Question

Water discharges from a cylindrical tank through an orifice, as shown in the figure. The flow is considered frictionless. Initially, the water level in the tank was \( h_1 = 2 \, \text{m} \). The diameter of the tank is \( D = 1 \, \text{m} \), while the diameter of the jet is \( d = 10 \, \text{cm} \), and the acceleration due to gravity is \( g = 10 \, \text{m/s}^2 \). The time taken (in seconds, up to one decimal place) for the water level in the tank to come down to \( h_2 = 1 \, \text{m} \) is \(\underline{\hspace{2cm}}\).
\includegraphics[width=0.5\linewidth]{image20.png}

Hint:

To find the time for the water level to change, use Torricelli's law for the discharge rate and apply the volume change equation.

Answer 1

Correct Answer: 17.5

Using Torricelli's law, the discharge rate through the orifice is given by:
\[ Q = C_d A \sqrt{2gh} \] where:
- \( C_d \) is the discharge coefficient (assumed to be 1 for frictionless flow),
- \( A = \frac{\pi d^2}{4} \) is the area of the jet,
- \( h \) is the height of the water in the tank.
The time for the water level to drop from \( h_1 \) to \( h_2 \) is:
\[ \text{time} = \frac{\Delta V}{Q} \] where \( \Delta V \) is the volume of water discharged, and \( Q \) is the discharge rate.
The change in volume is:
\[ \Delta V = A_{\text{tank}} (h_1 - h_2) = \pi \left(\frac{D}{2}\right)^2 (h_1 - h_2) \] Substituting the given values and solving for time:
\[ \Delta V = \pi \left(\frac{1}{2}\right)^2 (2 - 1) = 0.785 \, \text{m}^3 \] Now, calculate the discharge rate \( Q \):
\[ A = \frac{\pi (0.1)^2}{4} = 0.00785 \, \text{m}^2 \] \[ Q = 1 \times 0.00785 \times \sqrt{2 \times 10 \times 2} = 0.00785 \times 6.32 = 0.0496 \, \text{m}^3/\text{s} \] Finally, the time taken is:
\[ \text{time} = \frac{0.785}{0.0496} \approx 17.5 \, \text{seconds}. \] Thus, the time taken for the water level to come down to 1 m is \( \boxed{17.5} \, \text{seconds}. \)