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

Correct answer: C. 16 ~ms^ -1 v= 2 g h = 2 × 10 × 7 =12 ~ms^ -1 In this case when eye is inside water, x_ app. =μ x d x_ app. d t =μ · (d x)/(d t) or v_ app. =μ v=(4)/(3) × 12=16 ~ms^ -1

JEE Advanced 2009Paper 1

Exam: JEE Advanced
Year: 2009
Paper: Paper 1
Subject: Physics
Chapter: Ray Optics

JEE Advanced 2009 · Paper 1 · Physics · Ray Optics

A ball is dropped from a height of 20 ~m above the surface of water...

A ball is dropped from a height of

20 \mathrm{~m}

above the surface of water in a lake. The refractive index of water is

\frac{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 \mathrm{~m}

above the water surface, the fish sees the speed of ball as $A ball is dropped from a height of 20 ~m above the surface of water in a lake. The refractive index of water is (4)/(3)$

Options

A $9 \mathrm{~ms}^{-1}$
B $12 \mathrm{~ms}^{-1}$
C $16 \mathrm{~ms}^{-1}$
D $21.33 \mathrm{~ms}^{-1}$

Correct answer

C. $16 \mathrm{~ms}^{-1}$

Step-by-step solution

v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 7}=12 \mathrm{~ms}^{-1}

$A ball is dropped from a height of 20 ~m above the surface of water in a lake. The refractive index of water is (4)/(3)$
In this case when eye is inside water,

x_{\text{app. }}=\mu x \Rightarrow \frac{d x_{\text{app. }}}{d t}=\mu \cdot \frac{d x}{d t}

\quad v_{\text{app. }}=\mu v=\frac{4}{3} \times 12=16 \mathrm{~ms}^{-1}

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