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If a rubber ball is taken down to a 100 m deep lake its volume is decreased by $0.1 %$ . If $g = 10 m / \sec^{2}$ . Then the bulk modulus of elasticity is

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$10^{10} P a$

$10^{9} P a$

$10^{11} P a$

$3 \times 10^{9} P a$

answer is A.

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Detailed Solution

$k = \frac{ρ V}{Δ V} = \frac{h d g}{(\frac{Δ V}{V})} = \frac{100 \times 1000 \times 10}{0.1 %} = \frac{10^{6}}{\frac{0.1}{100}} = 10^{9} p a$

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