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If a rubber ball is taken at the depth of $200 \mathrm{~m}$ 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 $N / \mathrm{m}^2$ will be
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$2 \times 10^9$
$K=\frac{\Delta P}{\Delta V / V}=\frac{h \rho g}{\Delta V / V}=\frac{200 \times 10^3 \times 10}{0.1 / 100}=2 \times 10^9$
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