23-08-2025

4602CJA101001250013 JM

PART-1 : PHYSICS

SECTION-I

1) A stone of mass 1 kg is tied to the end of a string of 1 m long. It is whirled in a vertical circle. If
the velocity of the stone at the top be 4 m/s. What is the tension in the string?

(A) 6 N
(B) 16 N
(C) 5 N
(D) 10 N

2)

If the equation for the displacement of a particle moving on a circular path is given by
where is in radians and t in seconds, then the angular velocity of the particle after
2 s from its start is:-

(A) 8 rad/s
(B) 12 rad/s
(C) 24 rad/s
(D) 36 rad/s

3) A stone is projected with speed u and angle of projection is θ. Find radius of curvature at t = 0.

(A)

(B)

(C)

(D)

4) A road is banked at an angle of to the horizontal for negotiating a curve of radius . At

what velocity will a car experience no friction while negotiating the curve ? Take :-

(A) 54 km/hr
(B) 72 km/hr
(C) 36 km/hr
(D) 18 km/hr
5) A car is moving with constant speed on a road as shown in figure. The normal reaction by the road
on the car is NA, NB and NC when it is at the points A, B and C respectively :

(A) NA = NB = NC
(B) NA > NB > NC
(C) NA < NB < NC
(D) NC > NA > NB

6) A ball of mass m is hung on a thread. The thread is held taut and horizontal, and the ball is
released as shown. At what angle between the thread and vertical will the tension in thread be equal

to weight in magnitude ?

(A) 30o

(B)

(C)

(D) never

7) If a particle of mass m is moving in a horizontal circle of radius r with a centripetal force (–k / r2),
the total energy is :-

(A)

(B)

(C)

(D)

8) The centre of mass of the shaded portion of the disc is : (The mass is uniformly distributed in the

shaded portion) :-

(A)
to the left of A

(B)
to the left of A
(C)
to the right of A

(D)
to the right of A

9) In the given figure table is smooth and pulley and string are ideal then acceleration of centre of

mass of the two blocks system is :-

(A)

(B)

(C)

(D)

10) Two men of masses 80 kg and 60 kg are standing on a wooden plank of mass 100 kg, that has
been placed over a smooth surface. If both the men start moving towards each other with speeds
1m/s and 2 m/s respectively (w.r.t. plank) then find the velocity of the plank by which it starts

moving.

(A) (– 1/6) m/sec

(B) (– 1/3) m/sec
(C) (1/2) m/sec
(D) (–1/2) m/sec

11) For given F-t graph, Impulse in first 3 sec in (N.s.) :-

(A) 50 N–S
(B) 100 N–S
(C) 150 N–S
(D) 75 N–S

12) A stationary body explodes into two fragments of masses and If momentum of one
fragment is p, the minimum energy of explosion is :-

(A)

(B)

(C)

(D)

13) A wedge of mass M = 2m lies on a frictionless plane. A particle of mass m approaches the wedge
with speed v. There is no friction between the particle and the plane or between the particle and the
wedge. The maximum height climbed by the particle on the wedge is given by :-

(A)

(B)

(C)

(D)

14) A block of mass M with a semicircular track of radius R rests on a horizontal frictionless surface.
A uniform cylinder of radius r and mass m is released from rest from the top point A. The cylinder
slips on the semicircular frictionless track. The distance travelled by the block when the cylinder

reaches the point B is :

(A)
(B)

(C)

(D) None

15) A projectile is launched from the origin with speed v at an angle θ from the horizontal. At the
highest point in the trajectory, the projectile breaks into two pieces, A and B, of masses m and 2m,
respectively. Immediately after the breakup piece A is at rest relative to the ground. Neglect air
resistance. Which of the following sentences most accurately describes what happens next?

(A) Piece B will hit the ground first, since it is more massive.
Both pieces have zero vertical velocity immediately after the breakup, and therefore they hit the
(B)
ground at the same time.
Piece A will hit the ground first, because it will have a downward velocity immediately after the
(C)
breakup.
There is no way of knowing which piece will hit the ground first, because not enough
(D)
information is given about the breakup.

16) In the shown figure, the heavy block of mass 2 kg rests on the horizontal surface and the lighter
block of mass 1 kg is dropped from a height of 0.9 m. At the instant the string gets taut, the upward

speed (in m/s) of the heavy block is :

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

17) In the figure shown, a small ball hits obliquely a smooth and horizontal surface with speed u

whose x and y components are indicated. If the coefficient of restitution is , then its x and y
components vx and vy just after collision are respectively :-

(A) 4 m/s, 1 m/s

(B) 2 m/s, 1 m/s
(C) 2 m/s, 2 m/s
(D) 4 m/s, 2 m/s

18) Find the x co-ordinate of the cetre of mass of the bricks shown in figure. (length of each brick

:-

(A)

(B)

(C)

(D)

19) A block of mass m moving at a velocity v collides with another block of mass 2m at rest. The
lighter block comes to rest after collision. Find the coefficient of restitution:-

(A)

(B) 1

(C)

(D)

20) A ball of mass 1 kg strikes a heavy platform, elastically, moving upwards with a velocity of 5m/s.
The speed of the ball just before the collision is 10 m/s downwards. Then the impulse imparted by
the platform on the ball is :-

(A) 30
(B) 40
(C) 20
(D) 60

SECTION-II

1) A ball is projected with horizontal velocity at the bottom most point attached with
inextensible string of length R and fixed at O as shown. Tension in the string in horizontal position is

kmg. Find the value of k.

2) 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 at centre of larger disc find x-

coordinate of centre of mass of system. (in cm) :-

3) You are shown a photo of a car driven on a vertical inside wall of a huge cylinder with a radius
of 50 m. The coefficient of static friction between the car tires and the cylinder is μs = 0.8. The

minimum speed in (m/s), at which the car can be driven like that is 5x. Find x.
4) A body slides down an inclined surface which ends into a vertical loop of radius R = 40 cm.

What must be the height H, in metres, of the inclined surface for the
body not to fall at the uppermost point of the loop? Assume friction to be absent.

5) A bullet of mass m strikes a pendulum bob of mass 2m with velocity u as shown in figure. It
passes through and emerges out with a velocity u/2 from bob. The length of the pendulum bob is ℓ =
2m. If the minimum value of u for which the pendulum bob completes full circle is X m/s, then find X.

(Take g = 10 m/s2) :-

PART-2 : CHEMISTRY

SECTION-I

1) Three lines at three different values of constant pressure are given, which of the following

relation is correct for an ideal gas ?

(A) P1 = P2 = P3
(B) P1 > P2 > P3
(C) P3 > P2 > P1
(D) Can’t be predicted

2) The ratio of most probable velocity to the average velocity is?

(A)

(B)
(C)

(D)

3) Which of the following pairs will diffuse at same rate?

(A) NH3 ,PH3

(B) NO, CO
(C) NO,C2H6
(D) CO,NO2

4) 0.5 mole of each of H2, SO2 and CH4 are kept in a container. A hole was made in the container.
After 3 hrs, the order of partial pressures in the container will be?

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

5) Points I, II and III in the following plot respectively correspond to (Vmp : most probable velocity)

(A) Vmp of N2 (300K); Vmp of H2(300K); Vmp of O2(400K)

(B) Vmp of H2 (300K); Vmp of N2(300K); Vmp of O2(400K)
(C) Vmp of O2 (400K); Vmp of N2(300K); Vmp of H2(300K)
(D) Vmp of N2 (300K); Vmp of O2(400K); Vmp of H2(300K)

6) In a tube of length 5 m having 2 identical holes at the opposite ends. H2 & O2 are made to effuse
into the tube from opposite ends under identical conditions. Find the point where gases will meet for

the first time.

(A) 2m from H2 side

(B) 2m from O2 side
(C) 4m from H2 side
(D) 4m from O2 side

7)

The r.m.s. velocity of hydrogen is times the r.m.s velocity of nitrogen. If T is the temperature of
the gas, then :-

(A) T(H2) = T(N2)

(B) T(H2) > T(N2)
(C) T(H2) < T(N2)
(D)

8)

An open ended Hg manometer is used to measure the pressure exerted by a trapped gas as shown in
the figure. Atmospheric pressure is 749 m.m. of Hg. What is the pressure of the trapped gas :-

(A) 292 m.m. Hg

(B) 457 m.m. Hg
(C) 749 m.m. Hg
(D) 1041 m.m. Hg

9) At identical temperature and pressure, the rate of diffusion of hydrogen gas is times that of a
hydrocarbon having molecular formula CnH2n-2. What is the value of n ?

(A) 1
(B) 4
(C) 3
(D) 8

10) Select incorrect statement :

(A) We can condense vapour simply by applying pressure below critical temperature
To liquefy a gas, one must lower the temperature below it’s critical temperature (TC) and also
(B)
apply pressure
At critical temperature (TC), there is no distinction between liquid and vapour state. Hence,
(C)
density of the liquid is nearly equal to density of the vapour
(D) Whatever the pressure applied, a gas cannot be liquified below it’s critical temperature (TC)
11) If the Vander Waal's constant of two gases are given as, Identify the incorrect statement :-

a(Pa-dm6mol–2) b(dm3mol–1)

Gas A 120 0.02

Gas B 140.8 0.03

(A) a and b constant are independent of temperature
(B) Critical Volume of B > Critical Volume of A
(C) Critical Pressure of B < Critical Pressure of A
(D) Critical Temperature of B > Critical Temperature of A

12) Graph depicting correct behaviour of ideal gas & H2 gas will be (neglect a)

(A)

(B)

(C)

(D)

13) 'a' and 'b' are Vander Waals constants for gases. Chlorine is more easily liquefied than ethane
because :-
(A) a for Cl2 < a for C2H6 but b for Cl2 > b for C2H6
(B) a for Cl2 > a for C2H6 but b for Cl2 < b for C2H6
(C) a and b for Cl2 = a and b for C2H6
(D) a and b for Cl2 < a and b for C2H6

14) Critical temperature and critical pressure value of four gases are given :-

Critical Temp. Critical Pressure

Gas
(K) (atm)

P 5.1 2.2

Q 33 13

R 126 34

S 135 40
Which of the following gas(es) cannot be liquefied at a temperature 100 K and pressure 50 atm ?
(A) S only
(B) P only
(C) R and S
(D) P and Q

15) In which of the following, the oxidation number of oxygen has been arranged in increasing
order?

(A) BaO2 < KO2 < O3 < OF2

(B) OF2 < KO2 < BaO2 < O3
(C) BaO2 < O3 < OF2 < KO2
(D) KO2 < OF2 < O3 < BaO2

16) Which of the following can't act as oxidising agent ?

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

17) How many moles of KMnO4 are reduced by 1 mole of ferrous oxalate in acidic medium:-

(A)

(B)

(C)

(D)
18)

x mol of XeF4 quantitatively oxidized KI to I2 and liberated Xe, along with formation of KF. This
iodine required 20 ml of decinormal hypo solution for exact titration. The value of x is :-

(A) 0.5
(B) 1.0
(C) 2.0
(D) 5.0

19) The oxidation state of chromium in the final product formed by the reaction between KI and
acidified dichromate solution is :

(A) +3
(B) +6
(C) +5
(D) +4

20) Which of the following reactions are disproportionation reactions?

(A) Cu+ → Cu2+ + Cu
(B)
(C) 2KMnO4 → K2MnO4 + MnO2 + O2
(D)
Choose the correct answer from the options given below:

(A) (A), (B)

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

SECTION-II

1) 10 g CaCO3 were dissolved in 250 ml of 1 M HCI. What volume [ml] of 2 M KOH would be
required to neutralise excess HCI.

2) It requires 40 mL of 1M Ce4+ to titrate 20 mL of 1M Sn2+ to Sn4+. What is the oxidation state of the
Cerium in the product ?

3) An open flask contains air at 27°C. Calculate the temperature (in°C) at which it should be heated
so that 1/3rd of air measured at 27°C escapes out.

4) A real gas have vanderwaal constant as a and b which are related as a = 2.7b, find the

temperature (k) above which gas can not be liquified. [use R=0.08 ]
5) A glass bulb contains 10 gm O2 gas. Now, the bulb is completely evacuated and re-filled with CH4
gas at the same temperature and pressure. The mass of CH4 gas used in refilling is :

PART-3 : MATHEMATICS

SECTION-I

1)

In a triangle ABC, if b = ( – 1)a and ∠C = 30°,

then the value of (A – B) is equal to (All symbols used have usual meaning in a triangle.)

(A) 30°
(B) 45°
(C) 60°
(D) 75°

2) In a ΔABC, A : B : C = 3 : 5 : 4.
Then is equal to(All symbols used have usual meaning in a triangle.)

(A) 2b
(B) 2c
(C) 3b
(D) 3a

3)

If the sides of a triangle are sin α, cos α, ,

0<α< , the largest angle is

(A) 60°
(B) 90°
(C) 120°
(D) 150°

4)

In a triangle, the value of

is, where A, B, C are the angles of the triangle and a, b, c are

corresponding sides.
(Here s denotes the semi-perimeter).

(A) s2
(B)

(C) 2s2

(D)

5)

If where A, B, C are the angles of a triangle and a, b, c are the

corresponding sides respectively and then area of the triangle is

(A)

(B)

(C)

(D) 3

6)

In a if 9(a2 + b2) = 17c2.

then the value of the expression is

(A)

(B)

(C)

(D)

7) If in a triangle ABC,

.(All symbols used have usual meaning in a triangle.)

Then the value of ∠A =

(A) 90°
(B) 60°
(C) 30°
(D) None of these

8)

The coordinate axes are rotated through an angle of 135º in anti clock wise direction. If the
coordinates of a point P in the new system are known to be (4, –3), then the coordinates of P in the
original system are
(A)

(B)

(C)

(D)

9) The equation of the locus of all points equidistant from the point (4, 2) and the x-axis, is

(A) x2 + 8x + 4y – 20 = 0
(B) x2 – 8x – 4y + 20 = 0
(C) y2 – 4y – 8x + 20 = 0
(D) None of these

10)

The vertices A and D of square ABCD lie on the positive side of x- and y-axis respectively. If the
vertex C is the point (12, 17) then the co-ordinates of vertex B are

(A) (14, 16)

(B) (15, 3)
(C) (17, 5)
(D) (17, 12)

11) Let , be the circumcenter of a triangle with vertices and . Let

α denote the circumradius, β denote the area and γ denote the perimeter of the triangle. Then
is

(A) 60
(B) 53
(C) 62
(D) 30

12) If the equation of base of an equilateral triangle is 2x – y = 1 and the vertex is (–1, 2), then the
length of the side of the triangle is

(A)

(B)

(C)
(D)

13) The equation of the bisector of the acute angle between the lines 3x – 4y + 7 = 0 and 12x +
5y–2=0 is

(A) 21x + 77 y – 101 = 0

(B) 11x – 3y + 9 = 0
(C) 31x + 77y + 101 = 0
(D) 11x – 3y – 9 = 0

14) A line passes through the point of intersection of 2x + y = 5 and x + 3y + 8 = 0 and parallel to
the line 3x + 4y = 7 is

(A) 3x + 4y + 3 = 0
(B) 3x + 4y = 0
(C) 4x – 3y + 3 = 0
(D) 4x – 3y = 3

15) The reflection of A(3, 4) in the line x – 2y + 10 = 0 is the point B. The reflection of B in line x – y
+ 1 = 0 is the point C, then the circumcenter of the triangle ABC is

(A) (5, 6)
(B) (8, 9)
(C) (7, 8)
(D) None of these

16) For different values of a & b, the straight line

(a – 2b)x + (a + 3b)y + 3a + 4b = 0
will pass through a fixed point, that point is

(A) (1, 2)
(B) (–1, –2)
(C) (–2, –3)
(D) (2, 3)

17) A line L passes through the points (1, 1) and (0, 2) and another line M which is perpendicular to

L passes through the point The area of the triangle formed by these lines with y-axis is

(A)

(B)

(C)
(D)

18) Consider the family of lines (x + y –1) + λ(2x + 3y –5) = 0

and (3x + 2y – 4) + μ(x + 2y –6) = 0, equation of a straight line that belongs to both the families is

(A) x –2y –8 = 0
(B) x –2y + 8 = 0
(C) 2x + y – 8 = 0
(D) 2x – y – 8 = 0

19) If the points (1,2) and (3,4) were to be on the same side of the line 3x – 5y + a = 0, then

(A) 1 < a < 6

(B) 7 < a < 11
(C) a > 11
(D) a < 7 or a > 11

20) Let C be the centroid of the triangle with vertices (3, –1), (1,3) and (2,4). Let P be the point of
intersection of the lines x + 3y –1 =0 and 3x – y + 1 = 0. Then the line passing through the points C
and P also passes through the point

(A) (–9, –6)

(B) (–9, –7)
(C) (9, 7)
(D) (7, 6)

SECTION-II

1) A ray of the light is sent along the line x – 2y – 3 = 0. Upon reaching the line 3x – 2y – 5 = 0, the
ray is reflected. If the equation of the line containing the reflected ray is Ax + By – 31 = 0 find value
of A + B

2) If the straight line ax + by + p = 0 and are inclined at an angle and

concurrent with the straight line then is

3) If the area of parallelogram formed by lines

4y – 3x – a = 0, 3y – 4x + a = 0, 4y – 3x – 3a = 0

and 3y – 4x + 2a = 0 is then value of b is

4) If a, 2b, c are distinct numbers in A.P. and line ax + by + c = 0 always passes through a fixed
point (p, q), then value of |p + 2q| is
5) The number of possible straight lines passing through (2, 3) and forming a triangle with co-
ordinate axes, whose area is 12 sq. units is
 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. A C C C D C A A B A A C B B B B C A A A

SECTION-II

Q. 21 22 23 24 25
A. 7.00 5.00 5.00 1.00 40.00

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 C C A D C C D B D D A B D A B C A A A

SECTION-II

Q. 46 47 48 49 50
A. 25.00 3.00 177.00 10.00 5.00

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. C C C A C D A D B C B A B A B B C B D A

SECTION-II

Q. 71 72 73 74 75
A. 27.00 2.00 7.00 7.00 3.00
 SOLUTIONS

PART-1 : PHYSICS

1)

T = 16 – 10 = 6 N.

2)

3) At t = 0
a⊥ = g cos q,

R= =

4) For negotiating curve without friction

5)
where r is radius of curvature.

6)
7) mv2 = ∴ K.E. = mv2 =

P.E. =

∴ Total energy =

8)

A1 = πR2 A2 =

x1 = 0 x2 =

xcom =

9)

2mg – T = 2ma
T = ma

10)

Applying momentum conservation;

m/sec
11) Impulse = area of (f-t) graph with time axis

12)

Since the initial momentum of the system is zero. So, according to the conservation of linear
momentum, final momentum of the system is zero. Thus momentum of the second fragment is
also p.

Kinetic energy of first fragment

Kinetic energy of second fragment

Minimum energy of explosion

13)
Applying Linear momentum conservation

applying work energy theorem

Substituting from Eq. (i) in Eq. (ii), we have

14)

when ball reaches pt A

then block get shifted by x

∴ but since than there is no ext
force therefore com remain at its position
[(R–r) – x]m = Mx

∴x=
15) Since vertical velocity of particle before explassion is zero. So the vertical velocity after
explassion will also remain some i.e. zero.

16) Velocity of lighter block at the instant the string just gets taut.

Now by impulse−momentum theorem, let common speed be v1 then

(2 + 1)v1 = (1)v

17) vx = constant, vy = e × 4 = × 4 = 2 m/s

18)

19)
Momentum conservation
mv = 2mv′ ⇒ v′ = v/2

20)

21)

Centre of mass of a uniform solid cone of height h is at height from base therefore

0
h–z =

0
z =h–
 0
z =

22) Mass of removed part = , where M is the mass of disc.

Centre of mass (x-coordinate)

23) m/s

24)

25) Let V be velocity of mass 2m just after bullet passes through it. From momentum
conservation

mu + 0 = m + 2mV

V=
For completing circle

V> ; > ; u>4 > 40 m/s

PART-2 : CHEMISTRY

26) Charles Law [V α T]

27)

28) Molar masses of NO and C2H6 are same.

29) Rate of diffusion

30)

31)

32)

33)

Press. of the gas = 749 + 292 = 1041 mm Hg.

34)

35)

Above critical temperature gases can not be liquified whatever the pressures applied.
Below the critical temperature (TC) gases can be liquified

36) VC = 3b, PC = , TC =

37) For H2 z > 1, So graph (A) is correct.

38) Ease of liquification ∝ critical temperature.

39) R T = 126 K
S T = 135
The temperature above which gas cannot be liquefied both R and S have temperature greater
than 100 K.

40) BaO2 ⇒ O22– –1

KO2 ⇒ O2– –
O3 ⇒ O
OF2 ⇒ +2

41) In N–3, nitrogen is present in minimum O.N. & hence it cannot act as oxidising agent.

42) acc to law of Equivalence.

43)

Equivalent to XeF4 = Equivalent of I2

44) K2Cr2O7 + 6KI + 7H2SO4 → 4K2SO4 + Cr2(SO4)3 + 7H2O + 3I2.

45)

When a particular oxidation state becomes less stable relative to other oxidation state, one
lower, one higher, it is said to undergo disproportionation.
Cu+ → Cu2+ + Cu

46)

47) Ce+4 + Sn+2 → Sn+4 + Ce+x

40 × 1(4 – x) = 20 × 1 × 2
x = +3
 48) Let the initial number of moles of air at 27°C (300 K) be = n
Number of moles of air left when the air is heated T K

=n–
At constant pressure and constant volume,
n1T1 = n2T2

or n × 300 = ×T
or T = 450 K = (450 – 273) = 177°C

49)

50)

Same bulb is filled with CH4 gas at same temperature and pressure, hence P, V, T constant so
according to
PV = nRT

PART-3 : MATHEMATICS

51)

A – B = 60

52) ∵ A : B : C = 3 : 5 : 4
⇒
∴

(say)

53)

Largest side is
θc = 120°

54)

= s (s – a) + s (s – b) + s (s – c)
= 3s2 – s (a + b + c) = 3s2 – 2s2 = s2

55)

56)

57) We have
Multiplying both sides of abc, we get
⇒ 2bc cos A + ac cos B + 2ab cos C = a2 + b2
⇒ (b2 + c2 – a2) +

⇒ c2 + a2 – b2 = 2a2 – 2b2
⇒ b2 + c2 = a2
∴ ΔABC is right angled at A.
⇒ ∠A = 90°

58)

Let (x, y) be the cooordinates of P in the original system and (X, Y) be its coordiantes in the
new system
We have,
X = 4, Y = –3 and θ = 135°
∴ x = Xcosθ – Ysinθ, y = Xsinθ + Ycosθ
⇒ x = 4cos135° + 3sin135°,
 y = 4sin135° – 3cos135°

⇒ x = –2 + , y=2 +

⇒x=– , y=

Hence, the coordinates of P are

59) (x–4)2 + (y–2)2 = y2 ⇒ x2 – 8x – 4y + 20 = 0

60) Let the co-ordinates of B be (h, k) draw BL and CM perpendicular to the x-axis and the y-

axis. Therefore from figure co-ordinate of B (17, 5)

61) A(a, –2), B(a, 6), ,

AO = BO

a=8
AB = 8, AC = 6, BC = 10

62)

∵ tan60° =

⇒ ⇒

∴ BC = 2BD =
63) Bisectors of angles is given by

=± ....(1)
Now a1 a2 + b1b2 = (3) (– 12) + (– 4) (– 5)
= – 36 – 20 < 0
take '+' sign in eq. (1)
⇒ 11x – 3y + 9 = 0
Hence, 11x – 3y + 9 = 0 is the bisector of the acute angle between the given lines.

64) Equation of line is

(2x + y – 5) + λ(x + 3y + 8) = 0...(i)
x(2 + λ) + y(1 + 3λ) – 5 + 8 λ = 0

it's slope is = –
line (1) is parallel to 3x + 4y = 7

∴ – =–
⇒ 8 + 4λ = 3 + 9λ ⇒ 5λ = 5 ⇒ λ = 1
from eq. (i) 3x + 4y + 3 = 0

65)
Circumcenter is the point of intersection of the perpendicular bisectors of the sides.

66) a(x + y + 3) + b(–2x + 3y + 4) = 0

x + y + 3 + = 0 (1)
line (1) passes through intersection point of
x + y + 3 = 0 & –2x + 3y + 4 = 0
i.e. (–1, –2).

67)
Equation of the line L is

Equation of the line M is

If these lines meet y-axis at

Also, x-co-ordinate of their point of intersection R = 5/4.

Therefore, area of the ΔPQR

68)

The family of lines (x + y –1) + λ(2x + 3y – 5) = 0 passes through a point such that
x + y –1 = 0
2x + 3y – 5 = 0 i.e., (–2,3)
and family of lines
(3x + 2y –4) + μ(x + 2y –6) = 0
passes through a point such that 3x + 2y –4 = 0
and x + 2y – 6 = 0 i.e., (–1, 7/2)
∴ Equation of the straight line that belongs to both the families passes through (–2,3) and (-1,7
/ 2) is

69)

(3 – 10 + a)(9 – 20 + a) > 0

or (a –7) (a –11) > 0

∴ a ∈ (–∞, 7) ∪ (11, ∞)

70)

Centroid D(2,2)

Point of intersection of given lines;

Equation of line joining D and P

y–2= (x – 2)
⇒ 11y = 8x + 6
Only (–9, –6) satisfy this equation
71)

72) ax + by + p = 0

OM = perpendicular distance of O from

ax + by + p = 0

OM = = ON

= p.
⇒ a2 + b2 = 2

73) Use formula

74)
75) y – 3 = m (x – 2)
x = 0, y = 3 – 2m

y = 0, x = 2 –

Δ = |3 – 2m) = 12
2
(3 – 2m) = 24m
4m2 – 12m + 9 = –24m
4m2 + 12m + 9 = 0 (i)
2
4m – 12m + 9 = 24m
4m2 – 36m + 9 = 0 (ii)