JEE Advance - Physics (2009 - Paper 1 Offline - No. 8)
A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is 12.8 m above the water surface, the fish sees the speed of ball as (Take g = 10 m/s$$^2$$)
9 m/s
12 m/s
16 m/s
21.33 m/s
Explanation
The velocity of the ball when it has fallen through 7.2 m is
$$V = \sqrt {2g(7.2)} = 12$$ m/s
For a general height $$h$$ above the water surface, we have
$${{h'} \over h} = {4 \over 3} \Rightarrow h' = {4 \over 3}h$$
Therefore, $$V' = {4 \over 3}V = {4 \over 3} \times 12 = 16$$ m/s
where $$V'$$ is the velocity of ball with respect to the fish.
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