The reading of pressure metre attached with a closed pipe is $4.5 \times 10^4 \mathrm{~N} / \mathrm{m}^2$. On opening the valve, water starts flowing and the reading of pressure metre falls to $2.0 \times 10^4 \mathrm{~N} / \mathrm{m}^2$. The velocity of water is found to be $\sqrt{\mathrm{V}} \mathrm{m} / \mathrm{s}$. The value of $\mathrm{V}$ is__________
JEE MAIN 2024, Diffcult
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$ \text { Change in pressure }=\frac{1}{2} \rho \mathrm{v}^2 $
$ 4.5 \times 10^4-2.0 \times 10^4=\frac{1}{2} \times 10^3 \times \mathrm{v}^2 $
$ 2.5 \times 10^4=\frac{1}{2} \times 10^3 \times \mathrm{v}^2 $
$ \mathrm{v}^2=50 $
$ \mathrm{v}=\sqrt{50} $
$ \text { Velocity of water }=\sqrt{\mathrm{V}}=\sqrt{50} $
$ =\mathrm{V}=50$
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