28-05-2025

2001CJA101001250005 JA

PHYSICS

SECTION-I(i)

1) The relation between time and distance is , where and are constants. The
retardation is

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

2) A particle is thrown vertically upwards from the surface of the earth. Let TP be the time taken by
the particle to travel from a point P above the earth to its highest point and back to the point P.
Similarly, let TQ be the time taken by the particle to travel from another point Q above the earth to
its highest point and back to the same point Q. If the distance between the points P and Q is H, the
expression for acceleration due to gravity in terms of TP, TQ and H, is :

(A)

(B)

(C)

(D)

3) A large number of particles are moving each with speed v having directions of motion randomly
distributed. What is the average relative velocity between any two particles averaged over all the
pairs?

(A) v

(B)

(C)

(D) Zero
4) A ball is projected as shown in figure. The ball will return to point :

(A) O
(B) left to point O
(C) right to point O
(D) none of these

SECTION-I(ii)

1) Select CORRECT statement(s) for three vectors and .

(A) The above vectors can form triangle

(B) Component of along is 3

(C)
makes angle with y-axis
A vector having magnitude twice the vector and anti parallel to vector is
(D)

2) A particle moves along a straight line such that its position x changes with time t as
where a, b and c are constants. Then choose CORRECT alternative(s)

The acceleration varies as

(A)

(B) The acceleration will be zero if

(C) The acceleration will be zero if
(D) The acceleration can not be positive or negative.

3) A particle moving in a straight line has acceleration given by where k is a positive

constant and v is instantaneous velocity. Let v0 be the initial velocity, t be the total time of motion
and va be the average velocity in time t. Let s be total distance travelled. Choose the correct options.

(A)

(B)

(C)
(D)

SECTION-II(i)

Common Content for Question No. 1 to 2

A policeman is running behind a thief with constant velocity 20 m/s. Velocity of thief is 10 m/s.
Initially separation between police & thief is 1km. Police has a dog which runs at constant speed of
50 m/s. Dog runs from the police to the thief touches it and returns back to the police and then again
runs to the thief and returns to police & goes on repeating the trips till the police catches the thief.
Consider motion of all three along a straight line.

1) Total distance covered by the dog in that part of the trips when it was moving towards the thief.

2) Total distance covered by the dog in the time duration in which police catches the thief.

Common Content for Question No. 3 to 4

A moving particle is acted upon by three forces at different times to bring it to rest. Its velocity

versus time graph is given below

3) The average velocity for the first 12s is

4) The magnitude of average acceleration from t = 5s to t = 15s is

SECTION-II(ii)

1) A bird moves from point (1, –2) to (4, 2). If the speed of the bird is 10 m/s, then the velocity vector
of the bird is m/s then find product of a and b.

2) A ball is dropped from top of a building. If the ball crosses a window of height 45 m in 1 second.
Find the distance between top of building and upper end of window.
(g = 10 m/s2)
3) Two stones A and B are projected simultaneously as shown in figure. It has been observed that
both the stones reach the ground at the same place after 7 sec of their projection. Determine
difference in their vertical components of initial velocities in m/s. (g = 9.8 m/s2)

4) A ball is thrown from the ground to clear a wall 3 m high at a distance of 6 m and falls 18 m away

from the wall, the angle of projection of ball with horizontal is tan–1 . Find x.

5) In the figure shown, the two projectile are fired simultaneously. Find the minimum distance

between them during their flight.

6) Two bodies fall freely from different heights and reach the ground simultaneously. The time of
descent for the first body is t1 = 2s and for the second, t2 = 1s. At what height (in m) was the first
body situated when the other began to fall ? (Take g = 10 m/s2)

CHEMISTRY

SECTION-I(i)

1) A mixture of O2 and gas "Y" (mol. wt. 80) in the mole ratio a: b has a average molecular weight 40.
What would be average molecular weight, if the gases are mixed in the ratio b : a under identical
conditions ? (gases are non-reacting):

(A) 40
(B) 48
(C) 62
(D) 72

2) A sample of iron ore, weighing 0.700g, is dissolved in nitric acid. The solution is then diluted with
water, following with sufficient concentrated aqueous ammonia, to quantitative precipitation of iron
as Fe(OH)3. The precipitate is filtered, ignited and weighed as Fe2O3. If the mass of the ignited and
dried precipitate is 0.541 g, what is the mass percent of iron in the original iron ore sample (Fe =
56)

(A) 27.0 %
(B) 48.1 %
(C) 54.1%
(D) 81.1%

3) On subjecting 10 ml gaseous mixture of N2 and CO to repeated electric spark, 7 ml of O2 was

required for combustion. What was the mole percentage of CO in the mixture under identical
conditions? (Under the given conditions N2 on combustion forms NO)

(A) 45
(B) 70
(C) 40
(D) 60

4) Photoelectric emission is observed from a surface of metal when lights of frequency n1 and
n2 incident. If the ratio of maximum kinetic energy in two cases is K : 1 then (Assume n1 > n2)
threshold frequency is

(A) (K – 1) × (Kn2 – n1)

(B)

(C)

(D)

SECTION-I(ii)

1) (i) K4[Fe(CN)6] + 3H2SO4 2K2SO4 + FeSO4 + 6HCN (ii) 6HCN + 12H2O 6HCOOH +
6NH3
(iii) (a) 6NH3 + 3H2SO4 3(NH4)2SO4
(b) 6HCOOH 6CO + 6H2O
Above steps of reactions occur in a container starting with one mole of K4[Fe(CN)6], 5 mole of H2SO4
and enough water. Among reactions (a) and (b), (a) starts only after completion of (b).

st
(A) Limiting reagent in I reaction is H2SO4
st
(B) Limiting reagent in I reaction is K4[Fe(CN)6]
(C) 6 moles of CO, 2 moles of (NH4)2SO4 will be formed as final product.
(D) 6 moles of CO, 3 moles of (NH4)2SO4 will be formed as final product.

2) 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 (approx.) if 4 gm of NaOH is taken.

(B) 41.25 gm (NH4)2SO4 is required to produce 10.7 gm (approx.) NH4Cl.
(C) 25 gm NaOH is required to produce 10.7 gm (approx.) 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.

3) An α-particle having kinetic energy 4.0 MeV is projected towards tin nucleus (Z = 50). Select the
correct information(s) regarding the α-particle.
(e = –1.6×10–19 coul, k = 9×109 Nm2/coul2)

(A) Its distance of closest approach towards the nucleus is 3.6 × 10−14 m (approx.)
(B) Its potential energy at a distance of 9.0 × 10−14 m from the nucleus is 1.6 MeV (approx.)
(C) Its kinetic energy at a distance of 4.5 × 10−14 m from the nucleus is 0.8 MeV (approx.)
(D) At a moment, the distance between α-particle and the nucleus becomes 2.0 × 10−16 m (approx.)

SECTION-II(i)

Common Content for Question No. 1 to 2

81 gm mixture of MgCO3(s) and NH2COONH4(s) is heated to constant mass. If Mole % of MgCO3 in
the mixture is 50% then
Given that :
MgCO3(s) MgO(s) + CO2(g)
NH2COONH4(s) 2NH3(g) + CO2(g)

1) Mass of CO2 formed in gram is

2) Vapour density of gaseous mixture obtained on complete decompositon of the given mixture will
be

Common Content for Question No. 3 to 4

24 gm pure sample of magnesium is burned in air to form magnesium oxide and magnesium nitride.
When products are treated with excess of H2O, 3.4 gm of gaseous NH3 is generated according to
given reactions.
Mg + O2 → MgO
Mg + N2 → Mg3N2
Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3

3) Calculate the amount of Mg(OH)2 (in gm) produced in above reaction i.e., from Mg3N2.

4) Calculate the amount of magnesium oxide (in gm) in products when Mg is burned.

SECTION-II(ii)
1) Nitric acid can be produced in three steps process :
(i) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
(ii) 2NO(g) + O2(g) → 2NO2(g)
(iii) 3NO2(g) + H2O(ℓ) → 2HNO3(aq.) + NO(g)
% yield of Ist, IInd and IIIrd reaction are respectively 50%, 60% and 80%, and the volume of NH3(g) at
1 atm and 0°C required to produce 1575g of HNO3 is v (in litre). Then find v/100 ?

2) A 19.6 g of a given mixture of gaseous sample contains 2.8 g of A3 molecules, whose density is
0.75 g/litre ; 11.2 g of B2 molecules, whose density is 3g/litre and 5.6 g of C molecule, whose density
is 1.5 g/litre. All the density measurements are made at 0°C and 1 atm. Calculate the total number of
atoms (N) present in the given sample. Report your answer in . Assume, Avogadro's
number as

3) The 'roasting' of 100.0 g of a copper ore yielded 72 g pure copper. If the ore is composed of Cu2S
and CuS with 4 % inert impurity, calculate the percentage of Cu2S (by mass) in the ore. The
reactions are :
Cu2S + O2 2Cu + SO2 and CuS + O2 Cu + SO2
(Molar mass of Cu is 64)

4) A sample of carbon dioxide contains 12C and 14C in 3 : 1 mole ratio and 16O and 18O in 4 : 1 mole
ratio. How many moles of neutrons are present in 45.3 gm sample? Write your answer to the nearest
integer.

5) Similar to the % labelling of oleum, a mixture of H3PO4 and P4O10 is labelled as (100 + x) % where
x is the maximum mass of water which can react with P4O10 present in 100 gm mixture of H3PO4 and
P4O10. If such a mixture is labelled as 127 %. Find % by mass of P4O10 in the mixture.

6) 500 ml of '34.05 V' H2O2 solution is left open due to which some H2O2 decomposes, causes
evolution of 8 gram of O2 gas. The new volume strength of left H2O2 solution is ___.
[Volume of solution remains constant, and volume strength is defined at 1 bar and 0°C]
Express your answer to the nearest integer.

MATHEMATICS

SECTION-I(i)

1) The number of solution(s) of the equation

(A) 1
(B) 2
(C) 3
(D) 0
2) If complete solution set of the inequality (x2 + x – 2)(x2 + x – 16) + 40 ≥ 0 is x ∈ (–∞, –4] ∪ [a, b] ∪
[c, ∞) then value of a + b + c is equal to

(A) 4
(B) 3
(C) 2
(D) 7

3) Which of the following is a null set?

(A) A = {x : x > 1 and x < 1]

(B) B = {x : x + 3 = 3}
(C) C = {ϕ}
(D) D = {x : x ≥ 1 and x ≤ 1}

4) The value of , is

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

SECTION-I(ii)

1) In the equation 2ℓn(2x – 5) = ℓn2 + ℓn , the value of x can be -

(A)

(B)

(C)

(D)

2) If 2576a456b is divisible by 15, then possible value of (a – b) are

(a, b ∈ {1,2,3,....9} )

(A) 4
(B) 7
(C) 13
(D) 1
3) The product of the roots of the equation , is :

(A)

(B)

(C)

(D)

SECTION-II(i)

Common Content for Question No. 1 to 2

logaN = α1 + β1, logbN = α2 + β2, logcN = α3 + β3, where α1, α2, α3 are integers and β1, β2, β3 ∈ [0,1).

1) If for a = 3, c = 7, α1 = 5, α2 = 3, α3 = 2 and b is such that the largest integral value of N is 342,

then sum of integral values of b is (a < b < c)

2) The number of integral values of N if α2 = 3, α3 = 2 and b = 5, c = 7 is -

Common Content for Question No. 3 to 4

If x, y satisfy the equation , then

3) If x & y also satisfy xy = |x|, then the number of ordered pairs of (x, y), is -

4) The value of is

SECTION-II(ii)

1) The value of is equal to

2) If and kA = 9, then the value of is equal to

3) If , then value of x + y + z is (x, y, z ∈ R)

4) Number of solutions of equation is

5) Find the number of natural numbers satisfying the inequality,

(x – 8) (x – 10) (x – 12) ........... (x – 56) < 0 ?

6) If , then the value of is

 ANSWER KEYS

PHYSICS

SECTION-I(i)

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

SECTION-I(ii)

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

SECTION-II(i)

Q. 8 9 10 11
A. 3.5 5.00 5.00 0.5

SECTION-II(ii)

Q. 12 13 14 15 16 17
A. 48 80 7 3 10 15

CHEMISTRY

SECTION-I(i)

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

SECTION-I(ii)

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

SECTION-II(i)

Q. 25 26 27 28
A. 44.00 15.25 17.35 to 17.45 28.00

SECTION-II(ii)

Q. 29 30 31 32 33 34
A. 35 6 60 23 71 23

MATHEMATICS
 SECTION-I(i)

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

SECTION-I(ii)

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

SECTION-II(i)

Q. 42 43 44 45
A. 11.00 218.00 2.00 0.5

SECTION-II(ii)

Q. 46 47 48 49 50 51
A. 2 5 7 7 14 1
 SOLUTIONS

PHYSICS

1)

2)
Time taken from point P to point P

Time taken from point Q to point Q

and

3) Relative velocity, where v1 = v2 = v

If angle between them be q, then

Hence, average relative velocity

4) Here
⇒ Initial velocity & acceleration are opposite to each other.
⇒ Ball will return to point O.

5)

7)

⇒ ; VA = =

8) Time in which police catches thief,

Total distance covered by the dog

= 50 ×100
= 5000 m
= 5 km.
Now, suppose x distance is covered by dog in a trip when it is moving towards the thief and y
when moving towards the police.
Then
Σx + Σy = 5000
Σx - Σy = 2000
2Σx = 7000
Σx = 3500m = 3.5 km
and Σy = 1.5 km

9) Time in which police catches thief,

Total distance covered by the dog

= 50 ×100
= 5000 m
= 5 km.
Now, suppose x distance is covered by dog in a trip when it is moving towards the thief and y
when moving towards the police.
Then
Σx + Σy = 5000
Σx - Σy = 2000
2Σx = 7000
Σx = 3500m = 3.5 km
and Σy = 1.5 km

10) Average velocity for first

11) Average acceleration from t = 5 s to t = 15 s

12)

13)

Sn = 45 = 0 + (2n - 1)
9 = 2n - 1 ; n = 5

d= × 10 × 4 × 4 = 80

14) In time of flight i.e. 7 s, the vertical displacement of A is zero and that of B is 49 m so for
relative motion of B w.r.t. A
(u2sinθ2 – u1sinθ1) × 7 = 49 ⇒ u2sinθ2 – u1sinθ1 = 7 m/s
 15)
From equation of trajectory

16)
Taking origin at A and x axis along AB
Velocity of A w.r.t. B

So

CHEMISTRY

19)

Fe present in 0.541 gm Fe2O3 = 0.3787 gm Fe

20) Lest the mixture has ‘x’ mL CO and (10 – x) mL N2

Total O2 consumed = + 10 – x = 7

– = – 3 x = 6 mL

At similar conditions of ‘T’ and ‘P’ mole % same as volume %] Mole % = × 100 = 60%

22)

23)
⇒ = 1.6 times

24)

(a) (K.E.)initial = (P.E.)at distance of closest approach

or 4.0 MeV = K.

or, 4 × 106 × 1.6 × 10–19

= 9 × 109 ×

∴ Distance of closest approach, r = 3.6 × 10–14 m

(b) P.E. = K.

= 9 × 109 ×

= 1025 × (1.6 × 10–19)2 J

= = 1.6 MeV

(c) P.E. = K .

=
= 3.2 MeV

∴ K.E. of α-particle at this distance

= 4.0 –3.2 = 0.8 MeV

27)

2Mg + O2 → 2MgO
3Mg + N2 → Mg3N2
Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3

mole
 mole
2 moles of NH3 produced form 3 mole of Mg
0.2 moles of NH3 produced form 0.3 mole of Mg
0.3 mg produced 0.1 mole of Mg3N2
Remaining mole 1 – 0.3 = 0.7 mole Mg converted to 0.7 mole MgO = 0.7 × 40 = 28 g MgO
Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3

× 0.2 = 0.1 0.3 mol 0.2 mol

= 0.3 × 58
= 17.4 gm

28)

2Mg + O2 → 2MgO
3Mg + N2 → Mg3N2
Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3

mole

mole
(i) 2 moles of NH3 is produced from 3 mole of Mg
0.2 moles of NH3 is produced from 0.3 mole of Mg
(ii) 0.3 mole Mg produced = 0.1 mole of Mg3N2
Remaining mole of Mg = 1 – 0.3 = 0.7 mole
0.7 mole Mg converted to 0.7 mole MgO = 0.7 × 40 = 28 g MgO
Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3

× 0.2 = 0.1 0.3 mol 0.2 mol

= 0.3 × 58
= 17.4 gm

29)

Moles of NO2 required

Moles of NO required =

Moles of NH3 required =

= 156.25
Volume of NH3 at 1 atm and 0°C required is
= 156.25 × 22.4
= 3500 L

30) 2.8 g of molecule, A3 density = 0.75 g/L at NTP

vol occupied by
 No of mole of A3 at NTP

No of atoms at NTP

Similarly 11.2 g of B2, density 3g/L at NTP

No of B atoms at NTP =

and 5.6 g of C , density is 1.5 g

No of C atoms at NTP

Total number of A, B and C atoms in the mixture

31)

Mass of Cu2S & CuS = 100 – 4 = 96 gm

Let moles of Cu2S = a moles and CuS = b moles
a × 160 + b × 96 = 96
for Cu
[(a × 2) + b] × 64 = 72

a= ;b=

Mass of Cu2S =

32) Avg. molar mass of CO2

no. of moles of CO2 in sample =

Avg. number of neutrons per molecule

=
no. of moles of neutrons = 1 × 23.3 = 23.3

33)

P4O10 + 6H2O → 4H3PO4

27g

mol
 % P4O10 = 71 gm

34)

H2O2 → H2O + O2,

, moles evolved.
Moles of K2O2 decomposed = 0.5
Left moles of H2O2 = 1,

Mnew =
V.S. = '22.7 V'

MATHEMATICS

35)

(–ve) ≠ (+ve)

36)

Let x2 + x – 2 = t
t2 – 14t + 40 ≥ 0
(t – 4)(t – 10) ≥ 0
t≤4 or t ≥ 10
x2 + x – 2 ≤ 4 x2 + x – 2 ≥ 10
x2 + x – 6 ≤ 0 x2 + x – 12 ≥ 0
(x + 3)(x – 2) ≤ 0 (x + 4)(x – 3) ≥ 0
x ∈ [–3, 2] x ∈ (–∞, –4] ∪ [3, ∞)
x ∈ (–∞, –4] ∪ [–3, 2] ∪ [3, ∞)

37)

(1) A = {x ! x > 1 and x < 1}

these is no any number which is greater than as well as less than 1 so set A is an empty
set
(2) B = {0} as x + 3 = 3
⇒x=0
(3) C={ϕ} is a set having one element (ϕ)
So not an empty set
(4) D={1}

38) =
39) 2ℓn(2x – 5) = ℓn2 + ℓn

(2x – 5)2 =
Let 2x = t

(t – 5)2 =
t2 + 25 – 10t = 2t – 7
t2 – 12t + 32 = 0
t = 4, 8
2x = 4 and 2x = 5
x = 3, 2
(x = 2 rejected)

(B) =3

(C)

40)

Since the number is divisible by 5

∴ b = 0, 5
number is divisible by 3 iff a + b + 2 is divisible by 3
(i) If b = 0 then a = 1, 4, 7
(i) If b = 5 then a = 2, 5, 8

42) a = 3, α1 = 5 ⇒ log3N = 5 + β1
N ∈ [35, 36)
c = 7, α3 = 2 ⇒ log7N = 2 + β3
N ∈ [72, 73)
α2 = 3
⇒ N = [b3, b4)

N ∈ [35, 36)

N ∈ [72, 73)
N = [243, 343)
if b = 4 ⇒ N ∈ [243, 256)
⇒ rejected only b = 4, 5 and 6 can be accepted

43) b = 5, α2 = 3
⇒ N = [125, 625)
C = 7,a3 = 2
N ∈ [49,343)
N = [125,343)
Number of integral values = 218

44) ∵ x = 2y & xy = |x| ⇒ xx/2 = x

⇒ ⇒ x = 1 or 2

⇒ x=1⇒ or x = 2 ⇒ y = 1

⇒ (x, y) or (2, 1)

45)

Ans : 0.5

46) Let log23 = t

Let t2 + 3t = z

(z + 1) – z + 8 = 9
log3(9) = 2

47) A =
given kA = 9
⇒

Now

48)
log3(x + 2) – 1 = 0 ⇒ x + 2 = 3, y = –4, z = 10 ⇒ x+y+z=7

49)
(1) (log3x)2 = (log3x) + 2
Let log3x = t
t2 = t + 2
t = –1, t = 2

(2)
 |log2x| – 1 = ± 1
⇒ |log2x| = 2, 0
log2x = ±2, 0

x = 1,
(3) |log2x| – 1 = 0
log2x = ±1

x = 2,

51)