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Q.

A cubical ball is taken to a depth of 200 m inside a sea. The decrease in volume observed to be $0.1 %$ . The bulk modulus of the ball is -- $(g = 10 m s^{- 2})$ $(d e n s i t y o f w a t e r = 10^{3} k g / m^{3})$

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a

$2 \times 10^{6} P a$

b

$2 \times 10^{9} P a$

c

$1.2 \times 10^{9} P a$

d

$2 \times 10^{7} P a$

answer is C.

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Detailed Solution

$K = \frac{p V}{Δ V} = \frac{h d g V}{Δ V} = \frac{h d g}{(\frac{Δ V}{V})}$

$K = \frac{200 \times 10^{3} \times 10}{(\frac{0.1}{100})} = 2 \times 10^{9} p a$

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A cubical ball is taken to a depth of 200 m inside a sea. The decrease in volume observed to be 0.1% . The bulk modulus of the ball is -- (g=10ms−2) density of water =103 kg/m3