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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 $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ )

Option: 1
$11.6\: \mathrm{m/s}$

Option: 2
$10.8\: \mathrm{m/s}$

Option: 3
$17.8\: \mathrm{m/s}$

Option: 4
$14.4\: \mathrm{m/s}$

Answers (1)

$\mathrm{h_1 =15 \mathrm{~m} }$

$\mathrm{h_2 =5 \mathrm{~m} }$
From Eq of Continuity

$\mathrm{A_1 V_1 =A_2 V_2 }$

$\mathrm{V_1=\left(\frac{A_2}{A_1}\right) V_2}$

reference level

Applying Bernoulli's equation at $\mathrm{pt}$ (1) & (2)

$\mathrm{P_1+\frac{1}{2} \rho v_1^2+\rho g h_1=P_2+\frac{1}{2} \rho v_2^2+\rho g h_2}$

$\mathrm{\frac{5 \times 10^5}{\pi(1)^2}+0+\rho g h_1=1.01 \times 10^5+\frac{1 }{2}\rho v_{2}^2+\rho g\left ( h_{2} \right )}$

$\mathrm{v_1<<v_2 \quad\left(A_1 \gg A_2\right)}$

$\mathrm{\left(\frac{5}{\pi}-1.01\right) \times 10^5+10^5=500 v_2^2}$

$\mathrm{\frac{4.686}{\pi} \times 10^5=500 \mathrm{~V}_2^2}$

$\mathrm{\quad V_2=17.8 \mathrm{~m} / \mathrm{s} }$

Hence 3 is correct option.

Posted by

Sumit Saini

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