TS EAMCET · Physics · Mechanical Properties of Fluids
A large open tank has two holes in the wall. One is a square hole of side \(L\) at a depth \(y\) from the top and the other is a circular hole of radius \(R\) at a depth \(4 y\) from the top. When the tank is completely filled with water, the quantities of water flowing out per second from the two holes are the same. Then value of \(R\) is
- A \(\frac{L}{\sqrt{2 \pi}}\)
- B \(2 \pi L\)
- C \(L \sqrt{\frac{2}{\pi}}\)
- D \(\frac{L}{2 \pi}\)
Answer & Solution
Correct Answer
(A) \(\frac{L}{\sqrt{2 \pi}}\)
Step-by-step Solution
Detailed explanation
By the principle of continuity \(A_1 v_1=A_2 v_2\) According to question, \(A_1=L^2\) \(\begin{aligned} & v_1=\sqrt{2 g y} \\ & \text { and } \\ & A_2=\pi R^2 \\ & v_2=\sqrt{2 g 4 y} \\ & \therefore \quad L^2 \sqrt{2 g y}=\pi R^2 \sqrt{2 g 4 y} \\ & \end{aligned}\) or…
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