1

FIITJEE – All India Internal Test Series

PHYSICS, CHEMISTRY & MATHEMATICS
Pattern - APT-4 QP Code: AiiTS-1
Time Allotted: 3 Hours Maximum Marks: 186

▪ Please read the instructions carefully. You are allotted 5 minutes specifically for
this purpose.
▪ You are not allowed to leave the Examination Hall before the end of the test.

INSTRUCTIONS
BATCH – CFY 2023-2027

Caution: Question Paper CODE as given above MUST be correctly marked in the answer
OMR sheet before attempting the paper. Wrong CODE or no CODE will give wrong results.

A. General Instructions
1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.
2. This question paper contains Three Sections.
3. Section-I is Chemistry, Section-II is Mathematics and Section-III isPhysics.
4. Each Section is further divided into Two Parts: Part-A & B in the OMR.
5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be
provided for rough work.
6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic
devices, in any form, are not allowed.

B. Filling of OMR Sheet

1. Ensure matching of OMR sheet with the Question paper before you start marking your answers on
OMR sheet.
2. On the OMR sheet, darken the appropriate bubble with HB pencil for each character of your Enrolment
No. and write in ink your Name, Test Centre and other details at the designated places.
3. OMR sheet contains alphabets, numerals & special characters for marking answers.

C. Marking Scheme For All Two Parts.

(i) Part-A (01-08) – Contains seven (07) multiple choice questions which have One or More correct answer.
Full Marks: +4 If only the bubble(s) corresponding to all the correct options(s) is (are) darkened.
Partial Marks: +1 For darkening a bubble corresponding to each correct option, provided NO
incorrect option is darkened.
Zero Marks: 0 If none of the bubbles is darkened.
Negative Marks: −1 In all other cases.
(ii)Part-A (09-12) – Contains 2 paragraph questions which have ONLY ONE CORRECT answer Each question
carries +3 marks for correct answer and -1 marks for wrong answer.

(iii) Part-A (13-14) – This section contains Two (02) List-Match Sets, each List-Match set has Two (02) Multiple
Choice Questions. Each List-Match set has two lists: List-I and List-II. FOUR options are given in each
MultipleChoice Question based On List-I and List-II and ONLY ONE of these four options satisfies the
condition asked in the Multiple Choice Question. Each question carries +3 Marks for correct combination
chosen and -1 marks for wrong options chosen.

(ii)Part-C (01-04) contains five (04) Numerical based questions, the answer of which maybe positive or negative
numbers or decimals (e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30) and each question carries
+3 marks for correct answer and there will be no negative marking.

Name of the Candidate :____________________________________________

Batch :____________________ Date of Examination :___________________
Enrolment Number :_______________________________________________

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 2

SECTION-1 :CHEMISTRY
PART – A
(Multi Correct Choice Type)
This section contains 8 multiple choice questions. Each question has four choices (A), (B),
(C) and (D) out of which ONE OR MORE may be correct.

1. Kinetic energy of molecules is directly proportional to-

A. Pressure
B. Temperature
C. Both (A) and (B)
D. Atmospheric pressure
2. Which of the following statement is FALSE?
A. We perspire more on a sunny day rather than a cloudy day.
B. Perspiration is our body’s method of maintaining a constant temperature
C. The rate of evaporation decreases with increasing wind speed
D. Evaporation causes heating
3. Crystalline solids contain
A. Atoms
B. Molecules
C. Ions
D. All are correct
4. Gases show the following properties-
A. Fluidity
B. Compressibility
C. Rigidity
D. Diffusion
5. Which of the following change(s) is/are exothermic process(es)?
A. Solid →Gas
B. Gas →Solid
C. Liquid →Gas
D. Liquid →Solid
6. Which of the following state of matter exist at extremely high temperature-
A. Solid
B. BEC
C. Plasma
D. Gas
7. If we increase the temperature of a gas-
A. Pressure of gas increases
B. Kinetic energy of gas increases
C. More frequent collisions of gas molecules will occur.
D. None of the above
8. Latent heat of fusion for ice is-
A. 80 g/cal
B. 80 cal/g
C. 540 cal/g
D. None of these
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 3

(Paragraph Type)
This section contains 2 paragraphs. Based upon the paragraphs 2 multiple choice questions
have to be answered. Each of these questions has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.

Paragraph for question no. 9-10

The change of liquid into gaseous state below the boiling point is known as evaporation.
In evaporation a molecule of a liquid to escape to vapour state, it must overcome the
intermolecular forces attracting it.
9. Identify the factors not affecting evaporation.
A. Surface area
B. Humidity
C. Nature of liquid
D. Diffusion
10. Evaporation occurs when
A. A liquid molecule must have minimum kinetic energy to overcome intermolecular
forces.
B. Due to high potential energy
C. Both (a) and (b)
D. None of these

Paragraph for question no. 11-12

With increase in the temperature the kinetic energy of particles increases and this
increase results in vibration of particles and generation of heat. This heat keeps the
particles to overcome the attraction force operation among them. They leave their
position and start moving within the boundaries. When all the particles attain sufficient
energy, the state transforms to the state where the particles are with higher kinetic
energy, lesser attractive force and lesser density.
11. Increase in temperature leads to
A. Increase in kinetic energy
B. Condensation
C. Decrease in kinetic energy
D. freezing
12. Input of heat during state transition is consumed
A. To retain the previous state
B. To overcome the attraction force
C. By atmosphere
D. To evaporation

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 4

PART – B
Matrix Matching
This section contains 2 matrix match type. Each question has four statement (A,B,C & D)
given in Column I and four statements(P,Q,R,S and T) in Column-II. Any given statement in
Column I can the correct matching with ONE or MORE statements given in Column II

13. Matrix – 1

Column (A) Column (B)

A. Camphor P. Mass/Volume

B. Density Q. Piped Natural Gas

C. Matter R. Occupies Space and has mass

D. PNG S. Sublime

A. A-P,B-S,C-R,D-Q
B. A-P,B-Q,C-R,D-S
C. A-S,B-P,C-R,D-Q
D. A-Q,B-P,C-R,D-S

14. Matrix-2

Column (A) Column (B)

A. Solid P. Indefinite shape and definite Volume

B. Gas Q. Unlimited Free Surface

C. Liquid R. Exists at very high temperature

D. Plasma S. Fixed Shape

A. A-S,B-P,C-R,D-Q
B. A-S,B-Q,C-P,D-R
C. A-P,B-Q,C-R,D-S
D. A-Q,B-P,C-R,D-S

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 5

PART – C
(One Integer Value Correct Type)
This section contains 04 questions. Each question, when worked out will result in integer value either in
-ive or positive or decimal.

1. How much heat energy (in Joules) is required to change 10 mg of ice at 00 C to water at
400 C. [Latent heat of fusion = 3.34 J/Kg and specific heat of water = 4200 J/kgK]

2. 400 ml of a gas at 2270C is to be reduced to a volume of 300 ml. By what degree

Celsius, must the temperature be altered keeping pressure constant.

3. Convert 41 Fahrenheit to 0C.

4. A sample of helium occupies a volume of 3.6 L at -450C. What volume will it occupy at
450C?

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 6

SECTION-II :MATHEMATICS
PART – A
(Multi Correct Choice Type)
This section contains 8 multiple choice questions. Each question has four choices (A), (B),
(C) and (D) out of which ONE OR MORE may be correct.

2 1 3
1. 𝑥 = + 3+ 2 − 5+ 2. Then 𝑥 is
√5+√3 √ √ √ √
A. A rational number
B. An even number
C. A Fibonacci number
D. None of the above
5
4 5 4 5
2. 𝑎 · 𝑏 0.2 𝑐 = √ √(24 )3 − 5√8 + 2 √ √(23 )4. Here, 𝑎, 𝑏, 𝑐 can be
A. 𝑎 = −2, 𝑏 = 2, 𝑐 = 3
B. 𝑎 = −1, 𝑏 = 2, 𝑐 = 8
C. 𝑎 = −2, 𝑏 = 8, 𝑐 = 1
D. 𝑎 = −1, 𝑏 = 16, 𝑐 = 2
3. Let P(x) be a polynomial with integer coefficients such that for four distinct integers
𝛼, 𝛽, ɣ, µ; 𝑃(𝛼) = 𝑃(𝛽) = 𝑃(ɣ) = 𝑃(µ) = 7. If 𝑃(𝑘) = 𝑛 (𝑛 ≠ 7)for some integer k and n,
then n cannot be
A. 31
B. 29
C. 17
D. 11
4. Let 𝑔(𝑥) be a six degree monic polynomial such that𝑔(1) = 2, 𝑔(2) = 6, 𝑔(3) =
12, 𝑔(4) = 20, 𝑔(5) = 30 and 𝑔(6) = 42 If 𝑔(7) = 𝑛
A. n leaves remainder 6 when divided by 11
B. n leaves remainder 110 when divided by 111
C. (n +1) is divisible by 7
D. The last digit of n is 2
5. If ( x − 1) + ( y − 3) + ( z − 5 ) + ( t − 7 ) = 0 then xyzt + 16 is
2 2 2 2

A. A prime number
B. A perfect square number
C. Can be expressed as 1 + p + p2 + p3 + p4 for some prime p
D. Can be expressed as 1 + p! for some prime p
6. In a n-sided regular polygon, one interior angle is thrice of the exterior angle, then nis
A. An odd number
B. A multiple of 3
C. Square of an integer
D. Cube of an integer

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7. In rectangle ABCD, AC and BD intersect at O. If ∠AOB = 110°, and ∠ADO = x°, and p1
and p2 are two prime factors of x, with p1< p2, then
A. p1 is a root of 𝑥 2 – 5𝑥 + 6 = 0
B. p1 is a root of𝑥 2 – 12𝑥 + 35 = 0
C. p2 is a root of 𝑥 2 – 13𝑥 + 22 = 0
D. p2 is a root of 𝑥 2 – 9𝑥 + 14 = 0

8. Let 𝐺(𝑛) and 𝑆(𝑛) be two functions defined for positive integer n. 𝐺(𝑛) gives the greatest
divisor of n, whereas 𝑆(𝑛)gives the sum of the all divisors of n. Let 𝑅(𝑛) be another
𝐺(𝑛)
function defined as 𝑅(𝑛) = 𝑆(𝑛)
, then the correct statement(s) is/are
A. 𝑅(2003) < 𝑅(2007)
B. 𝑅(2017) > 𝑅(2019)
C. 𝑅(2011) > 𝑅(2021)
D. 𝑅(2023) < 𝑅(2027)

Part-B
(Paragraph Type)
This section contains 2 paragraphs. Based upon the paragraphs 2 multiple choice questions
have to be answered. Each of these questions has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.

Paragraph for question Nos. 9 to 10

Let P(x) be a polynomial with integer coefficients. If P(x) has some integer zeros then

9. P(2021)·P(2022)·P(2023) is divisible by
A. 2 but not 3
B. 3 but not 2
C. Neither by 2 nor by 3
D. Both by 2 and 3
10. Let P(α) and P(α+1) both be odd integers for some integer α, then α can be
E. Multiple of 3
F. Multiple of 4
G. Multiple of 5
H. None of these

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Paragraph for question nos. 11 to 12

If both a and b are rational numbers, find the value of a and b in each of the
following

11.
(A) a=5/2 and b=3/2
(B) a=7/2 and b=3/2
(C) a=5/2 and c=5/2
(D) a=7/2 and b=5/2

12.
(A) a=27/13 and b=14/13
(B) a=27/13 and b=16/13
(C) a=31/13 and b=14/13
(D) a=31/13 and b=16/13

PART – B
Matrix Matching
This section contains 2 matrix match type. Each question has four statement (A,B,C & D)
given in Column I and four statements (P,Q,R,S and T) in Column-II. Any given statement in
Column I can the correct matching with ONE or MORE statements given

13. Match the list

Column I Column II

(A) In a right triangle ABC, BC is (P) √3

hypotenuse and D is midpoint of BC.
𝐴𝐷
Then 𝐵𝐶 =
(B) 0.4 √3 (Q) 0.5√3
((640.3̅ ) ) = 2𝑎 , where 𝑎 =
(C) The equation 𝑥 2 − (2 + √3)𝑥 + (R) 0.8√3
(2 − √3) = 0 has roots α and β. Then
α2β + β2α =
(D) In triangle ABC, vertices have (S) 1
following coordinates:
A(sin45°,cos30°), B(sec60°, 0),
C(0,0). Then area of ABC =

A. A→Q, B→R, C→S, D→P

B. A→S, B→R, C→S, D→Q
C. A→Q, B→R, C→S, D→Q
D. A→S, B→Q, C→R, D→P
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 9

14. Match the following:

Column − I Column − II
(A) (P) 12.5
81 81 81
..........to  = ..........
64 64 64
(B) n  (Q) 9
2 − 4 
2 = 1024 then 3
n 4 
= .....................
(C) 10 6.25 (R) 81
= .......... 64
6.25 − 0.5
(D)
( ) (S) 6
36
3
The power of x in x

(A) (A) → (R), (B) → (Q), (C) → (P), (D) → (S)

B) (A) → Q), (B) → (R), (C) → (P), (D) → (S)
C) (A) → (R), (B) → (Q), (C) → (S), (D) → (P)
D) (A) → (R), (B) → (P), (C) → (Q), (D) → (S)

PART – C
Numerical Based
This section contains 4 numerical based questions, the answer may be positive or negative or
decimals (e. g 6.25, 7.00, −0.33, −.30, 30.27, − 127.30) and each question carries + 3 marks for
correct answer. There is not negative marking.

1. In a book, a statement is written as “G.C.D. of 483# and 210 is 1.” As it can be seen, the
unit digit of the first number is misprinted as #. Find the actual digit.

1 1
2. If x 4 + 4
= 47 , then the value of x2 + 2 is
x x

3. In the following figure, ABC is a triangle with AB = CD,

∠BAD = ∠DAC, and ∠ABC = 2∠ACB. If ∠BAC = x°, then
find the value of x.

4. A polynomial P(x) is given as, 𝑃(𝑥) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎2 𝑥 2 + 𝑎1 𝑥 + 1

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If P(1) = 23·P(0), then find the sum 𝑎𝑛 + 𝑎𝑛−1 + ⋯ + 𝑎2 + 𝑎1

SECTION-III :PHYSICS
PART – A
(Multi Correct Choice Type)
This section contains 8 multiple choice questions. Each question has four choices (A), (B),
(C) and (D) out of which ONE OR MORE may be correct.

1. Starting from rest a particle is first accelerated for time t1 with constant acceleration a1
and then stops in time t2 with constant retardation a2. Let v1 be the average velocity in
this case and s1 the total displacement. In the second case, it is accelerating for the
same time t1 with constant acceleration 2a1 and come to rest with constant retardation
a2 in time t3. If v2 is the average velocity in this case and s2 the total displacement, then
A. v2 = 2v1
B. 2v1 < v2 < 4v1
C. s2 =2 s1
D. 2s1 < s2 < 4s1

2. A particle is moving along a straight line. The displacement of the particle becomes zero
in a certain time (t>0). The particle does not undergo any collision.
A. The acceleration of the particle may be zero always
B. The acceleration of the particle may be uniform
C. The velocity of the particle must be zero at some instant
D. The acceleration of the particle must change its direction

3. For a moving particle, which of the following options may be correct?

A. |𝑽𝒂𝒗 | < 𝑣𝑎𝑣
B. |𝑽𝒂𝒗 | > 𝑣𝑎𝑣
C. |𝑽𝒂𝒗 | = 0 𝑎𝑛𝑑 𝑣𝑎𝑣 ≠ 0
D. |𝑽𝒂𝒗 | ≠ 0 𝑎𝑛𝑑 𝑣𝑎𝑣 = 0

Here 𝑽𝒂𝒗 is average velocity and 𝑣𝑎𝑣 is the average speed

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 11

4. Identify the correct graph representing motion of a particle along a straight line with
constant acceleration with zero initial velocity

A.

B.

C.

D.

5. The figure shows the velocity (v) of the particle plotted against time (t)

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A. the particle changes its direction of motion at some point

B. the acceleration of the particle remains constant
C. the displacement of the particle is zero
D. the intial and final speeds of the particle are the same

6. The speed of a train increases at a constant rate α from zero to v and then remains
constant for an interval and finally decreases to zero at a constant rate β. The total
distance travelled by the train is 𝑙. The time taken to complete the journey is t. Then
𝑙 (𝛼+𝛽)
A. 𝑡 = 𝛼𝛽
𝑙 𝑣 1 1
B. 𝑡 = + ( + )
𝑣 2 𝛼 𝛽
2𝑙𝛼𝛽
C. 𝑡 is minimum when 𝑣 = √(𝛼−𝛽)

2𝑙𝛼𝛽
D. 𝑡 is minimum when 𝑣 = √(𝛼+𝛽)

7. A car is moving with uniform acceleration along a straight line between along a straight
line between two stops X and Y are 2 m/s and 14 m/s. Which of the following statements
is(are) correct.
A. its speed at mid-point of XY is 10 m/s
B. its speed at a point A such that XA: AY = 1:3 is 5m/s
C. the time to go from X to the mid-point of XY is double that to go from mid-point to Y
D. the distance travelled in first half of the total time is half of the distance travelled in
the second half of the time
8. In the projectile motion shown is figure, given tAB=2s then which of the following is
correct : (g=10 ms−2)

A. particle is at point B at 3s
B. maximum height of projectile is 20m
C. initial vertical component of velocity is 20 m/s
D. horizonatal component of velocity is 20m/s

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(Paragraph Type)
This section contains 2 paragraphs. Based upon the paragraphs 2 multiple choice questions
have to be answered. Each of these questions has 4 choices (A), (B), (C) and (D) out of which
ONLY ONE is correct.

Paragraph for question Nos. 9 to 10

A body is allowed to fall from a height of 100 m. If the time taken for the first 50 m is t1
and for the remaining 50 m is t2.
9. Which is correct?
A. t1 = t2
B. t1 > t2
C. t1 < t2
D. Depends upon the mass

10. The ratio of times to reach the ground and to reach first half of the distance is
A. √3 ∶ 1
B. √2 ∶ 1
C. 5 : 2
D. 1 : √3

Paragraph for question nos. 11 to 12

The velocity- time graph of a particle in straight line motion is show in figure. The particle starts
its motion from origin.

11. The distance travelled by the particle in 8s is

A. 18 m
B. 16 m
C. 8 m
D. 6 m
12. The average acceleration from 2s to 6s
A. -2 m/s2
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 14

B. -3/2 m/s2
C. 2 m/s2
D. 3/2 m/s2

PART – B
Matrix Matching
This section contains 2 matrix match type. Each question has four statement (A,B,C & D)
given in Column I and four statements (P,Q,R,S and T) in Column-II. Any given statement in
Column I can the correct matching with ONE or MORE statements given

Matrix Match -1
13. The trajectories of the motion of 3 particles are shown in the figure. Match the entries of
column I with the entries of column II. Neglect air resistance.

Column I Column II
i. Time of flight is least for a. A

ii. Vertical component of velocity is greatest b. B

for
iii. Horizontal component of velocity is c. C
greatest for
iv. Launch speed is least for d. No appropriate
match

A. i – d, ii – d, iii – c, iv – a
B. i – c, ii – d, iii – c, iv – d
C. i – d, ii – d, iii – c, iv – d
D. i – a, ii – d, iii – c, iv – b

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Matrix Match -2
14. .The velocity - time graph of a particle moving along X-axis is shown in figure. Match the
entries of Column I with entries of Column II.

Column I Column II
A. For AB, particle is P. Moving in +ve x-direction with increasing speed

B. For BC, particle is Q. Moving in +ve x-direction with decreasing speed

C.For CD, particle is R. Moving in -ve x-direction with increasing speed
D. For DE, particle is S.Moving in +ve x-direction with decreasing speed

A. A-P, B-P, C-Q, D-R

B. A-R, B-S, C-Q, D-R
C. A-T, B-P, C-S, D-R
D. A-Q, B-P, C-S, D-R

PART – C
Numerical Based
This section contains 4 numerical based questions, the answer may be positive or negative or
decimals (e. g 6.25, 7.00, −0.33, −.30, 30.27, − 127.30) and each question carries + 3 marks for
correct answer. There is not negative marking.

1. An aeroplane is flying in a horizontal direction with a velocity 600 km/h at a height of

1960 m. When it is vertically above the point A on the ground, a body is dropped from it.
The body strikes the ground at point B. Calculate the distance AB in Km. ( take g = 9.8
m/s2)

2. Two particles A and B are thrown vertically upward with velocity, 5 m/s and 10 m/s,
respectively (g = 10 m/s2). Find separation between them after 1s.

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3. A ball is thrown downwards with a speed of 20m/s from the top of a building 150m high and
simultaneously another ball is thrown vertically upwards with a speed of 30m/s from the foot of
the building. Find the time (in sec) after which both the balls will meet. (g =10 m/s2)

4. A parachutist, after bailing out, falls 50 m without friction. When the parachute opens, it
decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height (in
m), did he bail out?

FIITJEECOMMON TEST
BATCHES: FOUR YEAR CRP (APT-4)

AiiTS-SET-A
ANSWER KEY
CHEMISTRY
PART-A
1. B 2. A, C, D 3. D 4. A, B, D
5. B, D 6. C 7. A, B, C 8. B
9. D 10. A 11. A 12. B
13. C 14. B
PART – C
1. 5.0 J 2. 1250C 3. 5 4. 5L

MATHEMATICS
PART-A
1. A, B 2. A, B, C, D 3. B, C, D 4. A, B, C
5. B, C, D 6. D 7. B, C 8. B, C, D
9. D 10. D 11. B 12. B
13. C 14. A

PART-C
1. 1.000 2. 7.000 3. 72.000 4. 22.000

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 17

PHYSICS
PART-A
1. A, D 2. B, C 3. A, C 4. A, D
5. A, B, C, D 6. B, D 7. A, C 8. A, B, C, D
9. B 10. B 11. A 12. B
13. A 14. A

PART – C
1. 3.33KM 2. 5.000 M 3. 3.000SEC 4 243.000 M

HINTS & SOLUTION

CHEMISTRY

1. Kinetic energy is directly proportional to temperature and is independent of pressure.

2. Because perspiration is our body’s method of maintaining a constant temperature.
3. Crystalline solids may contain atoms, molecules and ions.
4. Gases can diffuse, compress and fluid.
5. During the conversion of solid from liquid or gas, energy is liberated.
6. Plasma state is a state of matter which exist at extremely high temperature.
7. After increasing the temperature of a gas, its pressure, KE and collisions between
molecules will be increased.
8. Latent heat of fusion is 80 cal/g
9. The evaporation of a liquid depends mainly on temperature, surface area, humidity
and nature of liquid.
10. Some particles in a liquid always have more kinetic energy than the others which is
responsible for evaporation.
11. Increase in temperature leads to increase in KE
12. During state transition input of heat is consumed to overcome the attraction force

MATRIX MATCH

13. C
14. B

NUMERICALS

1. Mass of water = 10mg=10-5kg

Heat energy required = M.C.∆T + ML
= 10-5 x 4200 x (40-0) + 3.34 x 105 x 10-5
= 1.68 + 3.34 Joules
= 5.02 Joules
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 18

2. Given V1=400 ml, T1=2270C = (227 + 273)K = 500K, V2=300ml, T2=?

According to Charle’s Law
At constant pressure, V1/T1 = V2/T2
 400/500 = 300/T2
 T2 = 1500/4 = 375K
 375 – 273 = 1020C

Therefore alteration in temperature = (227-102)0C = 1250C

3. (F-32)/9 = C/5
41-32 = C/5
C= 50C
4. V1/T2 = V2/T2
3.6/228 = V2/318
V2=5L

MATHEMATICS

1. Rationalizing:
x = √5 - √3 + √3 - √2 - √5 + √2 = 0
Hence it is rational and even.
AB
2. Simplifying:
a·bc/5 = -2·81/5 = -28/5 = -2·23/5 = - 162/5
ABCD
3. f(x) = P(x) – 7 = (x – α) (x – β) (x – ɣ) (x – µ) Ø(x)
f(k) = (k – α) (k – β) (k – ɣ) (k – µ) Ø(k) = n – 7
∴(n – 7) must have at least four distinct integral factors (since, k – α, k – β, k – ɣ, k – µ all
are distinct)
BCD
4. g(x) = x(x+1) have roots 1, 2, 3, 4, 5
∴g(x) – x(x+1) = (x – 1)(x – 2)(x – 3)(x – 4)(x – 5)(x – 6)
→g(7) = 7·8 + 6! = 776
ABC

5. x = 1, y = 3, z = 5, t = 7
∴xyzt + 16 = 121
121 = (11)2 = 1 + 3 + 32 + 33 + 34 = 1 + 5!
BCD

6. Exterior angle = x, interior angle = 3x

→4x = 180
→8x = 360
Hence, n = 8 (Since sum of exterior angles = 360°)
D

7. ∠ADO = ∠AOB/2 = 55°

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 19

∴ p1 = 5, p2 = 11
BC

8. Certainly g(n) = n. So with increase in s(n), R(n) will decrease. Hence R(n) will be lesser
for a prime number than a nearer composite number.
2003, 2011, 2017 and 2027 all are primes.
BCD

9. P(x) = (x – n)·f(x), where n is integer.

P(2021) = (2021 – n)·f(2021), P(2022) = (2022 – n)·f(2022), P(2023) = (2023 –
n)·f(2023).
Now, (2021 – n), (2022 – n), (2023 – n) are three consecutive integers, so at least one of
them must be even and one must be divisible by 3.
D

10. Similarly for any integer α, either of (α – n) and (α + 1 – n) must be even.

D

11. B

12. B

13. AD2 = b2 + c2 – (a/2)2 = 3a2/4→AD/BC = 0.5√3

(641/3)0.4√3 = 40.4√3 = 20.8√3
αβ(α+β) = (2+√3)(2-√3) = 1
Base = sec60°, height = cos30°. ∴Area = (1/2)sec60° cos30°=0.5√3
C

14. B

Numerical
1. Since, G.C.D. is 1, the number has no common factor with 210 = 2·3·5·7
The number is not divisible by 2.Unit digit cannot be even, so remaining digits 1, 3, 5,
7, 9.
The number is not divisible by 3. Sum of the digits cannot be a multiple of 3, so
remaining digits 1, 5, 7
The number is not divisible by 5. Unit digit cannot be 5, so remaining digits 1, and 7.
4837 is divisible by 7.
∴The number is 4831 and the required digit is 1.
1

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2. 7

3. Let BE bisects ∠B

BEC is isosceles with BE = CE

∴In triangle ABE and CDE
CD = AB, ∠B = ∠C = α, CE = BE
Using S-A-S, ABE and CDE are congruent.
∴ AE = DE and ∠BAE = ∠CDE = 2β
∴ AED is isosceles with AE = DE and ∠ADE = ∠DAE = β
∴∠ADC = β + 2β = 3β
→3β = 2α + β (exterior angle of ABD)
→α=β
Also sum of angles 2α + 2β + α = 180°
→5β = 180°
→β = 36°
∴∠BAC = 72°
72

4. P(0) = 1
P(1) = an + … + a1 + 1 = 23
→an + … + a1 = 22

PHYSICS

1. Correct options are A) and D)

Velocity after acceleration by a for t1 time= a1.t1
It takes time t2 to retard to zero velocity, so, a1.t1 = a2.t2
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 21

𝑎
⟹ t2 = 𝑎1 𝑡1
2

Similarly for second case,

2𝑎
t3 = 𝑎 1 𝑡1
2

In first case,
1
Distance travelled during acceleration = 2 𝑎1 𝑡12
𝑎1 𝑡12
Distance travelled during retardation = 2𝑎2
1 𝑎12 𝑡12
⟹ s1=2 𝑎1 𝑡12 + 2𝑎2
𝑠1 𝑠1
𝑣1 = = 𝑎 =
𝑡1 +𝑡2 𝑡1 + 1 𝑡1
𝑎2

In second case,
Distance traveled during acceleration= 𝑎1 𝑡12
(2𝑎1 𝑡1 )2
Distance traveled during retardation= 2𝑎2
2𝑎12 𝑡12
⟹ 𝑠2 = 𝑎1 𝑡12 + 𝑎2
𝑠2 𝑠2
⟹ 𝑣2 = 𝑡 = 2𝑎1 𝑡1 = 𝑎1 𝑡1
1 +𝑡3 𝑡1 +
𝑎2

Hence v2=2v1 and 2s1<s2<4s1

2. Correct options are B) and C)

Since the particle is moving along a straight line and displacement is zero at certain time t>0,
so we can deduce that particle has to change its direction and should again pass through the
same point so that displacement becomes zero.
Thus the velocity of particle should be zero at some point so that particle can change its
direction and acceleration can be uniform i.e. opposing motion of the particle.
Acceleration of particle need not change its direction.
Acceleration of particle cannot be zero always because the particle has to change its direction
in order for the displacement to be zero. This all happens in the case of objects thrown up and
coming down on the surface of earth and the acceleration of the object is nothing but
acceleration due to gravity

3. Correct option is A and C

Displacement ≤ distance, therefore average velocity < average speed. Option A is correct

If the object was in motion and returns back to the intital position, then displacement = 0 but
distance ≠ 0. So average speed ≠ 0 but average velocity = 0. So option C is also correct.

4. Correct options are A and D

Acceleration is constant and u = 0
Now, v=u+at, so v=at
1 1
Now, 𝑥 = 𝑢𝑡 + 2 𝑎𝑡 2 , so 𝑥 = 2
𝑎𝑡 2
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i.e., v−t graph is a straight line passing through origin and x−t graph a parabola passing
through origin.

5. Correct options are A) , B) , C) and D)

the particle changes its direction of motion at t=T since the velocity becomes positive from
negative.
the acceleration of the particle remains constant since the slope is constant throughout the
curve.
the displacement of the particle is zero because the area under the curve of negative
velocity is equal to area under the curve of positive velocity.
the initial and final speeds of the particle are the same since speed is the magnitude of
velocity.

6. Correct options are B) and D)

v=αt1 ⇒ t1=αv
v = βt1 ⇒ t2 = βv
t o = t − t1 − t = (t − αv − βv)
Now ,
1 1
1 = 𝛼𝑡12 + 𝑣𝑡0 + 𝛽𝑡22
2 2

Substitutingvalue of t1 and t2 and to we get

𝑙 𝑣 1 1
E. 𝑡 = + ( + )
𝑣 2 𝛼 𝛽

For t to be minimum its first derivative with respect to velocity should be zero or,
1 𝛼+𝛽
0 = − 𝑣 2 + 2𝛼𝛽
2𝑙𝛼𝛽
𝑣 = √(𝛼+𝛽)
7. Correct options are A) and C) using third equation of motion,
let a be the constant acceleration ,s be the distance b/w x and y
v be velocity at y,u be velocity at x, V be velocity at the midpoint of line joining x and y.
v2=u2+2as
putting the given values:
as=96
now,
V2= u2+2a(s/2)
V=10m/s.
for other part using first equation of motion
time from x to the mid-point=(10−2)/a i.e.=8/a
time taken from midpoint to y=(14−10)/a i.e. =4/a
so,the c option is correct.

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8. Correct options are A) , B) , C) and D)

Let u and v be the horizontal and vertical components of the initial velocity vector. Now,
𝐴𝐵
u= , where AB=40m.
tAB

Therefore u=20m/s and as a result, tOA=1s. From the kinematic equations of motions,
s=vt−(1/2)gt2, where s=15m, we get v=20m/s.
And from 02=v2−2gh, we get, maximum height of projectile(h) = 20m.
tOB=tOA+tAB=3s.
1
9. 𝑠 = 2 𝑔𝑡12
10
or 𝑡1 =
√𝑔
1
and 100 = 2
𝑔𝑡 2
10√2
𝑡=
√𝑔
𝑡2 = 𝑡 − 𝑡1 = 0.4𝑡1
Ans. 𝑡1 > 𝑡2 , correct option is B

10√2 𝑔
10. 𝑡: 𝑡1 = ×√ = √2 , correct option is B
√𝑔 10
1 1
11. Distance covered = area of speed time graph = 2 × (4 + 2) × 4 + 2
(4 + 2) × 2 = 18m
𝑣2 −𝑣1 −2−4 3 −2
12. 𝑎𝑎𝑣 = 𝑡2 −𝑡1
= 6−2
= −2 𝑚𝑠

13. Correct option is A.

Since time of flight depends on uv and vertical height is same for all therefore, uvwill be
same for all. Hence, time of flight will be same.
so i- d and ii - d
iii. Since horizontal range is greatest for C therefore, uh will be greatest for (C)
iii - c
iv. Since uv is same for all but uh is least for (A) therefore, launch speed will be least for
iv -a
14. (A) – (P); (B) – (P); (C) – (Q); (D) – (R)
The slope of v-t graph gives acceleration. If velocity is positive and acceleration is
positive then speed will increase.
For part ‘BC’ acceleration is decreasing but positive so speed will increase.
If velocity is positive and acceleration is negative then speed will decrease.

Numerical type
1. The correct option is C. 3.33 km.
The velocity with which the body is released is 600 km/h (horizontal).
Vertical velocity of the body is zero. Therefore using the second equation of motion we
get,
1
⇒ℎ = 𝑢𝑡 + 𝑔𝑡 2
2
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1
⇒ℎ =0+ 𝑔𝑡 2
2
⇒ 𝑡 = √2𝑔ℎ
Horizontal displacement of the body
AB = Horizontal velocity × time available
AB=u x √2𝑔ℎ = 3.33km.
2. Using second equation of motion,
1
𝑠 = 𝑢𝑡 + 𝑎𝑡 2
2
Ball 'A' moves ⇒5m for 1 second;
Ball 'B' moves ⇒10 for 1 second;
After one second,
Separation is 5 m.
3. The distance covered is given as
s1=20t+5t2
s2=30t−5t2
s1+s2=150
⇒150=50t
⇒t=3s
4. Answer is 293 m
Step 1: Given
Distance travelled without friction = 50 m
Acceleration, a=−2m/s2
Speed just before reaching the ground = 3m/s

Step 2: Formula used

v2=u2+2as

Step 3 :Calculation
After bailing out from point A parachutist falls freely under gravity. The velocity acquired
by it will 'v'
From v2=u2+2as
=0+2×9.8×50 [As u=0,a=9.8ms-2,s=50m]
v2=980
v= √980 m/s
At point B, parachute opens and it moves with retardation of 2ms2 and reach at ground
(Point C) with velocity of 3ms
For the part 'BC' by applying the equation v2=u2+2as

v=3ms,u= √980 m/s

⇒(3)2=√980 x2 + 2×(−2)×h

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 25

⇒h ≅ 243m.
So, the total height by which parachutist bail out = 50 + 243 = 293m.

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