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A cylindrical tank has a hole of $1 \mathrm{~cm}^2$ in its bottom. If the water is allowed to flow into the tank from a tube above it at the rate of $70 \mathrm{~cm}^3 / \mathrm{sec}$. Then the maximum height up to which water can rise in the tank is
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The correct answer is:
$2.5 \mathrm{~cm}$
The height of water in the tank becomes maximum when the volume of water flowing into the tank per second becomes equal to the volume flowing out per second.
Volume of water flowing out per second
$=A v=A \sqrt{2 g h}$...(i)
Volume of water flowing in per second
$=70 \mathrm{~cm}^3 / \mathrm{sec}$ ...(ii)
From (i) and (ii) we get
$A \sqrt{2 g h}=70 \Rightarrow 1 \times \sqrt{2 g h}=70$
$\Rightarrow 1 \times \sqrt{2 \times 980 \times h}=70$
$\Rightarrow 1 \times \sqrt{2 \times 980 \times h}=70$
$\therefore h=\frac{4900}{1960}=2.5 \mathrm{~cm}$.
Volume of water flowing out per second
$=A v=A \sqrt{2 g h}$...(i)
Volume of water flowing in per second
$=70 \mathrm{~cm}^3 / \mathrm{sec}$ ...(ii)
From (i) and (ii) we get
$A \sqrt{2 g h}=70 \Rightarrow 1 \times \sqrt{2 g h}=70$
$\Rightarrow 1 \times \sqrt{2 \times 980 \times h}=70$
$\Rightarrow 1 \times \sqrt{2 \times 980 \times h}=70$
$\therefore h=\frac{4900}{1960}=2.5 \mathrm{~cm}$.
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