An $L-$ shaped glass tube is just immersed in flowing water towards tube as shown. If speed of water current is $V,$ then the height $h$ upto which water rises will be
A$\frac {V^2}{2g}$
B$\frac {g}{2V^2}$
C$\frac {V^2}{4g}$
D$\frac {2V^2}{g}$
Medium
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A$\frac {V^2}{2g}$
a $\mathrm{P}_{1}+\rho \mathrm{gh}_{1}+\frac{1}{2} \rho \mathrm{V}_{1}^{2}=\mathrm{P}_{2}+\rho \mathrm{gh}_{2}+\frac{1}{2} \rho \mathrm{V}_{2}^{2}$
$\Rightarrow \frac{1}{2} \rho V_{1}^{2}=\rho g h_{2}+0 \quad\left(V_{2}=0\right) \Rightarrow h_{2}=\frac{V^{2}}{2 g}$
$O R$
$KE$ per unit volume $=$ $P.E$ per unit volume $\Rightarrow \frac{1}{2} \rho \mathrm{V}^{2}=\rho \mathrm{gh} \quad \Rightarrow \quad \mathrm{h}=\frac{\mathrm{V}^{2}}{2 \mathrm{g}}$
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