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(D) The total pressure $P$ at a depth $h$ below the surface of a liquid is given by the formula: $P = P_0 + \rho gh$. Here,$P_0$ is the atmospheric pr

Vedclass · Accepted Answer

$2 	imes 10^5 \ Nm^{-2}$

Vedclass · Answer

(D) The total pressure $P$ at a depth $h$ below the surface of a liquid is given by the formula: $P = P_0 + ho gh$. Here,$P_0$ is the atmospheric pressure,$ho$ is the density of water $(1000 \ kg \ m^{-3})$,$g$ is the acceleration due to gravity $(10 \ ms^{-2})$,and $h$ is the depth $(10 \ m)$. Substituting the values: $P = 1.01 	imes 10^5 + (1000 	imes 10 	imes 10)$ $P = 1.01 	imes 10^5 + 10^5$ $P = 1.01 	imes 10^5 + 1.00 	imes 10^5 = 2.01 	imes 10^5 \ Nm^{-2}$. Rounding to the nearest significant value provided in the options,we get $2 	imes 10^5 \ Nm^{-2}$.

The pressure at a point $10 \ m$ below the surface of water is (take atmospheric pressure $P_0 = 1.01 \times 10^5 \ Nm^{-2}$ and $g = 10 \ ms^{-2}$):

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