Question

A large cylindrical tank of cross-sectional area 1m² is filled with water. It has a small hole at a height of 1m from the bottom. A movable piston of mass 5 kg is fitted on the top of the tank such that it can slide in the tank freely. A load of 45 kg is applied on the top of water by piston, as shown in figure. Find the value of v when piston is 7m above the bottom (g = 10 m/s²)

• A. 10 m/s
• B. 11 m/s
• C. 12 m/s
• D. 13 m/s

Answer 1

Answer: (B)

11 m/s

Answer 2

The total mass on the piston is $m_{total} = 5 \, \text{kg} + 45 \, \text{kg} = 50 \, \text{kg}$. The pressure at the water surface is $P_{surface} = \frac{m_{total} \times g}{A} = \frac{50 \, \text{kg} \times 10 \, \text{m/s}^2}{1 \, \text{m}^2} = 500 \, \text{Pa}$. The height of water above the hole is $h = 7 \, \text{m} - 1 \, \text{m} = 6 \, \text{m}$. Using Bernoulli's equation, $P_{surface} + \rho g h_{above\_hole} + \frac{1}{2} \rho v_{surface}^2 = P_{hole} + \rho g h_{hole} + \frac{1}{2} \rho v_{hole}^2$. Assuming $v_{surface} \approx 0$ and $P_{hole} = 0$ (gauge pressure), we get: $500 \, \text{Pa} + (1000 \, \text{kg/m}^3)(10 \, \text{m/s}^2)(6 \, \text{m}) = 0 + 0 + \frac{1}{2} (1000 \, \text{kg/m}^3) v^2$. $500 + 60000 = 500 v^2$. $60500 = 500 v^2$. $v^2 = 121$. $v = 11 \, \text{m/s}$.