A tank 1.5m stands on a trolly and is full of water. It has an orifice of diameter 0.1 m at 0.3m from the bottom of the tank. If the orifice is suddenly opened and coefficient of discharge of the orifice is 0.60, then the propelling force on the trolley will be
- A. 69.37 N
- B. 67.39 N
- C. 63.79 N
- D. 65.39 N
Answer: B. 67.39 N
Explanation: Head h = 1.5 - 0.3 = 1.2m. Velocity V = C_v \sqrt{2gh}. Actual velocity V_act = C_d \sqrt{2gh} / C_c? Propelling force F = \rho A V_{act}^2 / C_c? Standard formula F = \rho a V² where V is actual velocity. Or F = \dot{m} V. \dot{m} = \rho C_d a \sqrt{2gh}$. V = C_v \sqrt{2gh}$. If we assume C_d accounts for both, F = \rho a C_d^2 (2gh)/C_c? Using F = 2 \rho g h a C_d^2? Simple calculation: V = 0.6 \sqrt{2*10*1.2} = 2.94 m/s. F = \dot{m}V = (\rho A V) V = \rho A V²? No, flow rate is Q = Cd A \sqrt{2gh}. Force = Q * V. Here V is exit velocity. If Cd is given, V_actual? Let's try F = \rho A v² where v is jet velocity. $v = C_v \sqrt{2gh}$. Usually $C_d = C_c C_v$. If $C_c=1$, $C_d=C_v$. Then $F = 1000 * (\pi/4 * 0.1^2) * (0.6 \sqrt{20*1.2})^2 = 1000 * 0.00785 * 0.36 * 24 = 67.8$ N.