JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Geometrical Optics

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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 [Take

g = 10 m / s^{2}

(A)9 m/s
(B) $12 m / s$
(C)16 m/s✓ Correct
(D)21.33 m/s

Answer: (C)

Solution

The ball has fallen through a height of

20 - 12.8 = 7.2

m. Using

v^{2} = u^{2} + 2 g h

with

u = 0

g = 10

m/s², and

h = 7.2

v^{2} = 2 \times 10 \times 7.2 = 144

, so

v = 12

m/s. Due to refraction, the apparent position of the ball as seen by the fish is at a distance

x_{image} = μ \cdot x_{ball}

from the water surface, where

μ = 4/3

. Differentiating with respect to time:

v_{image} = μ \cdot v_{ball} = \frac{4}{3} \times 12 = 16

m/s.

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Single correct option

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JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Geometrical Optics

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