A) 10 m
B) 20 m
C) 60 m
D) 30 m
Correct Answer: B
Solution :
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 \[{{P}_{0}}+\frac{1}{2}h\rho g=\frac{2}{3}({{P}_{0}}+h\rho g)\] Þ \[\frac{1}{3}{{P}_{0}}=\frac{1}{6}h\rho g\] Þ \[h=\frac{2{{P}_{0}}}{\rho g}=\frac{2\times {{10}^{5}}}{{{10}^{3}}\times 10}=20m\].
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