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A bird ‘B’ in air is diving vertically downwards over a water tank with speed 5 cm/s. Base of the tank is silvered. A fish ‘F’ in the tank is rising vertically upwards along the same line with speed 2 cm/s. Water level is lowered at the rate of 2 cm/s. Take $μ = \frac{4}{3}$
Column – I Column – II
A) Speed of image of fish as seen by the bird directly in cm/s p) 4
B) Speed of image of fish formed after reflection in the mirror as Seen by the bird in cm/s q) 8
C) Speed of image of bird as seen by the fish looking upwards in cm/s r) 3
D) Speed of image of bird as seen by the fish looking downwards in cm/s s) 6
t) 9

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$A \to s, B \to r, C \to q, D \to p$

$A \to p, B \to r, C \to q, D \to s$

$A \to s, B \to r, C \to p, D \to q$

$A \to r, B \to s, C \to q, D \to p$

answer is B.

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Detailed Solution

A) $D = x_{1} + \frac{x_{2}}{μ}$

$\frac{d D}{d t} = \frac{d x_{1}}{d t} + \frac{1}{μ} \frac{d x_{2}}{d t}$

$= (- 3) + \frac{3}{4} (- 4) = - 6 c m / s$

B) $D = x_{1} + (\frac{x_{2} + 2 h}{μ})$

$\frac{d D}{d t} = \frac{d x_{1}}{d t} + \frac{1}{μ} (\frac{d x_{2}}{d t} + \frac{2 d h}{d t})$

$= (- 3) + \frac{3}{4} (- 4 + 2 \times 2) = - 3 c m / s$

B) $D = x_{2} + x_{1} μ$

$\frac{d D}{d t} = \frac{d x_{2}}{d t} + μ \frac{d x_{1}}{d t}$

$= (- 4) + \frac{4}{3} (- 3) = - 8 c m / s$

B) $D = (x_{1} μ + x_{2} + h) + h$

$\frac{d D}{d t} = μ \frac{d x_{1}}{d t} + \frac{d x_{2}}{d t} + 2 \frac{d h}{d t}$

$= \frac{4}{3} (- 3) + (- 4) + 2 (2) = - 4 c m / s$

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