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A cube is shifted to a depth of 200 m in a lake. The change in volume is $0.1 %$ . Then bulk modulus of the material is ( $g = 10 m s^{- 2}$ )

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$10^{9} P a$

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

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

None

answer is B.

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Detailed Solution

$B = - \frac{∆ P}{\frac{∆ V}{V}}$

$\frac{∆ V}{V} = 0.1 %$

$h = 200 m$

$B = \frac{h ρ g}{\frac{∆ v}{v}}$ $= \frac{200 \times 1000 \times 10 \times 100}{0.1}$

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

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