22-06-2025

1752CJA101021250018 JA

PART-1 : PHYSICS

SECTION–I (i)

1) A satellite of mass m is circulating around the earth with constant angular velocity. If radius of
the orbit is R0 and mass of the earth M, the ratio of angular momentum of the satellite about the
centre of the earth and its kinetic energy will be :-

(A)

(B)

(C)

(D)

2) A particle is projected from point A, that is at a distance 4R from the centre of the Earth, with
speed v1 in a direction making 30° with the line joining the centre of the Earth and point A, as
shown. Find the speed v1 of particle (in m/s) if particle passes grazing the surface of the earth.

Consider gravitational interaction only between these two. (use = 6.4 × 107 m2/s2)

(A)

(B) 800
(C)
(D) None of these

3) In the given circuit find the magnitude of current flowing through arm 'AB' :
(A) 4.5A
(B) 1.5 A
(C) 3.5A
(D) 4 A

4) Consider the potential at the corner of a uniformly charged cube of dimension L to be directly
proportional to ρL2, where ρ is volume charge density. What is the value of the ratio of the potential
at the centre to the potential at the corner of the cube :-

(A) 4 : 1
(B) 3 : 2
(C) 1 : 2
(D) 2 : 1

SECTION–I (ii)

1) A cuboid of dimension [a × a × b]. Charge q is placed at the centre of edge having length ‘b’. If

flux through face ‘ABCD’ is then select the correct statement(s) :-

(A)
Flux through the entire cuboid is
(B) Flux through the face ‘ABEH' is zero.

(C)
Flux through the face ‘BEFD’ is

(D)
Flux through the face ‘BEFD’ is

2) For which of the following continuous charge system magnitude of electric field at point P is
 0
where λ is linear charge density?

(A)

(B)

(C)

(D)

3) Regarding a satellite revolving around the earth with a time period of 18 hours, in the equatorial
plane, select the correct alternative(s)

(A) The time period of the orbital motion does not depend on the mass of the satellite
If the satellite revolves in the same sense as that of the rotation of earth, the time interval
(B)
between its consecutive appearances overhead a fixed point is 72 hrs
If the satellite revolves in the opposite sense as that of the rotation of earth, the time interval
(C)
between its consecutive appearances overhead a fixed point is nearly 1.3 hrs.
If the satellite revolves in the same sense as that of the rotation of earth, the time interval
(D)
between its consecutive appearances overhead a fixed point is infinite
4) A spherical planet of radius R has spherically symmetrical distribution of mass density, varying as
square of the distance from the centre, from zero at centre to maximum value ρ0 at its surface.

(A)
The value of escape velocity of a mass m at the surface of planet is .
The value of acceleration due to gravity 'g' varies inside the planet as cube of the distance from
(B)
centre.
The value of escape velocity from surface is same as the escape velocity from another planet of
(C)
same total mass & radius but having uniform mass density.
The energy required to impart escape velocity to particles of masses 'm' & '2m', at the surface
(D)
of planet, will be in ratio 1:2.

5) Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected
from the midpoint of the line joining their centers, perpendicular to the line. The gravitational
constant is G. The correct statement(s) is (are) :-

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is
(A)
.
The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is
(B)
.

(C)
The minimum initial velocity of the mass m, so that it reach 2L distance from masses is .
(D) The energy of the mass m and system remains constant.

6)

In the network shown in the figure. Choose the correct statement(s).

(A) Magnitude of potential difference between points H and B is 5V

(B) Magnitude of potential difference between points D and B is 1V
(C) Magnitude of potential difference between points D and E is 0.5V
(D) Magnitude of potential difference between points D and G is 2V

7) Given a square frame of diagonal length 2r made of insulating wires. There is a short dipole,
having dipole moment P, fixed in the plane of the figure lying at the center of the square, making an
angle θ as shown in figure. Four identical particles having charges of magnitude q each and
alternatively positive and negative sign are placed at the four corners of the square. Select the

correct alternative(s).
(A)
Electrostatic force on the system of four charges due to dipole is .

(B)
Electrostatics force on the system of four charges due to dipole is .
(C) Net torque on the system of four charges about the centre of the square due to dipole is zero.

(D)
Net torque on the system of four charges about the centre of the square due to dipole is .

8) A conducting wire has a non uniform cross section as shown in figure. A steady current flows

through it. Which of the following is/are true :

(A) Drift speed of electrons at A is less than at B.

(B) Drift speed of electron at A is more than at B.
(C) Magnitude of electric field at A is less than magnitude of electric field at B.
(D) Electric field is zero at every point, inside of wire.

SECTION–II

1) One end of a spring of negligible unstretched length and spring constant k is fixed at the origin (0,
0). A point particle of mass m carrying a positive charge q is attached at its other end. The entire
system is kept on a smooth horizontal surface. When a point dipole pointing towards the charge q
is fixed at the origin, the spring gets stretched to a length ℓ and attains a new equilibrium position
(see figure below). If the point mass is now displaced slightly by Δℓ << ℓ from its equilibrium
position and released, it is found to oscillate. It is found that equivalent force constant of the
oscillation is 'α' times of spring constant 'k' then what is the value of 'α' ?

2) A small body of mass ‘m’ is projected from the surface of a given planet as shown in the figure. If
small body has velocity which is just sufficient to get out from the gravity of planet then time taken

by small body to reach at height of 3R from the surface of planet is . Find the value of K.
3) If resistivity of the fustrum changes by relation ρ = Cr where C is a constant and r is the radius of

cross-section at a particular position. r = a and r = b = at ends A and B respectively. The

equivalent resistance between points A and B is R = . Then the value of is: ( is a

natural number)

4) The coefficient of static friction between two blocks is 0.5 and the table is smooth. The maximum
horizontal force that can be applied to move the blocks together is .......N.(take g = 10 ms–2)

5) A circular platform rotates around a vertical axis with angular velocity ω = 10 rad/s. On the
platform is a ball of mass 2 kg, attached to the long axis of the platform by a string of length 16 cm
(α = 30°). Find normal force exerted by the ball on the platform (in newton). Friction is absent.

6) A particle is projected horizontally with a speed u from the top of a plane inclined at an angle θ
with the horizontal. The particle will strike the plane at a distance tan θ sec θ from the point of
projection. Find the value of n.

PART-2 : CHEMISTRY

SECTION–I (i)

1) Given below are two statements : one is labelled as Assertion (A) and the other is labelled as
Reason (R).
Assertion (A) : Treatment of bromine water with propene yields 1-bromopropan-2-ol.
Reason (R) : Attack of water on bromonium ion follows Markovnikov rule and results in 1-
bromopropan-2-ol.
In the light of the above statements, choose the most appropriate answer from the options given
below :

(A) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
(B) (A) is false but (R) is true.
(C) Both (A) and (R) are true and (R) is the correct explanation of (A)
(D) (A) is true but (R) is false

2)
Product (A) and (B) are respectively

(A)

A is B is

A is B is
(B)
 A is B is
(C)

A is B is
(D)

3)

The product X is ?

(A)

(B)
(C)

(D)

4) In which of following species hypercongugation take place ?

(A) CH3 – BH2

(B)

(C)

(D)

SECTION–I (ii)

1) Select correct order of basic strength from the following :

(A)

(B)

(C)
(D)

2) Select the pair in which Ist is having higher heat of hydrogenation than IInd :

(A)
and

(B)
and
(C) CH3–CH = CH–CH3 and CH3–CH2–CH = CH2

(D)
and

3) Find out the major product 'X' and 'Y' in given reaction sequence.

(A) Product 'X' is CH3CH2–OH

(B) Product 'X' is CH3CH2–O–CH2–CH3
(C) Product 'Y' is CH3CH2–OH
(D) Product 'Y' is CH3CH2–O–CH2CH3

4) Product

(A)

(B)

(C)

(D)

5) Correct product formation takes place in which of the following reaction(s) ?

(A)

(B)

(C)

(D)

6) Match the reactions in List-I with the features of their products in List-II and choose the
INCORRECT option.

List-I List-II

Inversion of
(P) (1)
configuration

Retention of
(Q) (2)
configuration

Mixture of
(R) (3)
enantiomers

Mixture of
(S) (4) structural
isomers

Mixture of
(5)
diastereomers

(A) (P)-1; (Q)-2; (R)-5; (S)-3

(B) (P)-2; (Q)-1; (R)-3; (S)-5
(C) (P)-1; (Q)-2; (R)-5; (S)-4
(D) (P)-2; (Q)-4; (R)-3; (S)-5

7) the products are

(A)

(B)

(C) Et–I
(D) Et–OH

8) Which is/are the correct order?

(A) (Nucleophilicity in H2O)

(B)

< < (Leaving ability)

(C) F⊖ < Cl⊖ < Br⊖ < I⊖ (Nucleophilicity in DMSO)
⊖ ⊖ ⊖ ⊖
(D) F < Cl < Br < I (Nucleophilicity in H2O)

SECTION–II

1) A sample of water has hardness due to presence of only CaCl2. 10 kg of hard water required 10.6
gm of Na2CO3 for complete removal of hardness. What is degree of hardness of hard water sample
(in ppm CaCO3) ?
[Atomic mass : Na = 23, Ca = 40, Cl = 35.5]

2) How many of the following ethers can be formed by Williamson ether synthesis.

(i) (ii)

(iii) (iv)

(v) (vi)

(vii) (viii)

3) How many of the following alcohols given immediate turbidity with Lucass reagent (Conc. HCl /
ZnCl2)
(i) (ii) (iii) (iv)

(v) (vi) (vii)

(viii) (ix) (x)

4) 4 mole of a mixture of Mohr's salt and Fe2(SO4)3 requires 500 mL of 1

M K2Cr2O7 for complete oxidation in acidic medium. The mole % of the Mohr's salt in the mixture is :-

5) Calculate total number of compounds which give gas on reaction with aq. NaHCO3 ?

(a) (b) (c) Me–COOH

(d) (e) (f) (g)

(h) CH≡CH (i) CH2=CH2 (i)

6) How many orders is/are correct regarding SN1 reaction mechanism

(I) < <

(II) > >

(III) >

(IV) > > Ph – CH2 – Cl

(V) > >

PART-3 : MATHEMATICS

SECTION–I (i)

1) If the solution of equation be [a, b] then b – a is

(A) 1

(B)

(C)

(D)

2) Let be a function defined where and .

Number of function satisfies the condition

(A)
(B)
(C)
(D)

3) Let be such that then is

(A) 7
(B) 5
(C) 4
(D) 42

4) Let α,β be roots of quadratic equation x2 – 5x – 3 = 0. If an = 2αn + 3βn (n ∈ N), then

is equal to

(A)

(B)
(C) 1
(D) 5

SECTION–I (ii)

1) Which of following expression equivalent to

(A)

(B)

(C)

(D)

2) Which of the following limit expression is meaning less?

(A)

(B)

(C)

(D)

3) Consider
one more function g is defined such that and . Then

(A) Domain of is
(B) Domain of is

(C) Domain of is

Domain of is
(D)
​

4) Consider and . Let

 then

(A)
(B)
(C)

(D)

5) Let then Sn can take value's

(A) 1980
(B) 3422
(C) 2756
(D) 2970

6) Let and be the roots of with . For all positive integers n, define

, and . Then which of the following options

is/are correct?

(A)

(B)

(C)

(D)

7) Let and be two functions and gof: is defined. Then which of the
following statement(s) is(are) incorrect?

(A) If gof is onto then f must be onto

(B) If f is into and g is onto then gof must be onto function
(C) If gof is one-one then g is not necessarily one-one
(D) If f is injective and g is surjective then gof must be bijective mapping

8) Let is defined by
 is bijective function.

If where [.] is Greatest Integer function then

(A) f(0) = 2
(B) a + b + c = 1
(C) a, b & c must be integer
(D)

SECTION–II

1) The number of real solution of equation

2) Consider a quadratic function and satisfying the

following condition:
(a)
(b) b2 – a2 = 4ac
(c)

(d)
Then is

3) Let . Number of function are there satisfying is.

4) A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed
from the pack and the sum of the numbers on the remaining cards is 1000. If the smaller of the
numbers on the removed cards is k, then k – 20 =

5) Evaluate where [.] in G.I.F.

6) Consider the sequence, 489,448889,444888889,......., Tr

here denote '3r' digit number where the first r digit is 4, succedding 2r–1
digit is 8 and last digit is 9.

If then a+b+c is .
 ANSWER KEYS

PART-1 : PHYSICS

SECTION–I (i)

Q. 1 2 3 4
A. D A D D

SECTION–I (ii)

Q. 5 6 7 8 9 10 11 12
A. A,B,C A,B A,B B,C,D B,C,D A,B,C,D A,C B

SECTION–II

Q. 13 14 15 16 17 18
A. 4 7 18 15 4 2

PART-2 : CHEMISTRY

SECTION–I (i)

Q. 19 20 21 22
A. C B B A

SECTION–I (ii)

Q. 23 24 25 26 27 28 29 30
A. A,C,D A,B,D A,D B,C A,B,D A,C,D A,D A,B,D

SECTION–II

Q. 31 32 33 34 35 36
A. 1000 2 6 75 5 4

PART-3 : MATHEMATICS

SECTION–I (i)

Q. 37 38 39 40
A. A A B A

SECTION–I (ii)

Q. 41 42 43 44 45 46 47 48
A. A,C A,B,C A,C A,C,D A,C A,B,D A,B,D B,C
 SECTION–II

Q. 49 50 51 52 53 54
A. 0 4 26 20 1 9
 SOLUTIONS

PART-1 : PHYSICS

1)

where v =

2) Conserving angular momentum :

m(v1 cos60°)4R = mv2 R ⇒ = 2.

Conserving energy of the system :

3)

Current in branch AB 4 A

4)

VA ∝ ρL2

Divide cube into smaller cube of length

Therefore
VC ∝ 2ρL2

5)

Total flux through cube =

ϕABCD = ϕGFEH = & ϕBEFD = ϕCGFD & ϕABEH = ϕAHGC = 0

ϕABCD = ϕGFEH + ϕBEFD + ϕCGFD =

6)

Electric field at ∴

E|| =

and for charged circular arc E =

7) For B

For C

8)

ρ = kr2 =

(A) vesc =

M= =

so
(B) g = =
g ∝ r3

(C) as vesc = so will be same for other planet also.

(D) Energy required =

so

9) New Ans. (B,C,D) By Paras Jain Sir

10)

Correct Answer : (1), (2), (3), (4)

11)

Electric field diagram due to dipole

Force diagram on the system

Net force = =
Net force =
The net torque on the system = 0

12) by i = nAeVd
⇒ AB > AA ⇒ Vd|B < Vd|A

13) Initially at stretch of 'ℓ' .....(1)

after displacing by Δℓ (= x) we have

Fnet = Fspring – Fe = k(r + x) –

= k(ℓ + x) –

= kℓ + kx –

= kℓ + kx –

Fnet = kx + kℓ .

0
14) Speed of projection (v ) = ......(1)

......(2)

15) =

α = π, β = 3 ⇒

16)
F = 3a (For system) ….(i)
 (for 1kg block) ….(ii)
µ×1×g=a⇒5=a
F = 15N

17) T cos α = mω2 16 cos α

T sinα + N = mg

+N=20
N = 20 – 16 = 4

18)

Hence n = 2.

PART-2 : CHEMISTRY

19)

Correct Answer : (3)

20) Concept : It is based on pinacol-pinacolone rearrangement reaction. Product A is formed
by free radical initiated dimerisation of ketones followed by protonation of oxygen from the
solvent.
Product B is formed by dehydration to form more stable carbocation followed by
rearrangement of more electron dense alkyl group from adjacent carbon to gain more stable
carbocation by +M of OH group. Loss of H+ from the carbonyl oxygen will give ketone.

21) Correct option is (B)

NaNH2 is base which abstracted more acid hydrogen firstly then other hydrogen.
Acidic Hydrogen :- Hb > Hd > Hc > Ha

22) Solution/Explanation/Calculation:

• Sigma C–H bond electron pair overlap with vacant 2p-orbital of boron.
Hence, option (1) is correct.

23)

Correct Answer : (1), (3), (4)

24)

Correct Answer : (1), (2), (4)

25)

26)

27)

Correct Answer : (1), (2), (4)

28)
29)

30)

A, B, D (conceptual)

31) CaCl2 + Na2CO3 → CaCO3 + 2NaCl

0.1 mole
= 0.1 mole
gm eq. of CaCl2 = gm eq. of CaCO3

0.1 × 2 =
w = 10 gm

Degree of hardness =

32)
⇒ (i), (ii), can be formed by WE synthesis

33)

⇒Lucas reagent [conc. HCl + Anhy. ZnCl2] is used to distinguish 1°/2° and 3° Alcohol.
⇒ 3° Alcohol and Alcohol forming stable intermediate carbocation with Lucas reagent gives
turbidity immediately.
⇒ ∴ (i); (ii); (iv); (vi); (ix) & (x) gives turbidity immediately with Lucas reagent.

34) FeSO4 (NH4)2 SO4 . 6H2O + Fe2 (SO4)3

Mohr salt
x mol (4–x) mol
∵ (Eq) K2Cr2O7 = (Eq) Mohr solt
(Cr+6 → Cr+3) (Fe+2 → Fe+3)
N × V(L) = mol × n factor
1 × 6 × 0.5 = x×1
x = 3 mol

= 75%

35)

Correct Answer : (5)

36) Explain Question :

Asking for the reactivity of different halides towards SN1 reaction.
Concept :
Rate of SN1 ∝ Stability of carbocation formed in rds
∝ Leaving group ability (If carbocation stability is same)
Solution :
Correct stability of carbocations

(i)

(ii)
 (iii) [on the basis of leaving ability]
(iv)

(v)

SN1 reactivity is correct for option (i), (iii), (iv), (v)

Final Answer : (4)

PART-3 : MATHEMATICS

37)

Put

we get
38)

Total number of function is n + k – 1Ck

Here n = 20, k = 10

39)

40)

x2 – 5x – 3 = 0
an = 2αn + 3βn
⇒ an = 5an–1 + 3a–2
Put n = 20 a20 = 5a19 + 3a18
n = 19 a19 = 5a18

so

41)

Let
put

42)

Use fundamental of limit.

43)

(A) Domain of y = f(f(f(g(x))))

The innermost function is g(x). The domain of g(x) is .
The next function is f(g(x)). Since f(g(x))=x for all , the domain of f(g(x)) is .
The next function is f(f(g(x))), which is f(x) for . The domain of f(x) is , but we only
consider . The domain of f(f(g(x))) is .
The next function is f(f(f(g(x)))), which is f(f(x)) for . Since the range of f(x) for is
, the domain of f(f(f(x))) is .
(C) Domain of y = g(g(g(f(x))))

44)

Sketch the graph of f(x); x < 0 & g(x);

Now draw the graph of h(x) according to the definition

45)

Given

46)

Correct Answer: (A),B,(D)

47)
Counter example:

;
x
f(x) = e g(x) = sinx

; onto but f is into.

Similarly we can find counter example for B,D.

48)

Correct Answer : (2), (3)

49)

Draw the graph of both function:

So, number of solution is 0.

50)

f(x-1)=f(2-x)

f is symmetric about
(b) b2–a2 = 4ac
 Roots we 0 and 1.
f(x) = a(x) (x-1), by applying (c) and (d) we get a = –4

51)

f(f(x))=x can done two ways-

(1) f(f(x)) = x
(2) f(f(x))= y and f(f(y)) = x

52)

Clearly

from here we get n = 46

Here k = 40 so, k –20 = 20
53) where

So,

54)