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Q. A ball is dropped from a height of $20 \, m$ above the surface of the water in a lake. The refractive index of water is $\frac{4}{3}$ . A fish inside the lake, in the line of fall of ball is looking at the ball. At an instant, when the ball is $12.8 \, m$ above the water surface, the fish sees speed of ball as

NTA AbhyasNTA Abhyas 2022

Solution:

$\upsilon = \sqrt{2 \text{gh}} = \sqrt{2 \times 1 0 \times 7}$
$= 1 2 \text{ms}^{-1}$
In this case when eye is inside water,
Solution
$x_{\text{app.}} = \mu x$
$\therefore \frac{\text{d} x_{\text{app.}}}{\text{dt}} = \mu \cdot \frac{\text{d} x}{\text{dt}}$
or, $\upsilon_{\text{app.}} \mu \upsilon = \frac{4}{3} \times 1 2 = 1 6 \text{ m s }^{-1}$