MHT CET · Physics · Mechanical Properties of Fluids

A large vessel completely filled with water has two holes ' \(A\) ' and ' \(B\) ' at depths 'h' and '4h' from the top. Hole 'A' is a square of side 'L' and hole 'B' is circle of radius 'R'. If from both the holes same quanitity of water is flowing per second, then side of square hole is

A \(2 \pi R\)
B \(\sqrt{2 \pi \mathbf{R}}\)
C \(\sqrt{2 \pi} \cdot \mathrm{R}\)
D \(\frac{R}{2}\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(\sqrt{2 \pi} \cdot \mathrm{R}\)

Step-by-step Solution

Detailed explanation

The ratio of velocities of water is given by
\(\begin{aligned}
& \frac{V_{A}}{V_{B}}=\sqrt{\frac{h}{4 h}}=\frac{1}{2} \\
\therefore & V_{B}=2 V_{A}
\end{aligned}\)
Quantity of water flowing is same
\(\begin{aligned}
\therefore \mathrm{V}_{\mathrm{A}} \times \mathrm{L}^{2} &=\mathrm{V}_{\mathrm{B}} \times \pi \mathrm{R}^{2} \\
\therefore \mathrm{V}_{\mathrm{A}} \mathrm{L}^{2} &=2 \mathrm{~V}_{\mathrm{A}} \times \pi \mathrm{R}^{2} \\
\therefore \quad \mathrm{L}^{2} &=2 \pi \mathrm{R}^{2} \\
\mathrm{~L} &=\sqrt{2 \pi} \cdot \mathrm{R}
\end{aligned}\)

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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