Srinivas

For systematic learning, students can include AP Inter 1st Year English Grammar with Answers and Tenses in English Inter 1st Year in their daily study routine.

AP Inter 1st Year English Tenses with Answers

VERB:

A verb is a word that tells about an action, a state of being or existence, possession, or a change in state.

ఒక పనిని గురించి గాని, ఒక స్థితి గురించి గాని యాజమాన్యం గురించి గాని, స్థితిలో మార్పు గురించి గాని, తెలిపేది

e.g. : The boy cried.
I have a watch.
She is a nurse.
He became weak.

పై వాక్యాలలోని cried, have, is, became అను పదాలు పనిని గాని స్థితి గాని తెలుపుతున్నాయి. అందుచే అవి verbs.

TENSE

Tense : Tense is the form of the verb. It shows the time of the action or event. There are three tenses. They are :

1. Present Tense
2. Past Tense
3. Future Tense

The Use of the Tenses:

1. Simple Present Or Present Indefinite Tense :
i) It is used to express what actually happens at the time of speaking.
e.g.: Here comes John.

ii) It is used to express habitual action :
e.g. : I get up early in the morning.

iii) It is used to express general or universal truths :
e.g. : The earth is round. Stars twinkle.
Honesty is the best policy. Sugar is sweet.

iv) It is used to express future action that has already been planned :
e.g. : Our college reopens on 16th June.
He sails for England next Monday.

v) It is used to express historic present :
e.g.: i) Sivaji now sees the danger and immediately kills Afzul Khan with his lion-claws.
ii) Birbal now seizes the chance and cracks a joke on his rivals.

2. Present Continuous Tense :

i) It is used to express an action that is going on at the time of speaking :
e.g.: She is dancing – They are writing.
ii) It is used to express an action that will happen in the future.
e.g.: 1) I am going to Mysore tomorrow.

Note : The following verbs are not generally used in the Continuous Tense :
(ఈ క్రింది Verbs ను సామాన్యంగా Continuous Tense లో ఉపయోగించరాదు.)

• Verbs of sense of perception : hear, see, smell, notice, recognise, taste, feel.
• Verbs of appearing : look, seem, appear.
• Verbs of thinking : suppose, think, believe, realise, understand, know, imagine, mean, agree, consider, trust, remember, forget, expect, recall.
• Verbs of emotions : want, wish, desire, feel, like, love, hate, prefer, hope, refuse.

Miscellaneous :
own, possess, keep, concern, matter, owe, etc.
e.g. :He is looking fine. (Wrong)
He looks fine. (Correct)
I am believing you. (Wrong)
I believe you. (Correct)
She is understanding me. (Wrong)
She understands me. (Correct)
1 am hating you. (Wrong)
I hate you. (Correct)

3. Present Perfect Tense :
i) It is used to denote an action that has just been completed.
e.g.: The train has arrived.
They have finished the work.

ii) It is used to express a past action the result of which continues,
e.g.: I have not seen Ravi for many months.
He has been ill since Monday.

iii) It is used to refer to a past action in a more general way :
e.g.: Have you ever been to Simla ?

iv) A few adverbs or adverbial phrases are used with the Present Perfect Tense :
just, never, ever, so far, till, yet, already, since, today, this week, etc.
e.g.: I have just posted the letter.
So far he has not come.
He has not received the money yet.
This week there have been no rains.

4. Present Perfect Continuous Tense :

It is used to refer to an action that begins in the past and continues through a given period of time up to the present moment.
e.g. : I have been waiting here for Ramu for two hours.
I have been waiting here for Ramu since 3 p.m.
Note : ‘For’ denotes period of time.
Since’ denotes point of time.

Observe the following sentences :
i) He has been painting the door for 2 p.m. (Wrong)
He has been painting the door since 2 p.m. (Correct)

ii) The workers have been demanding more wages since ten days. (Wrong)
The workers have been demanding more wages for ten days. (Correct)

5. Simple Past Or Past Indefinite Tense :
i) It is used to express the action completed in the past. Adverbs and adverbial phrases expressing Past time are often used with this tense.
(భూతకాలంలో పని పూర్తయినట్లయితే దానిని తెలుపుటకు Simple Past లేక Past Indefinite Tense ను వాడెదరు. భూతకాలాన్ని తెలిపే క్రియ విశేషణ పదాలు యీ tense లో ప్రయోగింపబడతాయి.)
e.g.: She met me yesterday.
The clerk did his work hurriedly.

ii) It is used to express a habitual action in the past.
(భూతకాలంలో అలవాటుగా చేసే పనులను విశదపరచుటకు Simnple Past Tense వాడెదరు.)
e.g.: People performed child marriages in olden days. While I was in Chennai, 1 spoke to others in Tamil.

6. Past Continuous Tense :
It is used to express an action that was still going on in the past time.
(భూతకాలంలో పని కొనసాగింపును Past Continuous Tense తెలుపును.)
e.g. : I met Raghu while he was standing at the college gate.
We noticed some birds which were flying.

7. Past Perfect Tense :
It is used to express an action which had been completed at some point in the past time before another action was started.
(భూతకాలంలో ముందు జరిగిన పనిని Past Perfect Tense లోను రెండవ పనిని Simple Past Tense లోను చెప్పాలి.)
e.g.: The train had left before they reached the station.
I had finished my work when Alfred came in.
(పై వాక్యాలలోని మొదటి పనిని ఉదాహరణంగా తీసుకుందాం. ఇందులో జరిగిన రెండు పనులు సూచింపబడ్డాయి. The train left. They reached the station. ఇందులో ముందు జరిగినది The train left. దీనిని Past Perfect Tense లో చెప్పాలి. అప్పుడు వాక్యం The train had left before they reached the station అయింది.)

8. Past Perfect Continuous Tense :
It is used to show an action that had begun in the past and continued till some point of time in the past.
(భూతకాలంలోని ఒక పనిని ఎప్పుడో ప్రారంభించి, అదే పని కొంత కాలం వరకు కొనసాగుతున్నదని Past Perfect Continuous Tense లో తెలపాలి.)
e.g. : I had been watching TV for an hour when my uncle came to see me.
We had been playing hockey for three hours when it rained.
She had been passing the information for a long time before she was caught.

9. Simple Future Or Future Indefinite Tense:
It is used to express an action that will take place in future.
(భవిష్యత్తు కాలంలో జరగబోయే పనిని Simple Future Tense తెలుపును.)
e.g.: I shall meet you tomorrow.

10. Future Continuous Tense :
It is used to express an action going on at some point in future.
(భవిష్యత్తు కాలంలో కొనసాగే పనిని Future Continuous Tense తెలుపును.)
e.g.: She will be sleeping then.
We will be playing the match at this time tomorrow.
I shall be writing the examination at this time on Monday.

11. Future Perfect Tense :
It is used to express an action that will be completed at some point in future.
(భవిష్యత్తులో ఒక నిర్ణితాకాలానికి పని Future Perfect Tense పూర్తగునని విశదపరుచును
e.g. : I shall have done my work before you come.

12. Future Perfect Continuous Tense :
It is used to refer to an action that will be in progress at a point in the future after continuing for a given period.
e.g.: I shall have been completing thirty years of service by next March.
I shall have been writing the examination for two hours by this time on Monday.

→ Tenses in Conditional Clauses :

Clauses that show conditions are called conditional clauses. Sentences with conditional clauses use fixed patterns of Tenses.

Conditions are of three types :

1. Real conditions
2. Unreal or improbable conditions and
3. Past unfulfilled conditions.

The tense forms used in a sentence are determined by the type of the condition in that sentence.

Look at the following examples:

1. If it rains, I will not come to your home.
   (Real condition – Simple present in the conditional clause and Simple Future in the main clause)
2. If I were a bird, I would fly high in the sky.
   (Unreal condition – Simple Past (were-even with I) in the conditional clause and would +V’ in the main clause)
3. If he had worked hard, he would have passed the examination.
   (Past unfulfilled condition – Past Perfect in the conditional clause and would + have + Past Participle of verb in the main clause).

The information may easily be noted in the table form.

Exercises For Practice

1. Fill in the blanks with the correct form of the verb given in the brackets.

Question 1.
She usually __________ (cook) for the entire family.
Answer:
cooks

Question 2.
My brother often __________ (read) comics.
Answer:
reads

Question 3.
Students __________ (study) in the library daily.
Answer:
study

Question 4.
We __________ (go) to the market every Saturday.
Answer:
go

Question 5.
Raman __________ (go) to Tirupathi twice a year.
Answer:
goes

Question 6.
The hostelers __________ (watch) a movie every month.
Answer:
watch

Question 7.
Water __________ (boil) at 100°C.
Answer:
boils

Question 8.
__________ birds (fly) in the sky?
Answer:
Do, fly

2. Fill in the blanks with the correct form of the verb given in the brackets.

Question 1.
I __________ (wait) for the bus right now.
Answer:
am waiting

Question 2.
Hark ! The cuckoo __________ (sing) beautifully.
Answer:
is singing

Question 3.
Don’t disturb! She __________ (study) for her ensuing exams this week.
Answer:
is studying

Question 4.
Look ! They __________ (practise) dance for the upcoming performance.
Answer:
are practising

Question 5.
See! He __________ (ride) a bullock cart while holding a stick.
Answer:
is riding

Question 6.
I __________ (study) to become a doctor. Don’t disturb me.
Answer:
am studying

Question 7.
Hurry up! The train __________ (move) while the kids are playing.
Answer:
is moving

Question 8.
The birds __________ (fly) in the sky now.
Answer:
are flying

3. Fill in the blanks with the correct form of the verb given in the brackets.

Question 1.
I __________ already __________ (finish) my homework.
Answer:
have, finished

Question 2.
She __________ (visit) Paris three times.
Answer:
has visited

Question 3.
Gita’s English __________ (improve) since she joined college.
Answer:
has improved

Question 4.
Pavan __________ (meet) with an accident twice.
Answer:
has met

Question 5.
She __________ (be) to Delhi lately.
Answer:
has been

Question 6.
I __________ not __________ (see) him since last week.
Answer:
have, seen

Question 7.
The rain __________ not __________ (stop) yet.
Answer:
has, stopped

Question 8.
Rishi __________ just __________ (go) out to get something to eat.
Answer:
has, gone

4. Fill in the blanks with the correct form of the verb given in the brackets.

Question 1.
I ___________ (learn) English for five years.
Answer:
have been learning

Question 2.
She __________ (work) at this company since last year.
Answer:
has been working

Question 3.
They __________ (wait) for the bus for over an hour.
Answer:
have been waiting

Question 4.
Jaya __________ (draw) a picture for two hours.
Answer:
has been drawing

Question 5.
I __________ (water) the plants in the garden and I am tired now.
Answer:
have been watering

Question 6.
Jaya __________ (teach) English since she joined the profession.
Answer:
has been teaching

Question 7.
How long she __________ (practise) music?
Answer:
has been practising

Question 8.
Where you __________ (live) all these years?
Answer:
have been living

5. Fill in the blanks with the correct form of the verb given in the brackets.

Question 1.
I __________ (visit) my grandmother last weekend.
Answer:
visited

Question 2.
She __________ (finish) her homework before lunch.
Answer:
finished

Question 3.
Madhu __________ (go) to the beach yesterday.
Answer:
went

Question 4.
When _____ he __________ (write) the exam last year?
Answer:
did, write

Question 5.
__________ they __________ (plan) the meeting last month?
Answer:
Did, plan

Question 6.
They __________ (travel) to Japan two years ago.
Answer:
traveled

Question 7.
He __________ (forget) his wallet at home this morning.
Answer:
forgot

Question 8.
We __________ (talk) on the phone for thirty minutes last night.
Answer:
talked

6. Fill in the blanks with the correct form of the verb given in the brackets.

Question 1.
I __________ (cook) dinner when the phone rang.
Answer:
was cooking

Question 2.
She __________ (talk) to her father at this time yesterday.
Answer:
was talking

Question 3.
__________ you __________ (study) for the exam at 7 PM yesterday?
Answer:
Were, studying

Question 4.
They __________ (play) golf when it started raining.
Answer:
were playing

Question 5.
He __________ (work) on his project all afternoon yesterday.
Answer:
was working

Question 6.
Padmini __________ (watch) TV when the power was off.
Answer:
was watching

Question 7.
At midnight, we __________ still __________ (drive) through the desert that day.
Answer:
were, driving

Question 8.
When John __________ (sleep), someone stole his car.
Answer:
was sleeping

7. Fill in the blanks with the correct form of the verb given in the brackets.

Question 1.
By the time I arrived, they __________ already __________ (leave).
Answer:
had (already) left

Question 2.
We __________ never __________ (see) such a beautiful sunset before I went to Kochi.
Answer:
had (never) seen

Question 3.
After they __________ (eat), they went for a walk.
Answer:
had eaten

Question 4.
Only she understood the movie because she __________ (read) the book.
Answer:
had read

Question 5.
If I __________ (buy) lemons, I would have made lemon rice.
Answer:
had bought

Question 6.
I __________ (dream) of becoming an IAS officer, but I became a teacher.
Answer:
had dreamed

Question 7.
After they __________ (watch) the movie, they talked about it for hours.
Answer:
had watched

Question 8.
They __________ (vacate) the house before the owner came.
Answer:
had vacated

8. Fill in the blanks with the correct form of the verb given in the brackets.

Question 1.
I __________ (wait) for the bus for over 30 minutes when it finally arrived.
Answer:
had been waiting

Question 2.
By the time I called her, she (work) on the project for hours.
Answer:
had been working

Question 3.
They __________ (run) for an hour before they took a break.
Answer:
had been running

Question 4.
She __________ (study) all night for the exam and was exhausted.
Answer:
had been studying

Question 5.
He __________ (try) to fix the car for hours when I arrived.
Answer:
had been trying

9. Fill in the blanks with the correct form of the verb given in the brackets.

Question 1.
She __________ (buy) a new car next month.
Answer:
will buy

Question 2.
They __________ (start) the meeting at 10 AM tomorrow.
Answer:
will start

Question 3.
He __________ (finish) his homework after dinner.
Answer:
will finish

Question 4.
The team __________ (play) the championship match next month.
Answer:
will play

Question 5.
I __________ (help) you with your project later?
Answer:
Shall / Will, help

Question 6.
They __________ (build) a new library in the city next year.
Answer:
will build

Question 7.
She __________ (visit) her grandparents during the holidays.
Answer:
will visit

Question 8.
The weather forecast says it __________ (rain) tomorrow.
Answer:
will rain

10. Fill in the blanks with the correct form of the verb given in the brackets.

Question 1.
At this time tomorrow, I __________ (work) on my project.
Answer:
will be working

Question 2.
We __________ (have) dinner with friends at 7 PM tomorrow.
Answer:
will be having

Question 3.
He __________ (play) football in the park at this time next week.
Answer:
will be playing

Question 4.
I __________ (watch) TV when she arrives tonight.
Answer:
will be watching

Question 5.
She __________ (work) on her project all evening tomorrow.
Answer:
will be working

Question 6.
They __________ (wait) for us at the station when we arrive.
Answer:
will be waiting

11. Fill in the blanks with the correct form of the verb given in the brackets.

Question 1.
By next year, she __________ (complete) her degree in psychology.
Answer:
will have completed

Question 2.
By the time we arrive, the movie __________ (start).
Answer:
will have started

Question 3.
He __________ (travel) to five different countries by the end of summer.
Answer:
will have traveled

Question 4.
By the end of the week, we __________ (move) into our new house.
Answer:
will have moved

Question 5.
You __________ (save) enough money for your trip by next month.
Answer:
will have saved

12. Fill in the blanks with the correct form of the verb given in the brackets.

Question 1.
By next year, she __________ (study) at the university for three years.
Answer:
will have been studying

Question 2.
They __________ (wait) for us for over an hour by the time we arrive.
Answer:
will have been waiting

Question 3.
By 2035, scientists __________ (work) on space exploration for decades.
Answer:
will have been working

Question 4.
I __________ (train) for the marathon for two months by the end of the year.
Answer:
will have been training

Question 5.
By 10 PM, we __________ (celebrate) his birthday for several hours.
Answer:
will have been celebrating

13. Fill in the blanks with the correct form of the verb given in the brackets.

Question 1.
If we pay the bills, she __________ (go) home.
Answer:
will go

Question 2.
If you __________ (be) hungry, you may order online.
Answer:
are

Question 3.
If I have money, I __________ (visit) America.
Answer:
will visit

Question 4.
If Raju __________ (like) something, he will do anything.
Answer:
likes

14. Fill in the blanks with the correct form of the verb given in the brackets.

Question 1.
If I were the president, I __________ (declare) every Saturday a holiday.
Answer:
would declare

Question 2.
If they __________ (tire), they would take rest.
Answer:
were tired

Question 3.
If we had more time, we __________ (study) more.
Answer:
would study

Question 4.
If you __________ (play) the game before, you would be skilled.
Answer:
had played

15. Fill in the blanks with the correct form of the verb given in the brackets.

Question 1.
If we had paid more attention in class, we __________ (understand) the lesson.
Answer:
would have understood

Question 2.
If I __________ (see) him at the meeting, I would have asked him.
Answer:
had seen

Question 3.
If he had tried well, he __________ (get) the first rank.
Answer:
would have got

Question 4.
If she __________ (take) care of herself, the problem would never have been raised.
Answer:
had taken

Exercises

1. Fill in the blanks with the suitable forms of the verbs given in the brackets.

Question 1.
By the time you arrive, I __________ (finish) my homework.
Answer:
will have finished

Question 2.
Don’t disturb! She __________ (study) for exams.
is studynig

Answer:
Question 3.
Right now, they __________ watch a movie.
Answer:
are watching

Question 4.
Hurry up! We __________ (go) to watch a film.
Answer:
are going

Question 5.
By next year, I __________ (graduate) from college.
Answer:
will have graduated

Question 6.
When I was in college, I __________ (write) articles for the school newspaper.
Answer:
wrote

Question 7.
They __________ (work) on the project all week.
Answer:
have been working

Question 8.
I __________ (not / like) this food since I was a child.
Answer:
have not liked

Question 9.
He __________ already __________ (leave) by the time they called.
Answer:
had, left

Question 10.
By the time we reach the station, the train __________ (leave).
Answer:
will have left

Question 11.
He __________ (be) my friend for over 10 years now.
Answer:
has been

Question 12.
I __________ (try) to fix the car when the mechanic arrived.
Answer:
was trying

Question 13.
They __________ ( build) the bridge in 2020.
Answer:
built

Question 14.
She _____ (wait) for me for over an hour now.
Answer:
has been waiting

Question 15.
At this time tomorrow, I __________ (fly) to New York.
Answer:
will be flying

Question 16.
Venkat __________ (win) the medal last month.
Answer:
won

Question 17.
I __________ (eat) dinner when you called last night.
Answer:
was eating

Question 18.
She __________ (call) me as soon as she finishes the meeting.
Answer:
will call

Question 19.
He __________ (not / come) to the party if he had known about it.
Answer:
would not have come

Question 20.
I __________ (learn) Spanish for three months now.
Answer:
have been learning

Question 21.
I __________ never __________ (meet) anyone like her before.
Answer:
have, met

Question 22.
He __________ (wait) for an hour before the train arrived.
Answer:
had been waiting

Question 23.
We often __________ (finish) our projects in time.
Answer:
often finish

Question 24.
My father __________ (buy) a bike for me next month.
Answer:
will buy

Question 25.
I __________ (study) every evening.
Answer:
study

Question 26.
The movie __________ (begin) when we arrived.
Answer:
had begun

Question 27.
I __________ (not / see) her in the office this week.
Answer:
have not seen

Question 28.
When we were kids, we __________ (go) to the beach every summer.
Answer:
went

Question 29.
She always __________ (get) up late.
Answer:
gets

Question 30.
The bell __________ just __________ (strike).
Answer:
has, struck

2. Fill in the blanks with suitable forms of the verbs given in the brackets:

Question 1.
She never __________ (take) coffee.
Answer:
takes

Question 2.
Ramani __________ (buy) the painting long ago.
Answer:
bought
were sleeping

Question 3.
Yesterday, when I went home, my kids __________ (sleep).
Answer:
has been learning

Question 4.
She _______ (learn) classical music for a long time.
Answer:

Question 5.
If you had asked him, he __________ (help) you.
Answer:
would have helped

3. Fill in the blanks with suitable forms of the verbs given in the brackets:

Question 1.
Right now, I __________ (work) on a project.
Answer:
am working

Question 2.
She __________ (be) always my best friend.
Answer:
has been

Question 3.
Sujatha __________ (go) to her native place tomorrow.
Answer:
is going

Question 4.
At this time tomorrow, she __________ (write) the exam.
Answer:
will be writing

Question 5.
If I were a bird, I __________ (fly) v
Answer:
would fly

For systematic learning, students can include AP Inter 1st Year English Grammar with Answers and Synonyms and Antonyms in English Inter 1st Year in their daily study routine.

AP Inter 1st Year English Synonyms and Antonyms with Answers

Synonyms and antonyms are vital tools for expanding vocabulary, improving writing, and enhancing communication. Synonyms provide alternative words with similar meanings, while antonyms offer words with opposite meanings. Understanding and using them effectively allows for more precise and varied expression, enriching both written and spoken language.

ఎవరైనా ఫలానా వ్యక్తికి ఇంగ్లీషు బాగా వచ్చు, అని అంటే, ఆ వ్యక్తికి వివిధ రకాల వాక్యనిర్మాణం, పదాలకి అర్ధాలు, వాటి సమానార్ధాలు మరియు వాటి వ్యతిరేకార్ధాలు బాగా తెలుసన్నమాట. ఎన్ని ఎక్కువ పదాలు మనకి తెలిస్తే మరియు వాటిని వాక్యాలలో ఉపయోగించగలిగితే మనకి అంత ఇంగ్లీషు వచ్చన్నమాట. వాటిని ఇక్కడ నేర్చుకోండి.

Exercises for Practice

Question 1.
Fill in the blanks with the synonyms and antonyms of ‘Strength’.

Read the passage below and fill in the blanks with the correct words from the box.
Word Bank: (power, weakness, might, frailty, vigor, resilience, fragility, endurance)

To overcome challenges in life, one must rely on both physical and mental ____(1)____. However, there are times when ____(2)____ can hold us back, making it difficult to achieve our goals. People with great ____(3))____ are often admired for their ability to face hardships with courage.

On the other hand, ____(4)____ and ____(5)____ can make even simple tasks seem daunting. Developing ____(6)____ and ____(7)____ can help us stay strong during tough times. Remember, true strength is not just about ____(8)____; it is also about having the wisdom to adapt and persist.
Answer:
1. vigour
2. weakness
3. resilience
4. frailty
5. fragility
6. endurance
7. might
8. power

Question 2.
i) Choose the option that is a synonym of the given word or the key word.

1. glance
a) study
b) laze
c) look
Answer:
c) look

2. commence
a) begin
b) mirror
c) halt
Answer:
a) begin

3. acquisition
a) enthusiasm
b) imitate
c) gain
Answer:
c) gain

4. reflect
a) note
b) imitate
c) hinder
Answer:
b) imitate

5. wrath
a) rage
b) calmness
c) slow
Answer:
a) rage

ii) Choose the option that is an antonym of the given word the key word.

1. plod
a) endeavour
b) lounge
c) loss
Answer:
b) lounge

2. despair
a) gloom
b) struggle
c) joy
Answer:
c) joy

3. listlessness
a) enthusiasm
b) boredom
c) hopelessness
Answer:
a) enthusiasm

4. idly
a) casually
b) satisfy
c) actively
Answer:
c) actively

5. prowess
a) incompetence
b) brave
c) combativeness
Answer:
a) incompetence

Question 3.
Add suitable prefixes to the words given below and form their antonyms.

1. interested
Answer:
uninterested

2. visible
Answer:
invisible

3. sane
Answer:
insane

4. important
Answer:
unimportant

5. usual
Answer:
unusual

6. available
Answer:
unavailable

7. legible
Answer:
illegible

8. responsible
Answer:
irresponsible

9. possible
Answer:
impossible

10. trust
Answer:
distrust

11. qualified
Answer:
unqualified

12. appropriate
Answer:
inappropriate

Exercises

Question 1.
Choose the option that is a synonym of the given word.

1. symmetry
a) imbalance
b) equilibrium
c) indifference
Answer:
b) equilibrium

2. sparse
a) sprint
b) strong
c) scarce
Answer:
c) scarce

3. scary
a) calming
b) creepy
c) corrupt
Answer:
b) creepy

4. precious
a) cleanliness
b) treasured
c) liking
Answer:
b) treasured

5. revoke
a) repellent
b) approve
c) dismiss
Answer:
c) dismiss

6. crest
a) peak
b) aversion
c) brave
Answer:
a) peak

7. quaver
a) gruesome
b) prized
c) falter
Answer:
c) falter

8. dogma
a) adulterate
b) dignified
c) doctrine
Answer:
c) doctrine

9. meek
a) gentle
b) strangle
c) tranquil
Answer:
a) gentle

10. fascinating
a) alarming
b) charming
c) cheap
Answer:
b) charming

11. skirmish
a) harmony
b) bottom
c) quarrel
Answer:
c) quarrel

12. pugnacity
a) malice
b) nadir
c) combativeness
Answer:
c) combativeness

13. throttle
a) control
b) permit
c) esteem
Answer:
a) control

14. blot
a) hold
b) trust
c) blemish
Answer:
c) blemish

15. Slack
a) lax
b) brave
c) cure
Answer:
a) lax

2. Choose the option that is an antonym of the given word.

1. awful
a) pleasant
b) noisy
c) slow
Answer:
a) pleasant

2. hope
a) stain
b) talent
c) disbelief
Answer:
c) disbelief

3. clarity
a) accuracy
b) obscurity
c) mastery
Answer:
b) obscurity

4. mutilate
a) mend
b) tough
c) distort
Answer:
a) mend

5. slack
a) active
b) glee
c) lax
Answer:
a) active

6. sedate
a) abolish
b) malice
c) lively
Answer:
c) lively

7. contempt
a) disrespect
b) respect
c) retain
Answer:
b) respect

8. sterilize
a) adulterate
b) charming
c) ambiguity
Answer:
a) adulterate

9. delirium
a) slow
b) coherence
c) bottom
Answer:
b) coherence

10. remit
a) captivating
b) nadir
c) hold
Answer:
c) hold

11. blot
a) honour
b) aversion
c)encourage
Answer:
a) honour

12. throttle
a) possible
b) encourage
c) sane
Answer:
b) encourage

13. meek
a) desire
b) docile
c) bossy
Answer:
c) bossy

14. quaver
a) interested
b) composed
c) qualified
Answer:
b) composed

15. precious
a) worthless
b) calm
c) visible
Answer:
a) worthless

Question 3.
a) Choose the option that is a synonym of the given word.

1. dart
a) linger
b) dash
c) alarming
Answer:
b) dash

2. distort
a) alter
b) lax
c) smudge
Answer:
a) alter

3. acquisition
a) enthusiasm
b) imitate
c) gain
Answer:
c) gain

4. crest
a) peak
b) aversion
c) brave
Answer:
a) peak

b) Choose the option that is an antonym of the given word.

1. present
a) here
b) absent
c) light
Answer:
b) absent

2. plump
a) short
b) lean
c) fat
Answer:
b) lean

3. despise
a) like
b) need
c) hate
Answer:
a) like

4. sedate
a) abolish
b) malice
c) lively
Answer:
c) lively

Question 4.
a) Choose the option that is a synonym of the given word.

1. timid
a) frightened
b) bold
c) sober
Answer:
a) frightened

2. menace
a) safety
b) risk
c) week
Answer:
b) risk

3. slack
a) honesty
b) esteem
c) sluggis
Answer:
c) sluggis

4. precious
a) prized
b) cheap
c) steady
Answer:
a) prized

b) Choose the option that is an antonym of the given word.

1. throttle
a) encourage
b) control
c) dismiss
Answer:
a) encourage

2. remit
a) forward
b) retain
c) nadir
Answer:
b) retain

3. awful
a) gruesome
b) creepy
c) pleasant
Answer:
c) pleasant

4. revoke
a) abolish
b) approve
c) aversion
Answer:
b) approve

For systematic learning, students can include AP Inter 1st Year English Grammar with Answers and Silent Letters in English Inter 1st Year in their daily study routine.

AP Inter 1st Year English Silent Letters with Answers

Silent letters are a peculiar feature of English spelling. Letters are used in writing but the corresponding sounds are not uttered in pronunciation. This poses problems to learners both in spelling and pronunciation. A study says that 60% of the English words have silent letters!

Careful observation of both the spelling and pronunciation is the only way out to overcome this problem. There are, however, some generalizations about silent letters that will help learners understand this feature to some extent.

In the Intermediate Public Examination, normally, the words that appear in the prescribed selections with a silent consonant letter are set under this question.

Some Guidelines:

• ‘b’ before word-final ’t’ is silent – doubt debt.
• ‘b’ in final mb’ clusters is also silent. – comb womb tomb
• g’ before word final ’n’ is silent. – sign resign foreign campaign reign sovereign
• g’ in most ‘ough’, ‘augh’ combinations is silent. – though taught through thorough
  (But in some words with ’augh’ it may sound as /f/ as in ’laughter’)
• ‘h’ in word beginning position (not always) is silent. – honest honour
• ‘k’ in word beginning followed by n’ is silent. – knowledge know knit knee knave knuckle
• ‘l’ before word final ‘m’ is silent. – calm palm (but in realm – ‘l’ is not silent)
• Word final n’ preceded by m’ is silent, condemn column autumn
• ‘p’ in word beginning position followed by ‘s’ is silent, psychology psalm pseudo
• ‘p’ in words like receipt is also silent.
• ‘r’ followed by a consonant letter is silent, world card
• ‘t’ is silent in clusters like ‘astle’, ‘isten’, istle’ as in castle whistle listen
• w’ is silent in words like – write wrap know fawn lawn

(These guiding principles are by no way exhaustive. They help the learner to a great extent in identifying the silent letters. About vowels – the problem is not this simple or unambiguous. So no generalisation is given here.)

Some More Examples

Exercises

1. Identify the silent consonants in the following words.

1. thumb – b
2. thirst – r
3. muscle – c
4. iron – r
5. write – w
6. vehicle – h
7. numb – b
8. yolk – l
9. clutch – t
10. campaign – g
11. aisle – s
12. neighbour – gh
13. scent – c
14. almond – l
15. World – r
16. tackle – c
17. wrinkle – w
18. knot – k
19. bright – gh
20. journey – o
21. chalk – l
22. limb – b
23. weight – gh
24. wrath – w
25. folk – l
26. discipline – c
27. mortgage – t
28. fascinate – c
29. design – g
30. balm – l

31. thigh – gh
32. rustlings – t
33. knob – k
34. slaughter – gh
35. yacht – ch
36. heir – h
37. answer – w
38. cupboard – p
39. palm – l
40. thorn – r
41. Christmas – t
42. pledge – d
43. wrong – w
44. sword – w
45. knew – k
46. psychology – p
47. scene – c
48. debris – s
49. depot – t
50. alms – l
51. hour – h
52. salmon – l
53. knee – k
54. castle – t
55. Wednesday – d
56. wreath – w
57. judge – d
58. knock – k
59. honourable – h
60. rapport – t

2. Identify the silent consonants in the following words.

1. subtle – b
2. handsome – d
3. resign – g
4. knead – k
5. wreck – w
6. psyche – p

3. Identify the silent consonants in the following words.

1. soften – t
2. cupboard – p
3. honour – h
4. kneel – k
5. surprise – r
6. pneumonia – p

For systematic learning, students can include AP Inter 1st Year English Grammar with Answers and Prepositions in English Inter 1st Year in their daily study routine.

AP Inter 1st Year English Prepositions with Answers

A preposition is placed before nouns, pronouns, noun equivalents, noun phrases, or noun clauses. It shows its relation to some other word or words in the sentence.

(ఒక నామవాచకము, సర్వనామము, నామవాచక సమానాలు, నామవాచక పదసముదాయం లేక నామవాచక ఉపన్యాసాల ముందు విభక్తి ప్రత్యయం (preposition) ఉంటుంది. ఇది వాక్యంలోని ఇతర పదము లేక పదాలతో ఆ నామవాచకము యొక్క సంబంధాన్ని తెలుపుతుంది.)

e.g.: i) She is fond of sweets,
ii) He jumped into the river.

USING PREPOSITIONS:

A) In :

1. To denote place : in Canada; in a village.
2. To indicate time : in the morning.
3. To indicate position : in intensive care, in a jovial mood.
4. To indicate dress : in blue frock, in uniform.
5. To indicate surroundings : in open space, in prison.

B) At :

1. To denote state : India and Pakistan were at war in the early sixties.
2. To indicate a point of time : I shall meet you at 5 p.m.
3. To denote degree or price : We bought these apples at forty rupees a dozen.
4. To denote a place : She was at school then.
5. To tell the sense of being engaged : All the students are at play.
6. To denote aim : The hunter fired his gun at the bird.
7. With the names of small villages and towns : She lives at Kondapalli.

C) By:

1. To show nearness : There is a house by the stream.
2. To mean ‘during’ : By night or by day, he is a nuisance.
3. To mean ‘measure’ : He is taller by an inch.
4. To mean in the name of : I swore in the court by God.

D) On :

1. Place : My pen is on the table.
2. Time : I was on time to the examination.
3. Position : Sarala is on probation.
4. Concern : This is a book on law.
5. Objective : I am bent on visiting Mysore this summer.

E) Of :

1. Relationship : The results of S.S.C. are published today.
2. Phrases : A man of principles, a story of adventure.
3. Division : He took one metre of this cloth.
4. Subjective relation : The love of mother can never be repaid.
5. Cause : He died of sun-stroke.
6. Source : She comes of a rich family.
7. Quality : Shylock was a man of cruel character.
8. Contents : I received a bag full of gifts.
9. Peculiar constructions : How nice of you!

F) For:

1. In place of : Dalda is used for ghee.
2. On account of: We entered slowly for the fear of disturbing the baby.
3. Purpose : Let us pray for peace.
4. Destination : I am leaving for Chennai.
5. Period of time : They have been working for the last three hours.

G) After:

1. Resemblance : Sivaji takes after his mother.
2. Next : The dog ran after the hare.
3. Time : I usually return home after 5 p.m.

The following are some of the uses of prepositions of place:

: waiting at a bus stop, at the station, at the traffic lights, at a party, at a conference

: a spider on the wall, on the ceiling, on the road, on the floor, on the river.

: swimming in the river, in the bag, in the room, in Hyderabad, in the sky.

: to the station, to school, to the bank, to hospital, to my house.

: from the station, from school, from the bank, from home.

: into the house, into the room, into the bank, into the kitchen.

: out of the house, out of the room, out of the bank, out of the kitchen.

: over the head, over the building, over the house, over the tree.

: through the window, through glass, through the village.

: along the road, along the bridge, along the river.

: across the road, across the river, across the building.

Differences in using Prepositions

Commonly Used Prepositional Combinations

1. abide by: The students should abide by the college rules.
2. accuse of: The police have accused him of theft.
3. act up : Children usually act up at functions as they feel bored.
4. addicted to: Sheela is addicted to social media.
5. afraid of: Saritha is afraid of cockroaches.
6. agree to : Do you agree to the terms and conditions?
7. agree with : We all agree with the picnic proposal.
8. angry with (a person) : My father is angry with me.
9. angry at (something) : The principal is angry at the student’s rude behaviour.
10. apologise to (a person for something) : I apologised to my friend for forgetting her birthday.
11. apply to (a person for something) : Have you applied to the manager for a job?
12. ask for: The students asked the teacher for detailed notes.
13. averse to : I am averse to spicy food.
14. aware of: One should be aware of one’s strengths as well as weaknesses.
15. believe in : I believe in God.

16. beware of: Beware of pickpockets in crowded areas.
17. boast of: Do not boast of your wealth.
18. carry out: Many young persons come forward to carry out the older man’s plans for their village development.
19. congratulate on : I congratulate my brother on his success.
20. deal with: The new police officer initiated steps to deal with the deteriorating law and order situation.
21. depend on : You can always depend on your family.
22. die of: He died of a heart attack.
23. differ with (a person) : I differ with my colleague on that issue.
24. differ in (opinion) : The two experts differ in their opinion.
25. differ from : Southern India differs from Northern India in many ways.
26. famous for : The actor became famous for his role as Superman.
27. fond of: My sister is fond of classical music.
28. good at: She is good at solving puzzles.
29. grateful to : I am extremely grateful to my parents for their support.
30. guarantee for : There is no guarantee for this electronic device.
31. guilty of : The company was found guilty of environmental pollution.
32. inferior to : No one is inferior to anyone.
33. interest in : She has a keen interest in photography.
34. jealous of : Don’t be jealous of your friend’s success.
35. laugh at: Don’t laugh at people who are less fortunate.
36. listen to : Children should listen to their parents’ advice.
37. look after: Our English teacher looks after all our needs.
38. married to: She is married to her childhood friend.
39. pleased with : 1 am very pleased with the service at this hotel.
40. prefer to: I prefer reading books to watching TV.
41. prevent from: We should prevent children from playing with sharp objects.
42. proud of: Kavya is proud of her accomplishments.
43. rely on : You can rely on me to help you in an emergency.
44. response to : I received a positive response to my job application.
46. responsible to : The CEO is responsible to the board of directors.
47. search for: The police are searching for clues to solve the crime.
48. see off: We all went to the airport to see my son off, who was leaving for the U.S.

49. set up: My brother wants to set up his own business after he completes his MBA.
50. similar to : His writing style is similar to that of his favourite author.
51. succeed in : He succeeded in landing him in dream job.
52. suffer from : The patient is suffering from pneumonia.
53. superior to : Your design is superior to mine.
54. surprised at: We were surprised at the unexpected turn of events.
55. take off : The plane took off exactly at 7:05 AM.
56. thirst for: The girl has a great thirst for knowledge.
57. tolerance for: One should have tolerance for other cultures and religions.
58. turn off: All the machines were turned off as there was some technical problem.
59. worthy of : The film is worthy of a National Award.
60. yield to : Weak minds quickly yield to temptations, while strong minds never yield to any pressures.

Exercises for Practice

Question 1.
Fill in the blanks either with ‘since’ or ‘for’.

Raaji has had a number of job opportunities _____(1)_____ she left school. She worked in a fashion designing company _____(2)_____ two years. Then she went abroad. She is in Paris at this moment. She has been there _____(3)_____ 2009. She started a boutique, which has been running successfully _____(4)_____ ten years.
Answers:
1. since
2. for
3. since
4. for

Question 2.
Look at the picture given on Page No. 141 in the textbook and complete the sentences using ‘behind, opposite, next, in front of.

A serene lake lies _____(1)_____ them, its calm water reflecting the surrounding mountains. A dense forest stretches out _____(2)_____ them, its trees towering above. A village is situated _____(3)_____ the lake. An old-world little cafe is located _____(4)_____ to the village, offering a warm welcome to the tourists.
Answers:
1. in front of
2. behind
3. opposite
4. next

Question 3.
Fill in the blanks with prepositions of movement or place.

1. There is a fence ___________ (over/ around/ into) the garden.
Answer:
around

2. The park is ___________ (away/ on/ below) from here.
Answer:
away

3. He jumped ___________ (around/ through/into) the swimming pool.
Answer:
into

4. The plane flew ___________ (over/ in/ to) the mountains.
Answer:
over

5. The book is ___________ (on / in / down) the shelf.
Answer:
on

6. They walked ___________ (through / along / above) along the beach.
Answer:
along

7. The train passes ___________ (through / in / over) the tunnel.
Answer:
through

8. The car is parked ___________ (above / into / in front) of the crowd.
Answer:
in front of

9. The cafe is ___________ (among / between / on) the library and the bookstore.
Answer:
between

10. The Daintine is placed ___________ (above / UD / atonal the fireplace.
Answer:
above

11. Siya sat ___________ (over / beside / besides) Tia in the cinema hall.
Answer:
beside

12. She climbed ___________ (under / into / up) the stairs.
Answer:
up

13. The child hid ___________ (on / down / behind) the curtain.
Answer:
behind

14. She swam ___________ (into / on / across) the river.
Answer:
across

Question 4.
Choose the correct prepositions for the blanks.

1. She is proud ___________ (with / of / at) her wealth.
Answer:
of

2. I am very pleased ___________ (with / at / by) his behaviour.
Answer:
with

3. Himanish is married ___________ (by / to / with) Geethika.
Answer:
to

4. I am not jealous ___________ (of / on / to) his good luck.
Answer:
of

5. He was found guilty ___________ (for / of / at) his crime.
Answer:
of

6. Dheeraj is not afraid ___________ (of / for / to) failures.
Answer:
of

7. She is fond ___________ (for / of / to) traditional food.
Answer:
of

8. The painting was created ___________ (by / for / with) a local artist.
Answer:
by

9. The furniture was made ___________ (in / of / through) rose wood.
Answer:
of

Exercises

Question 1.
Choose the correct preposition from the options in brackets and fill in the blanks.

1. The principal congratulated the girl ___________ (on / for / to) her success.
Answer:
on

2. He is suffering ___________ (with / of / from) malaria.
Answer:
from

3. I prefer tea ___________ (for / over / to) coffee.
Answer:
to

4. He is angry ___________ (at / with / on) his son.
Answer:
with

5. Visakhapatnam is famous ___________ (of / for / with) beautiful beaches.
Answer:
for

6. Do you agree ___________ (with / to / at) this proposal?
Answer:
with

7. Don’t laugh ___________ (on / at / to) others’ mistakes.
Answer:
at

8. I am grateful ___________ (for / to / of) my professor for her invaluable guidance.
Answer:
to

9. Greeshmika is not interested ___________ (in / to / of) online games.
Answer:
in

10. I believe ___________ (at / in / with) hard work rather than luck.
Answer:
in

11. Nanny is very good ___________ (in / at / with) telling stories.
Answer:
at

12. My father always listens ___________ (of / off / to) Ilayaraja’s songs.
Answer:
to

13. MLAs and MPs must be responsible ___________ (for / of / to) the people who elected them.
Answer:
to

14. My parents got married ___________ (in / on / since) May.
Answer:
in

15. I had dinner ___________ (for / at / in) 7 PM.
Answer:
at

16. I’ve been playing ___________ (since / for / in) morning.
Answer:
since

17. I’m heading out ___________ (to / for / on) lunch.
Answer:
for

18. The park is ___________ (besides / between / among) the two buildings.
Answer:
between

19. She’s a great singer, ___________ (beside / besides / for) a good dancer.
Answer:
besides

20. The party’s ___________ (in / on / at) Friday night.
Answer:
on

21. The bridge was constructed ___________ (on / from / across) the river.
Answer:
across

22. I received a gift ___________ (from / to / on) my brother.
Answer:
from

23. The lights flickered ___________ (during / on / from) the storm.
Answer:
during

24. The swimmer dived ___________ (in / on / into) the pool for the race.
Answer:
into

25. The plane took ___________ (of / off / up) exactly on time.
Answer:
off

26. The meeting is scheduled from 2 pm ___________ (to / since / at) 3 pm.
Answer:
to

27. The new policy has sparked debate ___________ (among / between / in) the experts in the field of economics.
Answer:
among

28. She has been waiting ___________ (since / from / for) three hours for the flight in the airport.
Answer:
for

29. National Youth Day is observed ___________ (in / on / from) 12th January.
Answer:
on

30. The shops ___________ (along / over / under) the main road (on / across / from) the impact of social media on society.
Answer:
along

31. A stunning view of the city stretched ___________ (on / across / from) the horizon.
Answer:
across

32. She wrote a thought-provoking article ___________ (from / at / about) the impact of social media on society.
Answer:
about

33. Christ, the Redeemer statue is located ___________ (in / at / on) Brazil.
Answer:
in

34. The hikers made their way ___________ (over / through / at) the dense forest.
Answer:
through

35. The car was going ___________ (behind / under / over) the truck on the highway.
Answer:
behind

2. Fill in each blank with the correct preposition from options provided.

1. It closed its mouth ___________ (of / in / from) a death grin.
Answer:
in

2. I will not take ___________ (up / of / on) your time.
Answer:
up

3. But I will tell you what is not the matter ___________ (from / about / with) me.
Answer:
with

4. ___________ (In / On / At) a Thursday, the young man appeared at the eastern gate.
Answer:
On

5. Your buggy’s waiting ___________ (at / for / on) you, ain’t it ?
Answer:
for

6. Others have navigated ___________ (at / in / through) tough times.
Answer:
through

3. Fill in each blank with the correct preposition from the options provided.

1. The police arrived ___________ (at / in / from) the right moment.
Answer:
at

2. Always check your bill ___________ (after / before / during) paying money.
Answer:
before

3. The shop has been closed ___________ (from / for / since) two weeks.
Answer:
for

4. Children like ice creams ___________ (beside / besides / after) chocolates.
Answer:
besides

5. We go ___________ (to / for / from) market every Sunday.
Answer:
to

6. The chocolates are distributed ___________ (among / in / between) the children.
Answer:
among

For systematic learning, students can include AP Inter 1st Year English Grammar with Answers and Odd Sounds in English Inter 1st Year in their daily study routine.

AP Inter 1st Year English Odd Sounds with Answers

A set of three words are given. One or two letters which are common in all the three words are underlined. The underlined letter stands for one sound in two of the given words and for a different sound in the other word. The word with a different sound is to be written as the response.

Exercise For Pracitce

Exercises

Question 1.
Find the odd words in the given words keeping the underlined parts of the words.

Question 2.
Identify the odd words from the following words in respect of their underlined sounds and write them.

Question 3.
Identify the odd words from the following words in respect of their underlined sounds and write them.

For systematic learning, students can include AP Inter 1st Year English Grammar with Answers and Phonetic Transcription in English Inter 1st Year in their daily study routine.

AP Inter 1st Year English Phonetic Transcription with Answers

Meaningful speech sounds are the basic raw material for any language. These sounds are represented by symbols in writing. We refer to these symbols as the alphabet.

Most of the Indian languages have a fixed relationship between the sounds and their symbols. In other words, one symbol always stands for one sound and vice versa. Therefore, we find absolutely no problem while reading or writing Indian languages once we learn the alphabet of that language.

The same is not the case with English. One letter may stand for many sounds. Example c as / k / in car, as / s/ in century as /t ʃ/ in change. One sound is also represented by various letters.
Ex: / f / is represented by ph’ in philosophy, by ‘f in fan, by gh’ in rough. This results in a lot of problems for the learners, particularly for foreign learners in writing the spelling and pronouncing the written words.

A way out of this probelm is a set of forty-four symbols, called phonetic symbols. Each of these symbols stands for one sound only. Learning these symbols arms us with the necessary weapons to war against the problems in pronunciation and spelling.

Study the following forty four phonetic symbols carefully and learn to identify and use them. Phonetic symbols are always placed between two Slant lines.

Vowels

1) / i: / as in see
2) /ɪ/ as in sit
3) /e/ as in ten
4) / æ / as in bat
5) /ɑː/ as in arm
6) / ɒ/ as in got
7) /ɔː/ as in saw
8) /u/ as in put
9) /u:/ as in too
10) /ʌ/ as in cup
11) /ɜː/ as in fur
12) /ə/ as in ago
13) /eɪ/ as in page
14) /əʊ/ as in home
15) /aɪ/ as in five
16) /aʊ/ as in cow
17) /ɔɪ/ as in boil
18) /ɪə/ as in near
19) /eə/ as in hair
20) /ʊə/ as in poor

Consonants

1) / p / as in pen
2) / b / as in bin
3) /t / as in tea
4) / d / as in dog
5) / k/ as in cat
6) / g / as in goat
7) /t ʃ/ as in change
8) / d ʒ / as in July
9) / f / as in fall
10) /v/ as in voice
11) /θ/ as in thin
12) /ð/ as in then
13) / s / as in so
14) / z / as in zoo
15) / ʃ / as in she
16) / ʒ / as in vision
17) / h / as in how
18) / n / as in may
19) / n / as in now
20) / ŋ / as in sing
21) /l/ as in long
22) / r / as in ring
23) /j/ as in yes
24) /w / as in wall

Writing the symbols that represent the sounds in a word is called phonetic transcript. In many examinations phonetic transcription is given and the examinee is asked to write the spelling of those words.

Look at the following words and their phonetic transcription and you will certainly have an idea of it:

Word – Transcription

1. almond – /ɑːmənd/
2. asthma – /ˈæsmə/
3. business – /ˈbɪznɪs/
4. bury – /ˈber.i/
5. cache – /kæʃ/
6. colonel – /ˈkɜːnl/
7. coupon – /ˈkuːpɒn/
8. chassis – /ˈʃæsɪ/
9. chaos – /ˈkeɪɒs/
10. choir – /ˈkwaɪər/
11. cucumber – /ˈkjuːkʌm.bər/
12. data – /ˈdeɪ.tə/
13. dengue – /ˈdɛŋɡi/
14. determine – /dɪˈtɜːmɪn/
15. debris – /ˈdeɪb.riː/
16. discretion – /dɪˈskrɛʃᵊn/
17. epitome – /ɪˈpɪtəmi/
18. extempore – /eksˈtempərɪ/
19. etc./ et cetera – /ɪtˈsɛt.ər.ə/
20. flour – /flaʊə/
21. forte – /ˈfɔːteɪ/
22. gauge – /ɡeɪʤ/
23. genre – /ˈʒɒn.rə/
24. iron – /aɪən/
25. itinerary – /aɪˈtɪnərəri/
26. jewellery – /ˈʤuːəlri/
27. mischievous – /ˈmɪsʧɪvəs/
28. photography – /fəˈtɒɡrəfi/
29. picture – /ˈpɪkʧər/
30. plumber – /ˈplʌm.ər/
31. poem – /ˈpəʊɪm/
32. police – /pəˈliːs/
33. pizza – /ˈpiːt.sə/
34. receipt – /rɪˈsiːt/
35. saloon – /səˈluːn/
36. sour – /ˈsaʊər/
37. status – /ˈstaeɪ.təs/
38. subtle – /ˈsʌtl/
39. suite – /swiːt/
40. vehicle – /ˈviə.kəl/
41. vegetable – /ˈveʤɪtəbəl/
42. violence – /ˈvaɪə.ləns/
43. vineyard – /ˈvɪnjɑːd/
44. Wednesday – /ˈwɛnz.deɪ/
45. women – /ˈwɪm.ɪn/

Exercises

Study the following phonetic transcriptions and write them in ordinary spelling.

1. Exercise from UNIT – 1

Question 1.
/kəˈmɛns/
Answer:
commence

Question 2.
/ˈfænsi/
Answer:
fancy

Question 3.
/ˈeɪlmənt/
Answer:
ailment

Question 4.
/dɪˈspeə(r)/
Answer:
despair

Question 5.
/ɡlɑːns/
Answer:
glance

Question 6.
/plʌnʤ/
Answer:
plunge

Question 7.
/fɪəs/
Answer:
fierce

Question 8.
/ˈblʌʤənɪŋ/
Answer:
bludgeoning

Question 9.
/ˈhæmpə(r)/
Answer:
hamper

Question 10.
/ˈsɜːkəmstɑːns/
Answer:
circumstance

2. Exercise from UNIT – 2

Question 1.
/pʌɡˈnæsəti/
Answer:
pugnacity

Question 2.
/ɪˈlæbərəɪt/
Answer:
elaborate

Question 3.
/ˈʤɒsl/
Answer:
jostle

Question 4.
/ˈɜːrʧɪn/
Answer:
urchin

Question 5.
/skɜːʤ/
Answer:
scourge

Question 6.
/ˈmʌŋɡrəl/
Answer:
mongrel

Question 7.
/ˈtrɛbəl/
Answer:
treble

Question 8.
/ˈmjuːtɪleɪt/
Answer:
mutilate

Question 9.
/slæk/
Answer:
Slack

Question 10.
/ˈpɛrɪʃ/
Answer:
perish

3. Exercise from UNIT – 3

Question 1.
/ˈslɔːtər/
Answer:
slaughter

Question 2.
/ˈθɪkɪt/
Answer:
thicket

Question 3.
/ˈvænɪʃ/
Answer:
vanish

Question 4.
/ˈænsɛstə(r)/
Answer:
ancestor

Question 5.
/ˈæŋɡwɪʃ/
Answer:
anguish

Question 6.
/ˈpænɪk/
Answer:
panic

Question 7.
/rɪˈtriːvd/
Answer:
retrieved

Question 8.
/ˈsɛntəns/
Answer:
sentence

Question 9.
/ˈprəʊtɛst/
Answer:
protest

Question 10.
/vɔːlt/
Answer:
vault

4. Study the following phonetic transcriptions and write ANY FIVE of them in their ordinary spelling.

Question 1.
/ˈʤiːniəli/
Answer:
genially

Question 2.
/rɛtrɪˈbjuːʃən/
Answer:
retribution

Question 3.
/ʌnˈkɒŋkərəbəl/
Answer:
unconquerable

Question 4.
/ˈkæpʃən ɒv maɪ səʊl/
Answer:
caption of my soul

Question 5.
/səʊl/
Answer:
soul

Question 6.
/kənˈtɛmpt/
Answer:
contempt

Question 7.
/ˈpreəri/
Answer:
prairie

Question 8.
/bɪˈfɔːl/
Answer:
befall

Question 9.
/kəˈlɪɡrəfi/
Answer:
calligraphy

Question 10.
/’endəskəʊp/
Answer:
endoscope

5. Study the following phonetic transcriptions and write ANY FIVE of them in their ordinary spelling.

Question 1.
/əˈblaɪʤ/
Answer:
oblige

Question 2.
/skɜː(r)ʤ/
Answer:
scourge

Question 3.
/ˈeɪlmənt/
Answer:
ailment

Question 4.
/ɪnˈdjuːs/
Answer:
induce

Question 5.
/spɑːrs/
Answer:
sparse

Question 6.
/ˈʤɒsəl/
Answer:
jostle

Question 7.
/ˈɪnəsəns/
Answer:
innocence

Question 8.
/ˈbɜːnɪŋ/
Answer:
burning

Question 9.
/kənˈvʌlʃən/
Answer:
convulsion

Question 10.
/ˈθrɒtl/
Answer:
throttle

For systematic learning, students can include AP Inter 1st Year English Grammar with Answers and Parts of Speech in English Inter 1st Year in their daily study routine.

AP Inter 1st Year English Parts of Speech with Answers

There are eight parts of speech. They are

1. Noun,
2. Pronoun,
3. Verb,
4. Adjective,
5. Adverb,
6. Preposition,
7. Conjunction and
8. Interjection.

Noun

A noun is a word for a person, place, thing, or idea. Nouns are often used with an article (the, a, an), but not always. Proper nouns always start with a capital letter; common nouns do not. Nouns can be singular or plural, concrete or abstract. Nouns show possession by adding’s. Nouns can function in different roles within a sentence; for example, a noun can be a subject, direct object, indirect object, subject complement, or object of a preposition.

నామవాచకం అనేది ఒక వ్యక్తి, స్థలం, వస్తువు లేదా ఆలోచనను సూచించే పదం. నామవాచకాలు తరచుగా ఒక article (the, a, an)తో ఉపయోగించబడతాయి, కానీ ఎల్లప్పుడూ కాదు. సరైన నామవాచకాలు ఎల్లప్పుడూ capital అక్షరంతో ప్రారంభమవుతాయి. సాధారణ నామవాచకాలు అలా చేయవు. నామవాచకాలు ఏకవచనం లేదా బహువచనం, కాంక్రీటు లేదా వియుక్తంగా ఉండవచ్చు. నామవాచకాలు ‘s ని జోడించడం ద్వారా possession ని చూపుతాయి. నామవాచకాలు ఒక వాక్యంలో వేర్వేరు పాత్రలలో పనిచేయగలవు; ఉదాహరణకు, ఒక నామవాచకం ఒక కర్త, ప్రత్యక్ష వస్తువు, పరోక్ష వస్తువు, కర్త యొక్క వస్తువు కావచ్చు.

Common Nouns : man, mountain, state, ocean, country, building, cat, airline, etc.

Proper Nouns: Himalayas, Andhra Pradesh, Atlantic Ocean, Australia, Vijayawada, Moon, Sun, Ram, Raghu, Sita, etc. Abstract Nouns: temperature, wealth, happiness, pride, fear, religion, belief, history, communication, fever, headache, heat, etc.

Concrete Nouns: stone, water, wood, iron, glass, grass, bird, photograph, banana, hands, dog, suitcase, flowers, etc. Countable Nouns: coin, bed, cat, movie, train, country, book, phone, match, speaker, clock, pen, pencil, violin, etc.

Uncountable Nouns : money, milk, rice, snow, rain, water, food, music, oil, sugar, hair, etc.

Compound Nouns : headmaster, handkerchief, tablecloth, eyeglasses, seabird, classmate, daughter-in-law, pigtails, sunlight, snowfall, playground, doorbell, etc.

Collective Nouns: bunch, audience, flock, team, group, family, band, bundle, socks, trousers, herd, flock, group, procession, staff, jury, etc.

Pronoun

A pronoun is a word used in place of a noun. A pronoun is usually substituted for a specific noun. Pronouns are further defined by type: personal pronouns refer to specific persons or things; possessive pronouns indicate ownership; reflexive pronouns are used to emphasize another noun or pronoun; relative pronouns introduce a subordinate clause; and demonstrative pronouns identify, point to, or refer to nouns.

సర్వనామం అనేది నామవాచకం స్థానంలో ఉపయోగించే పదం. సర్వనామం సాధారణంగా ఒక నిర్దిష్ట నామవాచకానికి ప్రత్యామ్నాయంగా ఉంటుంది, దీనిని, ఆమె అనే సర్వనామం యొక్క పూర్వపదం అమ్మాయి. వ్యక్తిగత సర్వనామాలు నిర్దిష్ట వ్యక్తులు లేదా వస్తువులను సూచిస్తాయి.

Personal Pronouns:

Demonstrative Pronouns : This, that, these, those.

• This is a book.
• That is a book.
• These are my books.
• Those are your books.

Indefinite Pronouns : Anyone, something, all, some, both. Anyone can do it.

• Anybody can do it.
• Someone can do it.
• Somebody can do it.
• All can do it.
• Both are my sisters.

Interrogative Pronouns: What, whom, whose, who, which. What happened?

• With whom did you come ?
• Whose are these books?
• Who do you want?
• Which is your pen?

Distributive Pronouns : Each, everyone, either, neither. Each one of you will get a chance.

• Everyone of you will get a chance.
• Either of you can get a chance.
• Neither of you will get a chance.

Verb

The verb in a sentence expresses action or being. There is a main verb and sometimes one or more helping verbs. (“She can sing.” Sing is the main verb; can is the helping verb.) A verb must agree with its subject in number (both are singular or both are plural). Verbs also take different forms to express tense.

ఒక వాక్యంలోని క్రియ చర్య లేదా ఉనికిని వ్యక్తపరుస్తుంది. ఒక ప్రధాన క్రియ మరియు కొన్నిసార్లు ఒకటి లేదా అంతకంటే ఎక్కువ సహాయ క్రియలు ఉంటాయి. (“ఆమె పాడగలదు.” sing అనేది ప్రధాన క్రియ; can అనేది సహాయ క్రియ.) ఒక క్రియ దాని కర్తతో ఏకీభవించాలి (రెండూ ఏకవచనం లేదా రెండూ బహువచనం). క్రియలు కూడా కాలాన్ని వ్యక్తీకరించడానికి వేర్వేరు రూపాలను తీసుకుంటాయి.

• Action verbs : go, come, sit, stand, walk, jump, take, drive, do, drink, clean, cut, drive, eat, fly, go, live, make, play, read, run, shower, sleep, smile, stop, sweep, swim, think, throw, trip, wash, work, write etc. (There are thousands of these verbs.)
• Being verbs : am, is, are, was, were (be, been).
• Possessive verbs : have, has, had.
• Helping verbs : Am, is, are, was, were, have, has, had, will, would, shall, should, can, could, may, might, need, dare, ought, used, and must.

Adjective

An adjective is a word used to modify or describe a noun or a pronoun. It usually answers the question of which one, what kind, or how many. (Articles [a, an, the] are usually classified as adjectives.)
Examples : homeless, adorable, delightful, homely, adventurous, horrible, aggressive, determined, hungry, real, agreeable, etc.

విశేషణం అనేది నామవాచకం లేదా సర్వనామాన్ని సవరించడానికి లేదా వివరించడానికి ఉపయోగించే పదం. ఇది సాధారణంగా ఏది, ఏ రకం లేదా ఎన్ని అనే ప్రశ్నకు సమాధానం ఇస్తుంది. (articles [a, an, the] ను సాధారణంగా విశేషణాలుగా వర్గీకరిస్తారు.)

Kinds of adjectives:

a. Quantitative Adjectives : Quantitative adjectives provide information about the quantity of the nouns / pronouns. This type belongs to the question category of ‘how much’ and ‘how many.’
Examples:

I have just 200 rupees in my purse. (How much)
They have three children. (How many)
You should have completed the whole task. (How much)
There are a few students in the class. (How many)

b. Descriptive Adjectives : Descriptive adjectives are words which describe nouns and pronouns. Most of the adjectives belong to this type. These adjectives provide information and attribute to the nouns/pronouns they modify or describe. Descriptive adjectives are also called qualitative adjectives.

Examples:

• I have a fast car which can be used in races.
• I am hungry so let us go to any hotel.
• The mischievous children are making a lot of noise.
• I saw a flying Eagle.
• He bought an expensive old silver mirror.
• He gave me four fresh yellow roses.
• I want to buy a diamond ring.

c. Proper Adjectives: Proper adjectives are the adjective form of proper nouns. When proper nouns modify or describe other nouns/pronouns, they become proper adjectives. ‘Proper’ means ‘specific.’
Example:

• American cars are comparatively very strong.
• Chinese people are hard workers but their leaders are cunning.
• KFC restaurants are found only in big cities.
• Marxist philosophers don’t like capitalism.
• He belongs to the Communist party of India.
• Indian culture is getting diluted.

d. Demonstrative Adjectives: Demonstrative adjectives directly refer to something or someone. Demonstrative adjectives include the words: this, that, these, those.

Examples:

• These books are borrowed from our college library.
• This house is ours.
• That problem needs a very good solution.
• Those bridges were built by the British.

e. Possessive adjectives : Possessive adjectives indicate possession or ownership. It suggests the belongingness of something to someone/something. Some of the most used possessive adjectives are my, his, her, our, their, your.

Examples:

• Your cycle was stolen yesterday.
• We all liked his performance.
• The students of class submitted their assignments.
• I could not understand her intentions.
• My mind stops working when I think about the problem.
• Our intention is not to hurt anyone’s feelings.

f. Interrogative Adjectives : Interrogative adjectives ask a question. An interrogative adjective must be followed by a noun or a pronoun. The interrogative adjectives are: which, what, whose. These words will not be considered as adjectives if a noun does not follow right after them. ‘Whose’ also belongs to the possessive adjective type.

Examples:

• Whose book was that?
• Which pen do you like more?
• What books are you buying today?
• Which team scored the higher in the last match?
• What curry do you like on Sundays?
• Which song are you listening to?

g. Indefinite Adjectives : Indefinite adjective describes or modifies a noun unspecifically. They provide indefinite/ unspecific information about the noun. The common indefinite adjectives are few, many, much, most, all, any, each, every, either, nobody, several, some, etc.

Examples:

• I gave some ice cream to my friend.
• I want a few moments to take a decision.
• Several writers wrote about the recent incidents.
• Each student will have to submit homework tomorrow.

Adverb

An adverb describes or modifies a verb, an adjective, or another adverb, but never a noun. It usually answers the questions of when, where, how, why, under what conditions, or to what degree. Adverbs often end in ‘ly’.

ఒక క్రియా విశేషణం ఒక క్రియ, విశేషణం లేదా మరొక క్రియా విశేషణాన్ని వర్ణిస్తుంది లేదా మారుస్తుంది, కానీ ఎప్పుడూ నామవాచకంగా ఉండదు. ఇది సాధారణంగా ఎప్పుడు, ఎందుకు, ఎలా, ఎందుకు, ఏ పరిస్థులలో, లేదా ఎంతవరకు అనే ప్రశ్నలకు సమాధానం ఇస్తుంది. క్రియా విశేషణాలు తరుచుగా ly తో ముగుస్తాయి.

Adverbs of questions : How, when, where, why.
How did they do it ?
When did they do it ?
Where did they do it ?

Adverbs of Time : today, yesterday, later, now, then, last year, all day, for a while, sometimes,

Adverbs of Frequency: hourly, daily, weekly, monthly, yearly, every, second, once a minute, twice a year, once, twice, once or twice, three times, always, constantly, ever, frequently, generally, never, normally, occasionally, often, rarely, regularly, seldom, sometimes, usually, annually, daily, fortnightly, hourly, monthly, nightly, quarterly, weekly, yearly, yes, no, not, so, too, very, etc.

Preposition

A preposition is a word placed before a noun or pronoun to form a phrase modifying another word in the sentence. Therefore a preposition is always part of a prepositional phrase. The prepositional phrase almost always functions as an adjective or as an adverb.

సర్వనామం ముందు ఉంచిన పదం, వాక్యంలోని మరొక పదాన్ని సవరించి విభక్తి ప్రత్యయము ఒక పదబంధాన్ని ఏర్పరుస్తుంది. అందువల్ల, విభక్తి ప్రత్యయము రెండు పదాల మధ్య బంధాన్ని ఏర్పరుస్తుంది.

Examples: aboard, about, above, across, after, against, ahead of, along, amid, amidst, among, around, as, as far as, as of, aside from, at, athwart, atop barring, before, behind, below, beneath, beside, besides, between, beyond, but, by, despite, down, during, except, excluding, for, from, in, minus, near, next to, of, off, on, opposite, out, outside, over, since, through, till, to, toward, towards, under, underneath, unlike, until, up, upon, with, within, without, etc.

Conjunction

A conjunction joins words, phrases, or clauses, and indicates the relationship between the elements joined.

Coordinating conjunctions connect grammatically equal elements: and, but, or, nor, for, so, yet.

Subordinating conjunctions connect clauses that are not equal: because, although, while, since, etc.

There are other types of conjunctions as well.

ఒక సంయోగం పదాలు,పదబంధాలు లేదా నిబంధనలను కలుపుతుంది మరియు జతచేయబడిన elements మధ్య సంబంధాన్ని సూచిస్తుంది. సమన్వయ సంయోగాలు వ్యాకరణపరంగా సమాన అంశాలను కలుపుతాయి: and, but, or, nor, for, so, yet. సబార్డినేటింగ్ సంయోగాలు సమానం కానీ నిబంధనలు కలుపుతాయి: because, although, while, since, etc. ఇతర రకాల సంయోగాలు కూడా ఉన్నాయి.

Subordinating conjunctions: though, although, even though, even if, as, since, because, for, after, as if, as long as, as much as, as soon, as though, before, even if, if, lest, than, that, till, unless, until, when, where, how, what, why, whenever, wherever, while etc.

Coordinating conjunctions: but, yet, still, so, therefore, that’s why, and or, otherwise, then, either………… or, neither………… nor, not only……… but also etc.

Interjection

An interjection is a word used to express emotion. It is often followed by an exclamation point (!).

భావోద్వేగానికి వ్యక్తీకరించడానికి ఉపయోగించే పదం దాని తర్వాత తరచుగా ఆశ్చర్యార్థంకంగా గుర్తు (!) ఉంటుంది.

Example: Aha! Ahh! Ahoy! Alas! Boo! Bravo! Congratulations! Eh! Eureka! Goodbye! Goodness! Hey! Hurray! Oh! Oops! Ouch! Phew! uhhuh!

Wahoo! Wow! Yeah!, etc.

Examples

Question 1.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 2.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

(Note : In sentence 5, ‘out’ modifies the verb ‘went’ by indicating direction or movement away from a place, which is typical of an adverb of place.)

Exercises For Practice

Question 1.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 2.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 3.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 4.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 5.
Here’s a day from the life of Padmini. Can you identify at least 15 nouns from the given paragraph ?

Last weekend, Padmini visited the museum with her friends. They admired the paintings and marvelled at the sculptures in the art gallery. Later, they had lunch at a cozy cafe, where Padmini ordered a delicious sandwich and a cup of coffee. The group talked about their plans for the upcoming holiday and shared their ideas for a fun trip. Everyone enjoyed the afternoon, and they decided to visit the museum again next month.
Answers:

1. weekend
2. museum
3. friend
4. paintings
5. sculptures
6. gallery
7. lunch
8. cafe
9. sandwich
10. cup
11. coffee
12. group
13. plans
14. holiday
15. ideas
16. trip
17. afternoon
18. month

Question 6.
Fill in the blanks with correct plural form of the nouns given in the brackets.

1. The park was full of playful ___________. (child)
Answer:
children

2. We ran several ___________. (diagnosis)
Answer:
diagnoses

3. He explained the ___________ (focus) of the study.
Answer:
foci

4. My ___________ (foot) were sore after the run.
Answer:
feet

5. My dentist says I have healthy ___________. (tooth)
Answer:
teeth

6. A flock of ___________ (goose) flew across the sky.
Answer:
geese

7. All my ___________ (brother-in-law) are teachers.
Answer:
brothers-in-law

Question 7.
Fill in the blanks with appropriate Pronouns to complete the story.

One day, a young boy named Sam went to the park with ___________ (1) sister. ___________ (2) both loved playing on the swings and running around. While Sam was climbing a tree, ___________ (3)
saw a bird’s nest. ___________ (4) called out to ___________ (5) sister, “Look! There’s a nest up here!” ___________ (6) sister looked up and said. “Be careful, Sam. Don’t scare the birds!” Sam carefully climbed down and joined ___________ (7) on the bench. Together, ___________ (8) decided to draw the birds in ___________ (9) notebooks. When ___________ (10) finished, ___________ (11) showed the drawings to ___________ (12) parents, who were very impressed.
Answers:
1. his
2. They
3. he
4. He
5. his
6. His
7. her
8. they
9. their
10. they
11. they
12. their

Question 8.
Read each sentence carefully and underline or highlight the Adjectives.

1. The beautiful garden was filled with vibrant flowers.
Answer:
The beautiful garden was filled with vibrant flowers.

2. She wore a red dress and carried a small bag.
Answer:
She wore a red dress and carried a small bag.

3. The tall building cast a long shadow on the street.
Answer:
The tall building cast a long shadow on the street.

4. We had a delicious meal at the new restaurant.
Answer:
We had a delicious meal at the new restaurant.

5. His quick decision helped to avoid a serious mistake.
Answer:
His quick decision helped to avoid a serious mistake.

6. The young boy found a golden key in the ancient chest.
Answer:
The young boy found a golden key in the ancient chest.

7. It was a cold night with a bright moon in the sky.
Answer:
It was a cold night with a bright moon in the sky.

8. The teacher gave us a challenging yet interesting assignment.
Answer:
The teacher gave us a challenging yet interesting assignment.

9. I love my cozy, little home in the peaceful countryside.
Answer:
I love my cozy, little home in the peaceful countryside.

10. The angry crowd demanded immediate action from the authorities:
Answer:
The angry crowd demanded immediate action from the authorities.

Question 9.
Identify the Adverbs in the following sentences.

1. She sings melodiously during every performance.
Answer:
melodiously

2. The train arrived late because of the storm.
Answer:
late

3. He quickly finished his homework before heading out.
Answer:
quickly

4. The team played extremely well in the finals.
Answer:
extremely, well

5. They are always kind to their neighbours.
Answer:
always

6. The baby slept peacefully in her crib.
Answer:
peacefully

7. We will visit the museum tomorrow.
Answer:
tomorrow

8. He spoke very softly during the meeting.
Answer:
very, softly

9. The cat crept silently across the room.
Answer:
silently

10. I have never seen such a beautiful sunset.
Answer:
never

Question 10.
Underline the Conjunctions in the following sentences.

1. She wanted to go to the party, but she was feeling unwell.
Answer:
She wanted to go to the party, but she was feeling unwell.

2. I will call you when I reach home.
Answer:
I will call you when I reach home.

3. He likes both cats and dogs, but he cannot decide which to adopt.
Answer:
He likes both cats and dogs, but he cannot decide which to adopt.

4. They left early because they had a train to catch.
Answer:
They left early because they had a train to catch.

5. You can stay here or you can go with them.
Answer:
You can stay here or you can go with them.

Exercises

Question 1.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 2.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 3.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 4.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 5.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 6.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 7.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 8.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 9.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 10.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 11.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 12.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Regular practice with AP Inter 1st Year Physics Study Material Chapter 2 Motion in a Straight Line Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Physics 2nd Lesson Motion in a Straight Line Questions and Answers

I. Multiple Choice Questions

Question 1.
The velocity of a body depends on time according to the equation V = 10 + 2t². The body undergoing
(1) uniform acceleration
(2) uniform retardation
(3) non-uniform acceleration
(4) zero acceleration
Answer:
(3) non-uniform acceleration

Question 2.
Which of the following is true?
(1) Speed is always equal to magnitude of velocity.
(2) Speed is always greater than the magnitude of velocity.
(3) Velocity is always greater than the speed.
(4) Speed in general is either equal to or greater than the magnitude of velocity.
Answer:
(1) Speed is always equal to magnitude of velocity.

Question 3.
A bullet is dropped from the same time another bullet is fired horizontally. They will hit the ground
(1) one after the other
(2) simultaneously
(3) depends on the observer
(4) depends on the velocity of the fired bullet
Answer:
(2) simultaneously

Question 4.
If a car is moving with a uniform velocity of 60 km/h. Then its acceleration is
(1) 0km / h²
(2) 10km / h²
(3) 20km / h²
(4) 30km / h²
Answer:
(1) 0km / h²

Question 5.
A ball is thrown vertically upward with an initial velocity of 50 m/s. Its maximum height is (take g = 10/s²)
(1) 125m
(2) 250 m
(c) 500 m
(d) 1000 m
Answer:
(1) 125m

Question 6.
A particle moves in a straight line with an initial velocity of 10m/s and a constant acceleration of 2m/s². Its displacement after 5 seconds will be
(1) 50 m
(2) 75 m
(3) 150 m
(4) 200 m
Answer:
A = ut + \(\frac{1}{2}\)at² (or) x = v₀t + \(\frac{1}{2}[latex]at², ∴ x = (10 × 5) + [latex]\frac{1}{2}\)(2) × 5² = 50 + 25 = 75 m

Question 7.
A body starts from rest and moves with a uniform acceleration of 2m / s². Its velocity after 10 seconds will be
(1) 5 m/s
(2) 10 m/s.
(3) 15 m/s
(4) 20 m/s
Answer:
(4) 20 m/s
V = u + at = 0 + (2 × 10) = 20m/s

Question 8.
For the motion with uniform velocity, the slope of the velocity-time graph is equal to
(1) 1 m/s
(2) zero
(3) initial velocity
(4) final velocity
Answer:
(2) zero

Question 9.
If a cyclist takes two minutes to complete half revolution on a circular path of 120 m radius, what is the average speed?
(1) 1 m/s
(2) 2 m/s
(3) 3.14 m/s
(4) 4.14 m/s
Answer:
(3) 3.14 m/s
Average Speed = \(\frac{\text { Total distance }}{\text { Total Time }}=\frac{\pi \mathrm{r}}{2 \times 60}=\frac{3.14 \times 120}{2 \times 60}\) = 3.14 m/s

Question 10.
In a particular time, the net displacement travelled by the body is equal to
(1) the area which v -1 graph encloses with displacement axis
(2) the area which x -1 graph encloses with the time axis
(3) the area which v -1 graph encloses with time axis
(4) the area which a -1 graph encloses with displacement axis
Answer:
(3) the area which v -1 graph encloses with time axis

Question 11.
A particle moves along a straight line. At a time t, the distance x of the particle from the origin given by x = 20 + 12t – t³ where x is in meters and t in sec, the time taken for the particle fo coming to rest is
(1) 1 s
(2) 2 s
(3) 4 s
(4) 6 s
Answer:
(2) 2 s
x = 20 + 12t – t³, \(\frac{d x}{d t}\) = 12 – 3t² = V, 12 – 3t² = 0 ⇒ 3t² = 12 ⇒ t² = 4, t = 2s

Question 12.
A car travels half of the distance with speed v and the remaining distance with speed 2v. Its average speed is
(1) \(\frac{4 v}{3}\)
(2) \(\frac{3 v}{3}\)
(3) \(\frac{v}{3}\)
(4) \(\frac{2 v}{3}\)
Answer:
(1) \(\frac{4 v}{3}\)
Average Speed = \(\frac{2 u v}{u+v}=\frac{2 \times(v)(2 v)}{v+2 v}=\frac{2 \times v \times 2}{3}=\frac{4}{3} v\)

Question 13.
A car runs at constant speed on a circular track of radius 100 m, taking 62.8 seconds for every circular lap. The average velocity and average speed of each circular lap respectively is
(1) 10 m/s, 0 m/s
(2) 0m/s. 0 m/s
(3) 0 m/s, 10 m/s
(4) 10 m/s, 10 m/s
Answer:
(3) 0 m/s, 10 m/s
Hint: Average Velocity = 0
Average Speed = \(\frac{2 \pi \mathrm{r}}{62.8}=\frac{2 \times 3.14 \times 100}{62.8}=\frac{6.28 \times 100}{62.8}\) = 10m/s

Question 14.
The motion of the particle is given by the equation S = (2t³ + 5t² + 16t + 8)m. The value of acceleration of the particle at t = 1 sec is
(1) 20 m/s²
(2) 22 m/s²
(3) 24 m/s²
(4) 26 m/s²
Answer:
(2) 22 m/s²
S = 2t³ + 5t² + 16t + 8, v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 6t² + 10t + 16; a = \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 12t + 10
at t = 1, a = 12 + 10 = 22m/s²

Question 15.
The ratio of distances travelled by a freely falling body in the 1^st, 2^nd, 3^rd, 4^th second
(1) 1:2:3:4
(2) 1:4:9:16
(3) 1:3:5:7
(4) 1:1:1:1
Answer:
(3) 1:3:5:7
S_n = u + \(\frac{\mathrm{a}}{2}\)(2n – 1), S₁ = \(\frac{\mathrm{g}}{2}\)(2(1) – 1) = \(\)
S₂ = \(\frac{\mathrm{g}}{2}\)(2(2)- 1) = \(\frac{\mathrm{3g}}{2}\), S₃ = \(\frac{\mathrm{g}}{2}\)(2(3) – 1) = \(\frac{\mathrm{5g}}{2}\) ⇒ S₁ : S₂: S₃ : S₄ = 1 : 3 : 5 : 7

Question 16.
A stone falls freely under gravity. It covers distances h₁, h₂ and h₃ in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h₁, h₂ and h₃ is
(1) h₂ = 3h₁ and h₃ = 3h₂
(2) h₁ = h₂ = h₃
(3) h₁ = 2h₂ = 3 h₃
(4) h₁ = \(\frac{\mathrm{h}_2}{3}=\frac{\mathrm{h}_3}{5}\)
Answer:
(4) h₁ = \(\frac{\mathrm{h}_2}{3}=\frac{\mathrm{h}_3}{5}\)
h₁ = \(\frac{1}{2}\)g(5²) = \(\frac{1}{2}\)g(25), x₁ = \(\frac{1}{2}\)g(10²) = \(\frac{1}{2}\)g(100)
h₂ = x₁ – h₁ = \(\frac{1}{2}\)g (100 – 25) = \(\frac{1}{2}\)g(75), x₂ = \(\frac{1}{2}\)g (15²) = \(\frac{1}{2}\)g(225)
h₃ = x₂ – x₁ = \(\frac{1}{2}\)g(225 – 100) = \(\frac{1}{2}\)g(125), h, = \(\frac{\mathrm{h}_2}{3}=\frac{\mathrm{h}_3}{5}\) (or) 1:3:5:7

II. Fill in the Blanks

Question 1.
________ deals with the study of motion of objects without taking into account, factors which cause the motion.
Answer:
Kinematics

Question 2.
The shortest path travelled by a moving object in a given time from initial to final position is called _______.
Answer:
displacement

Question 3.
The distance covered by a body along a semicircular path of radius r is _______ .
Answer:
πr

Question 4.
The slope of velocity-time graph gives _______.
Answer:
acceleration

Question 5.
The slope of displacement-time graph gives _______.
Answer:
velocity

Question 6.
The area under acceleration-time graph gives _____.
Answer:
change in velocity

Question 7.
The acceleration of a body moving with uniform velocity is ________.
Answer:
zero

Question 8.
The velocity of a body at any instant of time is called _______.
Answer:
instantaneous velocity

Question 9.
The value of an acceleration of a freely falling body is ________.
Answer:
g = 9.8 m/s²]

Question 10.
The ratio of distances travelled by a freely falling body from the rest in the first three seconds is ______.
Answer:
1 : 3 : 5

III. One Word Answer Questions

Question 1.
What is uniform motion ?
Answer:
If an object moving along a straight line covers equal displacement in equal intervals of time, said to be in uniform motion along a straight line.

Question 2.
What is acceleration ?
Answer:
The rate of change of velocity is called acceleration.

Question 3.
What is average speed ?
Answer:
The ratio of total distance travelled by a body to total time is called average speed.

Question 4.
What is instantaneous velocity ?
Answer:
The velocity of a particle at a particular instant is called instantaneous velocity.

Question 5.
Draw position – time graph for an object at rest.
Answer:

Question 6.
What represents the area under the cure of velocity – time graph ?
Answer:
Area = v × t = displacement.

Question 7.
What is the maximum height reached by a body when it is vertically projected up with an initial velocity ‘u’ ?
Answer:
H_max = \(\frac{u^2}{2 g}\)

Question 8.
Mention one similarity between speed and velocity.
Answer:
Having same units and dimensions.

Question 9.
Draw velocity-time graph for a particle moving with uniform velocity.
Answer:

Question 10.
What is the time taken by the body, when it is freely dropped from a height “h” ?
Answer:
h = \(\frac{1}{2}\)gt²; t² = \(\frac{2 h}{g} \Rightarrow t=\sqrt{\frac{2 h}{g}}\)

IV. Very Short Answer Questions

Question 1.
The states of motion and rest are relative. Explain. [dlarch – 2026]
Answer:
REST: If the position of a body does not change with respect to surroundings, it is said to be at “Rest”.

MOTION : If the position of a body changes with respect to surroundings, it is said to be in “Motion”. By definitions, rest and motion are relative with respect to surroundings.

Question 2.
Is it necessary that the direction of acceleration and velocity of a body should be the same? Give an example.
Answer:
No. For a decelerating body velocity is in forward direction and acceleration will be in opposite direction.

Question 3.
Is it possible for a body to have acceleration but zero velocity? Give an example.
Answer:
In case of vertically projected body at maximum height its velocity v = 0. But acceleration due to gravity ‘g’ is not zero.
So even though velocity v = 0 ⇒ acceleration is not zero.

Question 4.
Which is greater, a velocity of 72kmph or 25m/s ?
Answer:
V₁ = 72 × \(\frac{5}{18}\) = 20m/s; V₂ = 25m/s ∴ 25m/s is greater than 72kmph.

Question 5.
Give an example of one-dimensional motion where a particle moving along a straight line comes to rest and moves backward.
Answer:

1. A body projected vertically upwards.
2. When the length of pendulum is high and amplitude is less then its motion is along a straight line. The pendulum will come to a stop at extreme position and moves back in forward direction (‘x’ + ve) periodically.

Question 6.
A vehicle travels half the distance L with speed Vj and the other half with speed v₂. What is the average speed ?
Answer:
The average speed of a vehicle for the two equal parts \(\frac{\mathrm{L}}{2}, \frac{\mathrm{~L}}{2}\) is
V_avg = \(\frac{\text { Total distance }}{\text { Total time }}=\frac{\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}}{\mathrm{t}_1+\mathrm{t}_2}\)
V_avg = \(\frac{\mathrm{L}}{\left(\frac{\frac{\mathrm{~L}}{2}}{\mathrm{~V}_1}+\frac{\frac{\mathrm{L}}{2}}{\mathrm{~V}_2}\right)}=\frac{\mathrm{L}}{\frac{\mathrm{~L}}{2}\left(\frac{1}{\mathrm{~V}_1}+\frac{1}{\mathrm{~V}_2}\right)}=\frac{1}{\frac{1}{2}\left(\frac{\mathrm{~V}_1+\mathrm{V}_2}{\mathrm{~V}_1 \mathrm{~V}_2}\right)}\) (∵ t = \(\frac{\mathrm{L} / 2}{\mathrm{~V}}\)) ⇒ V_avg = \(\frac{2 V_1 V_2}{V_1+V_2}\)

Question 7.
A body covers a distance of 5m along a semicircular path from its one end to another end. What is the ratio of its distance covered to its displacement ?
Answer:
πr = 5; r = 5/π

\(\frac{\text { Distance }}{\text { Displacement }}=\frac{\pi \mathrm{r}}{2 \mathrm{r}}=\frac{\pi}{2}=\frac{22}{7 \times 2}=\frac{11}{7}\)

Question 8.
Displacement travelled by a body is given by S = 2t + 5t² m. What is the acceleration of that body ?
Answer:
S = 2t + 5t²; V = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 2 + 10t; a= \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 10m/s²

Question 9.
An object falling through a fluid is observed to have an acceleration given by a = g – bv where g is the gravitational acceleration and b is a constant. After a long time it is observed to fall with a constant velocity. What would be the value of this constant velocity ?
Answer:
Acceleration, a- = g – bv when moving with constant velocity, a = 0 ⇒ 0 = g – bv
∴ Constant velocity, v = \(\frac{\mathrm{g}}{\mathrm{~b}}\) m/sec.

Question 10.
Under what circumstances is the relationship s = vt valid ?
Answer:
Body moving with uniform velocity (a = 0).

V. Short Answer Questions

Question 1.
Can the equations of kinematics be used when the acceleration varies with time ? If not, what form would these equations take ?
Answer:
The equations of motion are 1) v =u + at 2) s = ut + 1/2at² and 3) v² – u² = 2as. All these three equations applicable to a body moves with uniform acceleration ‘a’.

No, the equations are not applicable when the acceleration varies with time.

Question 2.
A particle moves in a straight line with uniform acceleration. Its velocity at time t = 0 is V₁ and at time t = t is v₂. The average velocity of the particle in this time interval is (v₁ + v₂) / 2. Is this correct ? Substantiate your answer.
Answer:
t₁ = 0 ⇒ u = v₁ ; t₂ = t ⇒ v = v₂; Average velocity = \(\frac{u+v}{2}=\frac{v_1+v_2}{2}\)

Question 3.
Can the velocity of an object be in a direction other than the direction of acceleration of the object ? If so, give an example.
Answer:
Yes. Velocity of a body and its acceleration may be in different directions.
Ex :
1) In case of vertically projected body ⇒ velocity of body is in the upward direction and acceleration is in downward direction.
2) When brakes are applied the velocity of body before coming to rest is opposite to retarding acceleration.

Question 4.
Obtain the equation of motion x = ut + \(\frac{1}{2}\)at² for constant acceleration using graphical method.
Answer:
The velocity – time graph of a body moving with initial velocity ‘u’ and with uniform acceleration ‘a’ as shown. Let V be the velocity of the body after a time t.
In v – t graph, area of velocity-time graph = total displacement travelled by it. Area under velocity – time graph = area of OABCD
∴ Area of Rectangular part OACD = Area of OACD + Area of ABC.
A₁ = OA × OD = ut …..(1)
Area of triangle ABC = A₂
A₂ = \(\frac{1}{2}\) × Base × height = \(\frac{1}{2}\) × AC × (BC) = \(\frac{1}{2}\)t(u – v)
But v – u = at; ∴ A₂ = \(\frac{1}{2}\)t. at = \(\frac{1}{2}\)at²
∴ Total area under graph = s = A₁+ A₂; s(n) = ut + \(\frac{1}{2}\)at².
∴ s = ut + \(\frac{1}{2}\)at² is graphically proved.

Question 5.
Explain the terms the average velocity and instantaneous velocity. When are they equal?
Answer:
Average velocity : It is the ratio of total displacement to total time taken. It is independent of path of the body.
∴ Average velocity = \(\frac{\mathrm{s}_2-\mathrm{s}_1}{\mathrm{t}_2-\mathrm{t}_1}\) (Or) V_avg = \(\frac{\mathrm{ds}}{\mathrm{dt}}\)
Velocity of a particle at a particular instant of time is known as instantaneous velocity. Here time interval is very small. V = \(\frac{\mathrm{dx}}{\mathrm{dt}}\)
Only in uniform motion, instantaneous velocity = average velocity. For all other cases instantaneous velocity may differ from average velocity.

Problems

Question 1.
A man walks on a straight road from his home to a market 2.5 km away with a speed of km h^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^-1. What is the (a) magnitude of average velocity mid (b) average speed of the man over the time interval 0 to 50 minutes ?
Solution:
Time taken by man to go from his home to market, t₁ = \(\frac{\text { distance }}{\text { speed }}=\frac{2.5}{5}=\frac{1}{2} \mathrm{~h}\)
Time taken by man to go from market to his home, t₁ = \(\frac{2.5}{7.5}=\frac{1}{3} h\)
∴ Total time taken = t₁ + t₂ = \(\frac{1}{2}+\frac{1}{3}=\frac{5}{6} h\) = 50min.
In time interval 0 to 50 min,
Total distance travelled = 2.5 + 2.5 = 5 km.; Total displacement = zero.

1. Average velocity = \(\frac{\text { displacement }}{\text { time }}\) = 0
2. Average speed = \(\frac{\text { distance }}{\text { time }}=\frac{5}{5 / 6}\) = 6 km/h

Question 2.
A car travels the first third of a distance with a speed of 10 kmph, the second third at 20 kmph and the last third at 60 kmph. What is its mean speed over the entire distance ?
Solution:
Total distance = s; distance travelled, S₁ = \(\frac{s}{3}\); velocity, v₁ = 10 kmph ; distance, s₂ = \(\frac{s}{3}\) velocity, v₂ = 20 kmph; distance, s₃ = \(\frac{s}{3}\); velocity, v₃ = 60 kmph
Average velocity = \(\frac{\text { total distance }}{\text { total time }}\)

Question 3.
A bullet moving with a speed of 150 m s^-1 strikes a tree and penetrates 3.5 cm before stopping. What is the magnitude of its retardation in the tree and the time taken for it to stop after striking the tree?
Solution:
Velocity of bullet, u = 150 m/s; Final velocity, v = 0 ; Distance travelled, s = 3.5 cm = 3.5 × 10^-2 m,
a) Acceleration, a = \(\frac{v^2-u^2}{2 s}=\frac{0^2-150^2}{2 \times 3.50 \times 10^{-2}}=\frac{-22500}{7 \times 10^{-2}}\)
= -3.214 × 10⁵ m/sec² (-ve sign for retardation)

b) Time taken to stop, t = \(\frac{v-u}{a}=\frac{-150}{-3.214 \times 10^5}\) = 4.67 × 10^-4 sec.

Question 4.
A motorist drives north for 30 min at 85 km/h and then stops for 15 min. He continues travelling north and covers 130 km in 2 hours. What is his total displacement and average velocity ?
Solution:
In first part:
Velocity, v₁ = 85 kmph ; Time, t₁ = 30 min ; Distance travelled, s₁ = v₁ t₁ = 85 × \(\frac{30}{60}\) = 42.5 km

In second part:
Distance travelled, s₂ = 0 ; Time, t₂ = 15.0 min.

In third part:
Distance travelled, s₃ = 130 km ; Time, t₃ = 2 hours = 120 min.
a) Total distance of the motorist, s = s₁ + s₂ + s₃ = 42.5 + 0 + 130 = 172.5 km

b) Total time travelled,
t = t₁ + t₂ + t₃ = 30 + 15 + 120 = 165 minutes = 2 hrs 45 minutes = 2\(\frac{3}{4}\) hrs. = \(\frac{11}{4}\) hrs.
∴ Average velocity, v_avg = \(\frac{\text { total displacement }}{\text { total time }}=\frac{172.5}{(11 / 4)}\) = 62.7 kmph.

Question 5.
A ball A is dropped from the top of a building and at the same time an identical ball B is thrown vertically upward from the ground. When the balls collide the speed of A is twice that of B. At what fraction of the height of the building did the collision occur ?
Solution:
Given at time of collision velocity of A = V_A = 2 × V_B (velocity of B)
Let the body be dropped from a height ‘h’. Let the two stones collide at x from the ground.
For the body dropped, s = h – x = \(\frac{1}{2}\)gt² → (1); For the body thrown up, x = ut – \(\frac{1}{2}\)gt² → (2)
For the body dropped, v = u + at ⇒ V_A = gt → (3)
For the body thrown up, v = u – gt ⇒ V_B = u – gt → (4)
Given V_A = 2V_B ⇒ gt = 2 (u – gt) or u = \(\frac{3gt}{2}\) → (5)
Divide equation (1) with equation (2)

∴ Fraction of height of collision = \(\frac{2}{3}\)

Question 6.
Drops of water fall at regular intervals from the roof of a building of height 16 m. The first drop strikes the ground at the same moment as the fifth drop leaves the roof. Find the distances between successive drops.
Solution:
Height of building, h = 16 m
∴ Time taken to reach the ground, t = \(\sqrt{\frac{2 h}{g}}\)
∴ Time taken to fall, t = \(\sqrt{\frac{2 \times 16}{9.8}}=\sqrt{\frac{32}{9.8}}=\sqrt{3.26}\) = 1.80

Number of drops, n = 5
∴ Number of intervals = n – 1 = 5 – 1 = 4
Time interval between drops = \(\frac{1.8}{4}\) = 0.45 sec
Time of travel of 1st drop, t₁ = 4 × 0.45 = 1.8
∴ Distance travelled by
1st drop, S₁ = \(\frac{1}{2}\)gt₁² = \(\frac{1}{2}\) × 9.8 × 1.8 × 1.8 = 16 m

For 2nd drop, t₂ = 3 × 0.45 = 1.35 sec., S₂ = \(\frac{1}{2}\)gt₂² ;
∴ S₂ = \(\frac{1}{2}\) × 9.8 × 1.35² = 4.9 × 1.822 = 8.9278 ≅ 9m

For 3rd drop, t₃ = 2 × 0.45 = 0.9 sec.
Distance, S₃ = \(\frac{1}{2}\)gt₃² = \(\frac{1}{2}\) × 9.8 × 0.9² = 3.97 ≅ 4m

For 4th drop, t₄ = 1 × 0.45 = 0.45 sec.
Distance travelled, S₄ = \(\frac{1}{2}\)gt₄² = \(\frac{1}{2}\) × 9.8 × (0.45)² ≅ 1

For 5 th drop, t₅ = 0 ⇒ S₅ = 0
Distance between 1^st and 2^nd drop S_{1, 2} = S₁ – S₂ = 16 – 9 = 7m
Distance between 2^nd and 3^rd drop S_{2, 3} = S₂ – S₃ = 9 – 4 = 5 m
Distance between 3^rd and 4^th drop S_{3, 4} = S₃ – S₄ = 4 – 1 = 3m
Distance between 4^th and 5^th drop S_{4, 5} = S₄ – S₅ = 1 – 0 = 1 m
∴ Distances between successive drops are 7m, 5m, 3m and 1m.

Question 7.
A food packet is dropped from an aeroplane, moving with a speed of 360 kmph in a horizontal direction, from a height of 500m. Find (i) its time of descent (ii) the horizontal distance between the point at which the food packet reaches the ground and the point above which it was dropped.
Solution:
Velocity of plane, V = 360 kmph = 360 × \(\frac{5}{18}\) = 100 m/s.
Height above the ground, h = 500 m; g = 10 m/s²

i) Time of descent, t = \(\) = 10sec.

ii) Horizontal distance between point of dropping and point where it reaches the ground = Range R.
∴ R = u \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\) = 100 \(\sqrt{\frac{2 \times 500}{10}}\) = 100 x 10 = 1000 m.

Motion in a Straight Line Class 11 NCERT Solutions

Question 1.
In which of the following examples of motion, can the body be considered approximately point object.
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on the top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
Answer:
(a) A railway carriage can be considered as a point object because the size of carriage is much smaller than the distance between the two stations.

(b) A monkey sitting on the top of the cyclist can be considered as a point object if the cyclist describes a circular path of radius which is sufficiently larger than the size of monkey.

(c) A spinning cricket ball cannot be considered as a point object because the size of ball is not negligible as compared to the distance through which the ball may turn on hitting the ground.
Remark: The spinning motion of cricket ball is due to its size. A point mass cannot have spin motion.

(d) A beaker slipping off the edge of a table cannot be considered as a point object because the size of beaker is not negligible as compared to the height of the table.

Question 2.
The position – time (x – t) graphs for two children A and B returning from their school O to their homes P and Q respectively as the shown in Fig.
Choose the correct entries in the brackets below :

(a) (A/B) lives closer to the school than (B/A).
(b) (A/B) starts from the school earlier than (B/A).
(c) (A/B) walks faster than (B/A).
(d) A and B reach home at the (same/different) time.
(e) (A/B) overtakes (B/A) on the road (once/twice).
Answer:
(a) As OP < OQ ; A lives closer to school than B.
(b) For A, x = 0 at t = 0 and for B, x = 0 at t = t₀ (let)
This implies that A starts from the school earlier than B.
(c) The speed = slope of x – t graph. As slope of B is greater than the slope of A, therefore, B walks faster than Answer:
(d) From graph, it is clear that the positions P and Q are attained at the same time, this implies that both A and B reach home at the same time.
(e) x – t graphs for A and B intersect each other only once, so B overtakes A on the road once.

Question 3.
A woman starts from her home at 9.00 am, walks with a speed of 5 kmh^-1 on a straight road upto her office 2.5 km away, stays at the office upto 5.00 pm and returns home by an auto with a speed of 25 kmh^-1. Choose suitable scales and plot the x-t graph of her motion.
Answer:
For motion from home to office :
Time to which woman leaves the office = 09.00 am
Speed = 5 km/h, distance traversed, s = 2.5 km
Time taken, t₁ = \(\frac{\text { Distance traversed }}{\text { Speed }}=\frac{2.5}{5}\) = 0.5 h = 30 min.
Time at which woman reaches office
= 9.00 am + 0.30 = 9.30 am
Stay at office : 9.30 am – 5.00 pm (position fixed)

For motion from office to home :
Time at which she leaves office = 5.00 pm
Speed of auto = 25 km/h, distance = 2.5 km
Time spent, t₂ = \(\frac{\text { Distance }}{\text { Speed }}=\frac{2.5}{25}\) = 0.1 h = 0.1 × 60 min = 6 min
Time at which she reaches home = 5.06 pm
The position-time (x – t) graph of motion of woman is as shown in the Fig.

Question 4.
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x – t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Answer:
The drunkard takes 5 steps forward and 3 steps backward. Each step is of 1 m length and takes 1 second. Therefore, the displacement of drunkard as a function of time is as shown below.

The position – time graph of motion of drunkard is as shown in the Fig.
From graph, it is clear that the drunkard will fall into the pit 37 seconds after the beginning.

Question 5.
A car moving along a straight highway with a speed of 126 kmir1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform); and how long does it take for the car to stop ?
Answer:
Initial speed of car, u = 126 km/h
1 km/h = \(\frac{1000 \mathrm{~m}}{3600 \mathrm{~s}}=\frac{5}{18}\) m/s ; u = 126 × \(\frac{5}{18}\) = 35 m/s
Final speed of car, v = 0, Distance traversed, s = 200 m
Using the formula v² = u² + 2as, we get
0 = (35)² + 2a × 200 ⇒ -400a = (35)² ⇒ a = \(\frac{(35)^2}{400}\) = -3.06 ms^-2
Retardation of Car = 3.06 ms^-2
If t is the time taken, then formula v = u + at gives
0 = 35 – 3.06 t ⇒ 3.06t = 35 ⇒ t = \(\frac{35}{3.06}\) = 11.4 s

Question 6.
A player throws a ball upwards with an initial speed of 29.4 ms1.
(a) What is the direction of acceleration during the upward motion of the ball ?
(b) What are the velocity and acceleration of the ball at the highest point of its motion ?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis and give the signs of position, velocity and acceleration of the ball during its upward and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands ?[Take g = 9.8 ms^-2 and neglect air resistance].
Answer:
(a) The ball moves under gravity throughout its motion. The acceleration due to gravity is always directed vertically downward. So, the direction of acceleration during the upward motion of the ball is vertically downward.
(b) At the highest point of motion, the velocity of ball is zero and the acceleration is 9.8 ms^-2 directed vertically downward.
(c) As the ball remains between highest point and earth, so during upward and downward motion of ball, x > 0 (or x is positive).
During vertically upward direction v is negative (v < 0) and during vertically downward direction v is positive (v > 0).
Acceleration is always g = + 9.8 ms^-2 (positive).
(d) Using formula v² = u² + 2as, we have
a = + g = 9.8 ms^-2, s = – H, v = 0 (at highest point)
u = – 29.4 ms^-1 (given)
0 = (-29.4)² – 2 × 9.8 × H ⇒ H = \(\frac{(29.4)^2}{2 \times 9.8}\) m = 44.1 m
Also from relation, s = ut + \(\frac{1}{2}\)at²,
we have s = 0, u = -29.4 m/s, a = 98m/s²
0 = -29.4 t + \(\frac{1}{2}\) × 9.8t² or \(\frac{1}{2}\) 9.8t² = 29.4 t ⇒ t = \(\frac{29.4}{4.9}\) = 6s.
∴ Height, (H) = 44.1m and time taken (t) = 6s.

Question 7.
Read each statement below carefully and state with reasons and examples, if it is true or false.
A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant.
(b) with zero speed may have non-zero velocity,
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.
Answer:
(a) True; at highest point of particle thrown in vertically upward direction, the speed is zero but acceleration is g = 9.8 ms^-2 vertically downward.
(b) False; because velocity = speed + direction. As speed is zero, velocity must be zero.
(c) True; as v = constant; acceleration a = \(\frac{d v}{d t}\) = 0
(d) False, when a body is thrown upward its speed decreases, but gravitational acceleration is downward (assumed positive direction).
However, if body continues to move along a straight line on a smooth horizontal surface, then positive acceleration will necessarily imply speeding up.

Question 8.
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one-tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Answer:
Let t₁ be the time taken by the ball to hit the floor.
Given, u = 0, h = 90 m, g = 9.8 m/s²
Formula, h = ut + \(\frac{1}{2}\)gt² gives
90 = 0 + \(\frac{1}{2}\)9.8t₁² ⇒ t₁² = \(\frac{90}{4.9}=\frac{900}{49}\) ⇒ t₁ = \(\frac{30}{7}\) = 4.3 s.
The speed of the ball when it hits the floor first time is given by
v₁ = u + gt₁ = 0 + 9.8 × \(\frac{30}{7}\) = 42 m/s
Now the ball bounces upward after collision. As the ball loses one-tenth of speed during collision, therefore, upward speed of ball just after collision
v₂ = \(\frac{9}{10}\) × 42 = 37.8 m/s
The ball now starts moving upward with retardation g = 9.8 m/s². The speed of ball at any instant t will be (u¹ = v₂ = 37.8 m/s).
∴ v = u¹ – gt = 37.8 – 9.8 t
The speed of ball becomes zero at highest point after time t₂¹ is given by
0 = 37.8 – 9.8 t₂¹ ⇒ t₂¹ = \(\frac{37.8}{9.8}\) = 3.85 s
Now the ball begins to fall downward and attains a speed of 37.8 m/s just before collision with the floor. As time of ascent = time of descent.

∴ Time taken by ball to hit the floor
t₂¹¹ = 3.85 s
Total time upto this instant
T = t₂ + t₂¹ + t₂¹¹ = 4.3 + 3.85 + 3.85 = 12 s
The speed of ball just after second collision
v₃ = \(\frac{9}{10}\) × 37.8 ≅ 34m/s
The graph of speed and time is as shown in the Fig.
Remark: Speed is never negative.

Question 9.
Explain clearly, with examples, the distinction between
a) magnitude of displacement (sometimes called distance) over an interval of time and the total length of the path covered by a particle over the same interval.
b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total paths length divided by the time interval].
Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only].
Answer:
For simplicity we suppose one-dimensional motion, where an object traverses from A to B and then returns back in total time t where AB = s.
(a) Displacement = 0 and total path length = AB + AB = 2s.
(b) Average velocity = \(\frac{\text { Displacement }}{\text { Time }}=\frac{0}{t}\) = 0
Average speed = \(\frac{\text { Total path length }}{\text { Time }}=\frac{2 \mathrm{~s}}{\mathrm{t}}\)

In both the cases (a) and (b), the second quantity (i.e., total path length and speed) turns out to be greater than the first (i.e., magnitude of displacement and velocity respectively). However, if the motion takes place along a straight line without change in direction, then equality sign will be true.
That is then, magnitude of displacement = Total path length
(i.e., motion along a straight line in one direction) and magnitude of velocity = Speed.

Question 10.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 kmh^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 kmh ‘. What is the
(a) Magnitude of average velocity, and
(b) The average speed of the man over the interval of time
(i) 0 to 30 min,
(ii) 0 to 50 min,
(iii) 0 to 40 min ?
Answer:
Distance of market from home = 2.5 km; Speed of man from home to market = 5 km/h
Time taken by man to reach market, t =Speed of man, while returning from the market = 7.5 km/h
∴ Time taken by man to return the home, t₁ = \(\frac{\text { Distance }}{\text { Speed }}=\frac{2.5}{5}=\frac{1}{2} \mathrm{~h}\) = 20 min
Speed of man, while returning from the market = 7.5 km/h
Time taken by man to return the home, t₂ = \(\frac{2.5}{7.5}=\frac{1}{3} h\) = 20 min
Total time of man returning to his home = 30 min + 20 min = 50 min

(i) Over the interval 0 to 30 min :
During this time, man goes from home to market; therefore, displacement, s = 2.5 km
Average velocity = \(\frac{\text { Displacement }}{\text { Time }}=\frac{2.5}{1 / 2}\) = 5km/h
Average speed = \(\frac{\text { Distance }}{\text { Time }}=\frac{2.5}{1 / 2}\) = 5 km/h
Clearly, during motion along a straight line (without turn), the average speed and average velocity are equal in magnitude.

(ii) Over the interval 0 to 50 min :
In 50 min the man goes from home to market and returns back to home.
Net displacement = zero
Average velocity = \(\frac{\text { Net displacement }}{\text { Total time }}=\frac{0}{50}\)
Average speed = \(\frac{\text { Total distance }}{\text { Total time }}=\frac{2.5 \mathrm{~km}+2.5 \mathrm{~km}}{(50 \mathrm{~min})}+\frac{5 \mathrm{~km}}{50 \times \frac{1}{60} \mathrm{~h}}\) = 6 km/h.

(iii) Over the interval 0 to 40 min :
During first 30 min man goes from home to market covering a distance of 2.5 km; in next 10 min the man is in path from market to home and covers a distance
= (75 km/h) × \(\left(\frac{10}{60} h\right)\) = 1.25 km
Displacement = 2.5 – 1.25 = 1.25 km
Distance = 2.5 + 1.25 = 3.75 km
Average velocity = \(\frac{\text { Displacement }}{\text { Time }}=\frac{1.25}{40 \times \frac{1}{60} \mathrm{~h}}\) = 1.875 km/h
Average speed = \(\frac{\text { Distance }}{\text { Time }}=\frac{3.75 \mathrm{~km}}{40 \times \frac{1}{60} \mathrm{~h}}\) = 5.625 km/h

Question 11.
In problems 9 and 10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why ?
Answer:
Instantaneous velocity, v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\)
Where ds is infinitesimal displacement in infinitesimal time interval.
For infinitesimal interval of time there is no change in direction so the magnitude of displacement is equal to the length of the path; hence magnitude of instantaneous velocity = instantaneous speed.

Question 12.
Look at the graphs (a) to (d) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Answer:
(a) It does not represent one-dimensional motion because according to this graph particle has two positions at the same time, which is impossible.

(b) It does not represent one dimensional motion because according to this graph the particle have equal positive and negative velocity at the same time, which is impossible.

(c) It cannot represent one-dimensional motion because according to graph speed is negative at certain instants, which is impossible (speed can never be negative).

(d) It cannot represent one-dimensional motion because according to this graph after a certain time the total path length begins to decrease with increase of time, which is impossible (total path length will always increase with increase of time).

Question 13.
Fig. shows the x – t graph of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0 ? If not suggest a suitable physical context for this graph.
Answer:
For t < 0, the particle is at rest; since, its position remains at x = 0. For t > 0, x – t graph is parabolic, it is possible if motion of particle is uniformly accelerated since x = ut + \(\frac{1}{2}\)at²
If u = 0, x = \(\frac{1}{2}\)at² or x ∝ t² (parabola)

Thus, it is not correct to say that the particle moves in a straight line for t < 0, because during t < 0, the particle does not move at all. However, for t > 0, the trajectory is parabolic which is possible if particle moves along a straight line with uniform acceleration. It is again not correct to say that particle moves on parabolic path for t > 0.

Physical context: The graph may represent the motion of a particle released from the top of a hill or tower.

Question 14.
Suggest a suitable physical situation for each of the following graphs.

Answer:
(a) The given x – t graph shows that initially x = 0, then ball gains constant velocity which then reverses in direction and finally its velocity becomes zero.
Physical situation : The ball is initially at rest on a smooth floor. It is kicked and given a constant velocity. Then, it strikes a wall, rebounds with reduced speed and strikes the opposite wall which stops it.

(b) The given velocity – time graph (v – t) shows that the velocity changes sign again and again with the passage of time and each time, it loses some speed.
Physical situation : A ball thrown up with some initial velocity rebounding from the floor with reduced speed after each collision.

(c) The given a – t graph shows that initially acceleration of object is zero. It gains acceleration for a very short duration and immediately it is decelerated and again, its acceleration is zero.
Physical situation: A cricket ball moving with a uniform speed is hit back with a bat for a very short duration.

Question 15.
Fig. gives the x – t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s and – 1.2 s.
Answer:
Position is positive when particle is above t – axis and negative if particle is below t – axis. Velocity is positive, when x increases with increase of time and velocity is negative when x decreases with increase of time. Acceleration is positive when the slope of x – t curve increases and negative when slope of x – t curve decreases.

At 0.3 s, x is negative, v is negative and a is positive.
At 1.2 s, x is positive, v is positive and a is negative.
At – 1.2 s, x is negative, v is positive and a is positive.

Question 16.
Fig. shows x – t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval.

Answer:
The average speed inx-t graph is equal to the slope of line joining the initial and final points in that interval. From the graph the slope is greatest in interval 3 and least in interval 2. Therefore, average speed is greatest in interval 3 and average speed is least in interval 2. Slope is positive in intervals 1 and 2 and negative in interval 3, so the average velocity is positive during intervals 1 and 2 and negative during interval 3.

Question 17.
Fig. shows a speed – time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which intervals is the average acceleration greatest in magnitude ? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ?

Answer:
The average acceleration is given by the slope of line joining the two points onv-f graph for given interval of time. From the graph, out of three intervals the slope is maximum in interval (2) but it is negative. Therefore, the magnitude of average acceleration is greatest in interval 2.
Average speed in a given interval = \(\frac{v_1+v_2}{2}\)
From the graph, it is clear that the magnitudes of v₁ and v₂ are maximum in interval 3, so average speed is maximum in interval 3.

Signs of v: The speed v is always positive. Therefore, it is positive in all the three intervals.
Signs of a: In interval T slope at v -1 graph is positive, therefore, acceleration is positive. In interval 2, the slope of v – t graph is negative, therefore, the acceleration is negative. In interval 3, the slope of v – t graph is positive, therefore, the acceleration is positive.

The instantaneous acceleration is equal to the slope of tangent on v – t graph at given instant. The tangents drawn on given v – t curve at points A, B, C and D are parallel to time axis, so slope is zero.
Hence, acceleration at points A, B, C and D are zero.

Question 18.
A police van moving on a highway with a speed of 30 km/h fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km/h, if the muzzle speed of the bullet is 150 m/s, with what speed does the bullet hit the thief’s car ?
(Note : Obtain that speed which is relevant for damaging thief’s car ?)
Answer:
Let v_p be the velocity of police van, v_T be the velocity of thief’s car and u_BP the muzzle velocity of bullet. Muzzle velocity means the velocity with respect to gun (or police van).
Given, v_P = 30 km/h = 30 × \(\frac{1100 \mathrm{~m}}{(60 \times 60) \mathrm{s}}=\frac{25}{3} \mathrm{~m} / \mathrm{s}\)
v_T = 192 km/h = 192 x \(\frac{1100 \mathrm{~m}}{(60 \times 60) \mathrm{s}}=\frac{160}{3} \mathrm{~m} / \mathrm{s}\)
u_B = 150 m/s
All velocities are in same direction, therefore, if u_B is the velocity of bullet with respect to earth, the velocity of bullet with respect to police van
u_BP = u_B + v_P = 150 + \(\frac{25}{3}=\frac{475}{3}\)m/s
Relative velocity of bullet with respect to thief’s car
u_BT = u_B – v_T = \(\frac{475}{3}-\frac{160}{3}=\frac{315}{3}\)= 105 m/s.

Regular practice with AP Inter 1st Year Chemistry Study Material Chapter 6 Equilibrium Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Chemistry 6th Lesson Equilibrium Questions and Answers

I. Multiple Choice Questions

Question 1.
The pH of a solution of hydrochloric acid is 4. The molarity of the solution is :
1) 4.0 M
2) 0.4 M
3) 0.0001 M
4) 0.04 M
Answer:
3) 0.0001 M
If [H⁺] = 0.0001M (Molarity), then
pH = -log₁₀ [H⁺] = -log 0.0001 = -log 10^-4 = 4.

Question 2.
The strong conjugate base in the following is
1) NO₃^–
2) Cl^–
3) SO₄^2-
4) CH₃COO^–
Answer:
4) CH₃COO^–
Strong conjugate base is CH₃COO^–, which is obtaining from weak acid.

Question 3.
When NH₄Cl is added to NH₄OH solution the dissociation of ammonium hydroxide is reduced. It is due to
1) common ion effect
2) hydrolysis
3) oxidation
4) reduction
Answer:
1) common ion effect
Due to common ion effect dissociation of NH₄OH is reduced.

Question 4.
The following reaction is at equilibrium.
Fe³⁺_(aq) + SCN^–_(aq) ⇌ [Fe(SCN)]²⁺_(aq)
In the above reaction, colour intensity of red colour can be increased by
l) addition of oxalic acid which reacts with Fe³⁺ ions.
2) addition of Hg²⁺ ions which reacts with SCN^– ions.
3) addition of KSCN.
4) red colour intensity cannot be changed.
Answer:
3) addition of KSCN.
\(\mathrm{Fe}_{(\mathrm{aq})}^{+3}+\mathrm{SCN}_{(\mathrm{aq})}^{-} \rightleftharpoons[\mathrm{Fe}(\mathrm{SCN})]_{(\mathrm{aq})}^{+2}\)
on adding KSCN colour intensity of red colour increased.

Question 5.
K_sp of a sparingly soluble salt AB₂ is 4 × 10^-12. The solubility of the salt is
1) 2 × 10^-6M
2) 4 × 10^-4M
3) 1 × 10^-12
4) 1 × 10^-4M
Solution:
4) 1 × 10^-4M
For AB₂ type of salt, K_sp = 4s³ ⇒ s² = \(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\)
⇒ s = \(\sqrt[3]{\frac{\mathrm{K}_{\mathrm{sp}}}{4}}=\sqrt[3]{\frac{4 \times 10^{-12}}{4}}\) = 10^-4m

Question 6.
An aqueous solution of ammonia consists of
1) H⁺ only
2) OH^– only
3) NH₄⁺ only
4) NH₄⁺ and OH^–
Answer:
Aqueous solution of ammonia contains NH₄⁺ and OH^– ions.
NH₄OH_(aq) → NH₄⁺ + OH^–

Question 7.
The pH of a 0.05 M of H₂SO₄ solution is
1) 0.05
2) 1
3) -1
4) 1.3010
Answer:
2) 1
For 0.05M of H₂SO₄
[H⁺] = 2 × 0.05(Di basic H₂SO₄) = 0.1
pH = -log₁₀[H⁺] = -log 0.1 = -log 10^-1 = 1

Question 8.
If the molar solubility of Zirconium phosphate Zr₃(PO₄)₄ is ‘S’. Solubility product K_sp of Zirconium phosphate is
1) 6912 (S)⁵
2) 6912 (S)⁷
3) 108 (S)⁷
4) 27 (S)⁵
Answer:
2) 6912 (S)⁷

Question 9.
For a reversible reaction, if the concentration of reactants is increased, the equilibrium constant of the reaction
1) Increases
2) Remains constant
3) Decreases
4) Depends on amount of reactant
Answer:
Equilibrium constant remains constant on increasing the concentration of reactants.

Question 10.
The pK_a of acetic acid and pK_b of ammonium hydroxide are 4.76 and 4.75 respectively.
1) 6.098
2) 7.005
3) 8
4) 5.0
Answer:
2) 7.005
CH₃COONH₄ is the salt formed by weak acid and weak base
∴ pH = 7 + — (PK_a – PK_b) = 7 + \(\frac{1}{2}\) (4.76 – 4.78) = 7 × \(\frac{1}{2}\) × 0.01 = 7.005

Question 11.
Which among the following salt solutions is basic in nature?
1) Ammonium chloride
2) Ammonium sulphate
3) Ammonium nitrate
4) Sodium acetate
Answer:
4) Sodium acetate
Sodium acetate is the salt obtained from sodium hydroxide (Strong base) and acetic acid (weak acid). So aq. CH₃COONa is in basic nature.

Question 12.
Observe the following equilibrium at T (Kelvin)
H_2(g) + I_2(g) ⇌ HI_(g)
Which one of the following does not disturb the above equilibrium at constant volume?
1) Addition of H_2(g)
2) Removal of HI_(g)
3) Addition of I_2(g)
4) Addition of He_(g)
Answer:
4) Addition of He_(g)
He is a Noble gas, it is chemically inert, so by adding inertgas the equilibrium does not disturbed.

Question 13.
Solution of 0.1 N NH₄OH and 0.1 N NH₄Cl has pH 9.25. Find out pK_b of NH₄OH.
1) 9.25
2) 4.75
3) 3.75
4) 8.25
Solution:
2) 4.75
P^OH = PK_b + log \(\frac{[\text { salt }]}{[\text { base }]}\) (basic buffer)
pH + P^OH = 14 ⇒ P^OH = 14 – 9.25 = 4.75
4.75 = PK_b + log \(\frac{0.1}{0.1}\) ⇒ PK_b = 4.75

Question 14.
For the following process, the expression for the K_c
Ni_(s) + 4CO_(g) ⇌ Ni(CO)_4(g)
1) K_c = \(\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]^2}{[\mathrm{Ni}][\mathrm{CO}]^4}\)
2) K_c = \(\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]}{[\mathrm{CO}]^4}\)
3) K_c = \(\frac{[\mathrm{Ni}][\mathrm{CO}]^4}{\left[\mathrm{Ni}(\mathrm{CO})_4\right]^{2-}}\)
4) K_c = \(\frac{[\mathrm{CO}]^4}{\left[\mathrm{Ni}(\mathrm{CO})_4\right]^2}\)
Answer:
2) K_c = \(\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]}{[\mathrm{CO}]^4}\)
K_c = \(\frac{\text { Molar concentrations of products }(\mathrm{g})}{\text { Molar concentrations of reactants }(\mathrm{g})}\) ⇒ K_c = \(\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]}{[\mathrm{CO}]^4}\)

Question 15.
K₁ and K₂ are equilibrium constants for reactions (i) & (ii). Then the relation between
K₁ and K₂
(i) H_2(g) + I_2(g) ⇌ 2HI(g)
(ii) HI_(g) ⇌ 1/2 H_2(g) + 1/2 I_2(g)
1) K₁ = K₂²
2) K₁ = \(\frac{1}{\mathrm{~K}_2}\)
3) K₁ = K₂⁰
4) K₁ = \(\frac{1}{K_2^2}\)
Solution:
4) K₁ = \(\frac{1}{K_2^2}\)
H_2(g) + I_2(g) ⇌ 2HI(g) …………… (1)
HI_(g) ⇌ \(\frac{1}{2}\) H_2(g) + \(\frac{1}{2}\) I_2(g) ………… (2)
K₁ = Equilibrium constant
K₂ = Equilibrium constant
K₂ = \(\frac{1}{\sqrt{\mathrm{~K}_1}}\) [eq.(2) is reverse reaction and halved]
Squaring on both sides, \(\mathrm{K}_2^2=\frac{1}{\mathrm{~K}_1}\)
⇒ K₁ = \(\frac{1}{K_2^2}\)

II. Fill in the Blanks

Question 1.
Acidity of BF₃ can be explained on the basis of __________ theory of acids and bases.
Answer:
Lewis acid base theory

Question 2.
The nature of aqueous solution of ammonium chloride (NH₄Cl) is ______.
Answer:
NH₄Cl is the salt of weak base (NH₄OH) and strong acid (HCl).
So aq. solution of NH₄Cl is acidic.

Question 3.
K_p for the following equation: ______.
CaCO_3(s) ⇌ CaO_(s) + CO_2(g)
Answer:
K_p = P_CO2

Question 4.
The molarity of pure water is ______.
Answer:
Molarity of pure water is = 55.5.M
M = \(\frac{\mathrm{W}}{\mathrm{GMW}} \times \frac{1000}{\mathrm{~V}(\mathrm{ml})}=\frac{1000}{18} \times \frac{1000}{1000}\) = 55.5 M

Question 5.
If K is in the range of ________ appreciable concentrations of both reactants and products are present.
Answer:
10^-3 to 10³

III. One Word Answer Questions

Question 1.
What is a bronsted base?
Answer:
The substance which can accept a proton is said to be bronsted base. Eg : Cl^–, HCO₃^–.

Question 2.
What is Lewis acid?
Answer:
Lewis acid is the substance which can accept a pair of electrons and form a coordinate covalent bond. Eg : BF₃, H⁺.

Question 3.
What is the value of K_w at 25°C? What are its units ?
Answer:
Ionic product of water K_w = 1 × 10^-14 ; Units are mole²/lit².

Question 4.
What is conjugate acid-base pair?
Answer:
An acid-base pair which differ by proton (H⁺) is called a conjugate acid base pair.
Eg : HCl and Cl^–.

Question 5.
Write the relation between Gibbs energy and equilibrium constant.
Answer:
∆G° : – 2.303 RT log K_c
R = gas constant ; T= temperature ; K_c = Equilibrium constant

IV. Very Short Answer Questions

Question 1.
State law of chemical equilibrium.
Answer:
The ratio of product of molar concentrations of products to the product of molar concentrations of reactants is a constant value. This constant is called equilibrium constant and this is called law of chemical equilibrium.
Eg : aA + bB ⇌ cc + dD (or) Equilibrium constant (K_c) = \(\frac{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{~A}]^{\mathrm{a}}[\mathrm{~B}]^{\mathrm{b}}}\)
Eg : H_2(g) + I_2(g) ⇌ (K_c) = \(\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}\)

Question 2.
What is homogeneous equilibrium? Give two examples.
Answer:
The equilibrium in which all the substances are present in the same phase is known as homogeneous equilibrium.
Eg: N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
CH₃COOH (l) + C₂H₅OH (l) ⇌ CH₃COO C₂H₅ (l) + H₂O (l)

Question 3.
What is heterogeneous equilibrium? Give two examples.
Answer:
The equilibrium in which the substances involved are present in different phases is called heterogeneous equilibrium.
CaCO₃ (s) ⇌ CO(s) + CO₂ (g) and H₂O (l) ⇌ H₂O (g)

Question 4.
What is meant by dynamic equilibrium?
Answer:
A state in a reversible reaction where the forward and backward reaction continue to takes place with same rate is known as dynamic equilibrium.

Question 5.
Give two chemical equilibrium reactions for which K_p > K_c
Answer:
In order that K_p > K_c, ∆n should be +ve
1) PCl₅ (g) ⇌ PCl₃ (g) + Cl₂(g); ∆n = (2 – 1)
2) N₂O₄(g) ⇌ 2NO₂ (g) ; ∆n = (2 – 1)

Question 6.
Give two chemical equilibrium reactions for which K_p <K_c.
Answer:
In order that K_p <K_c, ∆n should be -ve
1) N₂ (g) + 3H₂ (g) ⇌ 2NH₃(g) ; ∆n = (2 – 4)
2) 2SO₂ (g) + O₂ (g) ⇌ 2SO₃(g) ; ∆n = (2 – 3)

Question 7.
Write the equations for the conversion of K_c to K_p for the following reaction
CO_(g) + H₂O_(g) ⇌ CO_2(g) + H_2(g)
Answer:
CO_(g) + H₂O_(g) ⇌ CO_2(g) + H_2(g)
K_p = K_c (RT)^∆n
∆n = (1+1) – (1 + 1) = O
K_p = K_p (RT)
K_p = K_c

Question 8.
What are the factors which influence the chemical equilibrium?
Answer:
Molar concentration, temperature and pressure influence the equilibrium.

Question 9.
What is meant by ionic product of water?
Answer:
The product of concentration of H⁺ and OH^– ions in aqueous solution (or) in pure water is called ionic product of water.

Question 10.
Give two examples of salts whose aqueous solutions are acidic.
Answer:
NH₄Cl (Ammonium Chloride); Cu₂SO₄ (copper sulphate) and (NH₄)₂SO₄ (Ammonium sulphate) .
These aqueous solutions are acidic in nature.

V. Short Answer Questions

Question 1.
Derive the relation between K_p and K_c for the equilibrium reaction.
N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
Answer:
N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
K_c = \(=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}\) ; K_p = \(\frac{P_{\mathrm{NH}_3}^2}{P_{\mathrm{N}_2} \times \mathrm{P}_{\mathrm{H}_2}^3}\)
Partial pressures of P_N2, P_H2 and P_NH3 can be written as
P_N2 = \(\frac{\mathrm{n}_{\mathrm{N}_2} \mathrm{RT}}{\mathrm{~V}}\) = C_N2RT ; P_H2 = \(\frac{\mathrm{n}_{\mathrm{H}_2} \mathrm{RT}}{\mathrm{~V}}\) = C_H2RT and P_NH3 = \(\frac{\mathrm{n}_{\mathrm{NH}_3} \mathrm{RT}}{\mathrm{~V}}\) = C_NH3
Substituting these values in K_p equation, K_p = \(\frac{\left(\mathrm{C}_{\mathrm{NH}_3}\right)^2}{\left(\mathrm{C}_{\mathrm{N}_2}\right)\left(\mathrm{C}_{\mathrm{H}_2}\right)^3}\) × [RT]^-2
∴ K_p = K_c[RT]^-2 [∵ \(\frac{\mathrm{C}_{\mathrm{NH}_3}^2}{\left(\mathrm{C}_{\mathrm{N}_2}\right)\left(\mathrm{C}_{\mathrm{H}_2}\right)^3}\) = K_c]

Question 2.
State Le Chateliers’s Principle. Discuss the application of Le Chatelier’s principle for the industrial synthesis of Ammonia by Haber’s process.
Answer:
Le Chatelier’s principle : If a chemical reaction at equilibrium is subjected to a change in temperature, pressure or concentration that govern the chemical equilibrium, the equilibrium position shifts in the direction in which the change is reduced or nullified.
Application of Le Chatelier’s principle to by Haber’s process :
N_2(g) + 3H_2(g) ⇌ 2NH_3(g); ∆H = -92 kJ

Effect of pressure :
Increase of external pressure on the reaction at equilibrium favours the reaction in the direction in which the volume or the number of molecules decreases.
In the above reaction of formation of ammonia forward reaction involves decrease in volume or number of molecules. So high pressure is favourable for the formation of ammonia. Hence ammonia is manufactured at 200 atm pressure.

Effect of temperature :
In a reversible reaction, if the forward reaction is exothermic, the backward reaction will be endothermic.
High temperature favours the endothermic reaction while low temperature favours the exothermic reaction.

• Since the formation of ammonia is exothermic reaction, low temperature is favourable for the manufacture of ammonia.
• But at low temperature, the reaction is slow. So an optimum temperature of about 725 – 775 K is used.

Effect of catalyst:

• To increase the rate of formation of ammonia iron powder mixed with molybdenum powder is used as catalyst.
• Molybdenum acts as promoter to iron catalyst.

Effect of concentration :

• Increase in concentrations of reactants increases the rate of forward reaction.
• Removal of the products from reaction mixture decreases the rate of backward reaction.
• So more ammonia can be obtained by increasing the concentration of N₂ and H₂ and removing the NH₃ formed at regular intervals.
• Optimum conditions for the synthesis of ammonia are
  Pressure = 200 atm
  Temperature = 725 – 775 K
  Catalyst = Fe
  Promoter = Mo

Question 3.
State Le Chateliers’s Principle. Discuss the application of Le Chatelier’s principle for the manufacture of SO₃ by contact process.
Answer:
Le Chatelier’s principle : If a chemical reaction at equilibrium is subjected to a change in temperature, pressure or concentration that govern the chemical equilibrium, the equilibrium position shifts in the direction in which the change is reduced or nullified.

Application of Le Chatelier’s principle by contact process :
2SO_2(g) + O_2(g) ⇌ 2SO_3(g) + Heat; ∆H = -189 kJ

Effect of pressure :

• Increase of external pressure on the reaction at equilibrium favours the reaction in the direction in which the volume or the number of molecules decreases.
• Formation of SO₃ in the above reaction involves decrease in volume.
• According to Le Chatelier’s principle high pressure is favourable for the formation of SO₃.
• So, SO₃ is prepared at a pressure of 1.5 to 1.7 atmospheres.

Effect of temperature :

• In a reversible reaction if the forward reaction is exothermic, the backward reaction is endothermic.
• High temperatures are favourable for the endothermic reaction, while the low temperatures are favourable for the exothermic reactions.
• Since the formation of SO₃ is exothermic reaction, low temperatures are favourable. So, SO₃ is prepared at a temperature of about 673 K.
  Effect of catalyst: To increase the rate of formation of SO₃ catalyst like V₂O₅ or platinised asbestos is used.

Effect of concentration :

• Increase in the concentration of the reactants, increase the rate of forward reaction.
• Removal of products from reaction mixture decreases the rate of backward reaction.
• So more SO₃ can be obtained by increasing the concentrations of SO₂ and O₂ and by removing, the SO₃ is formed at regular intervals.
  Optimum conditions for the synthesis of ammonia are
  Pressure : 1.5 – 1.7 atmospheres Temperature : 673 K
  Catalyst: V₂O₅ (Vanadium pentoxide) or Platinised asbestos.

Question 4.
The species H₂O, HCO₃^–, HSO₄^– and NH₃ can act as both bronsted acids and bases. Give the corresponding conjugate acid and base for each of them.
Answer:

Question 5.
Write notes on
(i) Common ion effect.
(ii) Solubility product
Answer:
i) Common ion Effect : The decrease in the dissociation of the aq. solution of weak electrolyte by adding another strong electrolyte having one common ion is called common ion effect. Eg : To the saturated solution of AgCl, electrolyte NaCl is added to decrease the dissociation.
ii) Solubility Product (K_sp) : The solubility product is the product of molar equilibrium concentrations of cation and anion at a particular temperature in a saturated solution of sparingly soluble salt.
Applying the concept at equilibrium, A_xB_y(s) ⇌ xA^y+ + yB^x-
Applying the concept at equilibrium, K_sp = [A^y+]^x [B^x-]^y
AgCl ⇌ Ag⁺ + Cl^–
K_sp = [Ag⁺] [Cl^–]
⇒ K_sp = s × s [where s is the solubility]
⇒ K_sp = s²
Solubility s = \(\sqrt{\mathrm{K}_{\mathrm{sp}}}\)

Question 6.
Explain the concept of Bronsted-Lowry acids and bases. Illustrate with suitable examples.
Answer:
Bronsted – Lowry theory : Any substance that can lose a proton or protons is an acid.
Eg : HCl, H₂SO₄, H₃PO₄, CH₃COOH.

• Any substance that can gain a proton or protons is a base. Eg : NH₃, H₂O, OH^– etc.
• When an acid react with base neutralization takes place. As per Bronsted – Lowry theory proton transfer from an acid to a base is called neutralization.
  Eg : HCl + H₂O ⇌ H₃O⁺ + Cl^–
• The above reaction is reversible. In this reaction,
  1. HCl donates proton to water. So it is a bronsted acid.
  2. H₂O gains the proton. So it is a bronsted base.
  3. H₂O⁺ donates a proton to Cl^–. So it is bronsted acid.
  4. Cl^– can gain a proton. So it is a bronsted base.
• The acid base pair which differ by a single proton is said to be conjugate acid-base pair.
• In the above reaction HCl & Cl^– and H₃O⁺ and H₂O are conjugate acid-base pairs.
• Every bronsted acid has its conjugate base and every bronsted base has a conjugate protonic acid.
• According to bronsted – Lowry theory greater the tendency to donate proton stronger is the acid. Higher the tendency of the base to accept protons, stronger is the base.
• In the above reaction, HCl gives proton easily than H₃O⁺. So HCl is stronger acid than H20. H20 accepts the proton more easily than Cl^–. So H₂O is a stronger base than NH₃
• In all acid – base reactions, the reaction takes place in the direction of formation of weaker acid and a weaker base.
• A stronger acid has a weaker conjugate base and a stronger base has a weaker conjugate acid.

Advantages of Bronsted – Lowry theory :

• This theory explains the behaviour of acids and bases in both aqueous and non-aqueous solvents.
• It explains the behaviour of NH₃, CaO, etc. as bases and CO₂, SO₂, etc. as acids.

Drawbacks of Bronsted – Lowry theory :

• Proton donation or acceptance happens only in the presence of other substances.
• It could not explain the behaviour of electron deficient molecules like AlCl₃, BCl₃, etc. as acids.

Question 7.
Explain Lewis acids and bases with suitable examples. Classify the following species into Lewis acids and Lewis bases.
a) OH^–
b) F^–
c) H⁺
d) BCl₃.
Answer:
Lewis acid : The substance which can accept an electron pair to form a coordinate covalent bond is called Lewis acid.
Types of Lewis acids :
1) All +vely charged metal ions or cations act as Lewis acids.
Eg : Ag⁺, CO⁺³, Cu⁺², Fe⁺³, etc.
2) Electron deficient compounds whose central atom has an incomplete octet, possessing an empty orbital acts as Lewis acid. Eg : BF₃, BCl₃, AlCl₃, FeCl₃.
3) Compounds in which central atom has vacant’d’ orbitals and which can expand its octet. Eg : SiF₄, SnCl₄, SF₄, TeF₄, FeCl₃.
4) Molecules having multiple bonds between atoms of different electronegativities.
Eg : CO₂, SO₂, SO₃, NO₂, Cl₂O₇, P₄O₁₀.
5) Elements with an electron sextet. Eg : S, O.

Lewis base : The substance which donates electron pair to form a coordinate covalent bond is called Lewis base.
Types of Lewis bases :
1) All anions will act as Lewis bases. Eg : Cl^–, OH^–, CN^–, NH₂^–, F^–, SCN^–.
2) Molecules with one or two lone pairs on the central atom.
Eg : H₂O, NH₃, R – OH, R – NH₂,
R-O-R, C₅H₅N.
3) Molecules with multiple bonds. Eg : CO, NO, HC ☰ CH, H₂C ☰ CH₂
a) OH^– is Lewis base, b) F^– is Lewis base, c) H⁺ is Lewis acid, d) BCl₃ is Lewis acid.

Question 8.
Calculate the pH of
a. 10^-3 M of HCl
b. 10^-3 M of H₂SO₄ [log 2 = 0.3010]
Solution:
a) 10^-3M of HCl [H⁺] = 10^-3M
pH = – log [H⁺] = – Log 10^-3 = 3 log 10 = 3

b) 10^-3 M of H₂SO₄
[H⁺] = 2 × 10^-3
pH = -log [H⁺] = -log 2 × 10^-3 = – log 2 – log 10^-3
= – log 2 + 3 log 10 = -log 2 + 3
= 3 – log 2
= 3 – 0.3010 = 2.6990

Question 9.
Calculate the pH for
a. 0.001 M of NaOH
b. 0.01 M of Ca(OH)₂
Solution:
a) 0.001 M of NaOH
[OH⁺] = 0.001M
P^OH = -log [OH^–] = – log 10^-3 = 3 log 10 = 3
pH + POH =14
pH = 14 – 3 = 11.
∴ pH = 11

b) 0.01 M of Ca(OH)₂
[OH^–] = 0.01 × 2 = 2 × 10^-2
P^OH = – log [OH^–]
= – log 2 × 10^-2
= – log 2 – log 10^-2 = – log2 + 2 log 10
= 2 – log2 = 2 – 0.3010 = 1.6990
pH + P^OH = 14
pH = 14 – 1.699 = 12.3010

Question 10.
The pH of 0.1M solution of weak mono protic acid is 4.0. Calculate its [H⁺] and K_a.
Solution:
HA ⇌ H⁺ + A^–
pH = 4 ; [H⁺] = 10^-4
K_a = \(\frac{c \alpha^2}{1-\alpha}=\frac{(0.1)\left(10^{-4}\right)^2}{1}\) [Neglecting α in the denominator]
K_a = 10^-7

Question 11.
One litre of buffer solution contains 0.1 mole of acetic acid and 1 mole of sodium acetate. Find its pH if pK_a of CH₃COOH is 4.8.
Solution:
Henderson’s Equation, pH = PK_a + log \(\frac{[\text { Salt }]}{[\text { Acid }]}\)
Given [Salt] = 1M; [Acid] = 0.1M; PK_a = 4.8
= 4.8 + log \(\frac{1}{0.1}\)
= 4.8 + log 10
= 4.8 +1 = 5.8

VI. Long Answer Questions

Question 1.
What are equilibrium processes? Explain equilibrium in Physical and Chemical processes with examples.
Answer:

• In a reversible process, a state will be attained such that the rate of forward reaction is equal to the rate of backward reaction.
• Such a state is known as equilibrium state.
• Equilibrium processes are of two types. They are

1. Physical equilibrium process : In such a process, equilibrium occurs in a phase transformation process.
Ex: Solid ⇌ Liquid
Solid ⇌ Vapour
Vapour ⇌ Liquid

2. Chemical equilibrium process: If there exists an equilibrium in between reactants and products, such reactions are said to be chemical equilibrium process.
1) N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)
2) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

Question 2.
What is meant by dynamic equilibrium? Explain with suitable examples.
Answer:

• In a reversible process, a state will be attained such that the rate of forward reaction is equal to the rate of backward reaction. Such a state is known as equilibrium state.
• At equilibrium we find no change in the concentration of either reactants (or) products then we may feel that the reaction has stopped. But actually the reaction will be going on. Such a state of reaction is called dynamic equilibrium.

Equilibrium processes are of two types. They are :
1. Physical equilibrium process : In such a process, equilibrium occurs in a phase transformation process.
Ex : Solid ⇌ liquid
Solid ⇌ vapour
Vapour ⇌ liquid

2. Chemical equilibrium process: If there exists an equilibrium in between reactants and products, such reactions are said to be chemical equilibrium process.
1) N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)
2) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

The forward and backward reactions of a reversible reaction continuously takes place with equal rates simultaneously at the equilibrium stage also. Hence the equilibrium is called dynamic equilibrium.

At this stage, the reaction appears to be stopped.

All chemical equilibrium processes are dynamic equilibriums in nature.

Question 3.
What are the important features of equilibrium constant? Discuss any two applications of equilibrium constant.
Answer:

• The equilibrium constant has a definite value for every chemical reaction at a particular temperature. The value of equilibrium constant is independent of initial concentrations of reacting species.
• The value of equilibrium constant K changes with change in temperature.
• For a reversible reaction, the equilibrium constant for the backward reaction is inverse of the equilibrium constant for the forward reaction.
• The equilibrium constant is independent of the presence of catalyst.
• The equilibrium constant is expressed in terms of concentration, it has different units for different reactions.
• When addition of two equilibria leads to another equilibrium then the product of their equilibrium constants gives the equilibrium constant of the resultant equilibrium.
  Eg : N₂ + O₂ ⇌ 2NO ; K₁
  2NO + O₂ + O₂ ⇌ 2NO₂; K₂
  N₂ + 2O₂ ⇌ 2NO₂; K₃
  K = K₁ × K₂ × K₃

Applications of equilibrium constant:
i) Predicting of extent of reaction: The magnitude of equilibrium constant tells us about the extent to which the reactants are converted into the products before the equilibrium is attained.

ii) Predicting the direction of the reaction : From the knowledge of equilibrium con¬stant it is possible to predict the direction in which the net reaction is taking place at a given concentrations or partial pressures of reactants and products.

• If Q_c > K_c the reaction taking place in backward direction is more.
• If Q_c < K_c the reaction taking place in forward direction is more.
• If Q_c = K_c the reaction is at equilibrium. No reaction taking place.

Question 4.
What is Le Chatelier’s principle? Discuss briefly the factors which can influence the equilibrium.
Answer:
Le Chatelier’s principle : If a chemical reaction at equilibrium is subjected to a change in temperature, pressure or concentration that govern the chemical equilibrium, the equilibrium position shifts in the direction in which the change is reduced or nullified.
1) Effect of pressure : ,

• Pressure is inversely proportional to volume.
• When pressure is increased the equilibrium shifts in such a direction where the decrease in volume is observed.
• Similarly when pressure is decreased, the equilibrium shifts in such a direction where increase in volume is observed. This is applicable exclusively for gaseous reactions.

2) Effect of temperature :

• When temperature is increased, equilibrium shift in such a direction where the temperature change is nullified or equilibrium shifts towards the direction in which temperature is decreased or heat is absorbed. That is endothermic reaction is favoured.
• When temperature is decreased equilibrium shifts in such a direction where temperature is increased or heat is liberated. That is exothermic reaction is favoured.

3) Effect of concentration :

• Increase in concentration of the reactants favours forward reaction and decrease in concentration of the reactants favours the backward reaction.
• When products are added to the equilibrium reaction the concentration of the products increases and reaction proceeds in backward direction.
• Removal of products leads to decrease in concentration of the products and leads to forward reaction.

Question 5.
Describe the effect of:
a) addition of H₂
b) addition of CH₃OH
c) removal of CO
d) removal of CH₃OH on the equilibrium of the reaction 2H₂ (g) + CO (g) ⇌ CH₃OH (g)
Answer:
a) According to Le Chatelier’s principle increase in the concentration of reactant increase the forward reaction.
So the addition of H₂ results in the formation of more amount of CH₃OH.

b) According to Le Chatelier’s principle increase in the concentration of the product, shift the equilibrium in the backward direction. So the addition of CH₃OH (product) results in the decrease in concentration and CH₃OH and increase in the concentration of H₂ and CO.

c) According to Le Chatelier’s principle the decrease in the concentration of reactant equilibrium shift in the backward direction. So removal of CO (reactant) the equilibrium shift in the backward direction. So formation of CH₃OH decreases.

d) According to Le Chatelier’s principle the decrease in concentration of product shifts the equilibrium in the forward direction. So the removal of CH₃OH (product) shift the equilibrium in the forward direction. Thus more amount of CH₃OH is formed.

Question 6.
What is degree of ionization in respect of weak acids and weak bases? Derive the relationship between degree of ionization (α) and ionization constant (K_a) for the weak acid HX.
Answer:
Degree of ionization : It can be defined as the fraction of acid (a) base that undergoes ionization per one mole of acid or base. For strong acids or bases degree of ionization (a) is equal to ‘1’.
Let’s consider a weak acid HX with concentration ‘c’. It undergoes partial ionization.

where α is degree of ionization
Ionization constant K_a = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{X}^{-}\right]}{[\mathrm{HX}]}\)
⇒ K_a = \(\frac{(\mathrm{c} \alpha)(\mathrm{c} \alpha)}{\mathrm{c}(1-\alpha)}\) for weak acids α < < 1, so α is negligible.
⇒ K_a = \(\frac{c^2 \alpha^2}{c}\) ⇒ α = \(\sqrt{\frac{K_a}{c}}\)

Question 7.
Define pH. What is buffer solution? Derive Henderson – Hasselbalch equation for calculating the pH of an acid buffer solution.
Answer:
The pH of a solution may be defined as negative logarithm of the activity of hydrogen ion [H⁺]
pH = – log [H⁺]
Buffer solution : A buffer solution is the solution which can resist the change in pH on
addition of small amount of acid or base. The ability of a solution to resist change in pH on
addition of small amount of acid or base is called buffer solution.
Henderson – Hasselbalch equation :
HA + H₂ ⇌ H₃O⁺ + A^– Partial ionisation.
NaA → Na⁺ + A^– Complete ionisation
K_a = \(\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)
Since HA is weakly dissociated, the conc. HA ion solution can be taken equal to the initial conc. of the acid.
Since NaA is completely dissociated, conc. of A- can be taken equal to the conc. of NaA (salt)
K_a = \(\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right][\text {Salt }]}{[\text { Acid }]}\)
H₃O⁺ = \(\frac{\mathrm{K}_{\mathrm{a}} \text { [Acid] }}{[\text { Salt }]}\)
-log [H₃O⁺] = – logK_a – log\(\frac{\text { [Acid] }}{[\text { Salt }]}\)
pH = pKa + log \(\frac{\text { [Salt] }}{[\text { Acid }]}\)

Question 8.
Write notes on
(i) Common ion effect.
(ii) The relation between K_sp and solubility (S) of a sparingly soluble salt BaSO₄.
Answer:
i) Common ion effect: The solubility of an electrolyte in water decreases by the addition of another electrolyte which has one ion common with the electrolyte.

• For example the solubility of NaCl decreases on addition of HCl to the solution of NaCl in water which is known as salting out of water.
• Acetic acid is weak acid and it undergoes partial ionization as follows.
  CH₃COOH ⇌ CH₃COO^– + H⁺
• In case of such electrolytes there exists an equilibrium in between ions and molecular form. In such a condition, the addition of electrolyte with common ions shifts the equilibrium in backward direction which causes the decrease in solubility.

Significance :
1) Common ion effect is an important phenomenon in qualitative analysis.
2) The common ion effect principle is also used in controlling the H⁺ ion concentration in buffer solution.

ii) The relation between K and solubility (S) of a sparingly soluble salt BaSO₄ : The sparingly soluble salt like BaSO₄, when added to water form saturated solution. There exists an equilibrium between undissolved solid and the ions in saturated solution.
BaSO₄ ⇌ Ba²⁺ + SO₄^2-
The equilibrium constant, K = \(\frac{\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]}{\left[\mathrm{BaSO}_4\right]}\)
K = \(\frac{\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]}{\mathrm{K}_1}\) (∴ Activity of solid BaSO₄ is constant)

K × K₁ = [Ba²⁺][SO^2-] or K_sp = [Ba²⁺][SO₄^2-]
The new constant K_sp is called solubility product.

If S moles lit^-1 is the solubility of BaSO₄.

K_sp = s × s = s²
or solubility of BaSO₄ is s = \(\sqrt{\mathrm{K}_{\mathrm{sp}}}\) .

Equilibrium Class 11 NCERT Solutions

Question 1.
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature.
The volume of the container is suddenly increased.
i) What is the initial effect of the change on vapour pressure ?
ii) How do rates of evaporation and condensation change initially ?
iii) What happens when equilibrium is restored finally and what will be the final vapour pressure ?
Answer:
i) The vapour pressure will decrease due to increase in volume.
ii) Since, the temperature is fixed, the rate of evaporation will remain constant. However, the rate of condensation will be low initially due to lesser number of molecules per unit volume in the vapour phase.
iii) When equilibrium is restored, the final vapour pressure will be the same as original vapour pressure because at equilibrium the rates of evaporation and condensation are equal.

Question 2.
W hat is K_c for the following equilibrium when the equilibrium concentration of each substance is : [SO₂] = 0.60 M, [O₂] = 0.82 M and [SO₃] = 1.90 M ?
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Answer:
K_c = \(\frac{\left[\mathrm{SO}_3(\mathrm{~g})\right]^2}{\left[\mathrm{SO}_2(\mathrm{~g})\right]^2\left[\mathrm{O}_2(\mathrm{~g})\right]}=\frac{(1.90)^2}{(0.60)^2 \times(0.82)}\) = 12.229 L mol^-1

Question 3.
At a certain temperature and a total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms.
I_2(g) ⇌ 2I_(g)
Calculate K_p, for the equilibrium.
Answer:
Partial pressure of I atom (pI₁) = \(\frac{40}{100}\) × 10⁵
Partial pressure of I₂ (pI₂) = \(\frac{(100-40)}{100}\) × 10⁵
∴ K_p = \(\frac{\mathrm{p}_{\mathrm{I}}^2}{\mathrm{p}_{\mathrm{I}_2}}=\frac{\left(4.0 \times 10^4\right)^2}{6.0 \times 10^4}\)
= 2.67 × 10⁴ Pa

Question 4.
Write the expression for the equilibrium constant, K_c for each of the following reactions.
i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g)
ii) 2Cu(NO₃)₂(s) ⇌ 2Cu0(s) + 4NO₂(g) + O₂(g)
iii) CH₃COOC₂H₅(aq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(aq)
iv) Fe³⁺(aq) + 3OH^–(aq) ⇌ Fe(OH)₃(s)
v) I₂(s) + 5F₂(g) ⇌ 2IF₅(g)
Answer:
i) K_c = \(\frac{[\mathrm{NO}(\mathrm{~g})]^2\left[\mathrm{Cl}_2(\mathrm{~g})\right]}{[\mathrm{NOCl}(\mathrm{~g})]^2}\)
ii) K_c = [NO₂(g)]⁴[O₂(g)] {∵ [CuO(s) = [Cu(NO₃)₂(s)] = 1}
iii) K_c = \(\frac{\left[\mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq})\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{aq})\right]}{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(\mathrm{aq})\right]}\) (∵ [H₂O(l)] = 1)
iv) K_c = \(\frac{1}{\left[\mathrm{Fe}^{3+}(\mathrm{aq})\right]\left[\mathrm{OH}^{-}(\mathrm{aq})\right]^3}\) (∵ [Fe(OH)₃(s)] = 1)
v) K_c = \(\frac{\left[\mathrm{IF}_5\right]^2}{\left[\mathrm{~F}_2\right]^5}\) (∵ [I₂(s) = 1)

Question 5.
Find out the value of Kc for each of the following equilibria from the value of K_p.
i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) ;K_p = 1.8 × 10^-2 at 500 K
ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); K_p = 167 at 1073 K
Answer:
i) For the reaction, 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) ; ∆n = (2 + 1) – 2 = 1
K_p = K_c (RT)^∆n
∴ K_c = \(\frac{\mathrm{K}_{\mathrm{p}}}{(\mathrm{RT})^{\Delta \mathrm{n}}}=\frac{1.8 \times 10^{-2}}{(0.0831 \times 500)}\) = 4.33 × 10^-4 (∵ R = 0.0831 L bar mol^-1K^-1)

ii) For the reaction, CaCO₃(s) ⇌ CaO(s) + CO₂(g) ; ∆n = 1 – 0 = 1
∵ K_c = \(\frac{\mathrm{K}_{\mathrm{p}}}{(\mathrm{RT})^{\mathrm{An}}}=\frac{167}{(0.0831 \times 1073)}\) = 1.873

Question 6.
For the following equilibrium, K_c = 6.3 × 10¹⁴ at 1000 K.
NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reactions ?
Answer:
K_c(reverse) = \(\frac{1}{\mathrm{~K}_{\mathrm{c}(\text { forward })}}\) = \(\frac{1}{3.3 \times 10^{14}}\) = 1.59 × 10^-15

Question 7.
Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression ?
Answer:
[Pure solid] or [Pure liquid] = \(\frac{\text { No.of moles }}{\text { Volume (in } \mathrm{L})}\) = \(\frac{\text { Mass } / \text { molecular mass }}{\text { Volume }}\)
= \(\frac{\text { Density }}{\text { Molecular mass }}\) = Constant
(The density of a pure solid or a pure liquid is constant at constant temperature)

Question 8.
Reaction between N₂ and O₂ takes place as follows.
2N₂(g) + O₂(g) ⇌ 2N₂O(g)
If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a 10L reaction vessel and allowed to form N₂O at a temperature for which K_c = 2.0 × 10^-37, determine the composition of equilibrium mixture.
Answer:
Suppose x moles of N₂ react till equilibrium is attained.

Since, the value of Kc for the reaction (K_c = 2.0 × 10^-37) is very small, the value of x must be very small. Therefore,
[N₂(g)] = \(=\frac{0.482-x}{10}=\frac{0.482}{10}\) = 0.0482 mol L^-1
[O₂(g)] = \(\frac{0.933-x / 2}{10}=\frac{0.933}{10}\frac{0.933-x / 2}{10}=\frac{0.933}{10}\) = 0.0933 mol L^-1 and [N₂O(g)] = \(\frac{\mathrm{x}}{10}\)
From K_c = \(\frac{\left[\mathrm{N}_2 \mathrm{O}(\mathrm{~g})\right]^2}{\left[\mathrm{~N}_2(\mathrm{~g})\right]^2\left[\mathrm{O}_2(\mathrm{~g})\right]}\)
We have, 2.0 × 10^-37 = \(\frac{(x / 10)^2}{(0.0482)^2 \times 0.0933}\)
Which gives, x = 6.6 × 10^-20
∴ [N₂O(g)] = \(\frac{x}{10}=\frac{6.6 \times 10^{-20}}{10}\) = 6.6 × 10^-21

Question 9.
Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below.
2NO(g) + Br₂(g) ⇌ 2NOBr(g)
When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br₂.
Answer:

Given that x = 0.0518
No. of moles of NO(g) at equilibrium = 0.087 – x = 0.087 – 0.518 = 0.0352
and number of moles of Br₂(g) at equilibrium = 0.0437 – \(\frac{\mathrm{x}}{2}\) = 0.0437 – \(\frac{0.0518}{2}\) = 0.0178

Question 10.
At 450 K, K_p = 2.0 × 10¹⁰ bar for the given reaction at equilibrium.
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
What is K_c at this temperature ?
Answer:
For the given reaction, ∆n = 2 – (2 + 1) = -1
K_p = K_c(RT)^∆n
⇒ K_c = \(\frac{\mathrm{K}_{\mathrm{p}}}{(\mathrm{RT})^{\Delta \mathrm{n}}}=\frac{2.0 \times 10^{10}}{(0.0831 \times 450)^{-1}}\) L mol^-1, (∵ R = 0.0831 L bar K^-1 mol^-1)
= 2.0 × 10¹⁰ × 0.831 × 450 L mol^-1
= 7.48 × 10¹¹ L mol^-1

Question 11.
A sample of HI(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is K_p for the given equilibrium ?
2HI(g) ⇌ H₂(g) + I₂(g)
Answer:

[∵ (0.2 – 0.04) is the decrease in the pressure of HI]
K_p = \(\frac{\mathrm{p}_{\mathrm{H}_2} \times \mathrm{p}_{\mathrm{I}_2}}{\mathrm{p}_{\mathrm{HI}}^2}=\frac{0.08 \times 0.08}{(0.04)^2}\) = 4.0

Question 12.
A mixture of 1.57 mol of N₂, 1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 1.7 × 10². Is the reaction mixture at equilibrium ? If not, what is the direction of the net reaction ?
Answer:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Q_c = \(\frac{\left[\mathrm{NH}_3(\mathrm{~g})\right]^2}{\left[\mathrm{~N}_2(\mathrm{~g})\right]\left[\mathrm{H}_2(\mathrm{~g})\right]^3}=\frac{(8.13 / 20)^2}{\left(\frac{1.57}{20}\right) \times\left(\frac{1.92}{20}\right)^3}\) = 2.38 × 10³ (V = 20 L)
Given that K_c = 1.7 × 10²
Obviously, Q_c ≠ K_c
Therefore, the reaction is not in equilibrium because Q_c ≠ K_c. It will proceed in the reverse direction.

Question 13.
The equilibrium constant expression for a gas reaction is, K_c = \(\frac{\left[\mathrm{NH}_3\right]^4\left[\mathrm{O}_2\right]^5}{[\mathrm{NO}]^4\left[\mathrm{H}_2 \mathrm{O}\right]^6}\)
Write the balanced chemical equation corresponding to this expression.
Answer:
4NO(g) + 6H₂O(g) ⇌ 4NH₃(g) + 5O₂(g)

Question 14.
One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with CO according to the equation,
H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)
Calculate the equilibrium constant for the reaction.
Answer:
The given reaction is H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)
Initial number of moles of H₂O(g) = 1
Number of moles of H₂O(g) (reacted) = \(\frac{1 \times 40}{100}\) = 0.4
∴ Number of moles of H₂O(g) left (unreacted) = 1 – 0.4 = 0.6
From the given equation it is clear that 0.4 moles of H₂O(g) react with 0.4 moles of CO(g) to form 0.4 moles of H₂(g) and 0.4 moles of CO₂(g). The total volume is 10 litres.
∴ At equilibrium,
[H₂O(g)] = \(\frac{1-0.4}{10}\) = \(\frac{0.6}{10}\) = 0.06 mol L^-1 ; [CO(g)] = \(\frac{1-0.4}{10}\) = \(\frac{0.6}{10}\) = 0.06 mol L^-1
[H₂(g)] = \(\frac{0.4}{10}\) = 0.04 mol L^-1 ; [CO₂(g)] = \(\frac{0.4}{10}\) = 0.04 mol L^-1
Hence, K = \(\frac{\left[\mathrm{H}_2(\mathrm{~g})\right]\left[\mathrm{CO}_2(\mathrm{~g})\right]}{\left[\mathrm{H}_2 \mathrm{O}(\mathrm{~g})\right][\mathrm{CO}(\mathrm{~g})]}=\frac{0.04 \times 0.04}{0.06 \times 0.06}\) = 0 44

Question 15.
At 700 K, equilibrium constant for the reaction :
H₂(g) + I₂(g) ⇌ 2HI(g)
is 54.8. If 0.5 mol L_1 of HI(g) js present at equilibrium at 700 K, what are the concentration of H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K ?
Answer:
The given reaction can be written as 2HI(g) ⇌ H₂(g) + I₂(g)
For this reaction, K = \(\frac{1}{54.8}\) = 1.82 × 10^-2 (∵ because for the reverse reaction, K = 54.8)
As the number of moles of H₂ and N₂ are equal, their concentrations at equilibrium will also be equal.
Suppose [H₂(g)] = [I₂(g)] = x mol L^-1
Given that [HI(g)] = 0.5 mol L^-1
K = \(\frac{\left[\mathrm{H}_2(\mathrm{~g})\right]\left[\mathrm{I}_2(\mathrm{~g})\right]}{[\mathrm{HI}(\mathrm{~g})]^2}\) ⇒ 1.82 × 10^-2 = \(\frac{x \times x}{(0.5)^2}\) ⇒ x = [1.82 × 10^-2 × (0.5)²]^1/2 = 0.068 mol L^-1
Hence, at equilibrium, [H₂(g)] = [I₂(g)] = 0.068 mol L^-1

Question 16.
What is the equilibrium concentration of each of the substances in the equilibrium when the initial, concentration of ICl was 0.78 M ?
2ICl(g) ⇌ I₂(g) + Cl₂(g); K_c = 0.14
Answer:
Suppose at equilibrium, [I₂(g)] = [Cl₂(g>] = x mol L^-1

Given that, K_c = 0.14
K_c = \(\frac{\left[\mathrm{I}_2(\mathrm{~g})\right]\left[\mathrm{Cl}_2(\mathrm{~g})\right]}{[\mathrm{ICl}(\mathrm{~g})]^2}\)
⇒ 0.14 = \(\frac{x \times x}{(0.78-2 x)^2}\) ⇒ x² = 0.14(0.78 -2x)²
⇒ x = \(\sqrt{0.14}\) (0.78 – 2x) = 0.374(0.78 – 2x)
⇒ x = 0.292 – 0.748 x
⇒ 1.748 x = 0.292
x = \(\frac{0.292}{1.748}\) = 0.167
Hence, equilibrium concentrations will be as follows.
[I₂(g)] – [Cl₂(g)] = x = 0.167 M
[ICl(g)] = 0.78 – 2x = 0.78 – (2 × 0.167) = 0.446 M

Question 17.
K_p = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
C₂H₆(g) ⇌ C₂H₄(g) + H₂(g)
Answer:
Suppose the partial pressure of C₂H₆ decreases by p atm till equilibrium is attained.

K_p = \(\frac{\mathrm{p}_{\mathrm{C}_2 \mathrm{H}_4} \cdot \mathrm{p}_{\mathrm{H}_2}}{\mathrm{p}_{\mathrm{C}_2 \mathrm{H}_6}}\) = \(\frac{\mathrm{p} \times \mathrm{p}}{4.0-\mathrm{p}}\) ⇒ 0.04 = \(\frac{\mathrm{p}^2}{4.0-\mathrm{p}}\) = 0.16 – 0.04 p ⇒ p² + 0.04 p – 0.16 = 0
p = \(\frac{-0.04 \pm \sqrt{(0.04)^2-4 \times 1 \times(-0.16)}}{2 \times 1}=\frac{-0.04 \pm 0.801}{2}\)
Taking positive value, p = 0.38
∴ [C₂H₆(g)]_eq = 4.0 – p = 4.0 – 0.38 = 3.62 atm

Question 18.
Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as,
CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)
i) Write the concentration ratio (reaction quotient), Q_c, for this reaction (note : water is not in excess and is not a solvent in this reaction)
ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
Answer:
i) Q_c = \(\frac{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(l)\right]\left[\mathrm{H}_2 \mathrm{O}(l)\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}(l)\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)\right]}\)

ii)

If the volume of the reaction mixture is V L, we have at equilibrium,
[CH₃COOH(l)] = \(\frac{0.829}{\mathrm{V}}\) mol L^-1 ; [C₂H₅OH(l)] = \(\frac{0.009}{\mathrm{V}}\) mol L^-1
[CH₃COOC₂H₅(l)] = \(\frac{0.171}{\mathrm{V}}\) mol L^-1 ; [H₂O(l)] = \(\frac{0.171}{\mathrm{V}}\) mol L^-1
K_c = \(\frac{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(l)\right]\left[\mathrm{H}_2 \mathrm{O}(l)\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}(l)\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)\right]}=\frac{\frac{0.171}{\mathrm{~V}} \times \frac{0.171}{\mathrm{~V}}}{\frac{0.829}{\mathrm{~V}} \times \frac{0.009}{\mathrm{~V}}}=\frac{0.171 \times 0.171}{0.829 \times 0.009}\) = 3.92

iii)

∴ Q_c = \(\frac{\frac{0.214}{V} \times \frac{0.214}{V}}{\frac{0.786}{V} \times \frac{0.286}{V}}\) = 0.204
Since, Q_c < K_c, equilibrium has not been reached.

Question 19.
A sample of pure PCl₅, was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5 × 10^-1 mol L^-1. If value of K_c is 8.3 × 10^-5 what are the concentrations of PCl₃ and Cl₂ at equilibrium ?
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
Answer:
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
Since, the number of moles of PCl₃(g) and Cl₂(g) are equal, their concentrations at equilibrium will also be equal.
Suppose, at equilibrium, [PCl₃(g)] = [Cl₂(g)] = x mol L^-1
Given that at equilibrium, [PCl₅] = 0.5 × 10^-1 mol L^-1
∴ K_c = \(\frac{\left[\mathrm{PCl}_3(\mathrm{~g})\right]\left[\mathrm{Cl}_2(\mathrm{~g})\right]}{\left[\mathrm{PCl}_5(\mathrm{~g})\right]}\) ⇒ 8.3 × 10^-3 = \(\frac{\mathrm{x} \times \mathrm{x}}{0.5 \times 10^{-1}}\)
⇒ x = (8.3 × 10^-3 × 0.5 × 10^-1)^1/2 = 0.02
Hence, [PCl₃(g)] = [Cl₂(g)] = 0.02 mol L^-1.

Question 20.
One of the reaction that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO₂.
FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g); K_p = 0.265 atm at 1050 K
What are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial partial pressures are : P _CO = 1.4 atm and P_CO2 = 0.80 atm?
Answer:

Q_p = \(\frac{\mathrm{P}_{\mathrm{CO}_2}}{\mathrm{P}_{\mathrm{CO}}}=\frac{0.80}{1.4}\) = 0.571
Since, Q_p > K_p, the reaction will proceed in the reverse direction. On account of this partial pressure of CO₂ will decrease while that of CO will increase.
Suppose, the decrease in the partial pressure of CO₂ is p atm.
∴ At equilibrium, P_CO2 = (0.80 – p) atm and P_CO = (1.4 + p) atm
∵ K_p = \(\frac{\mathrm{P}_{\mathrm{CO}_2}}{\mathrm{P}_{\mathrm{CO}}}\), we have
0.265 = \(\frac{0.80-\mathrm{p}}{1.4+\mathrm{p}}\) ⇒ 0.265 (1.4 + p) = 0.80 – p
⇒ 0.371 + 0.265 p = 0.80 – p
⇒ 1.265 p = 0.429 ⇒ p = \(\frac{0.429}{1.265}\) = 0.339 atm
∴ P _CO at equilibrium = 1.4 + 0.339 = 1.739 atm.
and P_CO2 at equilibrium = 0.80 – 0.339 = 0.461 atm.

Question 21.
Equilibrium constant, K_c for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mole L^-1 N₂(g) 2.0 mol L^-1 H₂ and 0.5 mol L^-1 NH₃. Is the reaction at equilibrium ? If not in which direction does the reaction tend to proceed to reach equilibrium ?
Answer:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ; K_c = 0.061
For the given set of conditions, Q_c = \(\frac{\left[\mathrm{NH}_3(\mathrm{~g})\right]^2}{\left[\mathrm{~N}_2(\mathrm{~g})\right]\left[\mathrm{H}_2(\mathrm{~g})\right]^3}=\frac{(0.5)^2}{(3.00) \times(2.00)^3}\) = 0.0104
Since, Q_c < K_c, the reaction is not at equilibrium. It will proceed in the forward direction.

Question 22.
Bromine monochioride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:
2BrCl(g) ⇌ Br₂(g) + Cl₂(g)
for which K_c = 32 at 500 K. If initially pure BrCl is present at a concentration 3.3 × 10^-3 mol L^-1, what is its molar concentration in the mixture at equilibrium?
Answer:

K_c = \(\frac{\left[\mathrm{Br}_2(\mathrm{~g})\right]\left[\mathrm{C} l_2(\mathrm{~g})\right]}{[\mathrm{BrCl}(\mathrm{~g})]^2}\) ⇒ 32 = \(\frac{\frac{x}{2} \times \frac{x}{2}}{\left(3.30 \times 10^{-3}-x\right)^2}\)
Taking square root on both sides, we get
5.66 = \(\frac{\frac{x}{2}}{3.30 \times 10^{-3}-x}=\frac{x}{2\left(3.30 \times 10^{-3}-x\right)}\)
⇒ 0.037 – 11.32 x = x ⇒ (1 + 11.32)x = 0.037 ⇒ x = \(\frac{0.037}{1+11.32}\) = 3.0 × 10^-3
∴ Equilibrium centration of BrCl = 3.30 × 10^-3 – x
= 3.30 × 10^-3 – 3.0 × 10^-3
= 3.0 × 10^-4 mol L^-1