If a rubber ball is taken at the depth of 200 m in a pool, its volume decreases by 0.1 %. If the density of the water is $1 \times 10^{3} kg / m^{3} and g = 10 m / s^{2}$ , then the volume elasticity in $N / m^{2}$ will be

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$10^{8}$

2 $\times 10^{8}$

$10^{9}$

$2 \times 10^{9}$

answer is D.

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Detailed Solution

K = $\frac{∆ P}{\frac{∆ V}{V}} = \frac{hρg}{\frac{∆ V}{V}} = \frac{200 \times 10^{3} \times 10}{\frac{0.1}{100}} = 2 \times 10^{9}$

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