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```        . 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

(A) 9 m/s

(B) 12 m/s

(C) 16 m/s

(D) 21.33 m/s

```
7 years ago

510 Points
```										fish will see the image of ball after reraction through water surface...
using
n2/v - n1/u = n2-n1/R
R for plane surface is infinity & n2(refrative index of water) , n1(refractive index of air)
4/3v - 1/u = 0
v = 4u/3  ..........1
now differentiating eq 1 wrt t
(velocity of image)dv/dt = 4/3.(du/dt) (velocity of object)
now when ball has fallen 12.8m then its velocity is U..
so velocity of ball as seen by fish is (4/3) .U
V = 4/3. (U) =4/3 (sqrt2g(H-h))                                    (change in height (H-h) = 7.2m)
=4*12/3=16m/s
```
7 years ago
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