A cylindrical tank has a hole of area $2 \mathrm{~cm}^2$ at its bottom, If water is poured into the tank…
A cylindrical tank has a hole of area $2 \mathrm{~cm}^2$ at its bottom, If water is poured into the tank from a tube above it at the rate of $100 \mathrm{~cm}^3 \mathrm{~s}^{-1}$, then the maximum height upto which water can rise in the tank is (Acceleration due to gravity, $g=10 \mathrm{~ms}^{-2}$ )
2.5 cm
1.25 cm
5.5. cm
3.5 cm
Solution
Given, area of hole in tank, $A=2 \mathrm{~cm}^{-2}$
$
\begin{aligned}
& \Rightarrow \quad A=2 \times 10^{-4} \mathrm{~m}^{-2} \\
& \text { Volume flow rate }=100 \mathrm{~cm}^2 / \mathrm{sec} \\
& =100 \times 10^{-6} \mathrm{~m}^3 / \mathrm{sec}=10^{-4} \mathrm{~m}^2 \mathrm{sec}^{-1}
\end{aligned}
$
At maximum height $h$, velocity of water flowing through hole, $v=\sqrt{2 g h}$
$\therefore$ From principle of continuity of flow of liquid, volume flow rate $=A v$
$
\begin{array}{rlrl}
& & 10^{-2} & =A v=2 \times 10^{-4} \sqrt{2 g h}\{\because v=\sqrt{2 g h}\} \\
& \Rightarrow & \frac{10^{-4}}{2 \times 10^{-4}} & =\sqrt{2 g h} \\
& \Rightarrow & \frac{1}{2} & =\sqrt{2 g h} \\
& \Rightarrow & \left.\frac{1}{2}\right)^{-4} & =2 g h \\
\Rightarrow & \frac{1}{4} \times \frac{1}{2 g} & =h \\
\Rightarrow & h & =\frac{1}{80}=0.125 \mathrm{~m} & \\
\Rightarrow & h & =1.25 \mathrm{~cm}
\end{array}
$