Search any question & find its solution
Question:
Answered & Verified by Expert
A large tank filled with water to a height ' $h$ ' is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from $h$ to $\frac{h}{2}$ and from $\frac{h}{2}$ to zero is
Options:
Solution:
2606 Upvotes
Verified Answer
The correct answer is:
$\sqrt{2}-1$
Time taken for the level to fall from $H$ to $H^{\prime} \quad t=\frac{A}{A_0} \sqrt{\frac{2}{g}}[\sqrt{H}-\sqrt{H}]$
According to problem- the time taken for the level to fall from $h$ to $\frac{h}{2} \quad t_1=\frac{A}{A_0} \sqrt{\frac{2}{g}}\left[\sqrt{h}-\sqrt{\frac{h}{2}}\right]$
and similarly time taken for the level to fall from $\frac{h}{2}$ to zero $t_2=\frac{A}{A_0} \sqrt{\frac{2}{g}}\left[\sqrt{\frac{h}{2}}-0\right]$
$\therefore \frac{t_1}{t_2}=\frac{1-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}-0}=\sqrt{2}-1$.
According to problem- the time taken for the level to fall from $h$ to $\frac{h}{2} \quad t_1=\frac{A}{A_0} \sqrt{\frac{2}{g}}\left[\sqrt{h}-\sqrt{\frac{h}{2}}\right]$
and similarly time taken for the level to fall from $\frac{h}{2}$ to zero $t_2=\frac{A}{A_0} \sqrt{\frac{2}{g}}\left[\sqrt{\frac{h}{2}}-0\right]$
$\therefore \frac{t_1}{t_2}=\frac{1-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}-0}=\sqrt{2}-1$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.