Question

A tank is filled with a liquid upto a height H. A small hole is made at the bottom of this tank. Let $t_1$ be the time taken to empty first half of tank and $t_2$ the time taken to empty rest half of the tank. Then find $\frac{t_1}{t_2}$ (assume non viscous and in compressible fluid flowing in stream line manner)

• A. 0.414
• B. 0.212
• C. 0.313
• D. 0.515

Answer 1

The correct option is A 0.414

Time taken for the level to fall from H to 'H' $t=\frac{A}{A_0}\sqrt{\frac{2}{g}}[\sqrt{H}-\sqrt{H^{\prime }}]$
According to problem- the time taken for the level to fall from h to $\frac{h}{2}$ $t_1=\frac{A}{A_0}\sqrt{\frac{2}{g}}[\sqrt{h}-\sqrt{\frac{h}{2}}]$
and similarly time taken for the level to fall from $\frac{h}{2}$ to zero $t_2=\frac{A}{A_0}\sqrt{\frac{2}{g}}[\sqrt{\frac{h}{2}}-0]$
$\therefore$ $\frac{t_1}{t_2}=\frac{1-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}-0}=\sqrt{2}-1.$