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  4. If pressure at half of depth of a lake is equal to the (1/4) pressure at the bottom of the lake then what is the depth of the lake?

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Q. If pressure at half of depth of a lake is equal to the $\frac{1}{4}$ pressure at the bottom of the lake then what is the depth of the lake?

Mechanical Properties of Fluids

Solution:

Pressure at the bottom $= P _{ o }+ h \rho g$
Pressure at half the depth $= P _{ o }+\frac{ h }{2} \rho g$
$\therefore P_{0}+\frac{h}{2} \rho g=\frac{1}{4}\left(P_{0}+h \rho g\right)$
$4 P_{0}+4 \frac{h}{2} \rho g=P_{0}+h \rho g$
$4 P_{0}+2 h \rho g=P_{0}+h \rho g$
$ \Rightarrow 3 P_{0}=-h \rho g $
$\Rightarrow \frac{-3 P_{0}}{\rho g}=h$
$h =-\left(\frac{3 \times 10^{5}}{10^{3} \times 9.8}\right)$
$=-0.306 \times 10^{2}=-30.6 m$
(-ve sign indicates depth but only magnitude is considerd)