KCET · Physics · Mechanical Properties of Fluids

A closed water tank has cross-sectional area \(A\). It has a small hole at a depth of \(h\) from the free surface of water. The radius of the hole is \(r\) so that \(r \ll \sqrt{\frac{A}{\pi}}\). If \(p_o\) is the pressure inside the tank above water level and \(p_a\) is the atmospheric pressure, the rate of flow of the water coming out of the hole is ( \(\rho\) is density of water)

A \(\pi r^2 \sqrt{2 g h}\)
B \(\pi r^2 \sqrt{2 g h+\frac{2\left(p_o-p_a\right)}{\rho}}\)
C \(\pi r^2 \sqrt{2 g H}\)
D \(\pi r^2 \sqrt{g h+\frac{2\left(p_0-p_a\right)}{\rho}}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\pi r^2 \sqrt{2 g h+\frac{2\left(p_o-p_a\right)}{\rho}}\)

Step-by-step Solution

Detailed explanation

From Bernoulli's principle we have
\(\frac{1}{2} \rho v^2+p_0+\rho g h=\frac{1}{2} \rho v_A^2+p_A+\rho g(0) \)
\( p_0-p_A+\rho g h=\frac{1}{2} \rho v_A^2-\frac{1}{2} \rho v^2 \)
\( \Rightarrow v^2=\frac{2\left(p_0-p_A\right)}{\rho}+\frac{2 \rho g h}{P} \)
\( \Rightarrow v=\sqrt{\frac{2\left(p_0-p_A\right)}{\rho}+2 g h} (\because v=0 \text { at top }) \)
Water coming out of hole \(=\text {area} \times \text {velocity} \)
\( =\pi r^2 \sqrt{2 g h+\frac{2\left(p_0-p_A\right)}{\rho}}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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