Question 80 of 192

Consider a cylindrical tank of radius $1m$ is filled with water. The top surface of water is at $15m$ from the bottom of the cylinder. There is a hole on the wall of cylinder at a height of $5m$ from the bottom. A force of $5\times {10}^{5}N$ is applied an the top surface of water using a piston. The speed of efflux from the hole will be :
(given atmospheric pressure ${P}_{A}=1.01\times {10}^{5}Pa$, density of water ${\rho }_{w}=1000kg{m}^{-3}$ and gravitational acceleration $g=10m{s}^{-2}$)

A$11.6m{s}^{-1}$B$10.8m{s}^{-1}$C$17.8m{s}^{-1}$D$14.4m{s}^{-1}$

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Answer: (C) $17.8m{s}^{-1}$

Assuming there is no atmospheric pressure on piston, ${P}_{1}=\dfrac{F}{\pi {R}^{2}}$ and ${P}_{2}={P}_{0}$.
Applying Bernoulli's equation at point $1$ and $2$,
${p}_{1}+\rho g{h}_{1}+\dfrac{1}{2}\rho {v}_{1}^{2}={p}_{2}+\rho g{h}_{2}+\dfrac{1}{2}\rho {v}_{2}^{2}$
$\Rightarrow \left(\dfrac{F}{\pi {R}^{2}}\right)+\rho g\left({h}_{1}-{h}_{2}\right)={P}_{0}+\dfrac{1}{2}\rho {v}_{2}^{2}$
$\Rightarrow {v}_{2}=\sqrt{\dfrac{2F}{\pi {R}^{2}\rho }-\dfrac{2{P}_{0}}{\rho }+2g\left({h}_{1}-{h}_{2}\right)}$
${v}_{2}=\sqrt{\dfrac{2\times \left(5\times {10}^{5}\right)}{3.14\times {1}^{2}\times 1000}-\dfrac{2\times 1.01\times {10}^{5}}{1000}+2\times 10\times 10}$
$\approx 17.8m{s}^{-1}$