GATE 2024 CE (CE2) – Question 58
A 2 m × 1.5 m tank of 6 m height is provided with a 100 mm diameter orifice at the center of its base. The orifice is plugged and the tank is filled up to 5 m height. Consider the average value of discharge coefficient as 0.6 and acceleration due to gravity (g) as 10 m/s$^2$. After unplugging the orifice, the time (in seconds) taken for the water level to drop from 5 m to 3.5 m under free discharge condition is ____________ (rounded off to 2 decimal places).
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Correct answer: 103.47 to 104.51
Explanation
The time to fall from $h_1$ to $h_2$ is $t = \frac{2A_t}{C_da\sqrt{2g}}\left(\sqrt{h_1} - \sqrt{h_2}\right)$. The tank area is $A_t = 2 \times 1.5 = 3$ m² and the orifice area is $a = \frac{\pi}{4} \times 0.1^2 = 0.007854$ m². Then $t = \frac{2 \times 3}{0.6 \times 0.007854 \times \sqrt{20}}(\sqrt{5} - \sqrt{3.5}) = 284.68 \times 0.36524 = 103.99$ s.