An $ L-$ shaped glass tube is just immersed in flowing water such that its opening is pointing against flowing water. If the speed of water current is $v$, then
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(a) Here final velocity of water at the top of tube is zero after gaining a height.
At the lowest point in the tube height is taken zero.
From Bernoulli equation$:$
$P+0+\frac{1}{2} \rho v^{2}=P+\rho g h+0$
$\frac{1}{2} \rho v^{2}=\rho g h$
$h=\frac{v^{2}}{2 g}$
Thus, the final height attained by liquid in the tube is $\frac{v^{2}}{2 g}$
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