13-07-2025

1001CJA101016250015 JA

PART-1 : PHYSICS

SECTION-I

1) A force of 0.5 N is applied on the upper block as shown in figure. The coefficient of static friction
between the two blocks is 0.1 and that between the lower block and the surface is zero. The work
done by the lower block on the upper block for a displacement of 3 m of the upper block is :-

(A) 1 J
(B) –1 J
(C) 2 J
(D) –2 J

2) In the figure shown initially spring is in unstretched state & blocks are at rest. Now 100N force is
applied on block A & B as shown in figure. After some time velocity of 'A' becomes 2 m/s & that of 'B
' 4 m/s & block A displaced by amount 10 cm and spring is stretched by amount 30 cm. Then work

done by spring force on A will be :

(A) 9/3 J
(B) – 6 J
(C) 6 J
(D) None of these

3) Statement–1 : In circular motion, at any particular instant angle between velocity vector and
acceleration vector can be 0 < θ < 180.
Statement–2 : While in circular motion, speed may increase, decrease or remain constant.

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for
(B)
statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.

4) A projectile is launched from a horizontal ground with speed 5 m/s at a angle 37° with horizontal.
Radius of curvature of the projectile at its highest point of trajectory is (Take g = 10m/s2)

(A) 1m
(B) 1.8m
(C) 1.6m
(D) 2m

5) Small blocks A and B connected with a string are rotated with angular velocity ω about point O as
shown in the figure. Breaking strength of both the strings is 75N. The maximum angular velocity by

which system can be rotated on a horizontal smooth plane :-

(A) 4 rad/sec
(B) 5 rad/sec
(C) 6 rad/sec

(D)
rad/sec

6) A long horizontal rod has a bead which can slide along its length and is initially placed at a
distance L from one end A of the rod. The rod is set in angular motion with a constant angular
acceleration α about end A. If the coefficient of friction between the rod and bead is μ, and gravity is

neglected, then the time after which the bead starts slipping is :-

(A)

(B)

(C)

(D) infinitesimal

7) A block of mass m is pushed from the top of a fixed hemispherical mound of radius R as shown.
The coefficient of friction between the block and the mound is different at different values of angular
position of the block such that the horizontal speed remains constant. The block leaves the mound
when the angular position of the block with respect to vertical is 37°. What is the initial velocity of

the block? (Take : R = 0.5 m)

(A) 1.6 m/s

(B) 2.4 m/s
(C) 3.6 m/s
(D) 4.2 m/s

8) On a particle moving on a circular path with a constant speed v, light is thrown from a projectors
placed at the centre of the circular path. The shadow of the particle is formed on the wall. the

velocity of shadow up the wall is

(A) v sec2ϕ
(B) v cos2ϕ
(C) v cos ϕ
(D) none

9)

In the given figure,

List-II
List-I (values in appropriate
in SI unit)

(P) If F = 50 N, θ = 53° and (1)

then acceleration of block is

If F = 100 N, θ = 37° and µ = 1.5

(Q) (2)
then acceleration of block is

If F = 200 N, θ = 53° and µ = 1.0

(R) (3)
then acceleration of block is

If F = 110 N, θ = 90° and µ = 0.2

(S) (4)
then acceleration of block is
(A) P→2, Q→4, R→1, S→3
(B) P→4, Q→3, R→1, S→2
(C) P→2, Q→2, R→3, S→4
(D) P→3, Q→2, R→4, S→1

10) A small ball of mass m is released from rest at a height h1 above ground at time t = 0. At time t
= t0 the ball again comes to rest at a height h2 above ground. Consider the ground to be perfectly
rigid and neglect air friction. In the time interval from t = 0 to t = t0, pick up the incorrect
statement?
(A) Work done by gravity on ball is mg(h1 – h2)
(B) Work done by ground on ball for duration of contact is mgh2
(C) Average acceleration of the ball is zero
(D) Net work done on the ball by all forces except gravity is mg (h2 – h1)

11) A block of mass m is given initial speed 'u' over a wedge of mass 9m which is initially at rest. All
surfaces are smooth. Choose the incorrect statement. (Block never leaves contact with the wedge

during its upward journey).

(A)
Maximum height reached by the block is .

(B)
Final speed of the wedge (when block returns to the lowest position) is .

(C)
Final speed of the block (when block returns to the lowest position) is .
(D) Final speed of the block (when block returns to the lowest position) is towards right direction.

12) The ratio of distance carried away by the water current downstream in crossing a river by a
person making same angle with downstream and up stream are respectively as 3 : 2. Then of
ratio is {v is speed of river and u is speed of person in still water}

(A)

(B)

(C)

(D)

13) A ball is projected from a forward moving cart of sufficiently large size, with initial velocity
and angle of projection of θ relative to cart. Cart has acceleration and initial velocity . Santa
and Banta observed the trajectory from ground and cart respectively. Four trajectories are shown in
the figure and two of them are correct, in which one is observed by Santa and other by Banta. Mark
the CORRECT option-

(A) Trajectory observed by Santa is IV & Banta is I

(B) Trajectory observed by Santa is I & Banta is IV
(C) Trajectory observed by Santa is III & Banta is II
(D) Trajectory observed by Santa is II & Banta is III

14) A particle of mass m = 30 kg is released from rest at t = 0 from top of a high tower. In addition
to force of gravity drag force (air resistance) given by also acts on the particle where k = 15
–1
Nm s and Instantaneous velocity. Determine the speed of the particle after long time has
elapsed (t → ∞). (Take: g = 10 m/s2)

(A) 20 m/s
(B) 15 m/s
(C) 30 m/s
(D) 45 m/s

15) STATEMENT -1: Acceleration is Rate of change of velocity.

STATEMENT - 2: For an accelerating body its speed needs to be changing

STATEMENT -1 is True, STATEMENT- 2 is True; STATEMENT -2 is a correct explanation for

(A)
STATEMENT -1
STATEMENT -1 is True, STATEMENT- 2 is True; STATEMENT -2 is NOT a correct explanation
(B)
for STATEMENT -1
(C) STATEMENT-1 is True, STATEMENT-2 is False
(D) STATEMENT-1 is False, STATEMENT-2 is True

16) A particle is projected from origin with speed u and angle of projection θ. The minimum value of
speed of projection for which the projectile manages to pass through the point P(24, 7)?
(A)
(B) 21 m/s
(C) 27 m/s
(D)

17) A bead of mass m is located on a parabolic wire with its axis vertical and vertex directed towards
downward as in figure and whose equation is x2 = ay. If the coefficient of friction is µ, the highest

distance above the x-axis at which the particle will be in equilibrium is :

(A) μa
(B) μ2a

(C)

(D)

18) A man is coming down an incline of angle 30°. When He walks with speed m/s he has to
keep his umbrella vertical to protect himself to rain. The actual speed of rain is 5 m/s, At what angle
with vertical should he keep his umbrella when he is at rest so that he does not get drenched :-
(θ → angle from vertical)

tan θ =
(A)

(B) tan θ = 1
tan θ =
(C)

tan θ =
(D)

19) Two particles 1 and 2 move with constant velocities and . At the initial moment their radius
vectors are equal to and . How must these four vectors be inter-related for the particles to
collide?

(A)

(B)

(C)

(D) None

20) According to Maxwell-distribution law, the probability function representing the ratio of
molecules at a particular velocity to the total number of molecules is given by

Where m is the mass of the molecule,

v is the velocity of the molecule,
T is the temperature
k and k1 are constants.
The dimensional formulae of k1 is

(A) L2T–2
(B) L1T–1K–3/2
(C) L1T–1K+3/2
(D) L2T–1K+3/2

SECTION-II

1) Force acting on a particle in conservative field is given by If potential energy at

origin i.e., u(0, 0) = 0. Then determine u(1, 1) – u(3, 3).

2) A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power
0.5 W to the particle. If the initial speed (in m/s) of the particle is zero, the speed (in m/s) after 5 s is
____.

3) A rod of length L and mass M is located along the x-axis with ends at x = 0 and x = L. This rod
has a special property such that if we make a cut at any arbitrary x = a coordinate and consider the
rod between x = 0 and x = a, the centre of mass of this portion of the rod is located at a/4. If the

linear mass density λ(x) of the rod as a function of x is given as Find the value of
100n.

4) Starting from rest, a 5 kg object is acted upon by only one force as indicated in figure. Find the
total work done by the force (in J).

5) The following figure shows a disc of radius R = 20 cm with a portion of it removed symmetrically.
The removed part is a disc of radius R/2. The removed part is now placed in contact with the larger
disc as shown in figure. Disc has uniform mass distribution. With respect to origin as centre of larger

disc find x- coordinate of centre of mass of system (in cm).

PART-2 : CHEMISTRY

SECTION-I

1) Salt and water is formed by acid-base neutralisation reaction. Assume no dissociation of water in
following reactions and assume volume of final solution = 1 litre.
(Atomic mass of Cs = 133, I = 127, Rb = 85.5, Sr = 88)

List-I List-II

CsOH + HI → CsI + H2O Molarity of H+ in resulting

(P) (1)
0.2 mol 0.4 mol solution = 0.2 M.

RbOH + HNO3 → RbNO3 + H2O Molarity of cation in

(Q) (2)
0.4 mol 0.1 mol resulting solution = 0.4 M.

Sr(OH)2 + H2SO4 → SrSO4(aq) + 2H2O Molarity of cation in

(R) (3)
0.4 mol 0.4 mol resulting solution = 1.6 M.

Ba(OH)2 + 2HBr → BaBr2 + 2H2O Molarity of anion in

(S) (4)
1.6 mol 3.2 mol resulting solution = 0.4 M.
Choose the correct combination :
(A) P-3 ; Q-1,4 ; R-3 ; S-1,2
(B) P-1,2,4 ; Q-2,4 ; R-2,4 ; S-3
(C) P-1,2 ; Q-1,3 ; R-1,2 ; S-4
(D) P-1,4 ; Q-2, R-1,3 ; S-2

2) Statement -1 : Molality of pure ethanol is lesser than pure water.

Statement -2 : As density of ethanol is lesser than density of water.
 [Given : dethanol = 0.789 gm/ml; dwater = 1 gm/ml]

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for
(B)
statement-1.
(C) Statement-1 is false, statement-2 is true.
(D) Statement-1 is true, statement-2 is false.

3) A solution of glucose received from some research laboratory has been marked mole fraction x
and molality (m) at 10°C. When you will calculate its molality and mole fraction in your laboratory at
24°C you will find

(A) mole fraction (x) and molality (m)

(B) mole fraction (2x) and molality (2m)
(C) mole fraction (x/2) and molality (m/2)
(D) mole fraction (x) and (m ± dm) molality

4) 5 mole H2O2(ℓ) is placed in a container of volume 490 ml at 300 K, where it is completely

decomposed into H2O(ℓ) and O2(g). The pressure exerted by O2 gas formed is finally. (Given =
1 g/ml, R = 0.08 atm-L/mol-K)

(A) 150 atm

(B) 122.45 atm
(C) 153.75 atm
(D) 300 atm

5) A certain oxide of iron contains 2.5 gm oxygen for every 7 gm of iron. If it is regarded as a
mixture of FeO and Fe2O3 in the weight ratio of x : y ; find x : y.

(A) 9 : 20
(B) 9 : 40
(C) 9 : 10
(D) 2 : 1

6) A volume V of a gaseous hydrocarbon was exploded with an excess of oxygen. The

observed contraction was 2.5 V, and on treatment with potash, there was a further contraction of 2
V. What is the molecular formula of the hydrocarbon?

(A) C2H6
(B) C3H6
(C) C4H12
(D) C2H4

7) 50 ml of CO is mixed with 20 ml of oxygen and sparked. After reaction, the mixture is treated with
an aqueous KOH solution. Choose the incorrect option -

(A) The volume of the CO that reacts is 40 ml

(B) The volume of the CO2 formed is 40 ml
(C) The volume of the CO that remains after treatment with KOH is 20 ml.
(D) The volume of mixture obtained after reaction when treated with KOH solvent remains 10 ml

8) A hydrate of magnesium iodide has a formula MgI2.xH2O. If 1.055 g sample is heated to a constant
weight of 0.695 g. What is the value of x ?

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

9) 76 gram of a silver salt of dibasic acid on heating left a residue of 54 gram silver. Silver salt
contains Ag,C & O only and C & O in mole ratio of 1 : 2 then find the mass of CO2 gas liberated
during ignition of 76gm silver salt (Atomic mass : Ag = 108)

(A) 2.2 gm
(B) 0.22 gm
(C) 44 gm
(D) 22 gm

10) 25 ml of a solution of Na2CO3 having specific gravity of 1.25 required 28 ml of a solution of HCl
containing 109.5 gm of the acid per litre for complete neutralisation. Find the volume of 0.42 M
H2SO4 that will be completely neutralised by 125 g of Na2CO3 solution.

(A) 200 ml
(B) 400 litre
(C) 380 ml
(D) 400 ml

11) Assertion (A) : Among the two O-H bonds in H2O molecule, the energy required to break the
first O–H bond and the other O–H bond is the same.
Assertion (R) : This is because the electronic environment around oxygen is the same even after
breakage of one O–H bond.

(A) A and R both are true, and R is correct explanation of A.

(B) A and R both are true, but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A and R both are false.

12) Four orders are given according to their mentioned property.

Which of the following order is CORRECT?

(A) o-hydroxy benzaldehyde > p-hydroxy benzaldehyde : B.P.

(B) Water > Acetic acid : Molar entropy change of vaporization
(C) HF > H2O : Heat of vaporisation
(D) Fumaric acid > Maleic acid : acidic nature

13) Which of the following functional group is NOT present in the compound :-

(A) Ether
(B) Ketone
(C) Amine
(D) Anhydride

14) Which of the following are homologues?

(A) CH3CH2 – NH2 and CH3CH2–NH–CH3

(B) CH3–OH and CH3–CH2–OH
(C) CH3–CH2–OH and CH3CH2–O–CH3
(D) CH3–COOH and CH3–COOCH3

15) Examine the following structures :-

(i) ;

(ii)
Which of the following statement is correct :-

(A) (i) is tertiary alcohol while (ii) is tertiary amine

(B) (i) is primary alcohol while (ii) is primary amine
(C) (i) is tertiary alcohol while (ii) is primary amine
(D) (i) is primary alcohol while (ii) is tertiary amine

16) Which of the following order is CORRECT :-

(A) S > Se > Te > O : First Electron affinity

(B) < < < < : Hydration energy
(C) > > CO : Bond order of C–O Bond
(D) NaF > KF > RbF > CsF > LiF : Lattice energy
17) Which of the following order is INCORRECT :-

(A) H2O2 > O3 > O2 (O–O bond length)

(B) < (N–O bond order)
(C) O2F2 > H2O2 (O–O bond length)
(D) N2O > N2 (N–N bond length)

18) Select the INCORRECT match(s)?

Total number of shared oxygen in

Salt/silicate
anionic part of Salt/silicate

(a) Be3Al2(Si6O18) 6

(b) Zn3(Si2O7) 1

(c) Sodium tetrapolyphosphate 4

(d) Sodium trimetaborate 3

(A) a
(B) b
(C) c
(D) d

19) Select the CORRECT order sequence(s)

C–C bond length : CH2 = CH2 >

(A)

(B) Order of dipole moment : HF < HCl

(C) Boiling point : AsH3 > NH3
(D) Bond length (S = O) : SO2 > SO3

20) Number of statements which are correct.

(a) Graphite is elemental form of carbon
(b) White phosphorous is molecular solid
(c) Graphite is used as high temperature lubricant
(d) Hybridization of carbon in diamond and graphite is sp3,sp2 respectively
(e) C – C bond length in graphite is higher than diamond

(A) 2
(B) 3
(C) 4
(D) 5

SECTION-II
1) Kjeldahl's method was used for the estimation of nitrogen in an organic compound. The ammonia
evolved from 0.55 g of the compound neutralised 12.5 mL of 1 M H2SO4 solution. The percentage of
nitrogen in the compound is _________. (Nearest integer)

2) Molecular formula of compound is CaXBrY which contains 20% Ca and 80% Br by weight, if
molecular weight of compound is 200, then calculate (x + y).
(Atomic wt. Ca = 40, Br = 80)

3) Double bond equivalent of the following compound is

4)

How many functional groups have higher priority than (Aldehydic group) in IUPAC
nomenclature:

(i) –OH (ii) (iii) (iv) (Ketone) (v)

(vi) (vii) –NH2 (viii) –X (ix) –NO2

(x) –CN

5) From the following list, find the number of species which do not contain "X–O–X" linkage, where X
is central atom (B/S/P).
H3P3O9, H2S2O5, H2S2O8, H2SO5, Na2[B4O5(OH)4]. 8H2O, P4O10, [B2O4(OH)4]–2

PART-3 : MATHEMATICS

SECTION-I

1) Let a1, a2, ..... an be the terms of a G.P. whose common ratio is r. Let Sk denotes the sum of first k

terms of the G.P. then the value of in terms of Sm–1 and Sm is

(A) r Sm–1 Sm

(B)
(C)

(D) Sm-1 Sm

2) If k be the minimum value of expression (cos2 α – 2cos α )sec2 β + 9cosec2 β + 5sec2 β where α∈

[0, π] and β∈ , then k is

(A) 20
(B) 25
(C) 24
(D) 29

3) Assertion (A) : a + b + c + d + e = 8 and a2 + b2 + c2 + d2 + e2 = 16, where a, b, c, d are positive

real numbers, then .

Reason (R) :

(A) Both A and R are correct and R is the correct explanation of A

(B) Both A and R are correct but R is NOT the correct explanation of A
(C) A is correct but R is not correct
(D) A is not correct but R is correct

4) If then a1, a2, a3, a4 are in :-

(where a1, a2, a3, a4 are non-zero)

(A) A.P.
(B) G.P.
(C) H.P.
(D) None of these

5) Number of solution of the cos4 2α + 2 sin2 2α = 17 (cos α + sin α)8, α ∈ (0, 2π) is

(A) 10
(B) 6
(C) 4
(D) 2

6) The number of solutions of = –1 in (0, π) – is

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

7) If the value of α ∈ [0, 2π] such that the inequality sin + sin(α + x) ≥ 0 is true ∀ x ∈ R , is

equal to (where p and q are relatively prime positive integers), then the value of (p + q) is

(A) 3
(B) 5
(C) 7
(D) 9

8) If f(x) = 2 sin2 θ + 4 cos (x + θ) sin x sin θ + cos (2x + 2θ) then value of f2 (x) + f2 is

(A) 0
(B) 1
(C) –1
(D) x2

9) Match List-I with List-II and select the correct answer using the code given below the list.

List-I List-II

The number of solutions of 3[sinx + cosx] – 2[sin3x + cos3x] = 8 in x

(A) (I) 4
∈ [0, π] is

The number of solution(s) of sin4x – cos2x sinx + 2sin2x + sinx = 0

(B) (II) 5
in x ∈ [0,3π] is (are)

The number of ordered pairs which satisfy the equation

(C) (III) 0
x2 + 2xsin(xy) + 1 = 0 is (where y ∈ [0,2π])

The number of values of θ which satisfy the equation

(D) (IV) 2
sin3θ – sinθ = 4cos2θ – 2 ∀ θ ∈ [0, 2π] is
(A) (A) – II, (B) – IV, (C) – I, (D) – III
(B) (A) – III, (B) – II, (C) – IV, (D) – I
(C) (A) – I, (B) – II, (C) – III, (D) – IV
(D) (A) – III, (B) – I, (C) – IV, (D) – II

10) If f(x) = tan x + tan2 x tan 2x and g(n) = , then g(2013) – tan (22014) is equal to

(A) 0
(B) tan1
(C) –tan1
(D) 1

11) Let and then a2 +

c2 is equal to

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

12)

If in a right angled triangle ABC, and is finite, then tan(A – B) is equal to

(A)

(B)

(C)

(D)
–

13) If f(x) = x2 + ax + b where a, b ∈ R and f(f(x)) = 0 has roots 1 and 2. If k be the value of f(x) at x
= 0, then k is

(A)

(B) 2

(C)

(D) –2

14) Consider the equation x4 – ax3 – bx2 – cx – d = 0, a, b, c, d ∈ I+, a ≥ b ≥ c ≥ d then, the number of
integral solutions are

(A) 0
(B) 4
(C) 2
(D) none of these

15) Statement-I : If px2 + qx + r = 0 is quadratic equation (p, q, r ∈ R) such that its roots are α, β
and p + q + r < 0, p – q + r < 0 and r > 0, then [α] + [β] = – 1, (where [.] denotes the greatest
integer function)
Statement-II : If for any two real numbers a and b, polynomial f(x) is such that f(a) f(b) < 0, ⇒ f(x)
has at least one real root lying in (a, b).

(A) Both Statement-I and Statement-II are true

(B) Both Statement-I and Statement-II are false
(C) Statement-I is true but Statement-II is false
(D) Statement-I is false but Statement-II is true

16) Let x be then x is equal to :

(A) 10
(B) 8
(C) 4
(D) 2

17) lf (1 + 3 + 5 + ........ + p) + (1 + 3 + 5 + ........ + q) = (1 + 3 + 5 + ....... + r) where each set of

parentheses contains the sum of consecutive odd integers as shown, the smallest possible value of p
+ q + r, (where p > 6) is

(A) 12
(B) 21
(C) 45
(D) 54

18) Number of integers satisfying the inequalities

, is

(A) 5
(B) 6
(C) 7
(D) 8

19) The number of all polynomial whose coefficients are either 1 or –1 , that have only real roots are
:

(A) 6
(B) 8
(C) 10
(D) 12

20) If in triangle ABC, ∠C = 45° then value of sin2A + sin2B lies in the interval :-

(A) [0, 1]
(B) (0, 1)

(C)

(D) None of these

SECTION-II
1) If and tan α tan 2α + tan 2α tan 4α + tan 4α tan α = –k, then the value of k is :

2) Given that for a, b, c, d ∈ R, (where ac ≠ bd) if a sec(200°) – c tan(200°) = d and b sec(200°) + d

tan(200°) = c then cosec 200° then the value of p is

3) The number of solution in the interval [0, π] of the equation sin3 x⋅cos3 x = 0 is

4) Given that α, β, γ are the roots of cubic equation x3 – 3x2 + 2x + = 0. The value of α4 + β4 + γ4 is
equal to _________ .

5) If , where n ≥ 1, then is equal to _______

 ANSWER KEYS

PART-1 : PHYSICS

SECTION-I

Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A. B B A C B A A A C B D C C A C A C A B C

SECTION-II

Q. 21 22 23 24 25
A. 34 5 75 90 5

PART-2 : CHEMISTRY

SECTION-I

Q. 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
A. B B A A C A C D D D D B A B C A C C D C

SECTION-II

Q. 46 47 48 49 50
A. 64 3 11 5 4

PART-3 : MATHEMATICS

SECTION-I

Q. 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70
A. B B A C C B C B D C B D C A A D B B D C

SECTION-II

Q. 71 72 73 74 75
A. 7 2 3 9 7
 SOLUTIONS

PART-1 : PHYSICS

1) 0.5 = 3a ⇒ a = 5/30
f = 2a = 2(5/30) = 1/3N
fmax = 0.1 × 10 = 10N
∴ f = 1/3 N will act
work done by lower block
= –f(s)
= –1/3(3) = –1 Joule

2)

Wspring = 4 – 10 = – 6 J

3)

at could be in direction of velocity or in opposite direction while aC is always toward center.

4) Explanation :
A projectile is launched with a speed of 5 m/s at an angle of 370 with the horizontal. The task
is to find the radius of curvature of the projectile's path at its highest point of trajectory.

Concept :
At the highest point, the projectile's velocity is horizontal. The radius of curvature depends on
the horizontal velocity and the perpendicular acceleration (which is due to gravity).

Formula :

Brief Calculation
Step 1: Find the horizontal velocity v at the highest point.
At the highest point of the trajectory. the vertical velocity is zero, and only the horizontal
velocity remains. The horizontal velocity is:

m/s.
Step 2: Calculate the radius of curvature.
Now, using the formula for the radius of curvature:

= 1.6 m
5) Tension is more in inner string
T1 – T2 = I.ω2. 1
T2 = I.ω2. 2
∴ T1 = 3ω2 = 75
∴ ω = 5 rad/s

6)

mLω2 = µ(ma) = µmLα

⇒ ω2 = µα
By applying ω = ω0 + αt, we get ω = αt

∴ α2t2 = µα ⇒ t =

7)
N sin θ = µN cos θ
µ = tan θ

0
vx = v cos 37°
= 0.8 × 2 = 1.6 m/s

8)
y = R tan ϕ

, Vy = R sec2 ϕ(ω)

Vy = R sec2ϕ = V sec2ϕ
9)
If Fsinθ < mg then
Fsinθ + N = mg
N = mg – F sinθ
(f)max = µN

and acceleration =
if Fsinθ > mg then
N=0;f=0

10)
From the figure work done by gravity from Since initial and
final velocity of ball is zero its average acceleration will be zero
Since net work done is zero from time interval t = 0 to t = t0. Hence work done by forces
except gravity is mg (h2 – h1)

11) mu = 10 mv

v1 & v2 → Actual velocities of wedge & block respectively.

v1 + v2 = u ...(1)
mu = 9mu1 – mu2
⇒ u = 9(u – v2) – v2
⇒ u = 9u – 10v2
 (Maximum speed of the block)

(Maximum speed of the wedge)

12)

Downstream

Case II

13)

Trajectory is decided by relative acceleration.

14)
t→∞
15)

Statement-2 is not necessary for example in uniform circular motion speed is constant but
there is still acceleration (centripetal)

16)

17) For the sliding not to occur, we have tan θ ≤ µ

Since

18)
As per given condition
VR sin θ = 3

5 sin θ = 3 ⇒

∴ tan θ =

19) Position vector of particle-1 at time 't'

=
Position vector of particle-2 at time
't' →
For collision, their position vectors should become equal at some time 't'

t
So unit vectors of and should be same.

20) is unit less

is unit less

is unit less
 Hence

21)

u(x, y) = –x2 –y3

u(3, 4) = –32 – 33 = –36
Therefore, u(1, 1) – u(3, 4) = –2 – (–36) = 34.

22) As power is constant, P × t = ΔkE

2.5 = 0.1 v2
v = 5 m/s

23)

Correct answer is (75)

24) Change in velocity =

= 6 m/s

WF = ΔK.E. = (5) 62 = 90 J

25) Mass will be in the direct proportion of area.

So, if m is the mass of whole disc, then will be the mass of removed disc.

PART-2 : CHEMISTRY

26) (P) CsOH + HI → CsI + H2O

0.2 mole 0.4 mole – –
0 0.2 mole 0.2 mole
 Base is L.R., [H+] = 0.2 M [Cs+] = 0.2 M [I–] = 0.4 M
(Q) RbOH + HNO3 → RbNO3 + H2O
0.4 mole 0.1 mole – –
0.3 mole 0 0.1 mole
Acid is L.R. [OH ] = 0.3 M [Rb+] = 0.4 M [NO3–] = 0.1 M
–

(R) Sr(OH)2 + H2SO4 → SrSO4 + 2H2O

0.4 mole 0.4 mole – –
– – 0.4 mole
2+ 2–
[Sr ] = [SO4 ] = 0.4 M
(S) Ba(OH)2 + 2HBr → BaBr2 + 2H2O
1.6 mole 3.2 mole – –
– – 1.6 mole
2+ –
[Ba ] = 1.6 M [Br ] = 3.2 M

27) Molality of pure substance is depend on molecular mass.

∵m=

mα

28) Mole fraction & molality is independent of temperature.

29) Vgases = 400ml

30)
x x x

y 2y 3y
Given weight ratio
O : Fe
2.5 : 7

31)

The correct answer is option (A)

32)

The correct answer is option (C)

33) × x × 18 = (1.055 – 0.695 )
⇒ x = 8 Ans.

34) Ag2 A → Ag

A = 88

Ag2C2O4 → 2CO2
76 g

35)

Molarity of
1.68 M of moles of present in 1 litre mass of

Volume of required for neutralisation of

36) Energy require to break two OH bond successively will be different.

37) o-hydroxy benzaldehyde < p-hydroxy benzaldehyde : B.P.

HF < H2O : Heat of vaporisation
Fumaric acid < Maleic acid : acidic nature

38)

Ether is not present in the compound.

39)

(A) 1º & 2º amine

(C) Alcohol & Ether
(D) Acid & Ester

40)

3º alcohol and 1º amine.

41) will have highest hydration energy

• Lattice energy ∝

42) due to % p increases for O – F bond & % s increases for O–O bond.

43) Sodium tetrapoly phosphate

Na6P4O13

Anionic part has three oxygen shared in total.

44)

45) The correct answer is option (C)

46) Meq of H2SO4 used by NH3 = 12.5 × 1 × 2 = 25

% of N in the compound

= = 63.6
or
Meq. of H2SO4 = Meq. of NH3
12. 5 × 1 × 2 = 25 meq. of NH3
= 25 millimoles of NH3
So Millimoles of 'N' = 25
Moles of 'N' = 25 × 10–3
wt. of N = 14 × 25×10–3
 = 63.66

47) Ca →

Br →
⇒ CaBr2

48) DBE = Number of π-bond (unsaturation) + Number of cyclic ring present = 9 + 2 = 11 π

bond of cyanide group R–C≡C, aldehyde group & ketonic group are also
considered.

49)

(ii), (iii), (v), (vi) & (x)

50) H2S2O8, [B2O4 (OH)4]–2 H2SO5 have peroxy linkage H2S2O5 does not have any X – O – X
linkage.

PART-3 : MATHEMATICS

51)

52) k = cos2α – 2cosα+ 5 + sec2β +9cosec2β

k = ((cosα–1)2 + 4 × (1 + tan2 β) + 9 (1 + cot2 β)
k = ((cosα–1)2 + 4 + 9) + [[(cosα–1)2+4)tan2β + 90cos2β]
and use A.M. ≥ G.M. afte putting α = 0

kmin. = (4+ 9) + = 13 + 12 = 25
53)

64 + e2 – 16e ≤ 64 – 4e2
5e2 – 16e ≤ 0

0≤e≤

54) ....(1)

....(2)
∴ From (1) and (2),

we get
∴ a1, a2, a3, a4 are in H.P.

55) Let sin2α = t

(1 – t2)2 + 2t2 = 17(1 + t)4
8t4 + 34t3 + 51t2 + 34t + 8 = 0

Let
8x2 + 34x + 35 = 0
(2x + 5) (4x + 7) = 0

sin2α = (4 soln)

56) Let

⇒ t2 = 1 ⇒ 3t2 + 2t – 1 = 0 ⇒ t = –1,

57)

Now for x ∈ R hence (i) will be true.

if
possible k = –1

∴
58) f(x) = 2 sin2 θ + 4 cos (x + θ) sin x sinθ + cos (2x + 2θ)
= 2 sin2 θ + cos (2x + 2θ) + 2 cos (x + θ) cos (x – θ) – 2 cos2 (x + θ)
= 2 sin2 θ + 2 cos2 (x + θ) – 1 + 2 cos2 x – 2 sin2 θ – 2 cos2 (x + θ)
= cos 2x

59) (A) 3[sinx + cosx] – 2[sin3x + cos3x] = 8

3[sinx+cosx] – 2[(sinx + cosx)3
– 3sinxcosx(sinx + cosx)] = 8 ...(1)
We know
(sinx + cosx)2 = 1 + 2sinxcosx

⇒ sinxcosx =
Now put sinx + cosx = t in equation (1)

Now Do your self

(B) sin4x – cos2xsinx + 2sin2x + sinx = 0
⇒ sin4x + 2sin2x + sinx[1 – cos2x] = 0
⇒ sin4x + 2sin2x + sin3x = 0
⇒ sin2x[sin2x + 2 + sinx] = 0
⇒ sinx = 0
(C) x2 + 2xsin(xy) + 1 = 0
⇒ x2 + 1 = –2xsin(xy)

⇒ x + = –2sin(xy)
L.H.S. R.H.S
∈ (–∞, –2] ∪ [2, ∞) ∈ [–2, 2]
so equality can be established if
L.H.S. = R.H.S. = 2 or –2
If L.H.S. = R.H.S. = 2
⇒ x = 1 & –siny = 2

⇒ siny = –1 ⇒
similarly for x = –1

⇒ siny = 1 ⇒
(D) sin3θ – sinθ = 2[2cos2θ – 1]
⇒ 2cos2θsinθ = 2cos2θ
⇒ cos2θ[sinθ – 1] = 0
⇒ cos2θ = 0 or sinθ = 1

60) tan(2x–x)=
⇒tanx + tan2x tan2x = tan2x – tanx
f(x) = tan2x–tanx

g(n) =
g(n) = tan (2n+1)–tan1
g(2013)–tan(22014)= – tan1

61) (b + a cos θ)2 = 1 – a2 sin2 θ

(d + c cos θ)2 = 1 – c2 sin2 θ
(b + a cos θ)2 (d + c cos θ)2 = (–ac sin2 θ)2
(1 – a2 sin2 θ) (1 – c2 sin2 θ) = a2c2 sin4 θ
⇒ (a2 + c2) sin2 θ = 1

62)

is real

63) f(f(1)) = 0 ⇒ f(1 + a + b) = 0

f(f(2)) = 0 ⇒ f(4 + 2a + b) = 0
∴ 1 + a + b and 4 + 2a + b are roots of f(x)
∴ 3a + 2b + 5 = –a
⇒ 4a + 2b + 5 = 0 ...(1)
(a + b + 1)(2a + b + 4) = b

⇒ (a + b + 1) = b ⇒ (a + b + 1) =b
⇒ 3a + 3b + 3 = 2b
∴ f(0) = b
⇒ b = –3a – 3 ....(2)
From equation (1), we get
4a + 2(–3a – 3) + 5 = 0

⇒ –2a – 1 = 0 ⇒ a = –

∴ b = –3 –3=– , f(0) = b =

64) Suppose m is an integer root of x4 – ax3 – bx2 – cx – d = 0 as d ≠ 0

⇒m≠0
(I) m > 0
m4 – am3 – bm2 – cm = d
⇒ d = m(m3 – am2 – bm – c)
⇒d≥m
Also, m4 – am3 = bm2 + cm + d
m3(m – a) = bm2 + cm + d
⇒ m > a ⇒ contradiction
(II) m < 0
m = –n
⇒ n4 + an3 – bn2 + cn – d = 0
n4 + n2(an – b) + (cn – d) > 0
So, contradiction
Hence, equation has no integral solution

65) Given equation is px2 + qx + r = 0

Let, f(x) = px2 + qx + r
f(0) = r > 0
f(1) = p + q + r < 0
f(–1) = p – q + r < 0
Hence, one root lies in (–1, 0) and the other in (0, 1)
∴ [α] = – 1 and [β] = 0
⇒ [α] + [β] = – 1

66) Let a =
a – b = x, a3 – b3 = 20, ab = 2
x3 = (a – b)3 = a3 – b3 – 3ab(a – b) = 20 – 6x
⇒ x3 + 6x – 20 = 0
⇒ x = 2 is only one real solution.

67) We know that 1 + 3 + 5 + ......... + (2k – 1) = k2

Thus, the given equation can be written as

⇒ (p + 1)2 + (q + 1)2 = (r + 1)2
Therefore, (p + 1, q + 1, r + 1) forms a Pythagorean triplet, AS p > 6, p + 1 > 7.
The first Pythagorean triplet containing a number > 7 is (6, 8, 10).
∴ We may take p + 1 = 8, q + 1 = 6, r + 1 = 10
⇒ p + q + r = 21

68)

put log3 x – 1 = t2 log3 x = t2 + 1

2|t| – 3(1 + t2) + 4 > 0
2|t| – 3 – 3t2 + 4 > 0
3t2 – 3|t| + |t| – 1 < 0

x ∈ [3, 9) ⇒ 6 integers i.e. 3, 4, 5, 6, 7, 8

69)

If a polynomial a0xn + a1xn–1 + .......... + an is such a polynomial.

We can assume a0 = 1. Let r1, r2, .........rn are its roots
⇒ =

⇒ AM ≥ GM ⇒
a1, a2 = ±1, we must have a2 = –1, and n ≤ 3
⇒ Polynomials are
±(x – 1), ±(x + 1), ±(x2 + x – 1), ±(x2 – x – 1), ±(x3 + x2 – x –1), ±(x3 – x2 – x + 1)

70) Δ = sin2 A + sin2 (135° – A) = 1 + sin2A – sin2(45° – A)

= 1 + sin45° × sin (2A – 45°) = 1 + .sin (2A – 45°)

=Δ< and Δ >

0 < A < 135°
0 < 2A < 270°
–45° < 2A – 45° < 225°

⇒ .

71) cos (α + 2α + 4α) = cosα cos 2α cos 4α – sinα sin 2α cos 4α

– sinα cos 2α sin 4α – cosα sin 2α sin 4α
⇒ cos 7α = cosα cos2α cos4α (1 – tanα tan 2α – tan2α tan4α – tan4α tanα)
⇒ tanα tan2α + tan2α + tan 4α + tan 4α tan

α= ; where 7α = 2π

= –7

72) ..........(1)

..........(2)
a = d cos(200°) + c sin (200°) ..........(3)
b = c cos (200°) – d sin (200°) ..........(4)
∴ a2 + b2 = c2+ d2
∴ (a2 + b2 + c2 + d2) = 2(c2 + d2)
ac = dc cos (200°) + c2 sin (200°)
Also bd = cd cos(200°) – d2 sin (200°)
____________________________
ac – bd = (c2 + d2) sin (200°)
So
∴ p=2

73) Use sin3 x =

Equation becomes sin 4x = 0

x=

74) ∵ x3 – 3x2 + 2x + =0
∴ Σα = 3
Σαβ = 2

Σαβγ =
∵ Σα2 = (Σα)2 – 2 Σαβ
=9–4
=5

∴ α3 = 3α2 – 2α

∴ α4 = 3α3 – 2α2 – α = 7α2

= 35 – 20 – 6
=9

75) , find R1, R2, ..........., R20 it will be a telescopic

sum