Question

Find the depth of a lake at which the density of water is 1% greater than at the surface, if compressibility of water is $50\times {10}^{-6}/atm$

• A. 1 km
• B. 2 km
• C. 1.5 km
• D. 0.5 km

Answer 1

The correct option is B 2 km

${\rho }^{\prime }=\frac{\rho }{1-\frac{\Delta P}{B}}$

$\Rightarrow \frac{{\rho }^{\prime }}{\rho }\times 100=\frac{100}{1-\Delta P\times 50\times {10}^{-6}}$

$\Rightarrow \frac{101}{100}=\frac{1}{1-5\times h\times {10}^{-6}}$
$\Rightarrow 1-5\times h\times {10}^{-6}=\frac{100}{101}$
$\Rightarrow 5\times h\times {10}^{-6}=\frac{1}{101}$
$\Rightarrow h=\frac{{10}^6}{5\times 101}m\approx \frac{10000}{5}=2km$