाचक संज्ञा (Collective Noun) के साथ पूरे समूह का भाव व्यक्त करने के लिए एकवचन क्रिया (Verb in Singular) का प्रयोग होता है। परन्तु यदि समूह के व्यक्तियों (individuals) का बोध कराना हो तो बहुवचन क्रिया (Verb in Plural) का प्रयोग होता है; जैसे,
The Committee has chosen its Chairman.
The fleet has set sail.
The Council is one in its opinion for the choice of Chairman.
The Council (councillors) are not one in their opinion for the choice of Chairman.
There is a large number of boys in the class.
A number of players were playing foul.
The police were called out.

(10) कुछ संज्ञाएं देखने में बहुवचन लगती हैं, परन्तु वे अर्थ की दृष्टि से एकवचन होती हैं। ऐसी संज्ञाओं के साथ एकवचन क्रिया (Verb) लगती है; जैसे,
The news is good.
Physics is an easy subject.
The wages of sin is death.
Politics is a dirty game.

(11) कुछ संज्ञाएं देखने में एकवचन लगती हैं परन्तु अर्थ में बहुवचन होती हैं। ऐसी संज्ञाओं के साथ बहुवचन क्रिया (Verb in Plural) का प्रयोग होता है; जैसे,
The gentry were invited to the party.
These poultry are mine.
The cattle are grazing in the field.
People are shouting for nothing.

(12) क्रिया का मेल हमेशा वास्तविक Subject के साथ ही कराना चाहिये। यह जरूरी नहीं है कि क्रिया का निकटतम Subject ही वास्तविक हो। इसलिए वास्तविक Subject को ध्यान में रखना आवश्यक है; जैसे,
Each of the players was rewarded.
Neither of the women was tall.
Each one of plots is for sale.
The quality of these machines is good.
One of the boys was my friend.

(13) यदि Subject में दो Nouns या Pronouns ‘with’ या ‘as well as’ के साथ जुड़े हों, तो क्रिया उनमें से चहले Noun या Pronoun से मेल खती हहै; जैसे
Ram, with all the members of his family, was present.
He, as well as his friends, is present.
Mohan, and not his friend, has done this.
I, as well as he, have worked hard.

(14) जब किसी क्रिया का Subject कोई Relative Pronoun हो, तो क्रिया Relative Pronoun से ठीक पहले वाले Subject के अनुसार लगती है; जैसे
I, who am your guide, shall stand by you.
You, who are my guide, are expected to stand by me.
He is among the persons who are against me.

Exercises (Solved)

I. Say whether the Verbs in the following sentences are Transitive or Intransitive. If the Verb is Transitive name the Object.
(T = Transitive, I = Intransitive, O = Object)

1. He killed a snake.
2. The fire burns brightly.
3. Birds fly in the air.
4. I drink tea five times a day.
5. A blind man cannot see.
6. He came here last night.
7. She has lost her books.
8. He walked twenty miles.
9. A woman came to buy tea.
10. You should teach him a lesson.
Hints :
1. T, O = a snake
2. In.
3. In.
4. T, O = tea
5. In.
6. In.
7. T, O = her books
8. In.
9. T, O = tea.
10. T,O = him, a lesson.

II. Pick out the Auxiliary Verbs in the following sentences:

1. I am writing a novel.
2. He has done wrong.
3. He did not come in time.
4. He is treated badly.
5. I had heard of this before.
6. I shall leave for Mumbai tomorrow.
7. The boy said that he might have done so.
8. He must have gone home.
9. Did he ask you to write to him ?
10. I can fly an aeroplane.
Hints:
1. am
2. has
3. did
4. is
5. had
6. shall
7. might have
8. must have
9. Did
10. can.

III. Put the correct Verb in the blanks:

(a) 1. The tallest of these boys ……….. next door to me. (live, lives)
2. All the players in my team ……….. done well. (has, have)
3. The cost of all types of pens ………… gone up. (has, have)
4. The toys that were bought by my son ………… really useful. (are, is)
5. He ……….. regularly. (work, works)
Hints:
1. lives
2. have
3. has
4. are
5. works.

(b) 1. Slow and steady ………. the race. (win, wins)
2. Bread and butter ………. what they want. (is, are)
3. Time and tide ………… for none. (wait, waits)
4. Oil and water ……….. mix. (does not, do not)
5. Tobacco and alcohol ………. injurious to health. (is, are)
Hints:
1. wins
2. is
3. wait
4. do not
5. are.

(c) 1. Neither the captain nor the soldiers ………. been arrested. (has, have)
2. Neither he nor his servants ………… honest. (was, were)
3. Either Rajinder or his parents ……….. responsible for this. (was, were)
4. Neither the Principal nor the lecturers ………………. present at the meeting. (was, were)
5. He or his friends ……….. to blame. (was, were)
Hints:
1. have
2. were
3. were
4. were
5. were.

(e) 1. You, as well as he ……….. innocent. (is, are)
2. Not only the workman but the supervisor also …………………. been dismissed. (has, have)
3. He, as well as you, ………… innocent. (is, are)
4. The workmen, with their leader, ………… been arrested. (have, has)
5. The gallery, with its beautiful pictures, ………… a great attraction. (is, are)
Hints:
1. are
2. have
3. is
4. have
5. is.

(g) 1. None but the brave ………… the fair. (deserve, deserves)
2. Each day and each hour ………… its own importance. (has, have)
3. None of the ships ………… rescued from the storm. (was, were)
4. One of my friends ……….. the owner of this factory. (is, are)
5. Either of these two proposals ……….. acceptable to me. (is, are)
Hints:
1. deserve
2. has
3. was
4. is
5. is.

(h) 1. The United States ……….. a prosperous country. (is, are)
2. The news of the flood ……….. caused great anxiety. (has, have)
3. The Arabian Nights ………. interesting stories. (contain, contains)
4. Good crockery ………. expensive. (is, are)
5. The West Indies ………. a land of great cricketers. (is, are)

Punjab State Board PSEB 8th Class English Book Solutions English Vocabulary Prefixes and Suffixes Exercise Questions and Answers, Notes.

PSEB 8th Class English Vocabulary Prefixes and Suffixes

A- Prefixes

Word / Example:
in + active = inactive
in + accurate = inaccurate
in + capable = incapable
in + complete = incomplete
in + correct = incorrect
in + credible = incredible
in + curable = incurable
in + definite = indefinite
in + dependent = independent
in + direct = indirect

in + efficient = inefficient
in + finite = infinite
in + firm = infirm
in + flexible = inflexible
in.+ formal = informal
in + gratitude = ingratitude
in + hospitable = inhospitable
in + human = inhuman
in + secure = insecure
in + significance = insignificance
in + sincerity = insincerity
in + sufficient = insufficient
in + tangible = intangible
in + temperate = intemperate
in + tolerable = intolerable
in + visible = invisible
im + mature = immature
im + mobile = immobile
im + mobilize = inmodilize
im + moderate = immoderate
im + polite = impolite
im + possible = impossible
im + proper = improper
dis + advantage = inadvantage
dis + agree = disagree
dis + appear = disappear
dis + arm = disarm
dis + honour = dishonour
dis + inherit = disinherit
dis + loyal = irloyal
dis + obedient = ofirobedient
dis + orderly = orderly
dis + place = displace
dis + please = displease.
un + able = unable
un + accustomed = unaccustomed
un + affected = unaffected
un + armed = unarmed
un + avoidable = unavoidable
un + certain = uncertain
un + chartered = unchartered
un + common = uncommon
ir + reparable = irreparable
ir + resolute = irresolute
ir + responsible = irresponsible
ir + resistible = irresistible
ir + reverent = – irreverent
ir + revocable = irrevocable
non + aggression = nonaggression
non + aligned = nonaligned
non + cooperate = noncooperate
non + conformist = nonconformist
non + contentious = noncontentious
non + payment = nonpayment
non + poisonous = nonpoisonous
non + political = nonpolitical
non + resident = nonresident
non + religious = nowreligious
non + sense = nonsense
non + smoker = nonsmoker
non + taxable = nontaxable
non + vegetarian = nonvegetarian
non + violence = nonviolence
de + centralize = decentralize
de + increase = decrease
de + inflate = deflate
de + ascend = descend
il + legal = illegal
il + legible = illegible
il + literate = illiterate
il + literacy = illiteracy

B- suffixes

Word:
-able
Examples:
favour + able = favourable
objection + able = objectionable
reason + able = reasonable
remark + able = remarkable
value + able = valueable
prefer + able = preferable
knowledge + able = knowledgeable
notice + able = noticeable
like + able = likeable.

Word:
-ance
Examples:
accept + ance = acceptance
allow + ance = allowance
defy + ance = defiance
guide + ance = guidance
import + ance = importance
resist + ance = resistance
appear + ance = appearance
accept + ance = acceptance
perform + ance = performance.

Word:
-ment
Examples:
fulfill + ment = fulfillment
agree + ment = agreement
attach + ment = attachment
fulfil + ment = fulfilment
pay + ment = payment
refresh + ment = refreshment
settle + ment = settlement
advertise + ment = advertisement
entertain + ment = entertainment
misplace + ment = misplacement
replace + ment = replacement
refresh + ment = refreshment
pay + ment = payment

Word:
-fill
Examples:
law + ful = lawful
beauty + fill = beautiful
care + ful = careful
grace + fill = graceful
hope + fid = hopeful
use + ful = useful
success + fill = successful
power + fill = powerful
doubt + ful = doubtful
pain + fill = painful.

Word:
-er
Examples:
begin + er = beginner
build + er = builder
invade + er = invadeer
make + er = maker
command + er = commander
fast + er = faster
air + er = airer
act + er = acter
bike + er = biker
able + er = abler
old + er = older.

Word:
-less
Examples:
art + less = artless
hope + less = hopeless
noise + less = noiseless
taste + less = tasteless
weight + less = weightless
help + less = helpless
wire + less = wireless
need + less = needless
home + less = homeless
number + less = numberless
mother + less = motherless
driver + less = driverless
system + less = systemless
power + less = powerless
thank + less = thankless
pain + less = painless
fear + less = fearless.

Word:
-y
Examples:
cloud + y = cloudy
fault + y = faulty
sand + y = sandy
sleep + y = sleepy
rain + y = rainy
cream + y = creamy
noise + y = noisy.
bush + y = bushy
juice + y = juicy
snow + y = snowy
rock + y = rocky.

Word:
-en
Examples:
gold + en = golden
silk + en = silken
light + en = lighten
soft + en = soften
bright + en = brighten
straight + en = straighten
earth + en = earthen
wood + en = wooden
wool + en = woolen
dark + en = darken
sharp + en = sharpen

Word:
-ous
Examples:
grace + ous = grancious
nerve + ous = nervous
courage + ous = courageous
mystery + ous = mysterious
danger + ous = dangerous
teacher + ous = teacherous
continue + ous = continuous
luxury + ous = luxurious
glory + ous = glorious

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 10 Practical Geometry Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 10 Practical Geometry Ex 10.2

1. With the help of a ruler, construct line segments of given lengths:

Question (i)
5 cm
Solution:
Steps of Construction.

1. Place the ruler on a paper and hold it firmly.
2. Mark a point A with the pencil against 0 of the ruler and the point B against 5 cm mark of the ruler.

3. Join the two points A and B by moving the pencil along the ruler.

Thus, AB = 5 cm is the required line segment.

Question (ii)
6.5 cm
Solution:
Steps of Construction.

1. Place the ruler on a paper and hold it firmly.
2. Mark a point A with the pencil against 0 of the ruler and another point B against 6.5 cm mark of the ruler.

3. Join the two points A and B by moving the pencil along the ruler. Thus, AB = 6.5 cm is the required line segment.)

Question (iii)
5.2 cm
Solution:
Steps of Construction.

1. Place the ruler on a paper and hold it firmly.
2. Mark a point P with the pencil against 0 of the ruler and another point Q against 5.2 cm mark of the ruler.

3. Join the two points P and Q by moving the pencil along the ruler. Thus, PQ = 5.2 cm is the required line segment.

Question (iv)
6.8 cm
Solution:
Steps of Construction.

1. Place the ruler on a paper and hold it firmly.
2. Mark a point C with the pencil against o of the ruler and another point D against 6.8 cm mark of the ruler.
3. Join the two points C and D by moving the pencil along the ruler.

Thus CD = 6.5 cm is the required line segment.

Question (v)
9.7 cm
Solution:
Steps of Construction.

1. Place the ruler on a paper and hold it firmly.
2. Mark a point L with the pencil against zero mark of the ruler and another point M against 9.7 cm mark of the ruler.

3. Join the two points L and M by moving the pencil along the ruler.

4. Thus, LM = 9.7 cm is the required line segment.

Question (vi)
8.4 cm.
Solution:
Steps of Construction.

1. Place the ruler on a paper and hold it firmly.
2. Mark a point X with the pencil against zero of the ruler and another point Y against 8.4 cm mark of the ruler.

3. Join the two points X and Y by moving the pencil along the ruler.

4. Thus XY = 8.4 cm is the required line segment.

2. Draw line segments given in Question by using a ruler and compasses.
Solution:
(i) Steps of Construction.

1. Draw a line l and mark a point A on it.

2. Place the compasses pointer on the zero mark of the ruler. Open it to place the pencil point upto the 5 cm mark.
3. Now without changing the opening of compasses, place the pointer on A and draw an arc to cut the line l at point B.

4. Thus AB = 5 cm is the required line segment.

(ii) Steps of Construction.

1. Draw a line l and mark a point A on it.

2. Place the compasses pointer on the zero mark of the ruler. Open it to place the pencil point upto the 6.5 cm mark.
3. Now without changing the opening of compasses, place the pointer on A and draw an arc to cut the line l at point B.

4. Thus AB = 6.5 cm is the required line segment.

(iii) Steps of Construction.

1. Draw a line l and mark a point A on it.

2. Place the compasses pointer on the zero mark of the ruler.
Open it to place the pencil point up to the 5.2 cm mark.
3. Now without changing the opening of compasses, place the pointer on A and draw an arc to cut the line l at point B.

4. Thus AB = 5.2 cm is the required line segment.

(iv) Steps of Construction.

1. Draw a line l and mark a point P on it.

2. Place the compasses pointer on the zero mark of the ruler. Open it to place the pencil point upto the 6.8 cm mark.
3. Now without changing the opening of compasses, place the pointer on P and draw an arc to cut the line l at point Q.

4. Thus PQ = 6.8 cm is the required line segment.

(v) Steps of Construction.

1. Draw a line l and mark a point L on it.

2. Place the compasses pointer on the zero mark of the ruler. Open it to place the pencil point upto the 9.7 cm mark.
3. Now without changing the opening of compasses, place the pointer on L and draw an arc to cut the line l at point M.

4. Thus LM = 9.7 cm is the required line segment.

(vi) Steps of Construction.

1. Draw a line l and mark a point X on it.

2. Place the compasses pointer on the zero mark of the ruler. Open it to place the pencil point upto the 8.4 cm mark.
3. Now without changing the opening of compasses, place the pointer on X and draw an arc to cut the line l at point Y.

4. Thus XY = 8.4 cm is the required line segment.

3. Construct AB of length 8.4 cm. From it cut off AC of length 5.3 cm. Measure BC.
Solution:
Steps of Construction.
1. Draw a line l and mark a point A on it.

2. Place the compasses pointer on the zero mark of the ruler. Open it to place the pencil point upto the 8.4 cm mark.
3. Now without changing the opening of compasses, place the pointer on A and draw an arc to cut the line l at point B.
4. Thus AB = 8.4 cm.
5. Now open the compasses equal to AC = 5.3 cm.
6. Place the metal point of compasses on A. Then point with pencil point draw an arc, interecting the line l at C.
7. Now AC = 5.3 cm.
8. By measurement BC = 3.1 cm.

4. Draw two line segments AB and CD of lengths 8.4 cm and 4.5 cm respectively. Construct the line segments of the following lengths:

Question (i)
AB + CD
Solution:
AB + CD = 8.4 cm + 4.5 = 12.9 cm. We can construct a line segment AD of length 12.9 cm. Using a ruler and compasses.

Steps of Construction:

1. Draw line segment AB = 8.4 cm and segment CD = 4.5 cm line.

2. Draw a line l longer than combined length of AB and CD i. e. 12.9 cm. Mark a point A on it.

3. Take the compasses and measure AB. Without changing the opening of the compasses place its needle at A and draw an arc intersecting line l at B.

4. Again adjust the 8.4 cm compass and measure the line segment CD.
5. Without changing the opening of the compasses place the pointer at B on the line l and draw an arc cutting the line l at P.

6. Then AP is the required line segment whose length is equal to the sum of lengths of line segments AB and CD.
7. On measurement AP = 12.9 cm.

Verification:
AP= AB + BP
= AB + CD
= 8.4 cm + 4.5 cm
= 12.9 cm.

Question (ii)
AB – CD
Solution:
Steps of Construction:

1. Draw a line l and mark point A on it.
2. Take the compasses and measures AB. Without changing the opening of the compasses place its needle at A and draw an arc intersecting l at B.

3. Again adjust the compasses and measure the line segment CD.
4. Without changing the opening of the compasses place the pointer at B and draw an arc intersecting AB at Q.
5. AQ is the required line segment whose length is equal to difference of lengths of line segments AB and CD.

6. On measurement AQ = 3.9 cm.

Verification:
AQ = AB – OB
= AB – CD
= 8.4 cm – 4.5 cm
= 3.9 cm.

Question (iii)
2 CD.
Solution:
Steps of Construction.

1. Draw a line l and mark point P on it.

2. Open out the compasses and adjust measure CD without changing the opening of the compasses place the needle at point P and draw an arc intersecting line l at point X such that PX = CD.

3. Now again without changing the opening of compasses place the needle at point X and draw an arc cutting the line l at Q. Such that XQ = CD.

Thus, PQ is the required line segment which is equal to 2 CD.
4. Measure PQ, PQ = 9 cm.
Verification.
Now, PQ = PX + XQ
= CD + CD = 2CD
= 2 × 4.5 cm = 9 cm
Hence, PQ = 2 CD.

5. Draw two line segments PQ and RS of lengths 6.4 cm and 3.6 cm respectively. Construct the line segments of the following lengths:

Question (i)
PQ + RS
Solution:
Steps of Construction.

1. Draw line segment PQ = 6.4 cm and line segment RS = 3.6 cm.

2. Draw a line l longer than combined length of PQ and RS i. e. 10 cm. Take a point P on it.

3. Take the compasses and measure PQ. Without changing the opening of the compasses place its needle at P and draw an arc cutting line l at Q.

4. Again adjust the compasses and measure the line segment RS.
5. Without changing the opening of the compasses place the pointer at Q on the line l and draw an arc cutting the line l at R.

Thus PT is the required line segment whose length is equal to the sum of line segments PQ and RS.
7. Measure PT = 10 cm.

Verification:

PT = PQ + QT
= PQ + RS
= 6.4 cm + 3.6 cm
= 10 cm

Question (ii)
PQ – RS
Solution:
Steps of Construction:

1. Draw a line l and mark a point P on it.

2. Take the compasses and measure PQ. Without changing the opening of the compasses place its needle at P and draw an arc intersecting l at Q.

3. Again adjust the compasses and measure the line segment RS.
4. Without changing the opening of the compasses place the pointer at Q and draw an arc intersecting PQ at T.

5. PT is the required line segment whose length is equal to difference of lengths of line segments PQ and RS.
6. Measure PT, PT = 2.8 cm
Verification:
PT = PQ – QT
= PQ – RS
= 6.4 cm – 3.6 cm
= 2.8 cm.

Question (iii)
2 PQ
Solution:
Steps of Construction:

1. Draw a line l and mark a point A on it.

2. Take the compasses and measure PQ. Without changing the opening of the compasses place its needle at point A and draw an arc intersecting line l at point B such that AB = PQ.

3. Now again without changing the compasses, place the needle at point B and draw an arc cutting the line l at C such that BC = PQ.

4. Then AC is the required line segment whose length is equal to 2 PQ.
5. Measure AC, AC = 12.8 cm

Verification:
AC = AB + BC
= PQ + PQ
= 2PQ = 2 × 6.4 cm
= 12.8 cm.

Question (iv)
2 RS
Solution:
Steps of Construction

1. Draw a line l and mark a point A on it.

2. Take the compasses and measure RS. Without changing the compasses, place the needle at point A and draw an arc intersecting the line l at point B such that AB = RS.

3. Again without changing the opening of compasses, place the needle at point B and draw an arc cutting line l at C such that BC = RS.
4. Then AC is the required line segment whose length is equal to 2RS.

5. Measure AC, AC = 7.2 cm

Verification:
AC = AB + BC
= RS + RS
= 2 RS = 2 × 3.6 cm
= 7.2 cm.

Question (v)
3 RS.
Solution:
Steps of Construction.

1. Draw a line l and mark point A on it.

2. Take the compasses and measure RS. Without changing the compasses, place the needle at point A and draw an arc cutting the line l at point B such that AB = RS.

3. Again, without changing the opening of the compasses, place the needle at point B and draw an arc cutting line l at C such that BC = RS.

4. Once again, without changing the opening of the compasses, place the needle at point C and draw an arc cutting line l at D such that CD = RS. Then AD is the required line segment whose length is equal to 3 RS.

5. Measure AD, AD = 10.8 cm
Verification:
AD = AB + BC + CD
= RS + RS + RS
= 3RS = 3 × 3.6 cm
= 10.8 cm

6. Draw a line segment PQ of any length. Now without measuring it draw a copy of PQ.
Solution:
Steps of Construction.

1. Draw given line segment PQ

2. Draw a line l and mark a point A on it.

3. Take the compass and measure PQ. Without disturbing the compasses, place the needle of the compasses at point A on l and draw an arc, which intersect the line l at point B.

4. Then AB is the required line segment which is equal to the length of PQ. Thus AB = PQ

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.8 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

Question 1.
Find the volume of a sphere whose radius is
(i) 7 cm
Answer:
For the given sphere, radius r = 7 cm.
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 7 × 7 × 7 cm³
= \(\frac{4312}{3}\) cm³
= 1437\(\frac{1}{3}\) cm³

(ii) 0.63 m.
Answer:
For the given sphere, radius r = 0.63 m.
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.63 × 0.63 × 0.63 m³
= 1.05 m³ (approx.)

Question 2.
Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m
Answer:
Amount of water displaced by a solid spherical ball = Volume of spherical ball

(i) For the given spherical ball, diameter 28
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{28}{2}\) cm = 14 cm
Volume of spherical ball
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 14 × 14 × 14 cm³
= \(\frac{34496}{3}\) cm³
= 11498\(\frac{2}{3}\) cm³
Thus, the amount of water displaced by the given solid spherical ball is = 11498\(\frac{2}{3}\) cm³

(ii) For the given spherical ball, diameter 28
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{0.21}{2}\) m
Volume of spherical ball
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × \(\frac{0.21}{2}\) × \(\frac{0.21}{2}\) × \(\frac{0.21}{2}\) m³
= 11 × 0.01 × 0.21 × 0.21 m³
= 0.004851 m³
Thus, the amount of water displaced by the given solid spherical ball is 0.004851 m³.

Question 3.
The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9g per cm³?
Answer:
For the given spherical ball,
radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{4.2}{2}\) cm
= 2.1 cm
= \(\frac{21}{10}\) cm
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × \(\frac{21}{10}\) × \(\frac{21}{10}\) × \(\frac{21}{10}\) cm³
= 38.808 cm³
Now, the density of the metal of the ball is 8.9 g per cm³.
∴ Mass of the ball = Volume × Density
= 38.808 cm³ × 8.9 g/cm³
= 345.39 g (approx.)
Thus, the mass of the metallic ball is 345.39 g (approx.).

Question 4.
The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon ?
Answer:
As the diameter of the moon is one-fourth of the diameter of the earth, the radius of the moon is also one-fourth of the radius of the earth. In other words, the radius of the earth is four times the radius of the moon. Let, the radius of the moon be r and the radius of the earth be R.
Then, R = 4r
Now, \(=\frac{\text { volume of the moon }}{\text { volume of the earth }}\) = \(\frac{\frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi R^{3}}\)
= \(\left(\frac{r}{R}\right)^{3}\)
= \(\left(\frac{r}{4 r}\right)^{3}\)
= \(\left(\frac{1}{4}\right)^{3}\)
= \(\frac{1}{64}\)
∴ Volume of the moon
= \(\frac{1}{64}\) × Volume of the earth
Thus, the volume of the moon is \(\frac{1}{64}\) times the volume of the earth.

Question 5.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Answer:
For the hemispherical bowl,
radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{10.5}{2}\) cm
= 5.25 cm
= \(\frac{21}{4}\) cm
Capacity of the hemispherical bowl
= Volume of a hemisphere
= \(\frac{2}{3}\) πr³
= \(\frac{2}{3} \times \frac{22}{7} \times \frac{21}{4} \times \frac{21}{4} \times \frac{21}{4}\) cm³
= 303.19 cm³ (approx.)
= \(\frac{303.19}{1000}\) liters (approx.)
= 0.303 liters (approx)
Thus, the given hemispherical bowl can hold 0. 303 litres (approx.) of milk.

Question 6.
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Answer:
For the hemispherical tank, inner radius r = 1 m and the thickness of the iron sheet = 1 cm = 0.01 m.
∴ For the hemispherical tank, outer radius
R = 1 + 0.01 m = 1.01 m.
Volume of the iron used in the tank
= Volume of outer hemisphere – Volume of Inner hemisphere
= \(\frac{2}{3}\) πR³ – \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\) π (R³ – r³)
= \(\frac{2}{3}\) × \(\frac{22}{7}\) (1.01³ – 1³) m³
= \(\frac{44}{21}\) (1.030301 – 1) m³
Thus, the volume of the iron used to make the tank is 0.06349 m³ (approx.).

Question 7.
Find the volume of a sphere whose surface area is 154 cm².
Answer:
For the given sphere, surface area = 154 cm².
Surface area of a sphere = 4πr²
∴ 154 cm² = 4 × \(\frac{22}{7}\) × r²cm²
∴ r² = \(\frac{154 \times 7}{4 \times 22}\) cm²
∴ r² = \(\frac{49}{4}\) cm²
∴ r = \(\frac{7}{2}\)
Thus, the radius of the given sphere is \(\frac{7}{2}\) cm.
Volume of a sphere
= \(\frac{4}{3}\) πr³
= \(\frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}\) cm³
= \(\frac{539}{3}\) cm³
= 179\(\frac{2}{3}\) cm³
Thus, the volume of the given sphere is 179\(\frac{2}{3}\) cm³.

Question 8.
A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of ₹ 4989.60. If the s cost of whitewashing is ₹ 20 per square metre, find the
(i) inside surface area of s the dome.
Answer:
(i) Area of the region whitewashed at the
cost of ₹ 20 = 1 m²
∴ Area of the region whitewashed at the cost of ₹ 4989.60 = \(\frac{4989.60}{20}\) m² = 249.48 m²
Hence, the inner surface area of the dome is 249.48 m²

(ii) volume of the air inside the dome.
Answer:
Curved surface area of hemispherical dome = 2πr²
∴ 249.48 m² = 2 × \(\frac{22}{7}\) × r² m²
∴ r² = \(\frac{249.48 \times 7}{2 \times 22}\) m²
∴ r² = 39.69 m²
∴ r² = \(\sqrt{39.69}\) m
∴ r = 6.3 m
Thus. the radius of the hemispherical dome is 6.3m.
Volume of air inside the hemispherical dome = Volume of a hemisphere
= \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 6.3 × 6.3 × 6.3 m³
= 523.908 m³
= 5239 m³ (approx.)
Thus, the volume of the air inside the dome is 523.9 m³ (approx.).

Question 9.
Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’, Find the
(i) radius r’ of the new sphere,
Answer:
(i) 27 solid iron spheres of radius r are melted to form 1 iron sphere of radius r’.
∴ Volume of 1 sphere of radius r’
= Volume of 27 spheres of radius r
∴ \(\frac{4}{3}\) πr’³ = 27 × \(\frac{4}{3}\) πr’³
∴ r’³ = 27r³
∴ r’³ = (3r)³
∴ r’ = 3r

(ii) ratio of S and S’.
Answer:
The surface area of the sphere with radius r is S and the surface area of the sphere with radius r’ is S’.
Then,
\(\frac{\mathrm{s}}{\mathrm{S}^{\prime}}=\frac{4 \pi r^{2}}{4 \pi r^{\prime 2}}=\frac{r^{2}}{r^{\prime 2}}=\frac{r^{2}}{(3 r)^{2}}=\frac{r^{2}}{9 r^{2}}=\frac{1}{9}\) = 1 : 9
Thus, the ratio of S and S’ is 1 : 9.

Question 10.
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule ?
Answer:
For the spherical capsule, radius r = \(\frac{\text { diameter }}{2}\)
= \(\frac{3.5}{2}\) mm
= 1.75 mm
Capacity of the spherical capsule
= Volume of a sphere
= \(\frac{4}{3}\) πr’³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 1.75 × 1.75 × 1.75 mm³
= 22.46 mm³ (approx.)
Thus, 22.46 mm³ (approx.) medicine is needed to fill the given capsule.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7

Question 1.
Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm
Answer:
For the given cone, radius r = 6 cm and height h = 7 cm.
Volume of a cone
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 7 cm³
= 264 cm³

(ii) radius 3.5 cm, height 12 cm
Answer:
For the given cone, radius
r = 3.5 cm = \(\frac{7}{2}\) cm and height h = 12 cm.
Volume of a cone
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 12 cm³
= 154 cm³

Question 2.
Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
Answer:
For the given conical vessel, radius r = 7 cm and slant height l = 25 cm.
h = \(\sqrt{l^{2}-r^{2}}\)
= \(\sqrt{25^{2}-7^{2}}\)
= \(\sqrt{625-49}\)
= √576
∴ h = 24 cm
Capacity of conical vessel
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 7 × 7 × 24 cm³
= 1232 cm³
= \(\frac{1232}{1000}\) liters
= 1.232 litres

(ii) height 12 cm, slant height 13 cm
Answer:
For the given conical vessel, height
h = 12 cm and slant height l = 13 cm.
r = \(\sqrt{l^{2}-h^{2}}\)
= \(\sqrt{13^{2}-12^{2}}\)
= \(\sqrt{169-144}\)
= \(\sqrt{25}\)
∴ r = 5 cm
Capacity of conical vessel
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 5 × 5 × 12 cm³
= \(\frac{6600}{21}\)
= \(\frac{6600}{21 \times 1000}\) liters
= \(\frac{11}{35}\) liters

Question 3.
The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14.)
Answer:
For the given cone, height h = 15 cm and
volume = 1570 cm³.
Volume of a cone = \(\frac{1}{3}\) πr²h
∴ 1570 cm³ = \(\frac{1}{3}\) × 3.14 × r² × 15 cm³
∴ 1570 cm³ = 15.7 × r² cm³
∴ r² = \(\frac{1570}{15.7}\) cm²
∴ r² = 100 cm²
∴ r = 10 cm

Question 4.
If the volume of a right circular cone of height 9 cm is 48 πcm³, find the diameter of its base.
Answer:
For the given right circular cone, height h = 9 cm and volume = 48 πcm³
Volume of a cone = \(\frac{1}{3}\) πr²h
∴ 48π cm³ = \(\frac{1}{3}\) × π × r² × 9 cm³
∴ r² = \(\frac{48 \pi \times 3}{\pi \times 9}\) cm²
∴ r² = 16 cm²
∴ r² = 16 cm²
∴ r = 4 cm
Now, diameter = 2r = 2 × 4 cm = 8 cm
Thus, the diameter of the right circular cone is 8 cm.

Question 5.
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters ?
Answer:
For the given conical pit,
radius r = \(\frac{\text { diameter }}{\mathbf{2}}\) = \(\frac{3.5}{2}\) m = \(\frac{35}{2}\) m
and height (depth) h = 12 m
Capacity of the conical pit
= Volume of the cone
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{35}{20}\) × \(\frac{35}{20}\) × 12 m³
= 38.5 m³
= 38.5 kilioliters

Question 6.
The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find
(i) height of the cone,
(ii) slant height of the cone and
(iii) curved surface area of the cone.
Answer:
For the given right circular cone,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{28}{2}\) cm = 14 cm and
volume = 9856 cm³.

(i) Volume of a cone = \(\frac{1}{3}\) πr²h
∴ 9856 cm³ = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 14 × 14 × h
∴ h = \(\frac{9856 \times 3 \times 7}{22 \times 14 \times 14}\) cm
∴ h = 48 cm

(ii) l = \(\sqrt{h^{2}+r^{2}}\)
= \(\sqrt{48^{2}+14^{2}}\)
= \(\sqrt{2304+196}\)
= \(\sqrt{2500}\)
∴ l = 50 cm

(iii) Curved surface area of a cone
= πrl
= \(\frac{22}{7}\) × 14 × 50 cm²
= 2200 cm²

Question 7.
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm.. Find the volume of the solid so obtained.
Answer:
A right circular cone is received when ∆ ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm.
For the cone so obtained, radius r = 5 cm, height h = 12 cm and slant height l = 13 cm.
Volume of the cone obtained
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × π × 5 × 5 × 12 cm³
= 100π cm³

Question 8.
If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two ‘solids obtained in Questions 7 and 8.
Answer:
Now, if ∆ ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm, again a right circular cone is received.
For the cone so obtained, radiqs r = 12cm; height h = 5 cm and slant height l = 13 cm.
Volume of the cone obtained
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × π × 12 × 12 × 5 cm³
= 240π cm³
Ratio of the volumes of two cones obtained in question 7 and 8 = \(\frac{100 \pi}{240 \pi}\) = \(\frac{5}{12}\) = 5 : 12.

Question 9.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Answer:
For the conical heap of wheat,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{10.5}{2}\) m = \(\frac{105}{2}\) m and
height h = 3 m
Volume of the conical heap of wheat
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{105}{20}\) × \(\frac{105}{20}\) × 3 m³
= 86.625 m³
To cover the heap with canvas, the area of the canvas required will be equal to the curved surface area of the heap.
Now, l = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{\left(\frac{105}{20}\right)^{2}+(3)^{2}}\)
= \(\sqrt{(5.25)^{2}+9}\)
= \(\sqrt{27.5625+9}\)
= \(\sqrt{36.5625}\)
= 6.05 m (approx.)
Curved surface area of the conical heap
= πrl
= \(\frac{22}{7}\) × \(\frac{105}{20}\) × 6.05 m²
= 99.825 m²
Thus, 99.825 m² canvas is required to cover the conical heap of wheat to protect it from rain.

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6

Question 1.
The circumference of the base of a 7 cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm³ = 1l)
Answer:
For the given cylindrical vessel, height h = 25 cm and circumference of the base = 132 cm.
Circumference of base = 2πr
∴ 132 cm = 2 × \(\frac{22}{7}\) × r cm
∴ r = \(\frac{132 \times 7}{2 \times 22}\) cm
∴ r = 21 cm
Hence, for the cylindrical vessel, radius r = 21 cm
Capacity of cylindrical vessel
= Volume of cylinder
= πr²h
= \(\frac{22}{7}\) × 21 × 21 × 25 cm³
= 34650 cm³
= \(\frac{34650}{1000}\) liters
= 34.65 litres
Thus, the cylindrical vessel can hold 34.65 litres of water.

Question 2.
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.
Answer:
For the cylindrical wooden pipe,
outer radius R = \(\frac{\text { diameter }}{2}\) = \(\frac{28}{2}\) cm = 14 cm,
inner radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{24}{2}\) cm = 12 cm and
height (length) h = 35 cm.
Volume of the cylindrical wooden pipe
= Volume of outer cylinder – Volume of inner cylinder
= πR²h – πr²h
= πh (R² – r²)
= πh (R + r) (R – r)
= \(\frac{22}{7}\) × 35 × (14 + 12) (14 – 12) cm³
= 110 × 26 × 2 cm³
= 5720 cm³
Now, mass of 1 cm³ of wood = 0.6 g
∴ Mass of 5720 cm³ of wood = 5720 × 0.6 g
= 3432 g
= 3.432 kg
Thus, the mass of the given pipe is 3.432 kg.

Question 3.
A soft drink is available in two packs
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm.
Which container has greater capacity and by how much?
Answer:
(i) For the cuboidal container with rectangular base, length l = 5 cm; breadth b = 4 cm and height h = 15 cm.
Capacity of the cuboidal container
= Volume of cuboid
= l × b × h
= 5 × 4 × 15 cm³
= 300 cm³

(ii) For the cylindrical container,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{7}{2}\) cm and
height h = 10 cm.
Capacity of the cylindrical container = Volume of cylinder
= πr²h
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) × 10 cm³
= 385 cm³
Hence, the capacity of the cylindrical container is more than the cuboidal container by 385 – 300 = 85 cm³.

Question 4.
If the lateral surface area of a cylinder is 94.2 cm2 and its height is 5 cm, then find
(i) radius of its base and
(ii) its volume. (Use π = 3.14)
Answer:
(i) For the given cylinder, height h = 5 cm and lateral (curved) surface area = 94.2 cm².
Curved surface area of a cylinder = 2 πrh
∴ 94.2 cm² = 2 × 3.14 × r × 5 cm²
∴ r = \(\frac{94.2}{2 \times 3.14 \times 5}\) cm
∴ r = 3 cm
Thus, the radius of the cylinder is 3 cm.

(ii) Volume of a cylinder
= πr²h
= 3.14 × 3 × 3 × 5 cm³
= 141.3 cm³
Thus, the volume of the cylinder is 141.3 cm³.

Question 5.
It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m², find
(i) inner curved surface area of the vessel,
(ii) radius of the base and
(iii) capacity of the vessel.
Answer:
(i) Area of the region painted at the cost of ₹ 20 = 1 m²
∴ Area of the region painted at the cost of ₹ 2200 = \(\frac{2200}{20}\) m² = 110m²
Thus, the inner curved surface area of the vessel is 110 m².

(ii) For the cylindrical vessel, height (depth) h- 10 m and curved surface area = 110 m².
Curved surface area of cylindrical vessel = 2πrh
∴ 110 m² = 2 × \(\frac{22}{7}\) × r × 10 m²
∴ r = \(\frac{110 \times 7}{2 \times 22 \times 10}\) m
∴ r = \(\frac{7}{4}\) m
∴ r = 1.75 m
Thus, the radius of the cylindrical vessel is 1.75 m.

(iii) Capacity of the cylindrical vessel
= Volume of a cylinder
= πr²h
= \(\frac{22}{7}\) × 1.75 × 1.75 × 10 m³
= 96.25 m³
= 96.25 kilolitres
Thus, the capacity of the cylindrical vessel is 96.25 kilolitres.

Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it ?
Answer:
For the closed cylindrical vessel, height h = 1 m and capacity = 15.4 litres
∴ Volume of the vessel = 15.4 litres
= \(\frac{15.4}{1000}\) m³
= 0.0154 m³
Volume of cylindrical vessel = πr²h
∴ 0.0154 m³ = \(\frac{22}{7}\) × r² × 1 m³
∴ r² = \(\frac{154}{10000} \times \frac{7}{22}\) m²
∴ r² = \(\frac{49}{10000}\) m²
∴ r² = \(\frac{7}{100}\) m
∴ r = 0.07 m
Thus, the radius of the cylindrical vessel is 0. 07 m.
Area of the metal sheet required to make closed cylindrical vessel
= Total surface area of a cylinder
= 2πr (r + h)
= 2 × \(\frac{22}{7}\) × 0.07 (0.07 + 1) m²
= 0.44 × 1.07 m²
= 0.4708 m²
Thus, 0.4708 m² of metal sheet would be needed to make the closed cylindrical vessel.

Question 7.
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Answer:
For the solid cylinder of graphite,
radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{1}{2}\) mm = \(\frac{1}{20}\) cm and
height h = 14 cm.
Volume of cylinder of graphite
= πr²h
= \(\frac{22}{7}\) × \(\frac{1}{20}\) × \(\frac{1}{20}\) × 14 cm³ = 011 cm³
For the hollow cylinder of wood,
outer radius R = \(\frac{\text { diameter }}{2}\) = \(\frac{7}{2}\) mm = \(\frac{7}{20}\) cm,
inner radius r = \(\frac{1}{20}\) cm and height h = 14 cm.
Volume of hollow cylinder of wood
= πR²h – πr²h
= πh (R² – r²)
= πh (R + r) (R – r)
= \(\frac{22}{7}\) × 14 \(\left(\frac{7}{20}+\frac{1}{20}\right)\left(\frac{7}{20}-\frac{1}{20}\right)\) cm³
= 44 × \(\frac{8}{20}\) × \(\frac{8}{20}\) cm³
= \(\frac{528}{100}\) cm³
= 5.28 cm³
Thus, in the given pencil, the volume of wood is 5.28 cm³ and the volume of graphite is 0.11cm³.

Question 8.
A patient in a hospital is given soup daily in a cylindrical howl of diameter 7 cm. If the bowl is Oiled with soup to a height of 4 cm, how much soup the hospital has ‘ to prepare daily to serve 250 patients ?
Answer:
For the soup served in cylindrical bowl, radius r = \(\frac{\text { diameter }}{2}\) = \(\frac{7}{2}\) cm and height h = 4 cm.
Volume of soup served to one patient = Volume of a cylinder
= πr²h
= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 4 cm³
= 154 cm³
Thus, the volume of soup served to 1 patient =154 cm³
∴ The volume of soup served to 250 patients
= 154 × 250 cm³
= 38500 cm³
= \(\frac{38500}{1000}\) litres
= 38.5 litres
Thus, the hospital has to prepare 38500 cm³, i.e., 38.5 litres of soup daily.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 9 Understanding Elementary Shapes Ex 9.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 9 Understanding Elementary Shapes Ex 9.6

1. Give two examples of each of the following shapes from your surroundings:

Question (i)
Cube
Solution:
Cube. Examples:
(i) Dice,
(ii) Sugar cubes.

Question (ii)
Cuboid
Solution:
Cuboid. Examples:
(i) Matchbox,
(ii) Geometry box.

Question (iii)
Cone
Solution:
Cone: Examples:
(i) Ice cream cone,
(ii) Joker cap.

Question (iv)
Cylinder
Solution:
Cylinder. Examples:
(i) Drum,
(ii) Circular pipe.

Question (v)
Shpere.
Solution:
Shpere. Examples:
(i) Globe,
(ii) Ball.

2. Classify the following as plane figures and solid figures:

Question (i)
(i) Rectangle
(ii) Sphere
(iii) Cylinder
(iv) Circle
(v) Cube
(vi) Cuboid
(vii) Triangle
(viiii) Cone
(ix) Square
(x) Prism.
Solution:
Plane figures:
(i) Rectangle
(iv) Circle
(vii) Triangle
(ix) Square.

Solid figures:
(ii) Sphere
(iii) Cylinder
(v) Cube
(vi) Cuboid
(viii) Cone
(x) Prism.

3. Write the name of shapes in the base of the following solids:

Question (i)
Cube
Solution:
Square

Question (ii)
Cylinder
Solution:
Circle

Question (iii)
Tetrahedron
Solution:
Equilateral triangle

Question (iv)
Cuboid
Solution:
Rectangle

Question (v)
Square Pyramid.
Solution:
Square.

4. Fill in the table:

Shape	Number of Flat Faces	Number of Curved Faces	Number of Vertices	Number of Edges
(i) Cuboid
(ii) Cube
(iii) Cylinder
(iv) Cone
(v) Sphere
(vi) Triangular Prism
(vii) Square Pyramid
(viii) Tetrahedron

Solution:

Shape	Number of	Number of	Number of	Number of
	Flat Faces	Curved Facet!	Vertices	Edges
(i) Cuboid	6	Nil	8	12
(ii) Cube	6	Nil	8	12
(iii) Cylinder	2	1	Nil	2
(iv) Cone	1	1	1	1
(v) Sphere	Nil	1	Nil	Nil
(vi) Triangular Prism	5	Nil	6	9
(vii) Square Pyramid	5	Nil	5	8
(viii) Tetrahedron	4	Nil	4	6

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 15 Visualising Solid Shapes Ex 15.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.3

1. Count the number of cubes in each of the following figures :


Solution:
(i) 6
(ii) 21
(iii) 32
(iv) 13

2. If three cubes of dimensions 2 cm × 2 cm × 2 cm are placed side by side, what would be the dimensions of resulting cuboid ?

Solution:
Length 6 cm, breadth 2 cm and height 2 cm.

3. If we throw light an the following solids from the top name the shape of shadow obtained in each case and also give a rough sketch of the shadow.
(i) DVD player

Solution:

(ii) Sandwich

Solution:

(iii) Straw

Solution:

4. What cross-sections do you get when you given a.
(i) Vertical cut
(ii) Horizontal cut.
to the following solids ?
(a) A die
(b) A square pyramid
(c) A round melon
(d) A circular pipe
(e) A brick
(f) An ice cream cone.
Solution:
(a) Square, Square,
(b) Triangle, Square,
(c) Circle, circle
(d) Circle, Rectangle
(e) Rectangle, Rectangle
(f) Triangle, Circle.

5. If we throw light on following solids, from left name the shape of shadow in each and also give a rough sketch of the shadow.
(i)

Solution:

(ii)

Solution:

6. Here are the shadows of some 3-D objects, when seen under the lamp of an overhead projector. Identify the solids that match each shadow (There may be multiple answers for these)

Solution:
(i) Dice, chalk box etc
(ii) Book, Mobile, DVD Player etc
(iii) Cricket ball, Disc etc.
(iv) Birthday cap, Icecream cone etc.

7. Sketch the front, side and top view of the following figures :
(i)

Solution:

(ii)

Solution:

(iii)

Solution:

(iv)

Solution:

8. Multiple choice questions :

Question (i).
The number of cubes in the given structure is ?

(a) 12
(b) 10
(c) 9
(d) 8
Answer:
(a) 12

Question (ii).
The number of unit cubes to be added in above students to make a cuboid of dimensions 4 unit × 2 unit × 3 unit is ?
(a) 11
(b) 12
(c) 13
(d) 14
Answer:
(b) 12

Question (iii).
What cross-section is made by vertical cut in a cuboid
(a) Square
(b) Rectangle
(c) Circle
(d) Triangle
Answer:
(b) Rectangle

Question (iv).
What cross section is made by horizontal cut in a cone
(a) Triangle
(b) Circle
(c) Square
(d) Rectangle
Answer:
(b) Circle

Question (v).
Which solid cost a shadow of triangle under the effect of light
(a) Sphere
(b) Cylinder
(c) Cone
(d) Cube
Answer:
(c) Cone

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 15 Visualising Solid Shapes Ex 15.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2

1. Use isometric dot paper and make an isometric sketch of the following figures :
Question (i).

Solution:

Question (ii).

Solution:

Question (iii).

Solution:

Question (iv).

Solution:

2. Draw (i) an oblique sketch (ii) Isometric sketch for :
(a) A cube with a edge of 4 cm long
(b) A cuboid of length 6 cm, breadth 4 cm and height 3 cm
Solution:
(a) (i) Oblique sketch of cube with a edge 4 cm long.

(ii) Iso metric sketch for a cube with a edge of 4 cm long.

(b) (i) an oblique sketch for a cuboid of length 6 cm, breadth 4 cm and height 3 cm.

(ii) Isometric sketch for a cuboid of length 6 cm, breadth 4 cm and height 3 cm.

3. Two cubes each with edge 3 cm are placed side by side to form a cuboid, sketch oblique and isometric sketch of this cuboid.
Solution:

4. Draw an Isometric sketch of triangular pyramid with base as equilateral triangle of 6 cm and height 4 cm.
Solution:

5. Draw an Isometric sketch of square pyramid.
Solution:

6. Make an oblique sketch for each of the given Isometric shapes.

Question (i).

Solution:

Question (ii).

Solution:

7. Using an isometric dot paper draw the solid shape formed by the given net.

Solution:

8. Multiple choice questions:

Question (i).
An oblique sheet is made up of:
(a) Rectangles
(b) Squares
(c) Right angled triangles
(d) Equilateral triangles.
Answer:
(b) Squares

Question (ii).
An isometric sheet is made up of dots forming :
(a) Squares
(b) Rectangles
(c) Equilateral triangles
(d) Right angled triangle
Answer:
(c) Equilateral triangles

Question (iii).
An oblique sketch has :
(a) Proportional lengths
(b) Parallel lengths
(c) Non-Proportional lengths
(d) Perpendicular lengths.
Answer:
(c) Non-Proportional lengths

Question (iv).
An Isometric sketch has :
(a) Non proportional lengths
(b) Parallel lengths
(c) Perpendicular lengths
(d) Proportional lengths.
Answer:
(d) Proportional lengths.

Question (v).
Isometric sketches shows objects of:
(a) Two dimensions
(b) Shadows
(c) Three dimensions
(d) One dimension.
Answer:
(c) Three dimensions