17-10-2024

7501CJA101001240056 JA

PHYSICS

SECTION-I(i)

1) Q is a sleeve sliding on a fixed smooth vertical wire frame. P is the fixed end of a spring of natural

length . . If spring's constant is K = . Minimum speed required at lowest point to

complete circular path is :-

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

2) A particle of mass m = 0.1 kg is released from rest from a point A of a wedge of mass M = 2.4 kg
free to slide on a frictionless horizontal plane. The particle slides down the smooth face AB of the
wedge. When the velocity of the wedge is 0.2 m/s the velocity of the particle in m/s relative to the

wedge is

(A) 4.8
(B) 5
(C) 7.5
(D) 10

3) Mass m1 strikes m2 which is at rest. The ratio of masses m1/m2 for which they will collide again.
(Collisions between ball and wall are elastic. Coefficient of restitution between m1 and m2 is e and all
the surfaces are smooth)

(A)

(B)

(C)

(D) 1

4) A smooth chain AB of mass m rests against a surface in the form of a quarter of a circle of radius
R. If it is released from rest, the velocity of the chain after it comes over the horizontal part of the

surface is

(A)
(B)

(C)

(D)

SECTION-I(ii)

1) A uniform thin rod of mass m is placed on a horizontal surface. At one end vertical force F applied
by string, so that the center of the mass of the stick moves upward with acceleration a < g. The
normal force N of the ground on the other end of the stick just after the right end of the rod leaves
the surface can be
(A) 0.80 mg
(B) 0.74 mg
(C) 0.64 mg
(D) 0.54 mg

2) A rigid uniform bar AB of length L is slipping from its vertical position on a friqtionless floor (as
shown in the figure). At some instant of time, the angle made by the bar with the vertical is θ. Which
of the following statements about its motion is/are correct?

(A) Instantaneous torque about the point in contact with the floor is proportional to sin 0
(B) The midpoint of the bar will fall vertically downward
When the bar makes an angle θ with the vertical, the displacement of its midpoint from the
(C)
initial position is proportional to (1 – cos θ)
(D) The trajectory of the point A is a parabola

3) A thin ring is free to rotate in the vertical plane about O. Ring is released from rest at θ = 0, when
ring rotates through angle θ :

(A)
Angular acceleration of ring is
(B) Centripetal acceleration of ring is 2g sinθ

(C)
Net force by hinge is

(D)
Net force by hinge is

4) The disc of radius r is confined to roll without slipping at A and B. If the plates have the velocities
shown, then

(A) linear velocity v0 = v

(B)
angular velocity of disc is

(C)
angular velocity of disc is
(D) None of these

5) Two particles each of mass m are attached at end points of a massless rod AB of length ℓ. Rod is
hinged at point C as shown. Rod is released from rest from horizontal position. At the instant when
rod reaches its vertical position as shown, which of the following is/are CORRECT :-

(A) Speed of the particle at B is thrice the speed of particle at A

(B)
Net force on particle B is
(C) Angular accelration of the system is zero
(D) Both x & y components of hinge force are nonzero

6) Slender bar A is rigidly connected to a massless rod BC in Case 1 and two massless cords in Case
2 as shown. The vertical thickness of bar A is negligible compared to L. In both cases A is released

0 0

from rest at an angle θ = θ . When θ = 0° Mark

the correct statement(s) :-

(A) The kinetic energy will be the same in both the cases.
 The speed of the centers of gravity will be same
For Option (C) and (D) : Consider system to be at rest initially in vertical position. If
(B)
bullet D strikes A with a speed ν0 initially directed horizontally and becomes embedded
in it
(C) The speed of the center of gravity of A immediately after the impact in case-1 will be larger.
(D) The speed of the center of gravity of A immediately after the impact in case-2 will be larger.

7) A particle P1 moves with a constant velocity v along x-axis, starting from origin. Another particle
P2 chases particle P1 with constant speed v, starting from the point (0, d). Both motion begin
simultaneously. Select the correct alternative.

initial acceleration of P2 is
(A)

(B)
Ultimate separation between P1 and P2 is

(C)
initial acceleration of P2 is
Ultimate separation between P1 and P2 is
(D)

8) The radius of the tire of a car is R. The valve cap is at a distance r from the axis of the wheel. The
car starts from rest without skidding, at constant accleration. It's found that the valve cap has no

acceleration in the turn preceding the bottom most position. Select the correct alternatives.

(A)
The ratio

(B)
The car has travelled a distance upto this time
(C) The wheel of the car has turned by 0.5 rad upto this time

(D)
The car has travelled a distance upto this time.

SECTION-II

1) A uniform meter scale balances at the 40 cm mark when weights of 10 g and 20 g are suspended
from the 10 cm and 20 cm marks. The weight of the meter scale is __.

2) A symmetric lamina of mass M consists of a square shape with a semicircular section over each of
the edge of the square as in figure. The side of the square is 2 a. The moment of inertia of the lamina
about an axis through its centre of mass and perpendicular to the plane is 1.6 Ma2. The moment of
inertia of the lamina about the tangent AB in the plane of lamina is KMa2 then K is :
3) Potential energy of a particle moving on a circle in x-y plane is given by Where r is
the radius of circle and is the angle made by the radius vector of the particle with the positive x-

axis. Magnitude of x-component of force, when the particle is at given by Then

find the value of

4) A uniform beam rests on a rough horizontal floor at a point A and is held by a rope at a point B
(Fig.). The coefficient of friction is 0.2. The beam forms an angle α=45° with the floor. Determine
the angle of inclination of the rope to the horizontal when the beam just starts to slide. Fill tanf in

OMR sheet.

5) A bead can slide on a smooth straight wire and particle of mass m is attached to the bead by a
light string of length L. The particle is held in contact with the wire with the string taut and is then
let fall. If the bead has mass 2m. Then, when the string makes an angle θ = 60°. Determine the

distance which the bead slips on the string if L = 12 m.

6) Find the moment of inertia (in kg.m2) of a thin uniform square sheet of mass M = 3kg and side a =

2m about the axis AB which is in the plane of sheet :

CHEMISTRY

SECTION-I(i)

1) In a container initially only N2O4 is present. As shown in the diagram

If due to dissociation of N2O4 at constant temperature the difference

in the column of mercury becomes 7.6 cm then calculate % dissociation of N2O4.
Fill your answer as sum of digits (excluding decimal places) till you get the single digit
answer.

(A) 0
(B) 1
(C) 2
(D) 3

2) According to VSEPR theory, which of the following structure is incorrect ?

(A)

(B)
(C)

(D)

3) Which of the following is incorrect for real gas at high pressure and room temperature

(A) Z > 1
Z=1+
(B)

Z=1+
(C)

(D) Repulsive forces dominant

4) CH3OH(g) can be prepared using equation

; Kc = 4 × 102
At equilibrium, a 5 L flask contains equal moles of CO(g) and CH3OH(g). Hence number of moles of
H2 at equilibrium is

(A) 0.25 mol

(B) 0.10 mol
(C) 0.50 mol
(D) 0.125 mol

SECTION-I(ii)

1) Which of the following statements are correct ?

(A) It is not possible to compress a gas at a temperature below TC

At a temperature below TC, the molecules are close enough for the attractive forces to act and
(B)
condensation occurs
(C) No condensation takes place above TC
(D) Boyle’s temperature always greater than TC.

2) Which of the following species is/are linear ?

(A) C2H2
–
(B) N3
(C) SCN–
+
(D) NO2

3) Consider the following molecule If π–electron cloud of C1—C2 is present in

the plane of paper then incorrect statement is/are:

(A) Fluorine is perpendicular to the plane of paper

(B) Chlorine is present in the plane of paper
(C) σ–bond of C2 — C3 is perpendicular to the plane of paper
(D) π–electron cloud of C2 — C3 bond and Cl is present in same plane.

4) Choose the correct options:-

(A) Formal charge on 1st oxygen = 0

(B) Formal charge on 2nd oxygen = –1
(C) Formal charge on 3rd oxygen = –1
(D) Formal charge on P is O.

5) In an evacuated vessel of capacity 100 litres, 4 moles of Argon and 5 moles of PCl5 were
introduced and equilibrated at a temperature of 227 °C. At equilibrium, the total pressure of the
mixture was found to be 4.8 atm. Select the correct relation about the reaction.
PCl5(g) PCl3(g) + Cl2(g); at this temperature. (R = 0.08 L atm/mol.K)

(A) Degree of dissociation of PCl5 is 0.6

(B) KP = 1.8 atm
(C) Degree of dissociation of PCl5 is 0.5
(D) Total mole at Equilibrium = 12

6) Considering the following ionisation steps :

A(g) → A+ (g) + e¯ ΔH = 100 eV
2+
A(g) → A (g) + 2e¯ ΔH = 250 eV
Select the correct statements :

(A) IE1 of A(g) is 100 eV

+
(B) IE1 of A (g) is 150 eV
(C) IE2 of A(g) is 150 eV
(D) IE2 of A(g) is 250 eV
7) Consider the following Z vs P graph for two difference real gases P and Q at the same
temperature.

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

(A) Gas P has greater attraction forces in lower pressure region

(B) Gas Q has greater repulsion forces in higher pressure region
(C) Gas P shows lesser deviation from ideal gas
(D) Gas Q shows lesser deviation from ideal gas

8) The graph below shows the distribution of molecular speed of two ideal gases X and Y at 200K. on
the basis of the below graph identify the correct statements –

(A) If gas X is methane, then gas Y can be CO2

(B) Fraction of molecules of X must be greater than Y in all range of speed at 200K
(C) Under identical conditions, rate of effusion of Y is greater than that of X
(D) The molar kinetic energy of gas X at 200K is equal to the molar kinetic energy of Y at 200K

SECTION-II

1) In a molecule of NO2NH2 :
the total number of lone pairs = x N – H bond order = y What is the value of (2x – 3y) ?

2) 0.5 l of evacuated container is filled by gas upto 1 atm exactly, by connecting it to 20 litre cylinder

initially at 1.2 atm. How many evacuated containers can be filled.

3) Air is trapped in a horizontal gas tube by 36cm mercury column as shown below :
 If the tube is held vertical keeping the open end up, length of air
column shrink to 19cm. What is the length (in cm) by which the mercury column shifts down ?

4) How many of the following can react with bases :

Mn2O7, ZnO, Al2O3, BeO, SeO2, Cl2O7, Fe2O3, As2O3

5) Consider the following molecule

Calculate value of , here p and q are total number of dπ - pπ bonds and total number of sp3
hybridized atoms respectively in given molecule.

6) The gaseous reaction : A(g) + nB(g) mC(g) is represented by following curves

What is the value of 16Kp ?

MATHEMATICS

SECTION-I(i)

1) Let A ≡ (4,4), B ≡ (8,4), C ≡ (4,8). If P,Q,R are the midpoint of sides AB, BC & CA respectively &
(α, β) be the co-ordinates of orthocentre of ΔPQR, then the value of α + β is 6 -

(A) 8
(B) 6
(C) 12
(D) 16
2) Tangents are drawn to circle x2 + y2 = 4 at its distinct intersection points with the circle
. Find locus of point of intersection of tangents.

(A) 27x – 9y = 28
(B) 9x + 3y = 14
(C) 9x + 14y = 9
(D) 27x – 9y = 14

3) If the sum of values of in for which system of

equations and has a solution is then kπ is equal to

(A) 5/2
(B) 7/2
(C) 9/2
(D) 11/2

4) Locus of a point which moves in such a way that ratio of its distance from x axis to y axis is 3 : 2,
is S = 0 then which of the following point does not lie on S = 0

(A) (4, –6)

(B) (–3, –2)

(C)

(D) (6, 9)

SECTION-I(ii)

1) Let f(x) = . The minimum value of f(x) (given x > 1), is :

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

2) If the variable line L ≡ y = mx + c cuts the curve y = x2 – x at A and B such that angle AOB = 90°
(Where O is origin). Then

(A) m + c – 1 = 0
(B) m – c + 1 = 0
(C) Line 'L' passes through (1, 1)
(D) Line 'L' passes through (–1, –1)
3) Let the minimum value of be λ and the equation of the two
sides of a triangle are 4x + 5y – 20 = 0 and x – y + 4 = 0. If the circum-centre of the triangle be

, then

(A)
centroid of the triangle is
(B) orthocentre of the triangle is (1, 5)
(C) area of the triangle is 9 sq. units
(D) circum-radius of the triangle is units

4)

Consider the circle x2 + y2 – 10x – 6y + 30 = 0. Let O be the centre of the circle and tangent at A(7,
3) and B(5, 1) meet at C. Let S = 0 represents family of circles passing through A and B, then

(A) area of quadrilateral OACB = 4

(B) the radical axis for the family of circles S = 0 is x + y = 10
(C) the smallest possible circle of the family S = 0 is x2 + y2 – 12x – 4y + 38 = 0
(D) the coordinates of point C are (7, 1)

5) If C1, C2 be circles defined by (x – 10)2 + y2 = 36 and (x + 15)2 + y2 = 81, if ‘l’ is the length of
common tangent, then l can be :

(A) 20
(B)
(C)
(D)

6) If one vertex of a equilateral triangle of side length 2 lies at the origin and other lies on the line
then the co-ordinates of the third vertex can be

(A) (0, 2)

(B)

(C)

(D) (0, –2)

7) A circle is tangent to line 3x – 4y + 1 = 0 at (1, 1) and is normal to line 2x + y – 5 = 0. Choose

correct option(s)?

(A) Length of tangent to circle from (–1, 1) is 4 units

(B) Length of perpendicular from centre of circle to x-axis is 3 units.
(C) x-intercept of director circle of given circle is
(D) x-intercept of director circle of given circle is
8) If x,y,z are integers such that x + y + z = 0 and , then :

(A) x2 + y2 + z2 = 6
(B) x3 + y3 + z3 = 8
(C) xy + yz + zx = –3
(D) xyz = 8

SECTION-II

1) If , where x, y ∈ {–2, –1, 0, 1, 2}, then the number of ordered pairs of (x, y) is

2) In a sequence of circles C1, C2, C3, ..., Cn ; the centers lie along positive x-axis with abscissae
forming an arithmetic sequence of first term unity and common difference 3. The radius of these
circles are in geometric sequence with first term unity and common ratio 2. If the tangent lines with
slope m1 and m2 of C3 are intersected at the center of C5, then the value of |m1m2| is equal to
.

3) The radius of smallest circle touching the co-ordinate axes and passing through (8, 9) is :-

4)
As shown in the figure, a ray of light leaves the point (3,k) reflects off the y-axis towards the x-axis
and reflects off the x-axis to reach the point (15,8), now a ray of light leaves the points (15,8) reflects

off the y-axis towards the x-axis and reflects off the x-axis then value of is

5) Consider two points A(2, 4) & B(1, 2). Let P(a, a) be a point on the straight line L : y = x such that
AP + PB is minimum, then 'a' is equal to

6) Let τ be a circle with centre C(3, 5) and PA and PB are pair of tangents drawn from an external
point P(9, 11) to the circle τ, then the distance between the origin and the point inside the
quadrilateral ACBP which is equidistant from its four vertices is
 ANSWER KEYS

PHYSICS

SECTION-I(i)

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

SECTION-I(ii)

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

SECTION-II

Q. 13 14 15 16 17 18
A. 70.00 4.8 8.00 7.00 2 7.00

CHEMISTRY

SECTION-I(i)

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

SECTION-I(ii)

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

SECTION-II

Q. 31 32 33 34 35 36
A. 9.00 8.00 9.00 7.00 1.00 9.00

MATHEMATICS

SECTION-I(i)

Q. 37 38 39 40
A. C A C B

SECTION-I(ii)

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

Q. 49 50 51 52 53 54
A. 5.00 0.80 5.00 8.00 2.00 10
 SOLUTIONS

PHYSICS

1) –mg.2R + = Δx = 0 –

– 2mgR – =

u=

2) Mu = m (V cos 60 – u ) [from conservation of momentum]

2.4 × 0.2 = 0.1

V = 10 m/s

3)
Just before 1st collision
using momentum conservation
m1 u = – m1v1 + m2v2
⇒ m1u = m2v2 – m1v1

v2 + v1 = eu → (2)

v1 = eu – v2 = eu –

After collision with wall only dirn of velocity changes.

For again collision with balls.
⇒ e m2 – m1 > m1 + e m1

4)

Now,

or

5)

As it has tendancy to rotade about it's left end so

6)

(A)
(B) = vertical
(C)
(D) ellipse

7) mgR cosθ = 2 mR2α

α=

mg R sin θ =
g sin θ = Rω2

ω2 =

mg cosθ + Ry = mR =

Ry =

Rx – mg sin θ = mR ×
Rx = 2mg sin θ

8) vA = w0R – v0 = v
ω0R – v0 = v ...(i)
vB = ω0 R + v0 = 3v ...(ii)
From equation (i) and (ii)
2ω0R = 4V
⇒ ω0R = 2v

0
ω =
From (1) v0 = v

9)

=m×

In the given position α = 0 since Στcm = 0

10) For (A) since loss in potential energy in both case is same.
hence kinetic energy at θ0 = 0 will be same in both case.
For (D) in case (2) slender bar will have more speed because it has only translational kinetic
energy.

11) Let r be the distance between 1 and 2 at any time t.

...(1)

...(2)

Also,

Initially
Relative to 1, 2 moves such that its distance from 1 and line AB remains same so, it follows a
parabolic path.

12)
As shown in figure, anet = 0 when
.... (1)

Also,

13)
Balancing torques about O
Mg × .1 = 10 g × (.3) + 20g × (.2)
M = 70 gm

14) Using ⊥ axis theorem

ℓx = ℓy
2Ix = 1.6
Ix = .8 Ma2
IAB = Ix + M(2a)2
= 4.8 Ma2

15) Tangential component of force

Magnitude of the x-component of force

16)
(τA)N = 0

1.2N =

∴
(τB)N = 0

T(sinα – cosα) =

tanα – 1 = 6
tanα = 7
 17)

2M rightward

18) I = + = = 7.

CHEMISTRY

19)
N2O4 → 2NO2
1
1–P 2P
1 – p + 2P = 1.1
P = 0.1
α % = 10

21)

At high pressure
Z>1
 (Vm – b) = RT

At higher pressure can be neglected

P(Vm – b) = RT
PVm – Pb = RT

22)
teq a b a

23)

At a temperature below TC, the molecules are close enough for the attractive forces to act and
condensation occurs
No condensation takes place above TC
Boyle’s temperature always greater than TC.

24) (A) HC ≡ CH
(B) N ≡ N : → N– or N– = N+ = N–
(C) S = C = N–
(D) O = N+ = O

26)

Formal charge on each atom

27) nTotal =

nTotal = 5 – 5α + 5α + 5α + 4 = 12
α = 0.6

28)

29)

Greater the value of ‘a’ greater is the depth in curve in lower pressure region,
Greater the ‘b’ value, greater is the slope in high pressure region.

30)
For option (A) From Fig (i), if is clear that (umps)Y > (umps)X

or,
since both X and Y are at same temp.

∴ ......... (1)
since X is CH4 whose molar wass is 16.
∴ Y can't be CO2 because.
∴ (A) is wrong.
for option (B), see Fig (ii)
(ii)
in lower speed region shaded area in figure, fraction of molecules of X > fraction of molecules
of Y.
As shaded area is greater for X as compared to Y. In higher speed region, shaded area in fig.
fraction of molecules of Y > fraction of molecules of X. Therefore (B) is wrong.
for option 'C'
under identical conditions of P & T.

rate of effusion
∴ MY < XX (from (i))
∴ rate of effusion of Y > X.
∴ (C) is correct.
for option 'D'

Molar KE =
∴ Both X and Y are at same tempt. = 200k
∴ Molar (KE)X = Molar(KE)Y.
∴ (D) is correct.

31)
No. of lone pair = 6
NH-Bond order = 1
2 × 6 – 3 × 1 = 9.

32)

33)

P final = final height = 19cm

P initial = 1atm, initial length = hi cm
according to Boyle’s Law
Pivi = Pfvf
 hi = 28cm
The length by which the Hg column shifts down = hi - hf = 28 - 19 = 9cm

34)

Amphoteric oxide - ZnO, Al2O3, BeO, PbO2, As2O3

Acidic oxide - Mn2O7, Cl2O7
Basic oxide - Na2O
Amphoteric and acidic oxide reacts with Base.

35)
Total number of pπ – dπ bond = 6 = p
Total number of sp3 hybridised atom = 6 = q

36)
t=0 8M 12M
. 8 - x 12 - 2x mx
Given:
8-x=6
x=2M

12 - nx = 8
n(2) = 4
n=2

mx = 6
m(2) = 6
m=3
m+n=3+2=5

MATHEMATICS

37) Circumcentre of ΔABC

⇒ Orthocentre of ΔPQR

38) Equation of chord of contact w.r.t. P

xh + yk = 4 ...(1)
Equation of common chord
...(2)
Compare (i) and (ii)

∴ 27x – 9y = 28

39)

40)

Locus will be : 3|x| = 2|y|

41)

y= = x(1 – 0) – –

=x–1+ + = (x – 1) +
By applying A.M.–G.M. inequality, we have

y = (x – 1) + ≥ =4
(As x > 1) Ans.
42)
Homogenising, we get

= x2 – .x
2 2 2
y – mxy = cx – xy + mx
⇒ (m + c) x2 – (1 + m) xy – y2 = 0
⇒m+c=1
y = mx + c
y = mx + 1 – m ⇒ (1 – y) + m (x – 1) = 0
⇒ Line passes though (1, 1)

43)

Which represents the sum of the distances of from the points (0, 0) and (1, 3)
Again lies on the line x – y + 1 = 0 and points (0, 0) and (1, 3) lie on the opposite
side of x – y + 1 = 0

Let the equations of sides BC and AC are 4x + 5y – 20 = 0 and x – y + 4 = 0 respectively of

ΔABC.
∴ P(5, 0) lies on the side 4x + 5y – 20 = 0
⇒ P(5, 0) is the circum-centre
⇒ C(0, 4) ⇒ B(10, –4) and vertex A must be the ortho-centre.
Any point on AC is (α, α + 4)

⇒ Slope of AB × slope of AC = –1
44)

Coordinates of O are (5, 3) and radius = 2

Equation of tangent at A(7, 3) is 7x + 3y – 5(x + 7) – 3 (y + 3) + 30 = 0
i.e. 2x – 14 = 0 i.e. x = 7
Equation of tangent at B(5, 1) is 5x + y – 5(x + 5) – 3(y + 1) + 30 = 0
i.e. – 2y + 2 = 0 i.e. y = 1
coordinate of C are (7, 1)
area of OACB = 4
Equation of AB is x – y = 4 (radical axis)
Equation of the smallest circle is
(x – 7) (x – 5) + (y – 3) (y – 1) = 0
i.e. x2 + y2 – 12x – 4y + 38 = 0

45) Length of direct common tangent =

Length of transverse common tangent =

Where d = distance between the centres r1, r2 are radius of circles respectively

46) 4 possible triangle can be draw

47) (x – 1)2 + (y – 1)2 + λ(3x – 4y + 1) = 0

x2 + y2 + (3λ – 2)x – 2(2λ + 1)y + (2 + λ) = 0
–3λ + 2 + 2λ + 1 – 5 = 0
⇒ –λ – 2 = 0 ⇒ λ = –2

x2 + y2 – 8x + 6y = 0 (x – 4)2 + (y + 3)2 = 50
⇒

48) Given D.D2 = 64

⇒ (2xyz)3 = 64
⇒ xyz = 2
x = 2, y = –1, z = –1 or x = –2, y = 1, z = 1

49) Expand determiant

so x3 + y3 – 1 + 3xy = 0
⇒ x3 + y3 + (–1)3 – 3(x)(y)(–1) = 0
⇒ [x + y – 1][(x – y)2 + (x + 1)2 + (y + 1)2]=0
⇒ either x + y = 1 or x = y = –1
∴ number of ordered pair (x,y)
≡ (2,–1),(–1,2),(0,1), (1,0),(–1,–1)

50) Centre of circle C3 is (7, 0) and radius = 4

centre of circle C5 is (13, 0)

sinα =

slopes of tangents are

|m1m2| = = 0.8

51) Equation of circle touching both axes is

2 2 2
(x – r) + (y – r) = r
x2 + y2 – 2rx – 2ry + r2 = 0
64 + 81 – 16r – 18r + r2 = 0
r2 – 34r + 145 = 0
(r – 29)(r – 5) = 0
∴ r = 5, 29

52)

–15y + 16x = 3xy...(1)

15y – 8x = xy...(2)

(1) + (2) 8x = 3xy

∴
⇒

53) As shown in figure, Let A' be the image of point A in line L

∴ AP + PB = A'P + PB
For any point P on line L,
A'P + PB ≥ A'B

∴ Minimum value of A'P + PB is A' B

when A', P, B are collinear
Now, coordinates of A' will be reflection of A(2, 4) in the line L : y = x
⇒ A' ≡ (4, 2)
Equation of A'B : y = 2
∴ Point P(a, a) will be (2, 2)
∴ a=2

54) The point inside the quadrilateral

ACBP which is equidistant from all the four vertices is the centre M(6, 8) of the circle
discribed on PC as diameter.
Hence, distance from origin to the point M is = = 10.