03-08-2025

7501CJA101001250064 JA

PHYSICS

SECTION-I(i)

1) A gas bubble from an explosion under water oscillates with a period proportional to PadbEc where
P is the static pressure, d is the density of water and E is the energy of explosion. Then a, b and c
are :

(A)

(B)

(C)

(D) 1,1,1

2) A lift starts moving downward with acceleration 5 m/s2. At the same time a ball dropped at height
10 m from the floor of lift. The time after which it strike the floor is :-

(A) 2 sec
(B) 1 sec
(C) sec

(D)
sec

3) In the given diagram, if the string connecting B and C is cut then what will be the accelerations of

blocks A, B, C, D at that instant ?

(A) 0, g↓, g , 0

(B)
0, ,g ,0

(C)
0, , ,0

(D)
g↓, , g↓,
4) In a circular motion of a particle, the tangential acceleration is given by at = 9 m/s2. The radius of
circle is 4m. Initially particle was at rest. Time after which acceleration of the particle makes an
angle of 45° with centripetal acceleration is –

(A)

(B)

(C) 1 sec

(D)

SECTION-I(ii)

1) An ant travels along a long rod with a constant velocity relative to the rod starting from the
origin. The rod is kept initially along the positive x-axis. At t = 0, then rod also starts rotating with
an angular velocity ω (anticlockwise) in x-y plane about origin. Then :

(A) the position of the ant at any time t is

(B) the speed of the ant at any time t is

(C)
the magnitude of the tangential acceleration of the ant at any time t is .
(D) the speed of the ant at any time t is

2) Coefficient of friction between the two blocks Q and R is 0.3 whereas the surface AB is smooth

(A) Acceleration of P is 5.86 m/s2

(B) Tension T1 in the string is 17.7 N
(C) Tension T2 in the string is about 41.4 N
(D) Acceleration of P is 7.55 m/s2

3) A man is standing on a road and observes that rain is falling at angle 45° with the vertical. The
man starts running on the road with constant acceleration 0.5 m/s2. After a certain time from the
start of the motion, it appear to him the rain is still falling at angle 45° with the vertical, with speed
Motion of the man is in the same vertical plane in which the rain is falling. Then which of
the following statement(s) are true :

(A) It is not possible

(B) Speed of the rain relative to the ground is 2 m/s
(C) Speed of the man when he finds rain to be falling at angle 45° with the vertical, is 4 m/s.
The man has travelled a distance 16 m on the road by the time he again finds rain to be falling
(D)
at angle 45°.

SECTION-II(i)

Common Content for Question No. 1 to 2

All of you must have played with magnets at some point of time. Magnets interact due to magnetic
field whenever a charged particle. [Like electron and proton] enters in magnetic field, it might

experience a force. Formula of force is given by , here q is amount of charge on

particle, is its velocity and is magnetic field. Force on a charged particle comes from vector
cross product of velocity and magnetic field.

1) If a charged particle enters in a magnetic field with velocity . If acting on

particle is Find .

2) If another charged particle of mass 1 kg and charge 1C enters a magnetic field with
velocity find distance (m) travelled by particle in 5 sec.

Common Content for Question No. 3 to 4

Block B is placed on a smooth horizontal surface. Block A is placed on rough surface of block B with
coefficient of friction 0.60. The mass of A and B are 2 kg and 4 kg respectively. Find the frictional

force between A and B

3) The frictional force between A and B (in N) is ____________

4) Acceleration of upper block (in m/s2) is _______.

SECTION-II(ii)

1) As shown in figure below water is draining from a conical tank with 12 feet height and diameter 8
feet into a cylindrical tank that has a base with area 400 square feet. The depth h, in feet of the
water in the conical tank is changing at rate of 2 feet per second. The rate of change of height of
water in cylindrical tank (when h = 3ft) is ft/sec, then value of x is

2) A particle is moving with uniform acceleration along x-axis with initial velocity along positive x.

At the magnitude of displacement becomes the total distance travelled. By this time
the x coordinate of particle is still positive. The instant (in sec) at which displacement becomes zero
is :

3) A fixed rifle on ground is aimed at a point on a vertical wall 1440 m horizontally away and 1080
m high above the point of the rifle end. A bullet is fired at 150 m/s towards the target. 10 sec after
firing, the gravitational field vanishes. Find where the bullet will hit the vertical wall (in meter) from
the ground. [ g = 10 m/s2]

4) Find out tension (in N) in the string connecting blocks of masses 80 kg and 20 kg.

5) In the given figure, what is the minimum possible tension (in Newton) in the string 2 if 'θ' can be

varied. (Strings are ideal and take g = 10 m/s2)

6) A vehicle moves along a horizontal circular track. The coefficient of friction between the tyres and
the track (μ) varies directly as the nth root of the distance (r) of the vehicle from the centre of the
circular track. The variation of the maximum safe speed (Vmax) with the (r) is shown below on a log -

log scale. The value of 'n' is :

 CHEMISTRY

SECTION-I(i)

1)

In an atom, two electrons move around the nucleus in circular orbits of radii R and 4R, the ratio of
the time taken by them to complete one revolution is (Considering Bohr's atomic theory is
applicable)-

(A) 1 : 4
(B) 4 : 1
(C) 1 : 8
(D) 8 : 1

2) In a closed system : A( s) ⇌ 2B(g) + 3C(g) , if the partial pressure of C is doubled at equilibrium,

then partial pressure of B will be

(A) two times the original value

(B) one half of its original value

(C)
times the original value
(D) times the original value

3) Calculate the mass percent (w/w) of sulphuric acid in a solution prepared by dissolving 4 g of
sulphur trioxide in a 100 ml sulphuric acid solution containing 80 mass percent (w/w) of H2SO4 and
having a density of 1.96 g/ml. (molecular weight of H2SO4 = 98). Take reaction SO3 + H2O → H2SO4

(A) 80.8 %
(B) 88.85 %
(C) 41.65 %
(D) None of these

4) Cl2O7 gas decomposes as: Cl2O7 → Cl2 + O2 A partially decomposed gaseous mixture is allowed to
effuse through a pin-hole and the gas coming out initially was analyzed. The mole fraction of the O2
was found to be 0.60. The degree of dissociation of Cl2O7 will be:

(A) 0.1
(B) 0.2
(C) 0.42
(D) 0.6

SECTION-I(ii)
1) Select the correct statement(s) :

(A) Ratio of gm/litre & % w/v of a solution is independent of solute nature.

(B) Ratio of % w/v and molarity of a solution depends on solute nature.
(C) Ratio of % w/v and molarity of a solution depends on solvent nature
(D) Ratio of % w/v & ppm for any solution is same

2) With reference to the above graph, choose the correct alternatives

(A) PB > PA
(B) PA > PB
(C) Pressure first increases then decreases
(D) Pressure first decreases then increases

3) A sample of a mixture of CaCl2 and NaCl weighing 4.44 g was treated to precipitate all the Ca as
CaCO3, which was then heated and quantitatively converted to 1.12g of CaO. (At . wt. Ca
= 40, Na = 23, Cl = 35.5)

(A) Mixture contains 50% NaCl, by mass

(B) Mixture contains 60% CaCl2, by mass
(C) Mass of CaCl2 is 2.22 g
(D) Mass of CaCl2 1.11 g

SECTION-II(i)

Common Content for Question No. 1 to 2

Simultaneous equilibrium :
If in any container there are two or more equilibria existing simultaneously involving one or more
than one common species, then in both/all the equilibrium, the concentration of common species is
the total concentration of that species due to all the equilibria under consideration.
Now consider a set of reactions in which equilibrium is established simultaneously in a closed vessel
at T Kelvin.
A(s) ⇌ C(g) + D(g) ; KP = 144 atm2
B(s) ⇌ C(g) + E(g) ; KP = 25 atm2

1) If partial pressure of E(g) at equilibrium is x atm then value of 6.5 x is :

2) Partial pressure of C(g) at equilibrium (in atm) is :

Common Content for Question No. 3 to 4
Buffer solution is a solution of weak acid its conjugate base or a weak base and its conjugate acid. It
resists the change in pH when small amount of strong acid or base is added to it. Based on above
paragraph answer the following questions:

3) What will be the pH, when 24.4 g of benzoic acid is added at room temperature in 500 mL of 0.1
M NaOH solution (considering no change in volume on addition of benzoic acid in solution). [Given:
pKa of benzoic acid is 4.2 at 25°C. atomic mass C = 12, O =16, H = 1. log2 = 0.3, log3 = 0.5, logs =
0.7]

4) Calculate pH if 100 ml of 0.1 M NaOH is mixed with 100 ml of 0.1 M CH3COOH solution at 25oC
[Ka(CH3COOH) = 1.8 × 10–5, log 1.8 = 0.25, log(2) = 0.3, log 5 = 0.7]

SECTION-II(ii)

1) An electron is accelerated from rest and it has wavelength of 1.41 Å by how much amount
potential should be dropped so that wavelength associated with electron becomes 1.73 Å.

2) When 100 ml of a O2 – O3 mixture was passed through turpentine, there was reduction of volume
by 20 ml. If 100 ml of such a mixture is heated, what will be the increase in volume (in ml)?

3) 5 mL of a gaseous hydrocarbon was exposed to 30 mL of O2. The resultant gas, on cooling is found
to measure 25 mL, of which 10 mL was absorbed by NaOH and the remainder by pyrogallol. All
measurements are made at constant pressure and temperature. The sum of number of carbon and
hydrogen atom in a hydrocarbon molecule is

4) During "Sulphur" estimation, 160 mg of an organic compound gives 466 mg of barium sulphate.
The percentage of Sulphur in the given compound is ______ %.
(Given molar mass in g mol–1 of Ba : 137, S : 32, O : 16)

5) For one mole of a Vander Waals gas when b=0 and T=300K, the PV vs. 1/V plot is shown below.
The value of the Van der Waals constant is a (atm. litre2 mol-2). then what is the value of 10a?
6) Consider the statements:

(a) Critical temperature

(b) The value of compressibility factor (Z) at critical condition = (smallest fraction)
(Consider simplest ratio of Z & W)
Find the value of Z + W – x

MATHEMATICS

SECTION-I(i)

1) If are in H.P. and are in G.P., then

(A) 76
(B) 80
(C) 84
(D) None of these

2) ABC is a triangle such that sin(2A + B) = sin(C – A) = and A + C = 2B, then

Which of the following is correct ?

(A)

(B)

(C)

(D)

3) The minimum value of the function f(x) =

whenever it is defined is

(A) 4
(B) –2
(C) 0
(D) 2

4) If An = sin nθ . secn θ , Bn = cos nθ secnθ, then

(A) 0
(B) tan θ

(C)

(D)

SECTION-I(ii)

1) If , then:-

(A) P(n) > 1

(B) P(n) > 2
(C) P(n) < 1
(D) P(n) < 2

2) Let x1, x2, x3, x4 are four real and distinct roots of the equation x4 – 2x3 + (1 – 2k)x2 + 2kx = 0. If x1,
x2, x3, x4 are in arithmetic progression, then the possible value(s) of k can be

(A) 2
(B) 1

(C)

(D)

3) For each non-negative integer "n" function, "f" defined by

then which of the following are true?

(A)

(B)

(C)

(D)

SECTION-II(i)

Common Content for Question No. 1 to 2

There are 4n + 1 terms in a sequence of which first 2n + 1 are in A.P. and last 2n + 1 are in G.P. the

common difference of A.P. is 2 and common ratio of G.P. is . The middle terms of the A.P. is equal
to middle term of G.P. Let middle term of the sequence is Tm and Tm is the sum of infinite G.P. whose
sum of first two terms is and ratio of these two terms of GP is .

1) Middle term of the given sequence, i.e. Tm is equal to :

2) Number of terms in the given sequence is equal to -

Common Content for Question No. 3 to 4

Let a = cos 10°, b = cos 50° , c = cos 70°, d = sin 10º sin 50º sin 70º

3) The value of |a2 – bc + d| is equal to :

4) The value of (a2 + b2 + c2 + b + c – a) is equal to :

SECTION-II(ii)

1) The sum of first n terms of an A.P is given by Then

2) If (20)19 + 2(21)(20)18 + 3(21)2(20)17 + ….… +20(21)19 = k(20)19, then k is equal to_________.

3) Suppose that a,b,c,d are positive real number satisfying (a + c) (b + d) = ac + bd and

, then the smallest possible value of S is equal to

4) If ƒn(θ) = then ƒ29 is

5) If . Then (x + y) is
equal to

6) If = when p and q are in their lowest form,

then find (p + q).
 ANSWER KEYS

PHYSICS

SECTION-I(i)

Q. 1 2 3 4
A. A A B B

SECTION-I(ii)

Q. 5 6 7
A. A,B,C A,B,C C,D

SECTION-II(i)

Q. 8 9 10 11
A. 18.00 25.00 8 3

SECTION-II(ii)

Q. 12 13 14 15 16 17
A. 5 12 380 160 6 2

CHEMISTRY

SECTION-I(i)

Q. 18 19 20 21
A. C C A C

SECTION-I(ii)

Q. 22 23 24
A. A,B B,C A,C

SECTION-II(i)

Q. 25 26 27 28
A. 12.50 13.00 3.70 8.70 to 8.74

SECTION-II(ii)

Q. 29 30 31 32 33 34
A. 25 10 6 40 15 3

MATHEMATICS
 SECTION-I(i)

Q. 35 36 37 38
A. A A B C

SECTION-I(ii)

Q. 39 40 41
A. A,D B,C A,B

SECTION-II(i)

Q. 42 43 44 45
A. 6.85 or 6.86 13.00 0.87 to 0.88 1.50

SECTION-II(ii)

Q. 46 47 48 49 50 51
A. 2 400 8 4 10 7
 SOLUTIONS

PHYSICS

1) T ∝ Pa db Ec

a= ,d= ,c= .

2)
t = 2 sec

4) aT = aC ⇒ 9 = ⇒ V = 6m

6) 10 a = 10 g –T2
2 a = T1 – fK
3a = T2 – T1 – fK
⇒ a = 5.86;
T1 = 17.7 N and T2 = 41.4 N

8)

The correct answer is 18.00

9)

10)

If A & B move together

a←

f - 2 = 2a ....... (i)
a←
20 - f = 4a ...... (ii)

From (i) & (ii)

a = 3 m/s2
f=8N
Assumption is correct as fmax = 12 N

11)

a = 3 m/s2

12)

The correct answer is 5

13)
t1 + t2 = t

⇒ t12 + (t – t1)2 =

⇒ t1 = 6 for t =
so answer is 2t1 = 12

14)

First 10 sec y1 = 900 – 500 = 400 m

Next 2 sec y2 = –10 × 2 = –20
height = 380 m

15)

asystem =
 =
T = 80 × 2 = 160 N

16)

The correct answer is 6

17) Write equation of line using (y – y1) = m(x – x1)

⇒
Take anti log
Vmax = r3/4.e5/4
Friction force only centripetal force

CHEMISTRY

18)

The correct answer is 1 : 8

19)
.....(1)
ATQ,

.....(2)
Now form equation (1) & (2)

⇒ ⇒

20)

Use class note

21)

The degree of dissociation of Cl2O7 will be: 0.42

22) By theory and formula.

23)

The correct answer is (B), (C)

24)

CaCl2 → CaCO3 → CaO = 0.02 mol CaO

∴ Moles of CaCl2 = 0.02 Mol

Mass of CaCl2 = 0.02 × 111 = 2.22 g

∴ % of CaCl2 = × 100 = 50 %

25) A(s) ⇌ C(g) + D(g)

P1+P2 P1
B(s) ⇌ C(g) + E(g)
P1+P2 P2
P1(P1 + P2) = 144 .......(1)
P2(P1 + P2) = 25 .......(2)
add (1) & (2)
(P1 + P2)2 = 169
P1 + P2 = 13

PC = 13 atm, PD = atm., PE = atm

26) A(s) ⇌ C(g) + D(g)

P1+P2 P1
B(s) ⇌ C(g) + E(g)
P1+P2 P2
P1(P1 + P2) = 144 .......(1)
P2(P1 + P2) = 25 .......(2)
add (1) & (2)
(P1 + P2)2 = 169
P1 + P2 = 13

PC = 13 atm, PD = atm., PE = atm

27)
0.2 mol 0.5 × 0.1 mol 0
0.15 0 0.05

= 4.2 – log3 = 3.7

28) if 100 ml of 0.1 M NaOH is mixed with 100 ml of 0.1 M CH3COOH solution at 25oC then pH
equal to 8.00 to 8.10

29) 1 = V1 =

V1 = = 75 V and

V2 = = 50 V
Hence potential should be dropped by 25 V.

30)

The correct answer is 10

31)

Let the formula of hydrocarbon be CxHy.Its combustion reaction is given as

Initial vol. in ml 5 30 0 0

Final volume 0 5x 0
Out of 25 ml of resultant gas 10 ml was CO2 (absorbed by NaOH) and the remaining 15ml was
oxygen (absorbed by pyrogallol).
5x = 10 ; x = 2

On solving, y = 4
Formula of gaseous hydrocarbon is C2H4 .

32) Millimoles of BaSO4 = = 2m mol

%=
 33)

34) We know that

(b) Critical condition Z =

∴ z = 3, w = 8
z+w–x=3+8–8=3

MATHEMATICS

35) Clearly, will be in A.P.

Hence,

(say)
Now,

36) A + C = 2B ⇒ B = 60°

sin(C – A) = ⇒ C – A = 30° or 150°

sin(2A + B) =
⇒ 2A + B = 30° or 150° or 390°
∴ B = 60°, 2A + B = 150°, C – A = 30°
⇒ A = 45°, B = 60°, C = 75°

37)

38) Now =

=
39) (K is even or odd)

40) Given equation =0

⇒ x = 0 i.e. one root is zero
now x4 – 2x3 – (1 – 2k)x2 + 2kx = 0; x1x2x3x4 = 0
let roots be
a – 3d; a – d; a + d; a + 3d
sum = 4a = 2 ⇒ a = 1/2
and product = (a2 – d2)(a2 – 9d2) = 0
⇒ a2 = d2 or a2 = 9d2

⇒ d=± of d = ±

if a = and d =
roots are – 1, 0, 1 , 2

if a = ; d = – roots are 2, 1, 0, – 1

if a = ; d = roots are
Now, x – 2x + (1 – 2k)x + 2k ≡ x3 – 2x2 – x + 2
3 2

(deleting one zero root of f (x) = 0 and the remaining root being 2, 1, – 1)
comparing k = 1
if the roots are 1/3, 2/3, 1 then

also x3 – 2x2 + (1 – 2k)x + 2k ≡

comparing k = –
If a = 1/2 and d = 1/6 the roots are 0, 1/3, 2/3, 1

Hence, possible values of k are 1 and – ⇒ B, C

41)
42)

For Tm

Therefore,
…

Now, middle term of AP = Middle term of GP.

n=3
No. of terms = 4n + 1 = 13

43)

For Tm

Therefore,
…

Now, middle term of AP = Middle term of GP.

n=3
No. of terms = 4n + 1 = 13
44) [3 + cos 20° + cos 100° + cos 140°] [3 + cos 20° – (cos 80° + cos 40°)]

[3 + cos 20° – 2cos 60° cos 20°]

b + c = cos 50° + cos 70° = 2cos 60° cos 10° = cos 10° = a

45) [3 + cos 20° + cos 100° + cos 140°]

[3 + cos 20° – (cos 80° + cos 40°)]

[3 + cos 20° – 2cos 60° cos 20°]

b + c = cos 50° + cos 70° = 2cos 60° cos 10° = cos 10° = a

46) Denominator
So

47)

If (20)19 + 2(21)(20)18+3(21)2(20)17+…+ 20(21)19 = k(20)19 then k is

48)

Apply the AM-GM inequality twice as follows :

The above inequalities turn into equalities when a = c and b = d. Then the condition (a + c) (b
+ d) = ac + bd can be rewritten as 4ab = a2 + b2. So it is satisfied when a/b = 2 ± . Hence,
S attains value 8, e.g., when a = c = 1 and b = d = 2 + .

49) ƒn(θ) =

, n = 29

ƒ29 =4

50)
Consider, =

Now given expression reduces to -

51) sin3θ = (3 sin θ – sin 3θ)

= =
=

= ⇒ p + q = 7. Ans.