29-06-2025

1402CJA101031250002 JA

PART-1 : PHYSICS

SECTION-I(i)

1) In the given figure the object is placed at the center of curvature of the mirror. Find the

instantaneous velocity of image :-

(A)

(B)
(C) 5
(D)

2) A ray of light falls on a transparent sphere as shown in the figure. If the final ray emerges from
the sphere parallel to the horizontal diameter, then calculate the refractive index of the sphere.

Consider that the sphere is kept in air.

(A) 1.5
(B)

(C)

(D)

3) Figure shows, a glass prism ABC (refractive index 1.5) immersed in water (refractive index 4/3). A
ray of light incident normally on face AB. If it is totally reflected at face AC, then :-
(A)

(B)

(C)
sinθ =

(D) < sin θ <

4) A particle is projected at an angle θ above the horizontal with a speed u. After some time the
direction of its velocity makes an angle ϕ above the horizontal. The speed of the particle at this
instant is

(A)

(B)

(C)

(D)

5) Shots are fired from the top of a tower and from its bottom simultaneously at angles 30° and 60°
as shown. If horizontal distance of the point of collision is at a distance 'a' from the tower then

height of tower h is :

(A)

(B)
(C) 2a

(D)

6) A square platform of side length 8 m is situated in x–z plane such that it is at 16 m from the x–axis
and 8 m from the z-axis as shown in figure. A particle is projected with velocity m/s
relative to wind from origin and at the same instant the platform starts with acceleration

m/s2. Wind is blowing with velocity . (g = 10 m/s2)

List I List II

Least possible values of v2 (in m/s) so that particle hits the

(P) (1) 4
platform or edge of platform is

Least possible value of v1 (in m/s) so that particle hits the

(Q) (2) 6
platform or edge of platform is

If t is the time (in second) after particle hits the platform then
(R) (3) 8
2t is equal to

Value of displacement with respect to ground (in m) of the

(S) particle in y–direction, when v2 has its minimum possible value is (4) 20
(till particle hits the platform or edge of platform)

Codes :
P Q R S

(A) 2 4 3 1

(B) 2 1 3 4

(C) 2 3 4 1

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

SECTION-I(ii)

1) Two particles A & B projected along different directions from the same point P on the ground with
the same velocity of 70 m/s in the same vertical plane . They hit the ground at the same point Q such
that PQ = 480 m. Then : [g = 9.8m/s2]

(A) Ratio of their times of flight is 4 : 5

(B) Ratio of their maximum heights is 9 : 16
(C) Ratio of their minimum speeds during flights is 4 : 3
(D) The bisector of the angle between their directions of projection makes 45° with horizontal

2) A projectile is projected at an angle of elevation α. After t seconds, it appears to have an angle of

elevation β as seen from the point of projection. Which of the following are correct?

(A) the y-coordinate of the position of the projectile at time t is , where v is the
velocity of projection
the x-coordinate of the position of projectile at time t is , where v is the velocity of
(B)
projection

(C)
tan β = tan α – , where v is the velocity of projection

(D)
velocity of projection is

3) A cubical box dimension L = 5/4 m starts moving with an acceleration from the
state of rest. At the same time, a stone is thrown from the origin with velocity with
respect to earth. Acceleration due to gravity . The stone just touches the roof of box
and finally falls at the diagonally opposite point. then

(A)

(B) v2 = 5

(C)

(D)

4) ABCD is plane glass cube. A nearly horizontal beam of light enters the face AB at grazing

incidence. Then,

(A) sinθ = cot α, where α is critical angle

(B) Critical angle α < 45° , for TIR at second surface
(C) Refracting index for TIR at second surface
(D) None of these
5) Figure shows a convex lens cut symmetrically into two equal halves and separated laterally by a
distance h. A point object O placed symmetrically at a distance 30 cm, from the lens halves, forms
two images separated by a distance d. A plot of d versus h is shown in figure. The focal length of the

lens is :-

(A) 22.5 cm
(B) 40 cm
(C) 45 cm
(D) 20 cm

6) In the fig. shown consider the first reflection at the plane mirror and second at the convex mirror.

AB is object.

(A) the second image is real and inverted with magnification 1/5
(B) the second image is virtual and erect with magnification 1/5
(C) the second image moves towards the convex mirror
(D) the second image moves away from the convex mirror

SECTION-II

1) A curved thick glass surface is silvered at curved face & not silvered on plane surface. Object is
placed at A as shown in figure. Considering P (pole of the silvered surface) as origin. If x - co-
ordinate of final image is 'n' cm then find n.

2) A hemispherical bowl of radius 10 cm is filled with liquid of refractive index μ = 4/3. A glass plate
of refractive index 1.5 is placed on the top of bowl. If for the observer above the plate, the shift in
position of a point P on the bottom is 3 cm find the thickness (in cm) of glass plate.

3) M1 and M2 are two concave mirrors of the same focal length 10 cm. AB & CD are their principal
axes respectively. An object is kept on the line AB at distance 15 cm from M1. The distance between
the mirrors is 20 cm. Considering two successive reflections, first on M1 and then on M2. Find the

distance (in cm) of final image from the line AB.

4) A projectile is fired from horizontal ground with speed v and projection angle θ. When the
acceleration due to gravity is g, the range of the projectile is d. If at the highest point in its
trajectory, the projectile enters a different region where the effective acceleration due to gravity

is g' = , then the new range is d' = nd. The value of n is_______.

5) A ball is thrown from ground at an angle θ with horizontal and with an initial speed u0. For the
resulting projectile motion, the magnitude of average velocity of the ball up to the point when it hits
the ground for the first time is V1. After hitting the ground, ball rebounds at the same angle θ but
with a reduced speed of u0/α. Its motion continues for a long time as shown in figure. If the
magnitude of average velocity of the ball for entire duration of motion is 0.8 V1, the value of α

is______

6) The acceleration time graph of a particle is shown in the figure. What is the velocity (in m/s) of
particle at t = 8 s if its initial velocity is 3 m/s ?

PART-2 : CHEMISTRY

SECTION-I(i)

1) Order of Covalent bond;

A. KF > KI ; LiF > KF
B. KF < KI ; LiF > KF
C. SnCl4 > SnCl2 ; CuCl > NaCl
D. LiF > KF ; CuCl < NaCl
E. KF < KI ; CuCl > NaCl

(A) C, E only
(B) B, C only
(C) B, C, E only
(D) A, B only

2) Correct order of covalent character of alkaline earth metal chloride in

(A) BeCl2 < MgCl2 < CaCl2 < SrCl2

(B) BeCl2 < CaCl2 < SrCl2 < MgCl2
(C) BeCl2 > MgCl2 > CaCl2 > SrCl2
(D) SrCl2 > BeCl2 > CaCl2 > Mg Cl2

3) The correct order of the first ionization enthalpies is:

(A) Mn < Ti < Zn < Ni

(B) Ti < Mn < Ni < Zn
(C) Zn < Ni < Mn < Ti
(D) Ti < Mn < Zn < Ni

4) When an electron makes a transition from (n + 1) to n state, the frequency of emitted radiations
(ν) is related to 'n' according to (n >>1)

(A) ν α n–3
(B) ν α n2
(C) ν α n3
(D) ν α n2/3

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) The number of radial nodes of 3s, 3p and 3d electrons are respectively–

(A) 0,1,2
(B) 2, 1, 0
(C) 1, 3, 5
(D) 3, 2, 0

SECTION-I(ii)

1) The wave functions of 3s and 3pz orbitals are given by

3s =

=
from these we can conclude

(A) Total number of nodal surface for 3pz & 3s orbitals are equal
(B) The angular nodal surface of 3pz orbital has the equation = /2
(C) The radial nodal surfaces of 3s orbital and 3pz orbitals are at equal distance from the nucleus
(D) 3s electron have greater penetrating power into the nucleus in comparison to 3p, electrons

2) In which of these options do both constituents of the pair have the same magnetic moment -

(A) Zn2+ and Cu+

(B) Co2+ and Ni2+
(C) Mn4+ and Co2+
(D) Mg2+ and Sc+

3) If the wave number of 1st line of Balmer series of H-atom is ‘x’ then:

wave number of 1st line of lyman series of the He+ ion will be
(A)
 wave number of 1st line of lyman series of the He+ ion will be
(B)

(C)
the wave length of 2nd line of lyman series of H-atom is

(D)
the wave length of 2nd line of lyman series of H-atom is

4)

Which of the following order is/are INCORRECT for lattice energy of ionic solids -

(A) NaF > MgF2 > AlF3

(B) NaF > AlF3 > MgF2
(C) LiF > NaF > KF > RbF
(D) LiF < NaF < KF < RbF

5) Select the correct orders according to their indicated property -

(A) NaHCO3 < KHCO3 < RbHCO3 < CsHCO3 (solubility in water)
(B) NaF < MgO < ScN < TiC (lattice energy)
(C) MF2 < MCl2 < MBr2 < MI2 (covalent character)
– – –
(D) Cl3 < Br3 < I3 (stability order)

6) Which of the following orders are correct?

(A) Thermal stability BeSO4 < MgSO4 < CaSO4 < SrSO4
(B) Basic nature ZnO > BeO > MgO > CaO
(C) Solubility in water LiOH > NaOH > KOH > RbOH > CsOH
(D) Melting point NaCl > KCl > RbCl > CsCl > LiCl

SECTION-II

1) When a metal is exposed with light of wavelength λ, the maximum kinetic energy of electron
produced was found to be 2 eV. When the same metal was exposed with light of wavelength λ/2 , the
maximum kinetic energy of elelctron produced was 6 eV. What is the value of work function of metal
in eV.

2) An orbital having double dumbbell shape of an unielectronic specie is represented as

0
where and a represents Bohr’s radius. Calculate sum of angular & spherical nodes.

3) In a sample of H-atom electrons make transition from 5th excited state to ground state, producing
all possible types of photons, then number of lines in infrared region are

4) For 3s orbital of hydrogen atom, the normalised wave function is given by

The above mentioned orbital (3s) has two radial nodes at distances of '2a0' and 'xa0' from the
nucleus. Find the value of x / 4. [Given : ]

5) The Schrodinger wave equation for hydrogen atom is : ,

0 0
where a is Bohr’s radius. If the radial node of 2s be at r , then the value of is :

6) Find total number of orbitals in which electron density is observed along any of the axis (x, y or z).

PART-3 : MATHEMATICS

SECTION-I(i)

1) The minimum value of the expression is

(A) 6
(B) 7
(C) 8
(D) 9

2) If then x is equal to

(A) 110
(B) 50
(C) 40
(D) 55

3) Let fn(α) = . Then the value of is equal to

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

4)

The value of is equal to

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

5) Value of (1 + tan7°) (1 + tan13°) (1 + tan17°) (1 + tan32°) (1 + tan38°) (1 + tan28°) is equal to

(A) 4
(B) 8
(C) 16
(D) 32

6) The least value of cos2 θ – 6 sin θ. cos θ + 3 sin2 θ + 2 is -

(A)
(B)
(C) 0
(D) 4

SECTION-I(ii)

1) Which of the following is/are true ?

(A)

(B) sin50° – cos40° = 0

(C) sin2 – cos2 < 0
(D) sin2 – cos2 > 0

2) If , where a and b are natural numbers then is

divisible by

(A) 2
(B) 4
(C) 3
(D) 6
3) The value of expression (a, b ∈ N & co-prime), then

(A) a = 2
(B) b = 1
(C) a = 1
(D) b = 2

4) If , then which of the following is/are

correct ?

(A)
(wherever defined)
(B) Value of N when x = 2 is 2
(C) Value of N when x = 2 is 22008
(D) Number of solutions of |N| = 1 is one

2
5) If 'P' denotes the product of the roots of equation log10(10x ) – log10(100x) = 1 and 'Q' denotes the

value of then

(A) P + Q = 181
(B) P – Q = 19
(C) P × Q = 810

(D)

6) The equation 16 + log2x64 = 3 has :

(A) one irrational solution

(B) no prime solution
(C) two real solutions
(D) one integral solution

SECTION-II

1) Number of integral values of x for which is defined is

2) In a survey of 700 students in a college, 180 were listed as drinking Limca, 275 as drinking
Mirinda and 95 were listed as both drinking Limca as well as Mirinda. Find how many students
drinking neither Limca nor Mirinda?

3) Let A = {x : x2 – 4x + 3 < 0} and B = {x : 21– x + a ≤ 0, x2 – 2 (a + 7)x + 5 ≤ 0}

If A ⊆ B, then the number of integral values of 'a' is

4) In a survey of 220 students of a higher secondary Session school, it was found that at least 125
and most 130 students studied Mathematics; at least 85 and at most 95 studied Physics; at least 75
and at most 90 students Chemistry; 30 studied both Physics and Chemistry; 50 studied both
Chemistry and Mathematics; 40 studied both Mathematics and Physics and 10 studied none of these
subjects. Let m and n respectively be the least and the most number of students who studied all the
three subjects Then m + n is equal to _____________

5) A surrey shows that 61%. , 46%. and 29% of the eat "Pizza" ,"burger" and "sandwich" respectively
25%. people eat exactly two of the three item and 3% eat none. What percentage of people eat all
the three item?

6) Value of is
 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I(i)

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

SECTION-I(ii)

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

SECTION-II

Q. 13 14 15 16 17 18
A. 14.00 1.50 1.50 0.93 to 0.97 4.00 7

PART-2 : CHEMISTRY

SECTION-I(i)

Q. 19 20 21 22 23 24
A. C C B A A B

SECTION-I(ii)

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

SECTION-II

Q. 31 32 33 34 35 36
A. 2.00 5.00 6.00 1.75 2.00 6.00

PART-3 : MATHEMATICS

SECTION-I(i)

Q. 37 38 39 40 41 42
A. D D B B B B

SECTION-I(ii)

Q. 43 44 45 46 47 48
A. A,B,D A,B,C,D A,B A,B,D A,B,D A,B,C,D
 SECTION-II

Q. 49 50 51 52 53 54
A. 4.00 340.00 4.00 45.00 7.00 2.00
 SOLUTIONS

PART-1 : PHYSICS

1)
For x-direction
(m = –1)

For y-direction

= 5 m/s

2)

In the diagram, angle of emergence is 2r = 60°

∴ r = 30°
by Snell’s Law
1 × sin 60° = n sin30°; n =

3) Since TIR occurs at surface AC, thus

1.5 sin θ = ⇒ sin θ =

Therefore

4)

u cos θ = v cos ϕ

5)

For collision velocity in x direction vx is same

....(1)
Relative velocity along y direction

Relative accn = 0
S=h
For collision timing should be same

from (1)

6)

= Velocity of particle relative to platform

Time = = 4 sec.

8 ≤ V2 × 4 – × 2 × 42 ≤ 16
6 ≤ V2 ≤ 8
16 ≤ V1 × 4 ≤ 24
4 ≤ V1 ≤ 6

Y = 25 × 4 – × 10 × 42
= 100 – 80
= 20m

7) We are given u = 70, R = 480, g = 9.8

This implies,

Now

D is correct and θ1 and θ2 will be complimentary.

8) y-coordinate is

y = v sin α t [Therefore A is correct]

x –coordinate is
x = v cos α t [Therefore B is correct]
Equation of trajectory is

⇒ [C is correct]

Rearrenging

⇒ [∴ D is correct]

9) Max. height =

Time of flight

Now

10) Snell’s law at first surface,

1 sin (90°) = [sin (a)]n .... (i)

At second surface,
n sin (90 – α)= 1 sin θ ....(ii)
From Eqs. (i) and (ii), we get
sin θ = cot α
For θ = 90°, then (90° – α) will become critical angle.
sin 90° = cot α = 1
α = 45°
For TIR at second surface

11)

Here we use the concept of magnification to locate the position (height) of images.
Case (1) Erect image (+ve magnification)

Object height = , image height = 2h

⇒
Case (2) Inverted image (–ve magnification)

Object height = , image height = h

12) Since first reflection should be at plane mirror then image of AB would be A'B' which will
be equal distance from mirror
Now image A'B' will behave like object and image will be formed by convex mirror
f = +60 cm for convex mirror
= – 60 cm (distance of object from mirror)
= – 90 cm (distance of object from mirror)
Image coordinate's from pole of convex mirror = (30, 0) (36, 0)
Image will be virtual and erect.
size of image = 6 cm
size of object = 30 cm

magnification =

13) cm

cm.
for final image (refraction through plane surface) we can write.

x = 10 cm
so distance from P is 14 cm.

14)

Apparent depth

Shift

t1 = 1.5 cm

15) Reflection through M1

v = – 30 cm
Reflection through M2

∴ Distance of image from

AB = 3 - 1.5 = 1.5 cm

16)

; ;

d′ = New range =
d1 = vcosθ°t

n = 0.95

17)
Total time taken = t1 + t2 + t3 + ..............

= + ..........

Total time =
Total displacement = v1t1 + v2t2 +...........

=
On solving

<v> =
 18) Area in a-t graph = change in velocity
vf – 3 = 4 ⇒ vf = 7 m/s

PART-2 : CHEMISTRY

19) According to Fajan's Rule,

A. KF > KI - False ; LiF > KF - True
B. KF < KI - True ; LiF > KF - True
C. SnCl4 > SnCl2 ; True; CuCl > NaCl - True
D. LiF > KF - True ; CuCl < NaCl - False
E. KF < KI - True ; CuCl > NaCl - True

20)

Covalent character order acc. to Fajan’s rule for A.E.M.

i.e. BeCl2 > MgCl2 > CaCl2 > SrCl2

21)

Ti(22) : Ar4s23d2
Mn(25) : Ar4s23d5
Ni(28) : Ar4s23d8
Zn(30) : Ar4s23d10
On moving from left to right I.E. generally increases with atomic number exceptions which are
not mentioned also Zn has maximum.
∴ Ti < Mn < Ni < Zn

22) ν = C RHZ2

ν = C RHZ2
since n >>1 2n + 12n n + 1n

ν = C RHZ2
ν α n–3

23) Magnetic moment = = B.M.

∴ No. of unpaired electron = 4.
X26 : 1s22s22p63s23p63d64s2.
To get 4 unpaired electrons, outermost configuration will be 3d6.
∴ No. of electrons lost = 2 (from 4s2).
∴ n = 2.

24) Radial node = n – ℓ – 1.

25)

3s orbital has two radial node at the values of radius given by solutions of

3pz orbital has one radial nodal surface at = 0 & one angular node at = /2
for 3 pz , at r = 0 = 0 while for
3 s at r = 0 = maximum so, 3s has greater penetrating power than 3pz orbital

26)

Spin only magnetic moment

where n = no. of unpaired e–.

27)

Use

28)

29)

30) Basic nature CaO > MgO > ZnO > BeO
Solubility LiOH < NaOH < KOH < RbOH < CsOH
LiCl has less melting point due to more covalent nature

31)

32) From Ψ(r) 3 radial nodes also the orbital is d (∴ σ2)

∴ Total nodes = n – ℓ – 1 + ℓ = n – 1 = 5
ℓ = 2 & n – ℓ – 1 ∴ n = 6.
 33) infrared lines = total lines – visible lines – UV lines = – 4 – 5 = 15 – 9 = 6.
(visible lines = 4 6→2, 5→2, 4→2, 3→2) (UV lines = 5 6→1, 5→1, 4→1, 3→1, 2→1)

34) Radial distribution

At nodes, ψ3s = 0

35)

ψ=0

36)

PART-3 : MATHEMATICS

37)

Answer is - 9

38)

x = 1 + 2 + 3 + ..... + 10

39) fn(α) = tan nα.

40)

41)

Answer is - 8
42)

43)
sin50° – cos40° = 0
sin2 – cos2 > 0

44)

45)

a = 2 and b = 1

46)

(A)
(B) if x = 2 ⇒ N = 2

(D)

(Solve)

47)
Q = 27 × 3 = 81

48) + log2x 64 = 3
⇒4 + 6 log2x 2 = 3

⇒ + =3

⇒ + =3
but log2 x = t

∴ + =3
⇒ 2 + 2t + 6t = 3t + 3t2
⇒ 3t2 – 5t – 2 = 0
⇒ 3t2 – 5t – 2 = 0
⇒ (3t + 1)(t – 2) = 0

⇒t=– , t=2

⇒ log2 x = – log2 x = 2
⇒ x = 2–1/3 x=4

= .

49)

The number of integral values of x for which is defined is 4.

50)

Final answer is 340

51) x2 – 4 x + 3 < 0 ⇒ x ∈ (1, 3)

Since g(x) = 2x–1 + a is a decreasing function and B is a subset of A
⇒ g(1) ≤ 0 ⇒ 1 + a ≤ 0 ⇒ a ≤ – 1
Also for f(x) = x2 – 2(a + 7)x + 5. f(1) ≤ 0 and f(3) ≤ 0

⇒ a ≥ – 4 and a ≥ – ⇒ a ∈ [–4, –1]

52)

The final answer is 45

53)
we know that
61x + 46x + 29n – 25x – 2y = 97 x
2g = 14x
g = 7x

54) =1+1=2