The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by $0.02 %$ is ………m. (take density of sea water $= 10^{3} {kgm}^{- 3}$ . Bulk modulus of rubber $= 9 \times 10^{8} N m^{- 2}$ and $g = 10 m s^{- 2}$ )

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answer is 18.

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Detailed Solution

$B = \frac{s t r e s s}{s t r a i n} = \frac{ρ g h}{\frac{Δ v}{v}} \frac{Δ v}{v} \times 100 = \frac{ρ g h}{B} \times 100 0.02 = \frac{10^{3} \times 10 \times h \times 100}{9 \times 10^{8}} h = 18 m$

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