Consider a cylindrical tank of radius 1 m is filled with water. The top surface of water is at 15 m from the bottom of the cylinder. There is a hole on the wall of cylinder at a height of 5 m from the

Question

Consider a cylindrical tank of radius 1 m is filled with water. The top surface of water is at 15 m from the bottom of the cylinder. There is a hole on the wall of cylinder at a height of 5 m from the bottom. A force of 5 × 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 × 10 5 P a , density of water ρ w = 1000 kg m - 3 and gravitational acceleration g = 10 m s - 2 )

TestHub · Accepted Answer

Assuming there is no atmospheric pressure on piston, $P_{1} = \frac{F}{π R^{2}}$ and $P_{2} = P_{0}$ .
Applying Bernoulli's equation at point $1$ and $2$ ,
$p_{1} + ρ g h_{1} + \frac{1}{2} ρ v_{1}^{2} = p_{2} + ρ g h_{2} + \frac{1}{2} ρ v_{2}^{2}$
$\Rightarrow (\frac{F}{π R^{2}}) + ρ g (h_{1} - h_{2}) = P_{0} + \frac{1}{2} ρ v_{2}^{2}$
$\Rightarrow v_{2} = \sqrt{\frac{2 F}{π R^{2} ρ} - \frac{2 P_{0}}{ρ} + 2 g (h_{1} - h_{2})}$
$v_{2} = \sqrt{\frac{2 \times (5 \times 10^{5})}{3.14 \times 1^{2} \times 1000} - \frac{2 \times 1.01 \times 10^{5}}{1000} + 2 \times 10 \times 10}$
$\approx 17.8 m s^{- 1}$

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