JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
Consider a water tank as shown in the figure. It's cross-sectional area is \(0.4\, m ^{2}\). The tank has an opening \(B\) near the bottom whose cross-section area is \(1\, cm ^{2}\). A load of \(24\, kg\) is applied on the water at the top when the height of the water level is \(40\, cm\) above the bottom, the velocity of water coming out the opening \(B\) is \(v\, ms ^{-1}\). The value of \(v\), to the nearest integer, is ......\(m/s\). [Take value of \(g\) to be \(10 \,ms ^{-2}\) ]
- A \(3\)
- B \(6\)
- C \(9\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(3\)
Step-by-step Solution
Detailed explanation
\(m =24 \,kg\) \(A =0.4\, m ^{2}\) \(a =1 \,cm ^{2}\) \(H =40 \,cm\) Using Bernoulli's equation \(\Rightarrow\left( P _{0}+\frac{ mg }{ A }\right)+\rho gH + \frac{1}{2} \rho V _{1}^{2}\) \(=P_{0}+0+\frac{1}{2} \rho v^{2}.....(1)\) \(\Rightarrow\) Neglecting \(v _{1}\)…
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