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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
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Verified Answer
The correct answer is:
$\sqrt{2 \pi} \cdot \mathrm{R}$
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
$\begin{aligned}
\therefore \mathrm{V}_{\mathrm{A}} \times \mathrm{L}^{2} &=\mathrm{V}_{\mathrm{B}} \times \pi \mathrm{R}^{2} \\
\therefore \mathrm{V}_{\mathrm{A}} \mathrm{L}^{2} &=2 \mathrm{~V}_{\mathrm{A}} \times \pi \mathrm{R}^{2} \\
\therefore \quad \mathrm{L}^{2} &=2 \pi \mathrm{R}^{2} \\
\mathrm{~L} &=\sqrt{2 \pi} \cdot \mathrm{R}
\end{aligned}$
$\begin{aligned}
& \frac{V_{A}}{V_{B}}=\sqrt{\frac{h}{4 h}}=\frac{1}{2} \\
\therefore & V_{B}=2 V_{A}
\end{aligned}$
Quantity of water flowing is same
$\begin{aligned}
\therefore \mathrm{V}_{\mathrm{A}} \times \mathrm{L}^{2} &=\mathrm{V}_{\mathrm{B}} \times \pi \mathrm{R}^{2} \\
\therefore \mathrm{V}_{\mathrm{A}} \mathrm{L}^{2} &=2 \mathrm{~V}_{\mathrm{A}} \times \pi \mathrm{R}^{2} \\
\therefore \quad \mathrm{L}^{2} &=2 \pi \mathrm{R}^{2} \\
\mathrm{~L} &=\sqrt{2 \pi} \cdot \mathrm{R}
\end{aligned}$
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