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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 $R = \frac{200}{t} cm$ 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
Time Velocity of image Acceleration of image
A) 1 sec P) $\frac{400}{49} cm / s$ 1) $- \frac{64}{5} cm / s^{2}$
B) 2 sec Q) 16 cm/s 2) $- \frac{1600}{343} cm / s^{2}$
C) 3 sec R) $\frac{400}{9} cm / s$ 3) $- \frac{1600}{27} cm / s^{2}$
D)4 sec S)400 cm/s 4) –1600 cm/s²
Which of the following is correctly matched

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(A)(S)(4)

(A)(S)(1)

(A)(Q)(1)

(A)(Q)(4)

answer is A.

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Detailed Solution

$f = \frac{100}{t} \frac{1}{v} + \frac{1}{- 200} = \frac{- t}{100} \Rightarrow \frac{∣ 1}{v} = \frac{1}{200} - \frac{t}{100} = \frac{1 - 2 t}{200} v = \frac{200}{1 - 2 t} Velocity of image = \frac{dv}{dt} = \frac{(200) \times 2}{(1 - 2 t)^{2}} c m / s = \frac{400}{(1 - 2 t)^{2}} cm / s acceleration of image = \frac{d^{2} v}{{dt}^{2}} = \frac{2 (400) (2)}{(1 - 2 t)^{3}} = \frac{1600}{(1 - 2 t)^{3}} cm / s^{2}$
A) t = 1sec, velocity 400 cm/s, acceleration = -1600 cm/s²
B) $t = 2 \sec, velocity = \frac{400}{9} cm / s acceleration = - \frac{1600}{27} cm / s^{2}$
C) $t = 3 \sec, velocity = \frac{400}{25} = 16 cm / s acceleration = - \frac{1600}{125} = - \frac{64}{5} cm / s^{2}$
D) $t = 4 \sec, velocity = \frac{400}{49} cm / s, acceleration = - \frac{1600}{343} cm / s^{2}$

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