Chemistry and Physics With Solution

8/2/2019 Chemistry and Physics With Solution

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Rise AcademyA-2, Jawahar Nagar, KOTA (Raj.) -324005, 278-Rajeev Gandhi Nagar, Kota 1

Part I : CHEMISTRYSECTION - I (Total Marks : 21)

(Single Correct Answer Type)

This section contains 7 multiple choice questions. Each question has four choices

(A), (B), (C) and (D) out of which ONLY ONE is correct.

1. Bombardment of aluminum by -particle leads to its artificial disintegration into two ways, (i) and (ii) as shown.Products X, Y and Z respectivey are :

(A) Proton, neutron, position (B) Neutron, position, proton

(C) proton, positron, neutron (D) positron, proton, neutornAns. (A)

Sol. So, X11 is proton ( p11 ), Y

10 is neutron( n

10 ) and Z

01 is positron( e

01 ).

2. Dissoving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the

solution is :(A) 1.78 M (B) 2.00 M (C) 2.05M (D) 2.22 M

Ans. (C)

Sol. Moles of solute (urea) =60

120= 2

Mass of solution = 1000 + 120 = 1120 gm

Volume of solution =15.1

1120ml

So, molarity of solution = )lt(solutionofvolume

soluteofmoles= M05.2

1120

100015.12


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3. AgNO3(aq) was added to an aqueous KCl solution gradually and the conductivity of the solution was measured.

The plot of conductance ( ) versus the volume of AgNO3is

(A) (P) (B) (Q) (C) (R) (D) (S)

Ans. (D)

Sol. When AgNO3

(aq) was added to an aqueous KCl solution , due to Ag+ (aq)+ Cl- (aq) AgCl(s)reaction ,Cl- ions are replaced by NO-

3so no apprieciable change in of solution . After equivalant point Ag+

& NO3

- will be increased in solution so of solution increases.

4. Among the following compounds, the most acidic is :(A) p-nitrophenol (B) p-hydroxybenzoic acid

(C) o-hydroxybenzoic acid (D) p-toluic acidAns. C (o-hydroxybenzoic acid)

Sol. (A) p-nitrophenol (B) p-hydroxybenzoic acid

(C) o-hydroxybenzoic acid (D) p-toluic acid

Stability of conjugate base

< < Angular frequency for case B(D) Angular frequency for case A < Angular frequency for case B.

Sol. (A), (D)

Restoring torque in both cases will be equal but moment of inertia of rotating system will be less in case B as

disc is free to rotate about its centre, so time period will be less for case B and hence angular velocity will begreater.

33. A spherical metal shell A of radius RA

and a solid metal sphere B of radius RB

(< RA) are kept far apart and each

is given charge +Q. Now they are connected by a thin metal wire. Then :

(A) 0EinsideA (B) QA > QB (C)A

B

B

A

R

R

(D) surfaceOnB

surfaceanA EE

Sol. (A), (B), (C), (D)

0EinsideA

A B

A B

KQ KQ

R R

A A

B B

Q R

Q R

A A

A B A B

Q R

Q Q R R

QA R

A Q

A> Q

B

= 2Q

4 R


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A

1

R

A B

B A

R

R

Esurface

=0

So EA

< EB

34. An electron and a proton are moving straight parallel paths with same velocity. They enter a semi-infinite region

of uniform magnetic field perpendicular to the velocity. Which of the following statement(s) is/are true ?(A) They will never come out of the magnetic field region

(B) They will come out travelling along parallel paths(C) They will come out at the same time(D) They will come out at different times.

Sol. (B), (D)

T =2 m

qB

time period depends on mass charge ratio.

SECTION - III (Total Marks : 15)

(Paragraph Type)

This section contains 2 paragraphs. Based upon one of the paragraphs 3 multiple

choice questions and based on the other paragraf 2 multiple choice questions have

to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of

which ONLY ONE is correct.

Paragraph for Question Nos. 35 to 37

Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. momentum are changed.

here we consider some simple dynamical systems in one-dimension. For such systems, phase space is aplane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phasespace diagram is x(t) vs. p(t) curve in this plane. The arrow on the curve indicated the time flow. For example, the

phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We usethe sign convention in which position or momentum upwards (or to right is positive and downwards (or to left) is

negative.


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35. The phase space diagram for a ball thrown vertically up from ground is

(A) (B)

(C) (D)

Sol. (D)For a ball thrown upward momentum is +ve max at the intial position (Postion zero)

And then when it returns to the same position zero its momentum gets inverted (ie becomes negative)

36. The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two

circles represent the same oscillator but for different initial conditions, and E1

and E2

are the total mechanicalenergies respectively. Then

(A) E1

= 22E (B) E1 = 2E2 (C) E1 = 4E2 (D) E1 = 16 E2

Sol. (C)

We have

1

2m2 AA2 = E

E1

=1

2m2(2a) 2 E

2=

1

2m2 a2

E1

= 4 E2


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37. Consider the spring - mass system, with the mass submerged in water, as shown in the figure. The phase spacediagram for one cycle of this system is

(A) (B)

(C) (D)

Sol. (B)

Momentum is zero as the block starts from rest when the spring is compressed to its maximum (position ofblock will be taken as positive by the convention given) and then it comes down (momentum negative) ; the

block moves down till the spring is compressed again to its maximum (this time the position is negative) andfrom this position the block starts to move up(momentum positive). moreover there is loss of energy in process Final momentum is less than initial momentum

Paragraph for Question Nos. 38 and 39

A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids

containing fixed positive ions surrounded by free electrons can be treated as neutral plasema. Let N be thenumber density of free electrons, each of mass m When the electrons are subjected to an electric fieldthey are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the

electrons begin to oscillate about the positive ions with a natural angular frequency p

which is called theplasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an

angular frequency p where a part of the energy is absorbed and part of it is reflected As approaches pall the free electrons are set to resonace together and all the energy is reflected. This is the explanation of

high reflectivity of metals.

38. Taking the electronic charge as e and the permittivity as 0

, use dimensional analysis to determine thecorrect expression for

p.

(A)0

Ne

m (B)0m

Ne

(C)

2

0

Ne

m (D)02

m

Ne


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Sol. (C)

2 3 2 2 2 2

2 20

Ne L T MLT L

m M T

= 2T = 1T (same as )

39. Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons N ~

4 1027 m3. Take e0

~ 1011 and m = 1030, where these quantities are in proper SI units.

(A) 800 nm (B) 600 nm (D) 300 nm (D) 200 nm

Sol. (B)

=2

0

Ne

m

c = f

c =2

= 02mc(2 ) 2 cNe

By putting values, we have

= 600 nm

SECTION - IV (Total Marks : 28)

(Integer Answer Type)

This section contains 7 questions. The answer to each of the questions is a

single - digit integer, ranging from 0 to 9. The bubble corresponding to the correctanswer is to be darkned in the ORS.

40. Four point charges, each of +q, are rigidly fixed at the four corners of a square planar soap film of side The surface tension of the soap film is . The system of charges and planar film are in equilibrium, and a =

k

1/N2q

, where k is a constant. Then N is ]

Ans. 3

Sol.


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2ra =

2

2

2Kq

a+

2

2

2Kq

2a

1

2

a3 =2Kq

r

11

2 2

a =

1/32 1Kq 1

2 2

r

N = 3

41. A long circular tube of length 10 m and radius 0.3 m carries a current along its curved surface as shown. Awire-loop of resistance 0.005 ohm and of radius 0.1 m is placed inside the tube with its axis coinciding with

the axis of the tube. The current varies as = 0

cos (300 t) whre 0is constant. If the magnetic moment of

the loop is N 0

0sin (300 r), then N is

Ans. 6

Sol. M = i NA N = 1

M = iA

M =R

A =A

R

d

dt

=2A

R

dB

dtA = TTr2

B = . ni =.i

r = 0.1 m

dB

dt =

di

dt =

0

I (300) sin (300t) R = 0.005

M =2A

R

0

300 I

0sin (300t) = 10 m

M =

2A 300

R

.I0

sin(300t)

putting the values of A, R, we get the answer N = 6


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42. A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick applies

a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3 m/s2. The coefficient of

friction between the ground and the ring is large enough that rolling always occurs and the coefficient of

friction between the stick and the ring is (P/10). The value of P is

Ans. 4

Sol.

2 0.5 N(0.5) = 2(2) (0.5)20.3

0.5

2 2 = 1.20.8 = 2 = 0.4

43. Four solid spheres each of diameter 5 cm and mass 0.5 kg are placed with their centers at the corners of

a square of side 4 cm. The moment of inertia of the system about the diagonal of the square is N 104 kgm2, then N is

Ans. 9

Sol.

22 22 2 aI mR 2 2 mR m

5 5 2

a/ 2

2 2 24 4I mR mR ma5 5

=2 28 mR ma

5

=4 48 1 5 110 16 10

5 2 4 2

[1 + 8] 104 = 9 104


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44. The activity of a freshly prepared radioactive sample is 1010 disintegrations per second, whose mean life is

109s. The mass of an atom of this radioisotope is 1025 kg. The mass (in mg) of the radioactive sample is

Ans. 1

Sol.10dN 10

dt 91 10

m = 1025

dN Wdt

1010 = 91

N10

19 25N 10 10 m = 106 kg

= 1 mg

45. A block is moving on an inclined plane making an angle 45 with the horizontal and the coefficient of frictionis . Thje force required to just push it up the inclined plane is 3 times the force required to just prevent it

from sliding down. If we define N = 10 , then N isAns. 5Sol. F

1= 3 F

2

(mgsin + mgcos ) = 3 mg (mgsin mgcos )N F2

Nsin + cos = 3 sin 3 cos4cos = 2sin

4N2

10 (as = 45)

20N 5

4

46. Stell wire of length L at 40C is suspended from the ceiling and then a mass m is hung from its free end.The wire is cooled down from 40C to 30C to regain its original length L. The coefficient of linear thermalexpansion of the steel is 105/C, Youngs modulus of steel is 1011 N/m2 and radius of the wire is 1 mm.

Assume that L >> diameter of the wire. Then the value ofm

in kg is nearlyAns. 3

Sol.L

L T

L = L T

mg = K L

m =YA

Lg L

11 6 510 10 10 10m

10

40 C 30 C

m = 3.14 3


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Part III : MATHEMATICS

SECTION - I (Total Marks : 21)

(Single Correct Answer Type)

This section contains 7 multiple choice questions. Each question has four choices(A), (B), (C) and (D) out of which ONLY ONE is correct.

47. Let a i j k

, b i j k

and c i j k

be three vectors . A vector v

in the plane of a

and b

,

whose projection on c

is1

3is given by

(A) i 3 j 3k (B) 3i 3 j k (C) 3i j 3k (D) i 3 j 3k

Ans. (C)

Sol.47 Since a i j k

, b i j k

v a b

(where are real)

= i j k

Projection of v on c

=v. c ( ) ( ) ( ) 1

| c | 3 3

= 1

Hence a vector satisfying above condition would be (C) (for = 1, = 2).

Hence (C)

48. Let P : sin cos 2 cos and Q : sin cos 2 sin be two sets. Then

(A) P Q and Q P (B) Q P(C) P Q (D) P = Q

Ans. (D)

Sol.48 sin ( 2 1)cos (P) , cos ( 2 1)sin (Q)

(Q) ( 2 1)cos sin Hence P = Q

Hence (D)

49. Let the straight line x = b divide the area enclosed by y = (1 x)2, y = 0 and x = 0 into two parts R1

(0 x b) and R2 (b x 1) such that R

1 R

2=

1

4

. Then b equals

(A)3

4(B)

1

2(C)

1

3(D)

1

4

Ans. (B)


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Sol.49. R1 R

2=

1

4

=

b 12 2

0 b

1(x 1) dx (x 1) dx

4

= solving this we get1

b2

Hence (B)

50. Let and be the roots of x2 6x 2 = 0 , with . If an

= nn for n 1, then the value of10 8

9

a 2a

2a

is

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

Ans. (C)

Sol.50 10 69 28 = 0 {Since are roots}

10 69 28 = 0

Subtracting these two, we get, a10

6a9 2a

8= 0

10 8

9

a 2a3

2a

Hence (C)

51. A straight line L through the point (3, 2) is inclined at an angle 60 to the line 3x y 1 . If L also intersects

the x-axis, then the equation of L is

(A) y 3x 2 3 3 0 (B) y 3x 2 3 3 0

(C) 3y x 3 2 3 0 (D) 3y x 3 2 3 0

Ans. (B)

Sol.51 Given line makes an angle 120 with x-axis, the required line will make 120 60 (120 + 60 is rejected,

as it will not meet x-axis) with x-axis hence its equation is y + 2 = 3(x 3) .

Hence (B)

52. Let (x0

, y0) be the solution of the following equations

ln2 ln3

(2x) (3y)ln x lny3 2

Then x0

is

(A)1

6(B)

1

3(C)

1

2(D) 6

Ans. (C)


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Sol.52 Taking log to the base e in both equations

ln2 {ln2 + lnx} = ln3 {ln3 + lny} ...............(i)

lnx. ln3 = lny . ln2 ...............(ii)

Substituting lny from (ii) in (i)

we get ln2 2 ln2 3 =2 2ln 3 ln 2

.lnxln2

1x

2 Hence (C)

53. The vgalue of

ln3 2

2 2

ln2

xsinxdx

sinx sin(ln6 x ) is

(A)1 3

ln4 2

(B)1 3

ln2 2

(C)3

ln2

(D)1 3

ln6 2

Ans. (A)

Sol. 53 Substitute x2 = t and let the integral be I

ln3

ln2

1 sintI dt2 sin t sin(ln6 t)

ln3

ln2

1 sin(ln6 t)I dt

2 sin(ln6 t ) sin t

b b

a a

f(x) dx f (a b x) dx

ln3

ln2

12I dt

2 1 3I ln

4 2

Hence (A)_________________________________________________________________________________________________

SECTION - II (Total Marks : 16)

(Multiple Correct Answer Type)

This section contains 4 multiple choice questions. Each question has four choices

(A), (B), (C) and (D) out of which ONE or MORE may be correct.

____________________________________________________________________________

54. Let the eccentricity of the hyperbola

2 2

2 2

x y1

a b

be reciprocal to that of the ellipse x2 + 4y2 = 4. If the

hyperbola passes through a focus of the ellipse, then

(A) the equation of the hyperbola is2 2x y

13 2

(B) a focus of the hyperbola is (2, 0)

(C) the eccentricity of the hyperbola is5

3(D) the equation of the hyperbola is x2 3y2 = 3

Ans. (B, D)


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Sol.54 eellipse

=3

2 H

2e

3

focus of ellipse is 3, 0 , since hyperbola passes through focus

we get a2 = 3 b2 = 1 H2

as e 3

Hence (B, D)

55. Let f : R R be a function such that

f(x + y) = f(x) + f(y), x, y R

If f(x) is differentiable at x = 0, then

(A) f(x) is differentiable only in a finite interval containing zero

(B) f(x) is continuous x R

(C) f '(x) is constant x R

(D) f(x) is differentiable except at finitely many points

Ans. (B, C)

Sol.55 f(x+y) = f(x) + f(y) x, y R.

f '(0) exists, also f(0) = 0 {subsitute x = y = 0 in the relation}

h 0

f(x h) f(x)f '(x) lim

h

=

h 0

f(h)lim

h

Nowh 0

f(h) f(0)f '(0) lim

h

=

h 0

f(h)lim

h= (say)

Hence f '(x) = x R f(x) = x

So f(x) is diff. hence cont. x R

Hence (B, C)

56. The vecotr(s) which is/are coplanar with vectors i j 2k and i 2 j k and perpendicular to the vector

i j k is / are

(A) j k (B) i j (C) i j (D) j k

Ans. (A, D)

Sol.56 Required vector v (i j 2k) (i 2 j k) ( i j k)

4(i 2 j k) 4(i j 2k)

= j k

Hence (A, D)


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57. Let M and N be two 3 3 non-singular skew-symmetric matrices such that MN = NM. If PT denotes thetranspose of P, then M2N2 (MTN)1(MN1)T is equal to

(A) M2 (B) N2 (C) M2 (D) MN

Ans. (C)

Sol.57 MN = NM, MT = M & NT = N

M

2

N

2

(M

T

N)

1

(MN

1

)

T

= MM. NN . N

1

(M

T

)

1

. (N

1

)

T

. M

T

= M.M.N(N)1 (M)1(M). (as M & N are commutative)

= M2

(Note that the, per se the above condtions, given in this question cannot hold true as all skew-symmetric

matrice of odd order are singular)

Hence (C)

________________________________________________________________________________________________

SECTION - III (Total Marks : 15)

(Paragraph Type)

This section contains 2 paragraphs. Based upon one of the paragraphs 3 multiplechoice questions and based on the other paragraf 2 multiple choice questions have

to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of

which ONLY ONE is correct._________________________________________________________________________________________________

Paragraph for Question Nos. 58 to 60

Let a, b and c be three real numbers satisfying

[a b c]

1 9 7

8 2 77 3 7

= [0 0 0] ....(E)

58. If the point P(a, b, c), with reference to (E), lies on the plane 2x + y + z = 1, then the value of 7a + b + c is

(A) 0 (B) 12 (C) 7 (D) 6

Ans. (D)

59. Let be a solution of x3 1 = 0 with Im() > 0. If a = 2 with b and c satisfying (E), then the value of

a b c

3 1 3

is equal to

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

Ans. (A)


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60. Let b = 6 with a and c satisfying (E). If and are the roots of the quadratic equation ax2 + bx + c = 0, then

n

n 0

1 1

is

(A) 6 (B) 7 (C) 67

(D)

Ans. (B)

(Solution for Question Nos. 58 to 60)

Sol.58 [a b c]

1 9 7

8 2 7

7 3 7

[0 0 0]

X . A = B (say)

| A | =

1 9 7

8 2 7

7 3 7= 0

Now given equations are

a + 8b + 7c = 0 ....(1)

9a + 2b + 3c = 0 ....(2)

7a + 7b + 7c = 0 ....(3)

(3) a + b + c = 0Let a = + b + c = 0

From (1) b = ( + c) ....(4)a + 8b + 7c = 0

+ 8( c) + 7c = 078c + 7c = 0

c = 7From (4) b = 7

= 6Hence solution can be siven as (, 6, 7)Satisfying solution in 2x + y + z = 1

2 + 6 7 = 1 (a, b, c) (1, 6, 7)

7a + b + c = 7 + 6 7 = 6 Hence (D)

Sol.59 Now

If a = 2 then solution is

(2, 12, 14) a = 2, b = 12, c = 14

Hence,

a

3

+ b

1

+ c

3

= 2

3

+ 12

1

+ 14

3

= 3 + 1 + 32

= 3( + 2) + 1 = 3 + 1 = 2. Hence (A)


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Sol.60 If b = 6 then = 1 a = 1, b = 6 & c = 7

Hence ax2 + bx + c = x2 + 6x 7 = (x + 7) (x 1) = 7, = 1

1 1

= 17

+ 1 = 67

n

n 01 1

=

n

n 067

=

161 7

= 7 Hence (B)

Paragraph for Qeustion Nos. 61 and 62Let U1 and U2 be two urns such that U1 contains 3 white and 2 red balls, and U2 contains only 1 white ball. Afair coin is tossed. If head appears then 1 ball is drawn at random from U1 and put into U2. However, if tailappears then 2 balls are drawn at random from U1 and put into U2. Now 1 ball is drawn at random from U2.

61. The probability of the drawn ball from U2 being white is(A) 1330 (B)

2330 (C)

1930 (D)

1130

Ans. (B)62. Given that the drawn ball from U2 is white, the probability that head appeared on the coin is

(A) 1723 (B)1123 (C)

1523 (D)

1223

Ans. (D)(Solution for Question Nos. 61 to 62)


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there are total 10 possibilities

possibility (3) implies that first head occurs then a-Red ball is transferred to Urn U2and then White is drawn

from bag U2

and its probability is1

2

2

5

1

2=

1

10- - - - - - and so on.

Now

61. According to question Required probability is

=1

2

3

5 1 +

1

2

2

5

1

2+

1

2

1

10 1 +

1

2

6

10

2

3+

1

2

1

10

1

3

=3

10+

1

10+

1

20+

1

5+

1

60

=23

30Hence (B)

62. According to question required probability is

(1) (3)P (1) (3) (5) (7) =3 1

10 1023

30

=

12

23Hence (D)

________________________________________________________________________________________________

SECTION - IV (Total Marks : 28)

(Integer Answer Type)

This section contains 7 questions. The answer to each of the questions is a

single - digit integer, ranging from 0 to 9. The bubble corresponding to the correct

answer is to be darkned in the ORS.______________________________________________________________________________

63. The positive integer value of n > 3 satisfying the equation

1

sinn

=1

2sin

n

+1

3sin

n

is

Ans. (7)

63. Let n

=

given1

sin( ) =1

sin(2 ) +1

sin(3 )

1

sin( ) 1

sin(3 ) =1

sin(2 )

sin(3 ) sin( )

sin( ).sin(3 ) =

1sin(2 )


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2sin( )cos(2 )

sin( )sin(3 ) =

1sin(2 )

2sin (2) cos (2) = sin (3) sin (4) = sin (3) [As sin () 0] as n > 3 sin (4) sin (3) = 0

2sin( 2

) cos (

7

2

) = 0

sin2n

cos7

2n

= 0

cos7

2n

= 0 n = 7 Hence (7)

64. The minimum value of the sum of real numbers a5, a4, 3a3, 1, a8 and a10 with a > 0 is

Ans. (8)

64. Apply A.M. G.M. inequality on the numbers

5

1

a, 4

1

a, 3

1

a, 3

1

a, 3

1

a, 1, a8 and a10

8 10

5 4 3 3 31 1 1 1 1 1 a aa a a a a

8

188 10

5 4 3 3 31 1 1 1 1 1 a aa a a a a

Required sum8 Required sum 8 Minimum value = 8. Hence (8)

65. Consider the parabola y2 = 8x. Let 1

be the area of the triangle formed by the end points of its latus rectum

and the point P1

, 22

on the parabola, and 2

be the area of the triangle formed by drawing tangents at P

and at the end points of the latus rectum.

Then1

2

is

Ans. (2)

65. If points P : (at12

, 2at1), Q ; (at22

, 2at2) and R : (at32

, 2at3) are on parabola also Let A, B, C are points ofintersection of tangents at these points then A : (at

1t2, a(t

1+ t

2)), B : (at

2t3

a(t2

+ t3)) and C : (at

1t3, a(t

1+ t

3))

Now

Area of PQR = | 1

|

Where 1

=1

2

21 1

22 2

23 3

at 2at 1

at 2at 1

at 2at 1=

22a

2(t

1 t

2)(t

2 t

3)(t

1 t

3) ...(1)


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Similarly

Area of ABC = |2|

Where

2

=1

2

1 2 1 2

2 3 2 3

3 1 3 1

at t a(t t ) 1

at t a(t t ) 1

at t a(t t ) 1

=2a

2(t

1 t

2)(t

2 t

3)(t

1 t

3)

Hence from (1) and (2)

1

2

= 2.

Note that there is no use of the fact that two of the points on parabola are ends of latus rectum.

Hence (2)

66. Let () = sin1 sintan

cos2

, where

4

Chemistry and Physics With Solution

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