The pressure at half the depth of a lake is equal to two-third pressure at the bottom of the lake. So the depth 'h' of the lake is ( ρ = density of water in the lake, g = acceleration due to gravity, P₀ = atmospheric pressure). (a) (2P₀)/(ρ g) (b) (P₀)/(ρ g) (c) (2P₀)/(3ρ g) (d) (P₀)/(2ρ g)

Answer: (a). P₀ + ρ gh/2 = 2/3(P₀ + ρ g h) 3P₀ + 3/2ρ g h = 2P₀ + 2ρ g h ⇒ P₀ = 1/2ρ g h h = (2P₀)/(ρ g) So the answer is (a).

The pressure at half the depth of a lake is equal to two-third… | MHT CET 2026 (18 Apr, Shift 1)

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Pressure in a Fluid

The pressure at half the depth of a lake is equal to two-third pressure at the bottom of the lake. So the depth 'h' of the lake is ( $\rho$ = density of water in the lake, $g$ = acceleration due to gravity, $P_0$ = atmospheric pressure).

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Answer: (a)

$P_0 + \rho g\dfrac{h}{2} = \dfrac{2}{3}(P_0 + \rho g h)$
$3P_0 + \tfrac{3}{2}\rho g h = 2P_0 + 2\rho g h \Rightarrow P_0 = \tfrac{1}{2}\rho g h$
$h = \dfrac{2P_0}{\rho g}$
So the answer is (a).

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