question_answer

A cylindrical vessel of cross-section A contains water to a height h. There is a hole in the bottom of radius 'a'. The time in which it will be emptied is:

A) \[\frac{2A}{\pi {{a}^{2}}}\sqrt{\frac{h}{g}}\]              

B) \[\frac{\sqrt{2}A}{\pi {{a}^{2}}}\sqrt{\frac{h}{g}}\]

C) \[\frac{2\sqrt{2}A}{\pi {{a}^{2}}}\sqrt{\frac{h}{g}}\]               

D) \[\frac{A}{\sqrt{2}\pi {{a}^{2}}}\sqrt{\frac{h}{g}}\]

Correct Answer: B

Solution :

[b]

let the rate of falling water level be \[-\frac{dh}{dt}\]

Initially at \[t=0;h=h\]

Finally at \[t=t;h=0\]

Then, \[A\left( -\frac{dh}{dt} \right)=\pi {{a}^{2}}.v\]

\[dt=-\frac{A}{\pi {{a}^{2}}\sqrt{2gh}}dh\]

[\[\because \]velocity of efflux of liquid \[v=\sqrt{2gh}\] ]

Integrating both sides,

\[\int\limits_{0}^{t}{dt=-\frac{A}{\sqrt{2g}\pi {{a}^{2}}}\int\limits_{h}^{0}{{{h}^{-1/2}}}dh}\]

\[\left[ t \right]_{0}^{t}=-\frac{A}{\sqrt{2g}\pi {{a}^{2.}}}\left[ \frac{{{h}^{1/2}}}{1/2} \right]_{h}^{0}\]

\[t=\frac{\sqrt{2}A}{\pi {{a}^{2}}}\sqrt{\frac{h}{g}}\]