AP EAMCET · PHYSICS · Mechanical Properties of Fluids
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 \(=10 \mathrm{~ms}^{-2}\) )
- A \(14.14 \times 10^{-6} \mathrm{~m}^3 \mathrm{~s}^{-1}\)
- B \(2.1 \times 10^{-6} \mathrm{~m}^3 \mathrm{~s}^{-1}\)
- C \(1.77 \times 10^{-6} \mathrm{~m}^3 \mathrm{~s}^{-1}\)
- D \(0.177 \times 10^{-6} \mathrm{~m}^3 \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(C) \(1.77 \times 10^{-6} \mathrm{~m}^3 \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
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…
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