Fluid Mechanics

JEE Main level, numerical answer, medium

55 of 192

Question 55 of 192

The reading of pressure metre attached with a closed pipe is 4.5×104Nm−24.5\times {10}^{4}N{m}^{-2}. On opening the valve, water starts flowing and the reading of pressure metre falls to 2.0×104Nm−22.0\times {10}^{4}N{m}^{-2}. The velocity of water is found to be Vms−1\sqrt{V}m{s}^{-1}. The value of VV is _________.

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Answer: 50

The formula to calculate the change in pressure can be written as
P1−P2=12ρv2...(1){P}_{1}-{P}_{2}=\dfrac{1}{2}\rho {v}^{2}...\left(1\right)
From equation (1), it follows that
4.5×104−2.0×104=12×103×v2⇒2.5×104=12×103×v2⇒v2=50⇒v=504.5\times {10}^{4}-2.0\times {10}^{4}=\dfrac{1}{2}\times {10}^{3}\times {v}^{2} \Rightarrow 2.5\times {10}^{4}=\dfrac{1}{2}\times {10}^{3}\times {v}^{2} \Rightarrow {v}^{2}=50 \Rightarrow v=\sqrt{50}
Hence, V=50V=50.

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