A cylindrical tank having large diameter is filled with water to a height $H$. A hole of cross-sectional area $5 \mathrm{~cm}^2$ in the tank allows water to drain out. If the water drains out at the rate of $2 imes 10^{-3} \mathrm{~m}^3 \mathrm{~s}^{-1}$, then the value of $H$ is (acceleration…

Question

A cylindrical tank having large diameter is filled with water to a height $H$. A hole of cross-sectional area $5 \mathrm{~cm}^2$ in the tank allows water to drain out. If the water drains out at the rate of $2 	imes 10^{-3} \mathrm{~m}^3 \mathrm{~s}^{-1}$, then the value of $H$ is (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

Aicharya · Accepted Answer

The given situation is shown below

Velocity of efflux through a hole below depth $\mathrm{H}$ of fluid, $v=\sqrt{2 g H}$ Volume flow rate of fluid from hole of area $ \begin{aligned} & a=V=a v \\ & V=a \sqrt{2 g H} \end{aligned} $ Here $V=2 \times 10^{-3} \mathrm{~m}^3 / \mathrm{s}$ $ a=5 \mathrm{~cm}^2=5 \times 10^{-4} \mathrm{~m}^2 $ So from eq. (i), we get $ \begin{aligned} 2 \times 10^{-3} & =5 \times 10^{-4} \times \sqrt{2 \times 10 \times H} \\ \Rightarrow \quad H & =\frac{8}{10} \mathrm{~m}=80 \mathrm{~cm} \end{aligned} $

A cylindrical tank having large diameter is filled with water to a height $H$. A hole of cross-sectional…

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