A large open top water tank is completely filled with water. A small hole of diameter $4 \mathrm{~mm}$ is…

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 $=10 \mathrm{~ms}^{-2}$ )
  1. $14.14 \times 10^{-6} \mathrm{~m}^3 \mathrm{~s}^{-1}$
  2. $2.1 \times 10^{-6} \mathrm{~m}^3 \mathrm{~s}^{-1}$
  3. $1.77 \times 10^{-6} \mathrm{~m}^3 \mathrm{~s}^{-1}$
  4. $0.177 \times 10^{-6} \mathrm{~m}^3 \mathrm{~s}^{-1}$

Solution

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 $\begin{aligned} & Q=A v=\pi r^2 v \\ & =3.14 \times\left(\frac{4 \times 10^{-3}}{2}\right)^2 \times 14.14=177 \times 10^{-6} \mathrm{~m}^3 / \mathrm{s} \end{aligned}$

Asked in: AP EAMCET 2023 (17 May Shift 1)

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