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Bulk modulus of elastic of rubber is $10^{9} N m^{- 2}$ . if it is taken down to an 100 m deep lake, decreases in its volume is $(g = 10 m s^{- 2})$

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$0.2 %$

$0.1 %$

$2 %$

$1 %$

answer is A.

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Detailed Solution

$K = \frac{∆ P}{(\frac{∆ V}{V})} = \frac{- ∆ V}{V} = \frac{∆ p}{K} = \frac{10^{6}}{10^{9}} = 10^{- 3}$ $\frac{- ∆ V}{V} \times 100 = 10^{- 3} \times 100 = 0.1 %$

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