. 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

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

Approved