Question

Water stands at a depth $H$ in a tank, whose side walls are vertical as shown in diagram. A hole is made on one of walls at a depth $h$ below water surface. Water comes out from the hole. The maximum range will be equal to :

• A. $\frac{H}{2}$
• B. $H$
• C. $2H$
• D. $2h$

Answer 1

The correct option is B $H$

Time taken to fall of liquid ,$t=\sqrt{\frac{2(H-h)}{g}}$

Range=$\sqrt{2gh}\times \sqrt{\frac{2(H-h)}{g}}$

Range=$2\sqrt{h(H-h)}$

For range to be maximum $\frac{dR}{dh}=0$

$0=\frac{2\times (H-2h)}{2\sqrt{H.(H-h)}}$

$h=\frac{H}{2}$

Range=$2\sqrt{h(H-h)}$

$=2\sqrt{\frac{H}{2}\times \frac{H}{2}}$

$R_{max}=H$