AP EAMCET · PHYSICS · Mechanical Properties of Fluids
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 \(=10 \mathrm{~ms}^{-2}\) )
- A \(4 \mathrm{~mm}^2\)
- B \(1.5 \mathrm{~mm}^2\)
- C \(2.5 \mathrm{~mm}^2\)
- D \(2 \mathrm{~mm}^2\)
Answer & Solution
Correct Answer
(C) \(2.5 \mathrm{~mm}^2\)
Step-by-step Solution
Detailed explanation
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,…
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