Q.

An object O is kept on the principal axis of a concave mirror at a distance 200 cm from pole of the mirror. By some mechanism radius of curvature of mirror is changing with time as R=200tcm here t is in second. Column–I shows the time instant Column–II represents velocity of image at that time and Column –III represents acceleration of image at the time

Question Image
TimeVelocity of imageAcceleration of image
A) 1 secP) 40049cm/s1)645cm/s2
B) 2 secQ) 16 cm/s2) 1600343cm/s2
C) 3 secR)4009cm/s3)160027cm/s2
D)4 secS)400 cm/s4) –1600 cm/s2

Which of the following is correctly matched

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a

(A)(S)(4)

b

(A)(S)(1)

c

(A)(Q)(1)

d

(A)(Q)(4)

answer is A.

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Detailed Solution

f=100t 1v+1200=t1001v=1200t100=12t200 v=20012t Velocity of image =dvdt=(200)×2(12t)2 cm/s=400(12t)2cm/s acceleration of image =d2vdt2=2(400)(2)(12t)3=1600(12t)3cm/s2
A) t = 1sec, velocity 400 cm/s, acceleration = -1600 cm/s2
B) t=2sec, velocity =4009cm/s  acceleration =160027cm/s2
C) t=3sec, velocity =40025=16cm/s  acceleration =1600125=645cm/s2
D) t=4sec, velocity =40049cm/s, acceleration =1600343cm/s2

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