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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
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
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
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