JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 7)
Consider a cylindrical tank of radius $$1 \mathrm{~m}$$ is filled with water. The top surface of water is at $$15 \mathrm{~m}$$ from the bottom of the cylinder. There is a hole on the wall of cylinder at a height of $$5 \mathrm{~m}$$ from the bottom. A force of $$5 \times 10^{5} \mathrm{~N}$$ is applied an the top surface of water using a piston. The speed of ifflux from the hole will be : (given atmospheric pressure $$\mathrm{P}_{\mathrm{A}}=1.01 \times 10^{5} \mathrm{~Pa}$$, density of water $$\rho_{\mathrm{W}}=1000 \mathrm{~kg} / \mathrm{m}^{3}$$ and gravitational acceleration $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$$ )
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Explanation
By Bernoulli's theorem,
$${{5 \times {{10}^5}} \over {\pi {{(1)}^2}}} + \rho g(10) = 1.01 \times {10^5} + {1 \over 2}\rho {(v)^2}$$
$$ \Rightarrow {v^2} = 200 + {{{{10}^6}} \over {1000\pi }} - 202$$
$$ \Rightarrow v \simeq 17.8$$ m/s
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