The approximate depth of an ocean is 2700 m. The compressibility of water is and density of water is . What fractional compression of water will be obtained at the bottom of the ocean? [2015]

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Solution

Q.
The approximate depth of an ocean is 2700 m. The compressibility of water is $45.4 \times 10^{- 11} {Pa}^{- 1}$ and density of water is $10^{3} {kg/m}^{3}$ . What fractional compression of water will be obtained at the bottom of the ocean [2015]

1 $1.2 \times 10^{- 2}$

2 $1.4 \times 10^{- 2}$

3 $0.8 \times 10^{- 2}$

4 $1.0 \times 10^{- 2}$

Ans.
(1)

Depth of ocean, $d = 2700 m$

Density of water, $ρ = 10^{3} kg$ $m^{- 3}$

Compressibility of water, $K = 45.4 \times 10^{- 11} {Pa}^{- 1}$

$\frac{Δ V}{V} = ?$

Excess pressure at the bottom, $Δ P = ρ g d$

$= 10^{3} \times 10 \times 2700 = 27 \times 10^{6} Pa$

We know, $B = \frac{Δ P}{(Δ V / V)}$

$(\frac{Δ V}{V}) = \frac{Δ P}{B} = K \cdot Δ P$ $(∵ K = \frac{1}{B})$

$= 45.4 \times 10^{- 11} \times 27 \times 10^{6} = 1.2 \times 10^{- 2}$

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