A tank having cross sectional area A has a hole at the bottom of area of cross section $A_{1} = A / 1000$ . Bottom of the tank is a plane mirror. The tank contains water of refractive index 4/3. At the instant, when height of the water in the tank is h = 5 m, a fish is rising vertically in the tank with a velocity 3 cm/sec toward the surface. Find the velocity of the fish as observed by the observer looking directly at the fish (in cm/s). (Take g = $10 m / s^{2}$ ) [Take speed of liquid from hole is $\sqrt{2 g h}$ ]

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Detailed Solution

Velocity of efflux = $v = \sqrt{2 g \times 5} = 10 m / s$

Velocity of surface = $\frac{10 \times A / 1000}{A} = 1 c m / s$
The velocity of fish relative to the observer is
$v = \frac{d}{d t} (\frac{x}{μ} + y) = \frac{3}{4} \frac{d x}{d t} + \frac{d y}{d t} = \frac{3}{4} \times (- 4) + 1 = - 3 + 1 = - 2 c m / s$