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A large tank open to atmosphere at top and filled with water, develops a small hole in the side at a point $20 \mathrm{~m}$ below the water level. If the rate of flow of water from the hole is $3 \times 10^{-3} \mathrm{~m}^3 / \mathrm{min}$ then the area of hole is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
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The correct answer is:
$2.5 \mathrm{~mm}^2$
As we know that, rate of flow through a hole $=$
Velocity of efflux $\times$ Area of hole.
$\Rightarrow \quad \frac{\Delta V}{\Delta t}=v \times A$
$\Rightarrow \quad \frac{\Delta V}{\Delta t}=\sqrt{2 g h} \times A$...(i) $(\therefore v=\sqrt{2 g h})$
Here, $h=20 \mathrm{~m}$
$A=$ ?
Rated of flow of water,
$\frac{\Delta V}{\Delta t}=3 \times 10^{-3} \mathrm{~m}^3 / \mathrm{min}$
$=\frac{3}{60} \times 10^{-3} \mathrm{~m} / \mathrm{s}$
Substituting values in Eq. (i), we get
$\frac{3}{60} \times 10^{-3}=\sqrt{2 \times 10 \times 20} \times A$
$\Rightarrow \quad A=\frac{3 \times 10^{-3}}{60 \times \sqrt{2 \times 10 \times 20}}=25 \times 10^{-6} \mathrm{~m}^2$
$=2.5 \mathrm{~mm}^2$
Velocity of efflux $\times$ Area of hole.
$\Rightarrow \quad \frac{\Delta V}{\Delta t}=v \times A$
$\Rightarrow \quad \frac{\Delta V}{\Delta t}=\sqrt{2 g h} \times A$...(i) $(\therefore v=\sqrt{2 g h})$
Here, $h=20 \mathrm{~m}$
$A=$ ?
Rated of flow of water,
$\frac{\Delta V}{\Delta t}=3 \times 10^{-3} \mathrm{~m}^3 / \mathrm{min}$
$=\frac{3}{60} \times 10^{-3} \mathrm{~m} / \mathrm{s}$
Substituting values in Eq. (i), we get
$\frac{3}{60} \times 10^{-3}=\sqrt{2 \times 10 \times 20} \times A$
$\Rightarrow \quad A=\frac{3 \times 10^{-3}}{60 \times \sqrt{2 \times 10 \times 20}}=25 \times 10^{-6} \mathrm{~m}^2$
$=2.5 \mathrm{~mm}^2$
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