26-07-2025

4602CJA101001250012 JA

PART-A-PHYSICS

SECTION-I(i)

1) A block of mass m is attached with a massless spring of force constant k. The block is placed over

a rough inclined surface for which the coefficient of friction is , then minimum value of
M required to move the block up the plane is : (Neglect mass of string and pulley and friction in

pulley)

(A)
m

(B)
m
(C) 2m

(D)
m

2) P. E. of particle moving along x-axis is given as U = 2x3 + 3x2 – 12x + 6. At what position, particle
is in stable equilibrium :

(A) x = – 1
(B) x = +1
(C) x = – 2
(D) x = +2

3) A uniform chain is placed over a table such that part of it overhangs as shown. If chain
remains in equlibrium, then coefficient of static friction between chain and table will satisfy

following condition :-

(A) µ ≥ 1
(B) µ ≥ 0.75
(C) µ ≥ 0.5
(D) µ ≥ 0.3

4) In the shown system, pulleys and string are massless. One end of the string is pulled by the force

. Find the acceleration ( in m/s2) of the block. [ g = 10 m/s2]

(A) g/2
(B) g
(C) 2g
(D) 0

5) A car starts from rest with an acceleration of which decreases linearly with time to zero in
10 second; after which the car continues at a constant speed. Determine the time required for the
car to travel 400 m from the start.

(A) 15.2 s
(B) 11.67 s
(C) 16.67 s
(D) 26.67 s

6) A projectile is given an initial velocity of , where is along the ground & is along the
2
vertical. If g = 10 m/s , the equation of its trajectory is :-

(A) y = 2x – 5x2
(B) 2y = 3x – 5x2
(C) 4y = 6x – 5x2
(D) 4y = 3x – 5x2

SECTION-I(ii)

1) A boy in the elevator with open roof shoots a bullet in vertical upward direction from a height of
1.5 above the floor of the elevator. The initial speed of the bullet with respect to elevator is 15 m/s.
The bullet strikes the floor after 2 seconds. Then (Assuming g = 10 m/s2)

(A) Lift is moving with constant speed

(B) Lift is moving with upward acceleration of 5.75 m/s2
(C) Lift is moving with downward acceleration of 5.75 m/s2
(D) Distance travelled by bullet during its flight can not be calculated from the given data.
2) A man standing on the edge of the terrace of a high rise building throws a stone vertically up with
a speed of 20 m/s. Two seconds later an identical stone is thrown vertically downwards with the
same speed of 20 m/s. Then :

(A) the relative velocity between the two stones remain constant till one hits the ground.
(B) both will have the same kinetic energy when they hit the ground.
(C) the time interval between their hitting the ground is 2 seconds.
(D) none of these

3) Two particles are projected simultaneously as shown in figure. If collision take place between the

particles then choose the correct statement(s).

(A) Value of θ is π/4.

(B) If separation AB is 700 m then collision occur in 10 sec.
(C) If separation AB is 350 m then collision occur in 5 sec.
(D) Value of θ is π/2.

4) If in the given regular hexagon choose the correct option :-

(A)

(B)
(C)
(D)

5) The angle between two vectors is θ and the magnitude of is half of magnitude of . If
& then choose the correct statements :-

(A)
if c = then θ will be 90°

(B)
if c = then θ will be 60°
(C)
if c = then θ will be 45°

(D)
if c = then θ will be 180°

6) Two fixed frictionless inclined planes are shown in the figure. Two blocks A and B are placed on

the two planes (g = 10 m/s2).

(A) Direction of relative acceleration is horizontal

(B) Direction of relative acceleration is vertical
(C) Magnitude of relative acceleration is 2.8 m/s2
(D) Magnitude of relative acceleration is 2.45 m/s2

SECTION-III

1) In the given figure a wedge of mass M is kept on a horizontal smooth surface. Two blocks of equal
mass m are arranged as shown in figure. All surfaces are smooth. Find the value of acceleration a (in

m/s2), so that blocks A and B do not slip over the wedge. (g = 10 m/s2)

2) Two blocks of mass 2 kg and 4 kg are connected through a massless inextensible string.
Coefficient of friction between 2 kg block and ground is 0.4 and between 4 kg block and ground is
0.6. Two forces F1 = 10 N and F2 = 20 N are applied on the block as shown in figure. Friction force

(in N ) acting on 4 kg block minus 10 N is

3) A mass m is supported as shown in the figure by ideal strings connected to a rigid wall and to a
mass 3m at rest on a fixed horizontal surface. The string connected to larger mass is horizontal, that
connected to smaller mass is vertical and the one connected to wall makes an angle 60° with
horizontal. Then the minimum coefficient so static friction between the larger mass and the

horizontal surface that permits the system to remain in equilibrium in the situation shown is .
Find n.

4) A mass m is attached to two spring having spring constant k is in equilibrium as shown in the
figure. The work done to slowly lift the block upward at constant speed by a distance mg/4k, is

. Find n?

5) A ring of mass m slides on a smooth vertical rod. A light string is attached to the ring and is
passing over a smooth peg distant a from the rod, and at the other end of the string is a mass
M(>m). The ring is held on a level with the peg and released. The distance that the ring moves

before coming to rest is ________m. Take a = 90 cm, M = 5 kg and m = 4kg.

PART-B-CHEMISTRY

SECTION-I(i)

1) According to VBT, which of the following overlapping can result in π-type covalent bond in O2
molecule formation, when Z-axis is internuclear axis ?
(I) 2s – 2s (II) 2px – 2px (III) 1s – 1s (IV) 2py – 2py (V) 2pz – 2pz

(A) I, III
(B) II, V
(C) II, IV
(D) IV, V

2) Which of the following is correct order of bond angle ?

(A) H2O > OF2 > PH3 > H2S

(B) H2O > PH3 > OF2 > H2S
(C) H2O > OF2 > H2S > PH3
(D) H2O > H2S > OF2 > PH3

3) The hydrated salt Na2SO4.nH2O, undergoes 55% loss in weight on heating and becomes
anhydrous. The approximate value of n will be :

(A) 5
(B) 3
(C) 7
(D) 10

4) Difference in wavelength of two extreme lines of H-atom in Balmer region is (Where RH is Rydberg
constant) :-

(A) 7.2 / RH
(B) 0.25 / RH
(C) 3.2 / RH
(D) 4 /RH

5) Magnetic moment of Xn+ (Z = 26) is B.M. Hence number of unpaired electrons and value of n
respectively are -

(A) 4, 2
(B) 2, 4
(C) 3, 1
(D) 0, 2

6) 5.6 gm CaO react with 4.48 litre of CO2 at STP & form CaCO3. Calculate mass of CaCO3.

(A) 10 gm
(B) 20 gm
(C) 30 gm
(D) 40 gm

SECTION-I(ii)

1)

Select the correct statement(s).

(A) The value of electron gain enthalpy of an element can be -ve or +ve.
In the periodic table, metallic character of the elements increases down the group and
(B)
decreases across the period
(C) The Cl– & S2– are isoelectronic species and first one is smaller in size than the second
(D) Atomic size of Chlorine is bigger than Fluorine

2) Which of the following has (have) regular octahedral geometry :

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

3) Which one of the following molecules is expected to exhibit diamagnetic behaviour ?

(A) C2
(B) N2
(C) O2
(D) S2

4) Which among the following is/are correct about chromium ?

(A) Its outermost partially filled orbital electronic configuration is 3d54s1

(B) Total spin of chromium = 3
(C) Cr is paramagnetic and Cr3+ is diamagnetic
(D) Magnetic moment of chromium = BM

5)

Same mass of glucose (C6H12O6) and acetic acid (CH3COOH) contain :

(A) Same number of carbon atoms

(B) Same number of Hydrogen atoms
(C) Same number of oxygen atoms
(D) All of above

6) Which of the following is correctly plotted.

(A)
(B)

(C)

(D)

SECTION-III

1) 16 g of SOx gas occupies 5.6L at 1 atm and 273K. What will be the value of x ?

2)

A 50 watt bulb emits monochromatic red light of wavelength of 795 nm. The number of photons
emitted per second by the bulb is x × 1020. The value of x is _________.

[Given : h = 6.63 × 10–34 Js and c = 3.0 × 108 ms–1]

3) Among the triatomic molecules/ions, BeCl2, N3–, N2O, NO2+, O3, SCl2, ICl2–, I3– and XeF2, the total
number of linear molecule(s)/ion(s) where the hybridization of the central atom does not have
contribution from the d–orbital(s) is
[Atomic number : S = 16, Cl = 17, I = 53 and Xe = 54]

4)

Number of amphoteric compound among the following is __________

(i) BeO

(ii) BaO
 (iii) Be(OH)2

(iv) Sr(OH)2

5) According to MO theory, number of species/ions from the following having identical bond order is
______.
CN–, NO+, O2, O2+, O22+.

PART-C-MATHEMATICS

SECTION-I(i)

1) The complete solution set of the inequality is

(A) (–1, 3)

(B)

(C)

(D) (–1, 4)

2) The number of real solutions of the equation is

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

3)

If is an arthmetic progression with common difference 1 and

then find the value of

(A) 4
(B) 5
(C) 6
(D) 7

4) If the equation x2 –x–12 = 0 and kx2 + 10x + 3 = 0 may have one common root, then find the value
of k.

3 or
(A)
(B)
–3 or
3 or
(C)

(D)
–3 or

5) The number of elements in the set is

(A) 1
(B) 3
(C) 0
(D) infinite

6) The value of

(A)

(B)

(C)

(D) 1

SECTION-I(ii)

1) If the roots of the equation are equal in magnitude and opposite in sign, then

(A) p + q = r
(B) p + q = 2r

(C)

(D) sum of roots = 1

2) The solution(s) of the equation cos2x sin6x = cos3x sin5x in the interval [0, π] is/are

(A)

(B)

(C)
(D)

3) If 4cos2x – 8cosx + 3 0 for x ∈ [0,2π], then which of the following can be true

(A)

(B)

(C)

(D)

4) If terms then

(A)

(B) S16 = 2
(C) S33 = 3

(D)

5) If where p, q are relatively prime then

(A) p = 3
(B) q = 2
(C) p = 9
(D) q = 4

6)

Let f(x) = ax2 + bx + c, (a ≠ 0) and f(x) = 0 has no real root. If (a + 3b + 9c) > 0, then

(A) a + b + c > 0
(B) a > 0, c > 0
(C) a > 0, c < 0
(D) 4a + 16b + 64c > 0

SECTION-III

1)

The number of solution of the equation cos2x + sinx = in , is

2) The sum of the series is

3) If three successive terms of a G.P. with common ratio r(r > 1) are the lengths of the sides of a
triangle and [r] denotes the greatest integer less than or equal to r, then 3[r] + [–r] is equal to :

4) Two roots of the cubic equation x3 – 5x2 – ax + 45 = 0 are real and additive inverse of each other
then ‘a’ is

5) If x ∈ R, then least positive integral value of is

 ANSWER KEYS

PART-A-PHYSICS

SECTION-I(i)

Q. 1 2 3 4 5 6
A. A B C D C C

SECTION-I(ii)

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

SECTION-III

Q. 13 14 15 16 17
A. 5 8 3 4 4

PART-B-CHEMISTRY

SECTION-I(i)

Q. 18 19 20 21 22 23
A. C A D C A A

SECTION-I(ii)

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

SECTION-III

Q. 30 31 32 33 34
A. 2 2 4 2 3

PART-C-MATHEMATICS

SECTION-I(i)

Q. 35 36 37 38 39 40
A. C A C C A C

SECTION-I(ii)

Q. 41 42 43 44 45 46
A. B,C A,B,D A,B,C A,B,C,D C,D A,B,D
 SECTION-III

Q. 47 48 49 50 51
A. 8 3 1 9 2
 SOLUTIONS

PART-A-PHYSICS

1)

For mas m :
mg sin θ + fs, max = kx
mg sin θ + µmg cos θ = kx
for mass M :

Mgx =

2) U = 2x3 + 3x2 – 12x + 6

at x = 1 (stable equilibrium)

3)

Let µ be the minimum value of friction.

⇒µ ⇒µ=

4) a =0
5)

6)

vx = 2
vy = 3
x = 2t

y = 3t –

y=
2
4y = 6x – 5x

7) –1.5 = 15T (10 + a)T2. Where T = 2 sec ⇒ a = 5.75 m/s2

As we can not find velocity of bullet with respect to ground, we can not calculate distance from
the given data.

8) Relative Initial velocities

ur = 20 – (0) = 20 m/s
Relative acceleration
ar = 0
Relative velocity between them after time
vr = ur + ar.t
= 20m/s = constant ⇒ (A) is correct
⇒ Since they are thrown from same height
⇒ Speed is same after reaching ground
⇒ Same KE when they hit the ground
⇒ (B) is correct
The time taken by the first stone to come to same height from where it was thrown

= = 4s
Since, relative velocity is Constant between them. So time interval between their hitting the
ground = 2 s.
⇒ (C) is correct
Option (D) is obvious from conservation of energy.

9)
aA = aB Þ aAB = 0

Comp. of relative velocity perpendicular to line joining them is zero.

& θ = 45°

Time of collision

10) Using triangle law of vector addition.

(A)
(B)
(C)
(D)
11) c2 = a2 + b2 – 2ab cos θ ........... (1)
a=b ........... (2)

∴ c2 = – a2 cos2 θ
Now check for A,B,C,D

12)

13) mg – mg sin37° – ma cos37° = 0

14) For equilibrium,

10 = 8 + T …(i)
T + f2 = 20 …(ii)
⇒ f2 = 18N

15)
At the instant 3m is about to slip, tension in all the strings are as shown
∴ 3 μmg = T cos 60° …(1)
and mg = T sin 60° …(2)

∴
16)

Work is done against the gravity and spring forces.

Initially the system was in equilibrium

Top spring is stretched and bottom spring is compressed by

Now,

After lifting upwards,

in both springs

17)

Given a = 90 cm
M = 5 kg, m = 4 kg

Let use assume ring is fallen y below its initial.

position and block has raised by distance h. p.ε lost by ring = p.ε gain by block.
mgy = Mgh ... (i)
Suppose length of string between peg and the ring is L.

By (i) mg y = Mgh

h= , h = ky, , {k = 0.8}
So, L = a2 + y2
2
.... (a)
 h=y=L–a
h=L–a
h=L–a
L2 = (h + a)2 .... (b)
By (a) & (b)
a2 + y2 = h2 + 2ah + a2
y2 = k2y2 + 2aky

y2 =

y= =

y = 4m

PART-B-CHEMISTRY

18)

19)

B.A. ⇒ H2O > OF2 > PH3 > H2S

⇒ Drago’s Molecule 0 ≈ 90°
B.A. ∝ of central atom

20)
% loss in weight = 55

n = 9.67 ≈ 10

21)
= 3.2/RH

22)

The magnetic moment of 26Xn+ is BM

Electronic configuration of 26 x → [AR] 4s23d6
Formula of magnetic moment = BM
So, equate the given magnetic moment to the above formula and find 'n'

24 = n(n + 2)
24 = n2 + 2n
n2 + 2n - 24 = 0
(n + 6) (n - 4) = 0
n = -6, n = 4
n = 4 (Since, the number of unpaired electrons cannot be negative)
So, to above 4 unpaired electrons the electronic configuration of 26 Xn+ must be 3d6, and the
oxidation number of
x=2
n=+2
Ans.-(A)

23) CaO + CO2 → CaCO3

CaO is L.R.
1 mole CaO → 1 mole CaCO3
0.1 → 0.1 mole CaCO3
Mass = Mole × M. wt.
= 0.1 × 100 = 10 gm

24)

(A),(B) are general fact of periodic table.

(C) Cl⊝ < S2⊝ size
(D) Atomic size : Cl > F

25)

For regular octahedral geometry there should be 6σ bonds and no l.p. on central atom

26)

C2 = 12 electron
 All electron are paired.

27)

Total spin = Number of unpaired electrons × =

Spin multiplicity =

28)

Glucose molecular mass = 180

CH3COOH molecular mass = 60

∴ Same mass ⇒ a moles of CH3COOH & moles of glucose.

Hence each atom will be same.

29)

1s does not have any node, graph A is correct.

3p will have one radial node, graph B is correct.
3s will have two radial node, graph D is wrong.

30)
M = 64
64 = 32 + 16x
x=2

31)

Total energy per sec. = 50 J

n = 1998.49 × 1017 [ n = no. of photons per second]

= 1.998 × 1020

2 × 1020
= x × 1020
=2

32)

BeCl2 Linear sp hybridisation

sp hybridisation
sp linear
 sp2 Bent

sp3

linear sp3d

linear sp3d

linear

sp3d

33)

Both compounds BeO and Be(OH)2 are amphoteric in nature.

and both compounds BaO and Sr(OH)2 are basic in nature.
 34) CN–, NO+ and O22+. have bond order of 3.
O2 has bond order of 2.
O2+ has bond order of 2.5.
Therefore, 3 species have similar bond order.

PART-C-MATHEMATICS

35)

36) Now,
= (logex)2
∴ The equations is
(logex)2 = logex – (logex)2 +1
2t2 – t – 1 = 0 [where, t = logex]

⇒ t = 1, –
Loge x = 1
⇒x=e

But ⇒
Then, log2(logex) is not defined.
∴ There is only ne real solution.

37)

a1, a2, a3 ... A.P. Common diff. = d = 1

First term = a1 = a
a1 + a2 + .... + a98 = 137
(49) (2a + 97) = 137

2a =

a= ...(1)

=
38) Let α be a common root, then
α2 – α – 12 = 0, kα2 + 10α + 3 = 0.

∴
⇒ (12k + 3)2 = 117(10 + k)
⇒ (k – 3) (16k + 43) = 0

⇒ k = 3 or

39)
L.H.S ≤ 2. & R.H.S. ≥ 2
Hence L.H.S = 2 & R.H.S = 2

4x + 4-x = 2
Check x = 0 Possible hence only one solution.

40)

41)

Simplifying:

Roots are : α, –α

Product = pq – pr – qr
= pq – r (p + a)

42) cos 2x . sin 6x = cos 3x . sin 5x

2 sin 6x . cos 2x = 2 sin 5x . cos 3x
sin 8x + sin 4x = sin 8x + sin 2x
sin 4x – sin 2x = 0

x = nπ or

43)

4cos2x – 8cosx + 3 0
(2 cosx–3) (2 cosx –1) 0
⇒ 2cosx – 1 0 or 2 cosx – 3 < 0

44)

45)
P=9;q=4

46) ∵ a + 3b + 9c > 0

and f(x) has no real roots

⇒ f(x) > 0 ∀ x ∈ R, a > 0

f(1) > 0, f(0) > 0,

47)

1 – sin2x + sinx =
⇒ 9sin2x – 9sinx + 2 = 0

⇒
So, there are four principal solutions
in .

48) ..... (i)

Dividing (i) and (ii)

49) a, ar, ar2 → G.P.

Sum of any two sides > third side
a + ar > ar2, a + ar2 > ar, ar + ar2 > a
r2 – r – 1 < 0

...(1)
2
r –r+1>0
always true
r2 + r – 1 > 0

…(2)
Taking intersection of (1) , (2)

As r > 1

[r] = 1 [–r] = –2
3[r] + [–r] = 1.

50)

α–α+β=5⇒β=5
125 – 125 – 5a + 45 = 0 ⇒ a = 9

51)

Let
⇒ x2(y – 1) – 4x(y – 1) + 1 = 0
Case I : y – 1 = 0
⇒ y = 1 (not possible)
Case II : y – 1 ≠ 0
D>0
16(y – 1)2 – 4(y – 1) > 0
(y – 1)(4y – 5) > 0