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A large open top water tank is completely filled with water. A small hole of diameter $4 \mathrm{~mm}$ is made $10 \mathrm{~m}$ below the water level. The flow rate of water through the hole is
(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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(Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
$1.77 \times 10^{-6} \mathrm{~m}^3 \mathrm{~s}^{-1}$
From Bernoulli's equation
$\begin{aligned}
& \rho g h=\frac{1}{2} \rho v^2 \\
& \mathrm{v}=\sqrt{2 \mathrm{gh}}=\sqrt{2 \times 10 \times 10}=14.14 \mathrm{~m} / \mathrm{s}
\end{aligned}$
The flow rate of water through the hole is given by
$\begin{aligned}
& Q=A v=\pi r^2 v \\
& =3.14 \times\left(\frac{4 \times 10^{-3}}{2}\right)^2 \times 14.14=177 \times 10^{-6} \mathrm{~m}^3 / \mathrm{s}
\end{aligned}$
$\begin{aligned}
& \rho g h=\frac{1}{2} \rho v^2 \\
& \mathrm{v}=\sqrt{2 \mathrm{gh}}=\sqrt{2 \times 10 \times 10}=14.14 \mathrm{~m} / \mathrm{s}
\end{aligned}$
The flow rate of water through the hole is given by
$\begin{aligned}
& Q=A v=\pi r^2 v \\
& =3.14 \times\left(\frac{4 \times 10^{-3}}{2}\right)^2 \times 14.14=177 \times 10^{-6} \mathrm{~m}^3 / \mathrm{s}
\end{aligned}$
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