JEE MAIN - Physics (2025 - 3rd April Morning Shift - No. 7)
Consider a completely full cylindrical water tank of height 1.6 m and of cross-sectional area $0.5 \mathrm{~m}^2$. It has a small hole in its side at a height 90 cm from the bottom. Assume, the crosssectional area of the hole to be negligibly small as compared to that of the water tank. If a load 50 kg is applied at the top surface of the water in the tank then the velocity of the water coming out at the instant when the hole is opened is:
$$ \left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right) $$
Explanation
_3rd_April_Morning_Shift_en_7_1.png)
Apply Bernouli equation between points 1 & 2
$$ \begin{aligned} & P_1+\frac{1}{2} \rho v_1^2+\rho g h=P_2+\frac{1}{2} \rho v_2^2+0 \\ & P_0+\frac{m g}{A}+\rho g \frac{70}{100}=P_0+\frac{1}{2} \rho v_2^2 \\ & \frac{5000}{0.5}+10^3 \times 10 \frac{70}{100}=\frac{1}{2} \times 10^3 v_2^2 \\ & 10^3+10^3 \times 7=\frac{10^3}{2} v_2^2 \\ & v_2^2=16 \\ & v_2=4 \mathrm{~m} / \mathrm{s} \end{aligned}$$
As the tank area is large $v_1$ is negligible compared $$\text { to } v_2$$
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