20-10-2024

2001CJA101001240023 JA

PHYSICS

SECTION-I(i)

1) A particle is acted upon by a force of constant magnitude which is always perpendicular to the
velocity of the particle. The motion of the particle takes place in a plane. It follows that :–

(A) its velocity is constant

(B) its acceleration is constant
(C) its kinetic energy is constant
(D) it moves in a circular path

2) A simple pendulum of length L and mass (bob) M is oscillating in a plane about a vertical line
between angular limits –ϕ and ϕ. For an angular displacement θ, [|θ|<ϕ] the tension in the string and
velocity of the bob are T and v respectively. The following relations hold good under the above
conditions

(A)

(B)

(C) Tangential acceleration =

(D) T=Mg cosθ

3) The potential energy of a particle of mass 1 kg, free to move along x–axis under action of a
conservative force only, is given by U = 4x3 – 5x2, where x is in meters.

(A) The particle is at stable equilibrium at x = 0

(B) The particle is at unstable equilibrium at x = 5/6
If kinetic energy of particle at x = 5/6 m is 1J, the particle oscillates about its equilibrium
(C)
position
(D) The magnitude of acceleration of the particle at x = 1m is 2m/s2

4) A bullet of mass m = 1kg strikes a block of mass M =2kg kept on smooth surface and connected
to a light spring of stiffness
k =3N/m with a speed V0 = 3m/s. If the bullet gets embedded in the block then.

linear momentum of bullet and block system is not conserved during impact because spring
(A)
force is impulsive.
linear momentum of bullet and block system is conserved during impact because spring force is
(B)
non impulsive.
(C) Maximum compression in the spring is 2m.
(D) The maximum compression in the spring is 1m.

5) In the arrangement shown, horizontal surface is smooth, but friction is present between the block
and the surface of the wedge. Block is given velocity v0 at t = 0. After achieving height ‘h’ on the
wedge, block comes to rest with respect to wedge at t = t0. Then from t = 0 to t = t0 :-

(A) Work done by friction on the block is negative

(B) Work done by friction on the wedge is negative
(C) Work done by block on the wedge is positive
(D) Work done by wedge on the block is positive

6) In the figure shown a uniform rod of mass m and length l is hinged. The rod is released when the

rod makes angle θ = 60° with the vertical.

The angular acceleration of the rod just after release is

(A)

The normal reaction due to the hinge just after release is

(B)

The angular velocity of the rod at the instant it becomes vertical is

(C)

(D)
The normal reaction due to the hinge at the instant the rod becomes vertical is mg

SECTION-I(ii)

1) In all the situations shown in List I, the ideal spring is relaxed initially and all surfaces in contact
with the block are smooth. Match entries in List II for elongation of spring at equilibrium position of
the blocks. Blocks in all the cases start from rest. [g = 10 m/s2]

List - I List - I
 (I) (P) 2 cm

(II) (Q) 5 cm

(III) (R) 3 cm

(IV) (S) 1 cm

(A) I → R;II → P;III → R;IV → S

(B) I → S;II → P;III → Q;IV → R
(C) I → R;II → S;III → Q;IV → P
(D) I → P;II → Q;III → S;IV → R

2) Consider a body rolling on a horizontal surface as shown in figure. (Symbols have their usual

meaning)

List-I List-II

(I) Velocity is zero (P) Point A

 (II) Speed is maximum (Q) Point B

(III) 0 < speed < vcm (R) Point C

(IV) 1.3vcm < speed < 2vcm (S) Point E

(A) I → R;II → P;III → S;IV → Q

(B) I → R;II → S;III → Q;IV → P
(C) I → Q;II → S;III → P;IV → R
(D) I → R;II → P;III → Q;IV → S

3) A particle falls freely near the surface of the earth. Consider a fixed point O (not vertically below
the particle) on the ground. Consider the physical quantities in List-I of the particle about O. Match
them with their values mentioned in List-II.

List-I List-II

(I) Angular momentum of particle about O (P) Increasing

(II) Torque of gravitational force on particle about O (Q) Decreasing

(III) Moment of inertia of particle about O (R) Constant

(IV) Angular velocity of particle about O (S) First increasing and then constant
(A) I → Q;II → S;III → P;IV → R
(B) I → R;II → P;III → S;IV → Q
(C) I → P;II → R;III → Q;IV → P
(D) I → S;II → Q;III → P;IV → R

4) Body X may be a solid cylinder, solid or hollow sphere (having mass m and radius R). It is placed
on a horizontal smooth platform P, in a carriage moving with acceleration "a" horizontal leftwards. If
platform P is gently removed, body X rolls down without sliding the rough vertical wall of the

carriage.

List-I List-II

(I) X is hollow cylinder (P)

Downward acceleration w.r.t. carriage

(II) X is solid cylinder (Q)

Downward acceleration w.r.t. carriage

(III) X is hollow sphere (R)

Static friction
 (IV) X is solid sphere (S)
Static friction

(T)
Upward acceleration w.r.t. carriage
(A) I → Q;II → R;III → S;IV → P
(B) I → S;II → P;III → R;IV → Q
(C) I → Q;II → S;III → P;IV → R
(D) I → R;II → S;III → P;IV → Q

SECTION-II

1) An escalator is used to move 5 people (60 kg each) per minute from the first floor of a department
store to the second floor, 20 m above. Neglecting friction calculate the power (in kW) required.

2) In the ideal atwood machine arrangement shown, what is the change in kinetic energy (in Joule)

of the system in second of their motion starting from rest ?

3) In the shown arrangement all surfaces are smooth and system is released from rest. Mass of block
A and B are equal (m = 1 kg), mass of block C is M = 3 kg. After releasing from rest block B strikes
with the surface of C after travelling height h = 10 cm. Find the magnitude of displacement (in cm)

of the block C, just before, B strikes C.

4) A U shaped tube of mass 2m is placed on a horizontal surface. Two identical spheres each of
diameter d (just less than the inner diameter of tube) and mass m enter into the tube with velocity v
= 20 m/s as shown in the figure taking all collisions to be elastic and all surfaces smooth.
 Find the speed of the tube (in m/s) when the spheres come out
of it.

5) A thin circular coin of mass 5 gm and radius 4/3 cm is initially in a horizontal xy-plane. The coin is

tossed vertically up (+ direction) by applying an impulse of at a distance 2/3 cm

from its center. The coin spins about its diameter and moves along the + direction. By the time the
coin reaches back to its initial position, it completes n rotations. The value of n is ____.
[Given: The acceleration due to gravity g = 10 ms–2]

6) Consider a uniform cubical box of side a on a rough floor that is to be moved by applying
minimum possible force F at a point b above its centre of mass (see figure). If the coefficient of

friction is µ = 0.4, the maximum possible value of 100 × for a box not to topple before moving is

______ .

7) The density of newly discovered planet is twice that of the earth. The acceleration due to gravity
at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R
and the radius of the planet is R'. Then what is the value of R/R'?

8) A billiard ball of radius R, initially at rest over the rough surface is given a sharp impulse by a cue.

The cue is held horizontally a distance above the centre line as shown in figure. The ball leaves
 0
the cue with a speed v and eventually acquires a final speed . Find x.

CHEMISTRY

SECTION-I(i)

1) A open ended mercury manometer is used to measure the pressure exerted by a trapped gas as
shown in the figure. Initially manometer shows no difference in mercury level in both columns as
shown in diagram. After sparking 'A' dissociates according to following reaction

If pressure of Gas “A” decrease to 0.9 atm. Then :

(Assume temperature to be constant and is 300 K)

(A) total pressure increased to 1.3 atm

(B) total pressure increased by 0.3 atm
(C) total pressure increased by 22.3 cm fo Hg
(D) difference in mercury level is 228 mm

2) A 1 litre vessel contains 2 moles of a vanderwaal's gas.

Given data: a = 2.5 atm-L2 mole–2 , T = 240 K
b = 0.4 L-mole–1 , RT = 20 L-atm mole–1
Identify the correct options about the gas sample :

(A) Pressure of gas = 190 atm

(B) Compressibility factor = 4.75
(C) Attraction forces are dominant in the gaseous sample
(D) TB (Boyle temperature) = 75 K

3) Identify the correct statements when a fixed amount of ideal gas is heated in a container fitted
with a movable piston always operating at constant pressure.
(A) Average distance travelled between successive collisions will decreases.
Collisions frequency increases since speed of the molecules increases with increase in
(B)
temperature.
(C) Average relative speed of approach remains unaffected.
(D) Average angle of approach remains unaffected.

4) Which is/are incorrect for cyclic process as shown in figure - Choose the
incorrect statement

(A) Work is done by the system

(B) Heat is liberated by the system
(C) w = 0
(D) process is isoentropic as ΔS = 0

5) A reversible cyclic process for an ideal gas is shown below. Here, P , V and T are pressure ,
volume and temperature , respectively. The thermodynamic parameters q, w, H and U are heat,

work, enthalpy and internal energy, respectively. The correct

option(s) is (are)

(A) qAC = ΔUBC and wAB = P2 (V2 – V1)

(B) wBC = P2 (V2 – V1) and qBC = ΔHAC
(C) ΔHCA < ΔUCA and qAC = ΔUBC
(D) qBC = ΔHAC and ΔHCA > ΔUCA

6) Which of the following statement(s) is/are false?

(A) All adiabatic processes are isoentropic (or isentropic) processes

(B) When ; the reaction must be exothermic

(C)
at constant temperature for ideal gas

The heat of vaporisation of water at 100ºC is . When 9 gm of water vapour condenses

(D)
to liquid at 100ºC and 1 atm, then
 SECTION-I(ii)

1)

List-I List-II
Adiabatic expansion of an ideal gas against
(I) (P) ΔSsys > 0
vacuum
(II) Melting of ice at 1 atm & 10°C (Q) ΔU = 0
Adiabatic reversible compression involving an
(III) (R) ΔSuni = 0
ideal gas
(IV) Isothermal reversible expansion of an ideal gas (S) w=0
(T) ΔH = 0
(A) I → Q;II → P;III → R;IV → T
(B) I → P;II → Q;III → R;IV → S
(C) I → Q;II → R;III → T;IV → P
(D) I → P;II → R;III → T;IV → P

2) Match the following :

List-I List-II

(I) Work done in reversible adiabatic process (P) ΔStotal > 0

(II) Entropy of all perfectly crystalline substance (Q) Perfect differential

(III) Criteria of spontaneity (R) Zero when T → 0

(IV) ΔHsublimation (S)

(A) I → S;II → R;III → P;IV → Q

(B) I → S;II → P;III → Q;IV → R
(C) I → Q;II → R;III → P;IV → S
(D) I → P;II → Q;III → S;IV → R

3) One mole of N2(g) is taken in 1 litre empty container fitted with a movable piston at 300K. If it is
heated to 1200K at constant pressure then match the change (List-II) in parameters (List-I) of gas as
compared to initial state & select the correct code.

List-I List-II
(Parameter) (Change)

Z1 (number of collision made by

(I) (P) 1/8
a molecule per unit time)

(II) Z11 (collision frequency) (Q) 2

(III) λ (mean free path) (R) 1/2

(IV) Urms (root mean square speed) (S) 4

(A) I → R;II → P;III → S;IV → Q
(B) I → P;II → Q;III → S;IV → R
(C) I → R;II → S;III → P;IV → Q
(D) I → Q;II → P;III → S;IV → R

4) Match the description in List I with graph provided in List II. For n moles of ideal gas at
temperature 'T'.

List I List II

(I) (P)
vs P

(II) (Q)
vs V

(III) (R)
vs P–2

(IV) (S)
vs log P

(A) I → P;II → R;III → S;IV → Q

(B) I → Q;II → S;III → R;IV → P
(C) I → S;II → R;III → Q;IV → P
(D) I → R;II → S;III → Q;IV → P

SECTION-II

1) One litre of oxygen gas is passed through a ozonizer & the final volume of the mixture becomes
820 ml. If this mixture is passed through oil of turpentine. Find final volume (in ml) of gas remaining.

2) 190 mL of N2 was collected in a jar over water at same temperature, water level inside and
outside the jar standing at the same height (As shown in the fig.). If barometer reads 740 mm Hg
and aqueous tension at the temperature of the experiment is 20 mm Hg, the volume of the gas (in

mL) at 1 atm pressure and at the same temperature would be :

3) A perfectly expandable balloon filled with helium gas at 27°C and a pressure of 720 mm of Hg has
a volume of 100 L. The balloon rises to an altitude where the pressure is 420 mm of Hg and the
temperature -63°C. What is the change in the volume (in L) of the balloon ?

4) For the reaction 2NH3(g) → N2 (g) + 3H2 (g). What is the % of NH3 converted if the mixture
diffuses twice as fast as that of SO2 under similar conditions

5) When the formation of one mole HBr(g) takes place reversibly from gaseous H2 and liq. Br2 at
300K and 1 atm increase in entropy of the surrounding is 170 J/K. What is change in entropy (J/K) of
the system if reaction 2HBr(g) → H2(g) + Br2(ℓ) is carried out reversibly at 300K in rigid closed
container. [Given : R =8.3 J/mol-K]

6) The change in entropy when two moles of a monoatomic perfect gas is compressed to half its
volume and simultaneously heated to twice its initial temperature is ________. [in cal /K]
[ln 2 = 0.69, R = 2cal /mole -K] [Nearest Integer]

7) How many of the following statement(s) is/are Incorrect?

(1) Final temperature in adiabatic free expansion is less than final temperature in isothermal free
expansion if carried out from same initial state involving an ideal gas.
(2) An adiabatic process is always an isoentropic process.
(3) More heat is absorbed by an ideal gas if subjected to isothermal reversible expansion process
than an irreversible isothermal expansion process between same states.
(4) During isothermal expansion of any substance internal energy will remain constant.
(5) On isothermal expansion of an ideal gas upto same final volume from same initial state, final
pressure will be more for irreversible process than reversible process.
(6) Work is always calculated by the expression

8) For the reaction at 300 K

A(g) + B(g) —→ C (g)
ΔU = –3.0 kcal ; ΔS = – 10.0 cal/K
value of ΔG is –y ×103 cal. What is the value of y?

MATHEMATICS

SECTION-I(i)

1) If px4 + qx3 + rx2 + sx + t = then

which of following is/are correct.

(A) t = 21
(B) p + q + r + s + t = 100
(C) p + r = 30
(D) r = 34

2) The parameter on which the value of the

determinant . depends upon is-

(A) a
(B) p
(C) d
(D) x

3) In an A.P., let Tr denote rth term from beginning, , , then :

(A) T1 = common difference

(B)

(C)

(D)

4) If a, b, c are in A.P. and a2, b2, c2 are in G.P. the common ratio of G.P. can be

(A) 1
(B)
(C)
(D) -1

5) Consider two sequence P : p1,p2,p3..... and Q : q1,q2,q3..... such that pn = 73 – 2n ∀n∈N and sum of
first n terms of sequence Q is Sn = 3n2 + 2n ∀n∈N, then :-

(A)

(B) q10 = 59

(C)

(D)

6) If Sn = then S10 is less than :

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

SECTION-I(ii)

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

List-I List-II

Number of common integers in two arithmetic progression

(I) (P) 25
1,3,5,7,...157 and 2,5,8,11...161 is

If a and b are roots of 11x2 – 4x – 2 = 0, then value of 10(1 + a + a2

(II) (Q) 26
+ ...∞) (1 + b + b2 + ...∞) is

If the sum of first 20 terms is same as the sum of first 10 terms in

(III) (R) 22
an arithmetic progression then the sum of first 30 terms is

(IV) If sum is (where a and b are (S) 20

coprime) then value of 20|b – a| is

(T) 0
(A) I → Q;II → T;III → R;IV → S
(B) I → Q;II → R;III → T;IV → S
(C) I → P;II → S;III → T;IV → Q
(D) I → P;II → S;III → R;IV → T

2)

List–I List–II

Two adjacent sides of a

parallogram are 4x + 5y = 0
and 7x + 2y = 0 and one diagonal
(I) (P) 1
is 11x + 7y = 9 and other diagonal
is ax + by + c = 0, then a + b + c
is equal to

If line 2x – by + 1 = 0 intersects
the curve 2x2 – by2 + (2b – 1)xy – x
– by = 0 at points A & B and
(II) (Q) 0
AB subtends a right angle at
origin, then value of b + b2
is equal to
 A line passes through point (3, 4)
and the point of intersection of the
lines 4x + 3y = 12 and 3x + 4y =
(III) (R) 5
12 and length of intercepts on the
co-ordinate axes are a and b, then
ab is equal to

A light ray emerging from the point

source placed at P(2, 3) is reflected
at a point 'Q' on the y-axis and then
(IV) (S) 4
passes through the point R(5, 10).
If co-ordinates of Q are (a, b), then
a + b is
(A) I → Q;II → S;III → P;IV → R
(B) I → Q;II → S;III → R;IV → P
(C) I → S;II → Q;III → P;IV → R
(D) I → Q;II → P;III → S;IV → R

3)

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

List-I List-II

Let and Ai, Bi, Ci are co-factor

(I) (P) 1

of ai, bi, ci respectively and then

is

If x > 0 such that

(II) (Q) 2
then x is

Number of solution(s) of the equation

(III) (R) 3
sinx + cosx – 1 = sin2x – 2 is [0, 3π] is

(IV) (S) 4
Minimum value of is
(x, y, z > 0)

(T) 7
(A) I → P;II → R;III → Q;IV → T
(B) I → Q;II → T;III → S;IV → P
(C) I → P;II → T;III → R;IV → P
(D) I → Q;II → S;III → R;IV → P

4) Consider a triangle ABC having vertices A(3, 4), B(4, 3), C(5, 0). Let the orthocentre,
circumcentre, centroid of the triangle are (a, b), (c, d), (e, ƒ ) respectively and image of the point
(–10, –5) in the line x + y – 2 = 0 is (g, h), then on the basis of above match the following

List-I List-II

(I) a + b is equal to (P) 19

(II) c + d is equal to (Q) 0

(III) e + 3ƒ is equal to (R) a+g

(IV) g + h is equal to (S) b–g

(T) 11
(A) I → P,R;II → Q,S;III → T;IV → P,R
(B) I → PQ;II → TS;III → QR;IV → PS
(C) I → QS;II → RP;III → QR;IV → ST
(D) I → RT;II → PS;III → QR;IV → PT

SECTION-II

1) Let ABC be a triangle and A ≡ (1, 2), y = x be the perpendicular bisector of AB and x – 2y + 1 = 0
be the angle bisector of ∠C. If the equation of BC is given by ax + by – 5 = 0, then the value of a – 2b
is :

2) The line x + y = p meet x-axis and y-axis at A and B respectively. A triangle APQ is inscribed in
the triangle OAB, O being origin, with right angle at Q. P and Q lies, respectively on OB and AB. If

area of triangle APQ is of area of then is equal to

3) The equations of the two adjacent sides of a rhombus formed in first quadrant are represented by
7x2 – 8xy + y2 = 0, then slope of it's longer diagonal is :

4) If S, S1, S2 be the circles of radii 5, 3, 2 respectively. If S1 and S2 touch externally and they touch
internally with S. The radius of circle S3 which touches externally with S1 and S2 and internally with
S is

5) A circle passes through the vertex C of a rectangle ABCD and touches its sides AB and AD at M
and N, respectively. If the distance from C of the line segment MN is equal to 5 units and the area of
the rectangle ABCD is S sq. units, then the values of is _______.

6) One of the diameter of the circle circumscribing the rectangle ABCD is 4y = x + 7. If A and B are
the points (–3, 4) and (5, 4) respectively, then the area of rectangle is
7) If all the roots of the equation x1008 + a1x1007 + a2x1006 +.....+ a1006x2 – 1008x + 1 = 0 are ∈ R+, then

value of is

8) If harmonic means (H1, H2, H3,.........,H8) are inserted between two positive number a and b (a < b)
such that arithmetic mean of a and b is 5/4 times equal to geometric mean of a and b,
then is equal to
 ANSWER KEYS

PHYSICS

SECTION-I(i)

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

SECTION-I(ii)

Q. 7 8 9 10
A. A A C A

SECTION-II

Q. 11 12 13 14 15 16 17 18
A. 1.00 5.00 2.00 20.00 30.00 75.00 2.00 5.00

CHEMISTRY

SECTION-I(i)

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

SECTION-I(ii)

Q. 25 26 27 28
A. A A A C

SECTION-II

Q. 29 30 31 32 33 34 35 36
A. 460.00 180.00 20.00 6.25 348.00 1.00 5.00 0.60

MATHEMATICS

SECTION-I(i)

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

SECTION-I(ii)

Q. 43 44 45 46
A. B A C A
 SECTION-II

Q. 47 48 49 50 51 52 53 54
A. 5.00 3.00 2.00 1.57 or 1.58 5.00 32.00 503.50 2.00
 SOLUTIONS

PHYSICS

1) If force is alueags ⊥ to the velocity the path of the particle will be circular.

2) T – Mg cosθ =
af = g sinθ

4)

Velocity of the system just after the collision

mvo = (m + M) V ⇒ V=

Using work energy theorem.

DK = WAII = Wg + WN + WS (Assume friction force is absent) ......(i)

0– (m + M) V2 = 0 + 0 – K X2max

= KX2max ⇒ Xmax = =

5) Since kinetic energy of block decreases.

Since kinetic energy of wedge increases.
6)

(A) mg sin 60° = α

α=

(B) N2 – mg cos60° = 0 Þ N2 =

mg sin 60 – N1 = m

mg – . = N1

N1 =

N= = =

(C) mg (1 – cos 60°) = =

w=

(D) N – mg =

N = mg

11)

12) aof block =

v=

∴ ΔKE =
 13)

Displacement of mass A is 10 cm
m × 10 = (m + m + 3m) × x
10m = 5xm
x = 2cm

16)
F = µmg ...(1)

F = mg ...(2)

µmg = mg ×

µ=

0.4 = µ =
0.8b + 0.4a = a
0.8 b = 0.6 a

CHEMISTRY

20)

21)

Fixed amount ⇒⇒ n = const.

Given P = const.
(A) Average distance travelled by a molecule between successive collision = mean free path (l).

∵
on heating temperature will increase and other parameters (s and p) are constant.
∴ λ increases ⇒ (A) is wrong.
for option 'B'
collision frequency = z11 = Total number of bimolecular collisions per unit time per unit
volume.

∴ (at const. P)
Clearly as tempt increases z11 decreases.
⇒ (B) is incorrect.
for option (C),

Average relative speed of approach =

∴ As tempt. increases, average relative speed of approach increases.
∴ option (C) is incorrect.
Average angle of approach is considered to be 90º which remains unafferted on increasing
tempt.
∴ option (D) is correct.

24) (a) Reversible adiabatic process is iso-entropic process

(b) When then process may be endothermic/exothermic
(c) , only PV work consider

(d)

25)

In free adiabatic expansion q = 0, w = 0, ΔH = 0, ΔU = 0

Water(s)⇌ Water(l)
ΔSsys > 0
Adiabatic reversible compression in ideal gas
ΔSuni = 0
Isothermal reversible expansion of ideal gas
ΔSsys > 0 ΔU = 0, ΔSuni = 0, ΔH = 0

27) , , λ ∝ T, Urms ∝

30)

33)

ΔSsurr. = –

170 = –
q = qp = –51 kJ
ΔH = ΔU + ΔngRT

– 51 = ΔU + × 8.3 × 10–3 × 300

–51 = ΔU + 1.245
 ΔU = –52.245
2HBr(g) + 104.5 → H2(g) + Br2(l)

ΔS =

34) ΔS =

Here n = 2 CV = R (for monoatomic gas)

ΔS = = 3Rln2 + 2R[– ln2] = Rln2

= 2ln2 = 2 × 0.69 = 1.38cal /K

35) only (3) is correct

MATHEMATICS

41) Pn = 73 – 2n ⇒ seq 71, 69, 67, 65,......

Sn = 3n2 + 2n ⇒ seq 5, 11, 17, 23, ......

∴
q10 = 5 + 9(6) = 59

43) (P) first A.P.

1, 3, 5, 7 …….. 157
Common diff. (d1) = 2
second A.P.
2, 5, 8, 11 …… 161
Common diff. (d2) = 3
Common A.P. have common diff = LCM (d1, d2)
d=6
Common A.P. will be
5, 11, 17, 23 ……. 155
⇒ 155 = 5 + (n – 1) 6
⇒ (n – 1)6 = 150
⇒ (n – 1) = 25
⇒ n = 26
∴ (2) will be correct

(Q) 11x2 – 4x – 2 = 0

,
⇒ 10(1 + a + a2…∞) (1 + b + b2…∞)

⇒ ⇒
∴ (3) will match
(R) Sum of first 20 terms = Sum of first 10 terms

⇒ 4a + 38d = 2a + 9d
⇒ 2a + 29d = 0

⇒
⇒ S30 = 15[2a + 29d] ⇒ 15 × 0 = 0
⇒ S30 = 0
∴ (1), (2), (3), (4), (5) will match

(S)

∴ a = 9999, b = 10000
20|b – a| = 20|10000 – 9999| = 20 × 1 = 20
∴ (4) will match.

44) (P) Let the lines 4x + 5y = 0 and 7x + 2y = 0 represents the sides AB & AD of the
parallelogram ABCD, then the
verti ces of

A, B, D are (0,0), and respectively the mid point of BD is

∴ the equation of the line passing through and (0, 0) will be x – y = 0 which is the
required equation of the other
diagonal
So a = 1, b = –1, c = 0
∴a+b+c=0
(Q) Joint equation of lines OA & OB, O being the origin will be
2x2 – by2 + (2b – 1) xy – (x + by)(–2x+by) = 0
⇒ 4x2 – (b + b2)y2 + (3b – 1)xy = 0
If these lines are perpendicular then
4 – b – b2 = 0 ⇒ b + b2 = 4
(R) Equation of line passing through inter section of 4x + 3y = 12 and 3x + 4y = 12 will be
(4x + 3y – 12) + λ(3x + 4y – 12) = 0
If passes through (3, 4) ⇒ (12 + l(13)) = 0

⇒λ=–
∴ Equation of the required line
16x – 9y – 12 = 0

length of intercepts on x and y axes are and so ab = 1

45) (I) Δ` = Δ2 ⇒

(II)

⇒x=7
(III) (sinx + cosx) – 1 = (sinx + cosx)2 – 1 – 2
sinx + cosx = t ⇒ t2 – t – 2 = 0
t = 2 or t = –1
sinx + cosx = 2 is rejected &
sinx + cosx = –1
⇒ sinx = –1 or cosx = –1

, x = π, 3π
total solutions are 3

(IV) (A.M. ≥ G.M.)

multiply all of them

46) It is obvious that the circumcentre is (0, 0) centroid is

Now let the co-ordinates of orthocentre be (a,b), then (a, b) ——— (4, 7/3) ——— (0,0)
2 : 1
Which gives a = 12 and b = 7
To find the image

⇒ g + 10 = h + 5 = 17
⇒ g = 7, h = 12
Now we have
a = 12, b = 7, c = 0, d = 0, e = 4, ƒ = 7/3,
g = 7, h = 12

47) Here, B is the image of A w.r.t. line y = x

∴ B ≡ (2, 1) and C is the image of A w.r.t. line x – 2y + 1 = 0

If C ≡ , then or and

Equation of BC is
 or 3x – y – 5 = 0 ( Eq. of BC is ax + by – 5 = 0)
Here a = 3, b = – 1 ∴ a – 2b = 5

49) Adjacent sides

50)
For diagram we get
(r + 2)2.2 + (r + 3)2.3
= 2.32 + 3.22 + 5(5 – r)2
⇒ 76r = 120

⇒
51)

52)

Let O be the centre of the circle. M is the mid point of AB. Then
OM ⊥ AB
Let OM when produced meets the circle at P and Q.

∴ PQ is a diameter perpendicular to AB and passing through M.

Slope of AB = =0
∴ PQ is a line parallel to y- axis passing through (1, 4).
∴ Its equation is x = 1 … (1)
Also equation of one of the diameter given is
4y = x + 7 … (2)
Solving (1) and (2), we get coordinates of centre O
O (1, 2) Also let coordinates of D be (α, β)
Then O is mid point of BD, therefore

⇒ α = –3, β = 0
∴ D (–3, 0)
Using the distance formula, we get

∴ Area of rectangle ABCD = AB × AD = 8 × 4 = 32 square units.

53) From the relation between the zeroes and the coefficient we obtain
and x1x2......x1008 = 1

Hence,
So se have the equality case in the A.M. – G.M. inequality
So x1 = x2 ...... = x1008 = 1

So

54)