08-07-2025

6901CJA10153125MJ003SB JA

PHYSICS

SECTION-I(i)

1) The angle between vector and is :

(A)

(B)

(C)

(D)

2)

What is the dimensional formula of in the equation Where letters have

their usual meaning. Where P → pressure; V → volume; T → temperature.

(A)

(B)

(C)

(D)

3) The relation between time t and distance x is given as t = αx2 + βx where α and β are constant.
The retardation will be :

(A) 2αv3
(B) 2βv3
(C) 2αβv3
(D) 2β2v3

4) Figure shows a sine curve, as the displacement time curve of a particle executing rectilinear

motion. The velocity of particle at

(A)
(B)

(C)

(D) None of these

SECTION-I(ii)

1) If and are two vectors then the unit vectors

(A)
Perpendicular to is

(B)
Parallel to is

(C)
Perpendicular to is

(D)
Parallel to is

2) Which of the following statement(s) is/are correct ?

(A)

(B)
lies in the plane determined by
If the sum and difference of the two vectors are at right angles then the vectors are equal in
(C)
magnitude
(D) denotes the area of parallelogram formed by .

3) Which of the following is/are correct?

(A)
at x = 1
(B)

Average value of (1 + sin2x) for 0 ≤ x ≤ is

(C)

(D)

the value of y is

4)

In a new system of units, the unit of mass is 1000 kg (1 metric ton), unit of length is 1000m (1km)
and unit of time is 3600 s(1hr). Select correct statement(s).

(A) The numerical value of 500 m in this new system is 0.5.

(B) The numerical value of 7200 s in this new system is 2.
(C) The numerical value of 1/36 J in this new system is 3.6 × 10–4.
(D) The numerical value of 1/36N in this new system is 0.36.

5) The equation of stationary wave is :

Which of the following is NOT correct. Where length, t → time, x → distance, y → distance.

(A) The dimensions of nt is [L]

(B)
The dimensions of n is
(C) The dimensions of is [T]
(D) The dimensions of x is [L]

6) The velocity, acceleration and force in two systems of units are related as under

(i) V' = V (ii) a' = (xy)a (iii) F' = F

x and y are dimensionless constants. Which of the following are correct ?

(A)
Length standards of the two systems are related by L' = L

(B)
Mass standards of the two systems are related by M' = M

(C)
Time standards of the two systems are related by T' = T
(D)
Momentum standards of the two systems are related by P' = P

7) A particle is moving along a straight line. Its velocity varies as v = 6 – 2t where v is in m/s and t in
seconds. If the difference between distance covered and magnitude of displacement in first 4
seconds is 2n, then find n.

(A) 1
(B) 0
(C) 2
(D) 1/2

8) A man in a balloon rising vertically with an acceleration of 4.9 m/s2 releases a ball 2 sec after the
balloon is let go from the ground. The greatest height above the ground reached by the ball is

(A) 14.7 m
(B) 19.6 m
(C) 9.8 m
(D) 24.5 m

SECTION-II

1) A stone is dropped into a quiet lake and waves move in circles spreading out radially at the speed

of 0.5 m/s. At the instant when the radius of the circular wave is m, how fast is the enclosed area
(in m2/s) increasing?

2) A function f in terms of ‘x’ is defined as :

f(x) = x2 + 2x + 1
where as a function g in terms of y is defined as g(y) = (2y + 1)

then the value of is :

3) Volume of a gas undergoing certain process varies with temperature as shown

0 0
where T is initial temperature, V is initial volume. If apparatus can
measure temperature with least count of 0·2 K, find the percentage error in calculation of volume at

0 0
temp T = (V and T are accurately known)

4)

Water drops fall from a tap on the floor 5 m below at regular intervals of time, the first drop striking
the floor when the fifth drop begins to fall. The height (in m) of the third drop from the ground, at
the instant when the first drop strikes the ground, will be (g = 10 ms–2)

5) A stone dropped from the top of a tower travels part of height of tower during the last second
of fall. If height of tower is 5Z, find Z. (g = 10 m/s2)

6) A bird flies for 4s with a velocity |t–2|m/s in a straight line where time t is in seconds. What is the
distance covered in meters by it ?

CHEMISTRY

SECTION-I(i)

1) 30 ml of 2M HCl solution is diluted by 70 ml of H2O. Molarity of final solution is–

(A)

(B) 0.6 M

(C)

(D) 1.6 M

2) If the de-Broglie wavelength of the fourth Bohr orbit of hydrogen atom is 4Å, the circumference of
the orbit will be :-

0
(A) 4A
(B) 4 nm
0
(C) 16 A
(D) 16 nm

3) The electronic configuration of an element is 1s2 2s2 2p6 3s2 3p4. The atomic number and the group
number of the element ‘X’ which is just below the above element in the periodic table are
respectively.

(A) 24 & 6
(B) 24 & 15
(C) 34 & 16
(D) 34 & 8

4) What is the ratio of time periods (T1/T2) in second orbit of hydrogen atom to third orbit of He+ ion?

(A)
(B)

(C)

(D)

SECTION-I(ii)

1) Consider the following reactions

Choose correct option(s). Assuming all other reactant required are in sufficient amount.

(A) Produced mass of NH4Cl is 2.71 gm if 4 gm of NaOH is taken.

(B) 41.25 gm (NH4)2SO4 is required to produce 10.7 gm NH4Cl.
(C) 25 gm NaOH is required to produce 10.7 gm NH4Cl.
Produced moles of NH4Cl (in IInd reaction) are 1.6 times of produced moles of Na2SO4 (in Ist
(D)
reaction), if 4 gm of NaOH is taken.

2) Select the correct statement(s) regarding 3py orbital

(A) Total number of nodes are 2

(B) Number of maxima in curve 4πr2R2 vs r are two
(C) Qunantum no. n, and m for orbital may be 3, 1, -1 respectively
(D) The magnetic quantum number may have a positive value

3) When a mixture consisting 10 moles of CO and 15 moles of O2 is sparked and CO2 is produced
then, which of the following statement is/are correct :

(A) Ratio of number of reacted moles of CO and O2 is 2 : 1

(B) Ratio of number of moles of CO and O2 remains unreacted is 6 : 1

(C)
Mole fraction of CO2 in resultant gaseous mixture is
(D) Vapour density of resultant gaseous mixture is 19

4) Which of the following orders of atomic / Ionic radius is/are correct ?

(A) B < Ga < Aℓ

(B) Sc > Cu < Zn
(C) C < O < N
(D) Al+3 < Al+2 < Al+1

5) Which is/are correct statements about 1.7 g of NH3 :

(A) It contain 0.3 mol H – atom

(B) it contain 2.408 × 1023 atoms
(C) Mass % of hydrogen is 17.65%
(D) It contains 0.3 mol N-atom

6) Which is/are correct statement.

The difference in angular momentum associated with the electron present in consecutive orbits
(A)
of H–atom is (n–1)
Energy difference between energy levels will be changed if, P.E at infinity assigned value other
(B)
than zero.
(C) Frequency of spectral line in a H–atom is in the order of (2 → 1) < (3 → 1) < (4 → 1)
(D) On moving away from the nucleus, kinetic energy of electron decrease.

7) If n + =5, then

(A) Number of possible orbitals are 4

Maximum number of electrons that can be filled with ms = +1/2 is 6 in an atom with above
(B)
condition
(C) for iron (Z = 26) atom, number of electron with above condition must be 1 only
(D) For Germanium (Z = 32), number of electron with above condition can be 5

8) Which of the following Ist ionisation energy order is/are correct

(A) Be < B
(B) N < O
(C) Mg > Na
(D) P > S

SECTION-II

1) An electron makes a transition from third excited state to ground state in a hydrogen atom. Find
the total possible number of radiations in visible region.

2) A certain metal when illuminated alternatively by light of λ1 = 155 nm and λ2 = 310 nm emit
photoelectrons of maximum kinetic energies in the ratio 5 : 1. The work function of the metal in eV is
[Take hc = 1240 eV nm]

3) For an element the successive ionisation energy values (in eV/atom), are given below.
14.534, 29.601, 47.448, 77.472, 97.888, 552.057, 667.029
Find the number of valence shell electrons in that element.

4) Mass % of oxygen in monovalent metal carbonate is 48%. If the number of atoms of metal present
in 5mg of this metal carbonate sample is y × 1019 then value of y is (NA = 6 × 1023).
5) Average atomic mass of magnesium is 24.31 a.m.u. This magnesium is composed of 79 mole % of
24
Mg and remaining 21 mole % of 25Mg and 26Mg. Calculate mole % of 26Mg. Report your answer
after multiplying by 0.1.

6) Consider a helium (He) atom that absorbs a photon of wavelength 330 nm. The change in the
velocity (in cm s–1) of He atom after the photon absorption is ____.
(Assume : Momentum is conserved when photon is absorbed.
Use : Plank constant = 6.6 × 10–34 J s, Avogadro number = 6 × 1023 mol–1, Molar mass of He = 4 g
mol–1)

MATHEMATICS

SECTION-I(i)

1) The numbers of pairs (a, b) of real numbers, such that whenever α is a root of the equation x2 + ax
+ b = 0 , α2 – 2 is also a root of this equation, is :

(A) 6
(B) 2
(C) 4
(D) 8

2) The equation has

(A) No solution
(B) One solution
(C) Two solution
(D) More than two solutions

3) If , then the value of x is equal to -

(A)

(B)

(C)

(D)

4) If α and β are solutions of equation , then -

(A) α + β = 1
(B) α + β = 0
(C)

(D) α + β = log23

SECTION-I(ii)

1) If the quadratic equation ax2 + bx + c = 0 (a > 0) has sec2θ and cosec2θ as its roots then which of
the following must hold good ?

(A) b + c = 0
(B) b2 − 4ac ≥ 0
(C) c ≥ 4a
(D) 4a + b ≥ 0

2) If Pn = cosnx + sinnx, then the value of 2P6 – 3P4 + 1 is equal to

(A)

(B)

(C) sin(1000π)
(D) cos(13π)

3) a, b, c ∈ R, a ≠ 0 and the quadratic equation ax2 + bx + c = 0 has no real roots, then -

(A) c(a + b + c) > 0

(B) a(a + b + c) > 0
(C) b(a + b + c) > 0
(D) c(a + b + c) < 0

4) If p and q are the roots of equation x2 + px + q = 0, then the possible values of p are

(A) 1
(B) 0

(C)

(D) –2

5) Let S = cos2α + cos2β ∀ α, β ∈ R, then which of the following is not the value of cos(α + β)cos(α –
β)

(A) S – 1

(B)

(C) S2
(D) 1 – S
6) If cos(A – B) = , and tanA tanB = 2, then

(A)

(B)

(C)

(D) None of these

7)

If , t > 0, then which of the following is/are true-

(A)

(B)

(C)

(D)

8)

For quadratic equation x2 + (1 – 2a)x + a2 – a = 0, correct statement(s) is/are -

(A) One root is positive & one negative, if a ∈ (0,1)

(B) Both roots lie on (–1,1), if a ∈ (0,1)
(C) Roots are integers a∈I
(D) Difference of roots is always constant for all real values of 'a'

SECTION-II

1) If x1, x2 are roots of the equation such that 10x1 = x2. If k denotes
the sum of all possible values of λ then value of 2k is :-

2) Number of real solution(s) of equation (2x – 1)x2 +

3) The smallest integral value of α for which the inequality 1+ log5(1 + x2) < log5(αx2 + 4x + α) is
true for all x ∈ R is

4) If least common multiple of quadratic expressions x2 – ax + b and x2 – cx + 2a is x3 – 6x2 + 11x – 6

and greatest common divisor is (x – 2), then a + b – c is equal to

5) Let a = sin 10°, b = sin 50°, c = sin 70° then is equal to

6) If and Let are the roots of then

value of is
 ANSWER KEYS

PHYSICS

SECTION-I(i)

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

SECTION-I(ii)

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

SECTION-II

Q. 13 14 15 16 17 18
A. 4.00 25.00 0.00 3.75 9.00 4.00

CHEMISTRY

SECTION-I(i)

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

SECTION-I(ii)

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

SECTION-II

Q. 31 32 33 34 35 36
A. 2.00 3.00 5.00 6.00 1.00 30.00

MATHEMATICS

SECTION-I(i)

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

SECTION-I(ii)

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

Q. 49 50 51 52 53 54
A. 5.00 3.00 7.00 0.00 0.75 1600
 SOLUTIONS

PHYSICS

1)
Angle between and , θ = 60°
Angle between and

2) Solution :

3)

t = ax2 + bx
Differentiate w.r.t. time

1 = 2αxv + βv
DIfferentiate again

0 = a (2αx + β) + 2αv2

a = –2αv3

4)

5) For two vectors which are perpendicular then their Dot product becomes zero.
6) Conceptual

7) For A: (4x2 + 2x)|x = 1

= 2x + 2|x = 1 = 10

For B:

For C:

Average value of function

For D: y = mx + C

m = tan 30° =

y= x+C

0= +C

C=

8)

For any physical quantity; numerical value × unit = constant

For (A) n1u1 = n2u2

⇒ (500)

= = 0.5

For (B)
=

For (C)

=
= 3.6 × 10–4

For (D)

= = 0.36

9) The dimensions of is [T]

10) L1T–1 = [LT–1] ...(i)

L T = xy [LT–2]
1 –1–2
...(ii)

M1L1T1–2 = [MLT–2] ...(iii)

(iii)/(ii)

M' = [M]

(i)/(ii) T' = T

So, and a ~ b

11) Displacement =

Distance =
= 10 m
∴ Distance – displacement = 10 – 8 = 2

12)

v = 0 + 4.9 × 2 = 9.8 m/s

Total height = 4.9 + 9.8 = 14.7 m

13) Area A = πr2

m2/s

14)

f′(x) = 2x + 2

f′′ (x) = 2

∴ g(f′′(x)) = g (2) = 2 (2) + 1 = 5

∴ = 5 [3 – (– 2)] = 5 × 5 = 25.

15) dv =

At T =
∴ % error = 0

16)
time taken by first drop to strike the ground.
Let the time interval b/w two drops be T
4T = 1
T = 0.25 sec.
Therefore third drop falls from the tap after 0.5 sec. of the first are
When 1st drop is about to strike the ground third drop has travelled for 0.5 seconds

= 1.25 m
Height from the ground = 5 – 1.25
= 3.75 m
 17)

.....(i)

.....(ii)
Now,
V2 = V1 + gt

h = 45 m
5z = 45
z=9

18)

Turning point at 2 sec

between time t = 0 to t = 2

between t = 2 to t = 4 sec

CHEMISTRY

19) M1V = M2V2

20) n = 4 λ = 4 Å
circumference ⇒ 2πr = nλ
=4×4
 = 16 Å

21)

Se → Z = 34 and group number = 16.

22)

23)

⇒ = 1.6 times
nNaOH > NHCl always in above reaction
If % yield 100%
Then nNaOH = NHCl ⇒ mass of NH4Cl = 32 × 10–3 × 53.5 = 1.712 g

24)

n = 3, ℓ = 1
Maxima : 3 – 1 = 2
Total nodes = 3 – 1 = 2

25)
10 15
LR is CO(g)
(A) reacted moles of CO : O2
10 : 5
(B) unreacted moles of CO : O2
0 : 10
(C) Final gas mixture has 10 moles CO2(g) and 10 moles O2(g)

(D) Vapour density of resultant gas mixture =

26) C > N > O

27)

Mole of NH3 = = 0.1

Mole H atom = 0.3
Total atoms = 0.4 × 6.023 × 1023
= 2.408 × 1023 atoms

%H= × 100 = 17.65%

28) (A) The difference in angular momentum associated with the electron present in

consecutive orbital of H-atom is (According to Bohr's postulate) Hence, the option A is

incorrect.
(B) Energy difference between energy levels will remain same even if P.E. at infinity assigned
value other than zero. Hence, the option B is incorrect.
(C) Frequency of spectral line in a H-atom is in the order of (2→1) < (3→1)<(4→1). This is same
order as that of the energy. Hence, the option C is correct.
(D) The kinetic energy of an electron is inversely proportional to its radius. On moving away
from the nucleus, kinetic energy of electron decreases. Hence, the option D is correct.

29)

If n + =5,
m = +1 or -1

n=4 l=1 4p

n = 3 l = 2 3d
no. of posible orbitals = 4

30)

Answer - Option (3,4 )

Explanation - Let's analyze each option for the ionization energy order:
The species with the lowest nuclear charge and hence the lowest ionization energy is S2-, since
its electrons are least tightly bound by the nucleus.

31) 3rd excited state ⇒ n2 = 4

n2 = 4 ⇒ n1 = 1

⇒ Total of
⇒ Out of 6 radiations, 2 radiations fall in Balmer series or visible region.

32)

We know

ϕ = 3 eV.

33) IE6 value increases suddenly so

No. of valence electrons = 5

34)

let the monovalent metal be : M, the carbonate will be : M2CO3

let atomic wt. of metal = x

% O = 48 = × 100 ⇒ x = 20
Molar mass of M2CO3 = 100

Moles of metal carbonate =

Molar of metal atom = × 2 = 10–4

No. of metal atoms = 10–4 × 6.022 × 1023

35) Let mole % of 26Mg be x.

∴ = 24.31
x = 10%

Answer = 1

36)

​
MATHEMATICS

37) Consider the equation x2 + ax + b = 0

If has two roots (not necessarily real α & β)
Either α = β or α ≠ β
Case (1) If α = β, then it is repeated root. Given that α2 – 2 is also a root
So, α = α2 – 2 ⇒ (α + 1)(α – 2) = 0
⇒ α = – 1 or α = 2
When α = –1 then (a, b) = (2, 1)
α = 2 then (a, b) = (–4, 4)
Case (2) If α ≠ β Then
(I) α = α2 – 2 and β = β2 – 2
Here (α, β) = (2, –1) or (–1, 2)
Hence (a, b) = (–(α + β), αβ)
= (–1, –2)
(II) α = β2 – 2 and β = α2 – 2
Then α – β = β2 – α2 = (β – α) (β + α)
Since α ≠ β we get α + β = β2 + α2 – 4
α + β = (α + β)2 – 2αβ – 4
Thus –1 = 1 – 2 αβ – 4 which implies
αβ = –1 Therefore (a, b) = (–(α + β), αβ)
= (1, –1)
(III) α = α2 – 2 = β2 – 2 and α ≠ β
⇒α=–β
Thus α = 2, β = –2
α = – 1, β = 1
Therefore (a, b) = (0, –4) & (0, –1)
(IV) b = α2 – 2 = β2 – 2 and α ≠ β is same as (III)
Therefore we get 6 pairs of (a, b)
Which are (2, 1), (–4, 4), (–1, –2), (1, –1) (0, –4)

38)
squaring
2x – = 4x – 1
= 2x – 1 ; Again squaring
4(x – 1) = 4x2 + 1 – 4x
2

x= ⇒ Rejected

39)
(log23)xlog22 = (log32)x(log23)

= (log23)2x = (log23)1

⇒
40)
3x(2x – 3x–1) – 2x–1(2x – 3x–1) = 0
(3x – 2x–1)(2x – 3x – 1) = 0
3x = 2x–1, 2x = 3x–1
also

41) sum = product and roots are real

b+c=0 b2 − 4ac ≥ 0 A,B,C]

42) 2P6 – 3P4 + 1 = 2(cos6x + sin6x) – 3(cos4x + sin4x) + 1

= 2(1–3sin2xcos2x) – 3(1–2sin2xcos2x)+1= 0

43) ax2 + bx + c = 0 has no real roots

So the sign of ax2 + bx + c will be either positive
or negative will depend on sign of a
If a is positive then the expression is positive
If a is negative then the expression is negative
so a(a + b + c) > 0 {put x =1 in the expression}
when x = 0 then the expression has the same sign as c
so c(a + b + c) > 0{put x = 1 in the expression}

44)

p and q are the root of the equation x2 + px + q = 0

∴ p2 + p2 + q = 0 and q2 + pq + q = 0
∴ 2p2 + q = 0 and q(p+q+1)=0
∴ 2p2 + q = 0 and (q = 0 or q = – p – 1)
If q = 0 then from 2p2 + q = 0 we get p = 0
If q = – p – 1 then 2p2 – p – 1 = 0
∴ (2p+1)(p–1)=0

∴ p = 0, 1 or is possible

45) (a) S – 1 = cos2α + cos2β – 1

= cos2β – sin2α
= cos(β + α)cos(β – α)
46) cos (A – B) tan A tan B = 2
tan A . tan B = 2

⇒ cos A cos B

sinA sinB = 2 cos A cos B =

47) (t > 0)

48)

x2 + (1 – 2a)x + a2 – a = 0
x2 – 2ax + x + a2 – a = 0
(x – a)2 + (x – a) = 0
(x – a)(x – a + 1) = 0
x = a, a – 1

(A) a – 1 < 0 & a > 0 ⇒ a ∈ (0, 1)

(B) a – 1 > –1 & a < 1 ⇒ a ∈ (0, 1)
(C) if a ∈ int ⇒ a – 1 & a both are integers
(D) difference of roots 1.

49)

∴ y2 – (λ + 1)y + λ = 0

{10x1 = x2}
∴ y2 = 2y1
Now one root is double the other
2(b2) = 9ac (condition)
2(λ + 1)2 = 9λ
2λ2 – 5λ + 2 = 0

Sum of possible values of λ is

Hence

50) Given equation is

put x2 –1 = t
∴ (2x – 1)t + 2x(2t – 1) = 0
it is possible only when x = 0 or t = 0
∴ x = 1, –1, 0.

51) We have (5 – α)x2 – 4x + (5 – α) < 0 ; ∀ x ∈ R

∴ 5 – α < 0 Disc < 0
α > 5 & 16 ≤ 4(α – 5)2

52) L.C.M = (x – 1) (x – 2) (x – 3) given

H.C.F. = (x – 2) given
so the quadratic expressions are
(x – 1) (x – 2) & (x – 2) (x – 3)
⇒ x2 – 3x + 2 & x2 – 5x + 6
⇒ x2 – ax + b & x2 – cx + 2a
⇒ a = 3, b = 2, c = 5

53) abc = sin10° . sin50° . sin70°

= sin10° . sin(60° – 10°) . sin(60° + 10°)

= sin(3 × 10°) =

So,

54)