If pressure at half the depth of a lake is equal to $2/3$ pressure at the bottom of the lake then what is the depth of the lake ........ $m$
Diffcult
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(b)Pressure at bottom of the lake = ${P_0} + h\rho g$
Pressure at half the depth of a lake $ = {P_0} + \frac{h}{2}\rho g$
According to given condition
${P_0} + \frac{1}{2}h\rho g = \frac{2}{3}({P_0} + h\rho g)$ ==> $\frac{1}{3}{P_0} = \frac{1}{6}h\rho g$
==> $h = \frac{{2{P_0}}}{{\rho g}} = \frac{{2 \times {{10}^5}}}{{{{10}^3} \times 10}} = 20m$.
Pressure at half the depth of a lake $ = {P_0} + \frac{h}{2}\rho g$
According to given condition
${P_0} + \frac{1}{2}h\rho g = \frac{2}{3}({P_0} + h\rho g)$ ==> $\frac{1}{3}{P_0} = \frac{1}{6}h\rho g$
==> $h = \frac{{2{P_0}}}{{\rho g}} = \frac{{2 \times {{10}^5}}}{{{{10}^3} \times 10}} = 20m$.
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