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 $= {10}^{3}kg{m}^{- 3}$ , Bulk modulus of rubber $= 9 \times {10}^{8}N{m}^{- 2}$ , and $g = 10m{s}^{- 2}$ )

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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 $= {10}^{3}kg{m}^{- 3}$ , Bulk modulus of rubber $= 9 \times {10}^{8}N{m}^{- 2}$ , and $g = 10m{s}^{- 2}$ )

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Correct answer: 0

Bulk modulus is given by, B = (- ∆ P)/((∆ V)/(V)) ⇒ ∆ P = - B(∆ V)/(V) ⇒ ρgh = - B(∆ V)/(V) ⇒ 10 ^ 3 × 10 × h = - 9 × 10 ^ 8 × (- (0 . 02)/(100)) ⇒ h = 18m

The depth below the surface of sea to which a rubber ball be taken...

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