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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
$\frac{{{V^2}}}{{2g}}$
$\frac{g}{{2{V^2}}}$
$\frac{{{V^2}}}{{4g}}$
$\frac{{2{V^2}}}{g}$
Solution
$\mathrm{P}_{1}+\rho \mathrm{gh}_{1}+\frac{1}{2} \rho \mathrm{V}_{1}^{2}=\mathrm{P}_{2}+\rho \mathrm{gh}_{2}+\frac{1}{2} \rho \mathrm{V}_{2}^{2}$
$\Rightarrow \frac{1}{2} \rho \mathrm{V}_{1}^{2}=\rho \mathrm{gh}_{2}+0 \quad\left(\mathrm{V}_{2}=0\right) \Rightarrow \mathrm{h}_{2}=\frac{\mathrm{V}^{2}}{2 \mathrm{g}}$
$O R$
$KE$ per unit volume $=$ $P.E$ per unit volume
$\Rightarrow \frac{1}{2} \rho \mathrm{V}^{2}=\rho \mathrm{gh} \quad \Rightarrow \quad \mathrm{h}=\frac{\mathrm{V}^{2}}{2 \mathrm{g}}$
