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CGP EDU Academic Team
Published on: September 12, 2026
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 $1 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}$ and $g = 10 \mathrm{m}/\mathrm{s}^2$ , then the volume elasticity in $\mathrm{N/m^{2}}$ will be
Text Solution
Verified by ExpertsThe correct answer is:
D
$\mathrm{K} = \frac{\Delta \mathrm{P}}{\Delta \mathrm{V}/\mathrm{V}} = \frac{\mathrm{h} \rho \mathrm{g}}{\Delta \mathrm{V}/\mathrm{V}} = \frac{200 \times 10^{3} \times 10}{0.1/100} = 2 \times 10^{9}$
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