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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
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The correct answer is:
$20 \mathrm{~m}$
Pressure at half the depth $=P_0+\frac{h}{2} d g$
Pressure at the bottom $=P_0+h d g$
According to given condition
$P_0+\frac{h}{2} d g=\frac{2}{3}\left(P_0+h d g\right)$
$\Rightarrow 3 P_0+\frac{3 h}{2} d g=2 P_0+2 h d g$
$\Rightarrow h=\frac{2 P_o}{d g}=\frac{2 \times 10^5}{10^3 \times 10}=20 \mathrm{~m}$
Pressure at the bottom $=P_0+h d g$
According to given condition
$P_0+\frac{h}{2} d g=\frac{2}{3}\left(P_0+h d g\right)$
$\Rightarrow 3 P_0+\frac{3 h}{2} d g=2 P_0+2 h d g$
$\Rightarrow h=\frac{2 P_o}{d g}=\frac{2 \times 10^5}{10^3 \times 10}=20 \mathrm{~m}$
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