Question

A hemispherical tank of diameter 4m is filled with water. A very small hole is punched at 1m above the bottom as shown in figure. Find x (distance at which water strikes the surface)

• A. 0.81 m
• B. 1.81 m
• C. 2.81 m
• D. 3.81 m

Answer 1

The correct option is C

2.81 m

velocity of water coming out from hole

$v=\sqrt{2gh}=\sqrt{2gl}$

$y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$1=\left(\sqrt{2g}sin{30}^{\circ }\right)t+\frac{1}{2}\times 10t^2$

$1=\sqrt{2g}\frac{t}{2}+5t^2$

$5t^2+2.24t-1=0$

$t=\frac{-2.24\pm \sqrt{(2.24{)}^2+20}}{10}=0.28s$

$x_1=v_{x}t=\sqrt{2g}cos{30}^{\circ }\times 0.28=1.08s$

$x_2=\sqrt{2^2-1^2}=\sqrt{3}$

Velocity of water coming out from hole , $x=x_1+x_2=1.08+1.73=2.81m$