Question

Water is filled in a container of uniform area of cross section 'A'. A hole of cross sectional area ${}^{\prime }{a}^{\prime }(a<<A)$ is made in the container at a height of 20 m above the base. Water streams out and hits a small block on the surface at some distance from the container. The block is moved on the surface in such a way that the stream always hits the block. The initial velocity of the block (in m/s) is
$(\text{Given}\frac{a}{A}=\frac{1}{20}.\text{Take}g=10{\text{ms}}^{-2})$.

Answer 1

Velocity of efflux $v=\sqrt{2gy}$
Range $x=\sqrt{2gy}\times \sqrt{\frac{2h}{g}}$
The velocity of the block must be $\left(\frac{dx}{dt}\right)$.
$V_b=\frac{dx}{dt}=\sqrt{\frac{2h}{g}}\times \sqrt{2g}\times \frac{1}{2\sqrt{y}}\frac{dy}{dt}$
$V_b=\frac{\sqrt{h}}{\sqrt{y}}.\frac{dy}{dt}\dots \dots \left(i\right)$
Using equation of continuity,
$\frac{Ady}{dt}=a\sqrt{2gy}\dots \dots \left(ii\right)$
From equation (i) and (ii),
$V_b=\sqrt{\frac{h}{y}}\times \frac{a}{A}\sqrt{2gy}$
$=\sqrt{2gh}\times \frac{a}{A}=20\times \frac{1}{20}=1{\text{ms}}^{-1}$