Question

A cylinder containing water stands on a table of height $H$. A small hole is punched in the side of cylinder at its base. The stream of water strikes the ground at a horizontal distance $R$ from the table. Then the depth of water in the cylinder is:

• A. $H$
• B. $R$
• C. $\sqrt{\left(RH\right)}$
• D. $R^2/4H$

Answer 1

The correct option is D $R^2/4H$

A hole is punched at a height $H$ of table. Let ${}^{\prime }{d}^{\prime }$ be the depth of hole.

u: velocity at which water comes out

From toricells Law, Velocity of water ejecting out is given by:

$u^2=2gd\to \left(2\right)[d:depthofhole]$

The water stream follows a Horizuntala projectile, whose range ${}^{\prime }{R}^{\prime }$ is given by:

$R=u\sqrt{\frac{2H}{g}}\to \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$R=\sqrt{2gd.\frac{2H}{g}}\Rightarrow d=\frac{R^2}{4H}.$