A) \[\sqrt{3}\]
B) \[\sqrt{2}\]
C) \[\frac{1}{\sqrt{2}}\]
D) \[\frac{1}{\sqrt{3}}\]
Correct Answer: D
Solution :
Let h is the height of water, A is the area of tank and \[{{A}_{0}}\]be the area of hole orifice which is in the bottom. Time taken by the tank to be emptied is given by \[t=\frac{A}{{{A}_{0}}}\sqrt{\frac{2h}{g}}\] ?(i) Let \[{{t}_{1}}\] is the time for tank to be emptied upto \[\frac{1}{4}\]height is given by \[{{t}_{1}}=\frac{A}{{{A}_{0}}}\sqrt{\frac{2}{g}\left( \frac{h}{4} \right)}\] ?(ii) \[{{t}_{2}}\]is the time, for emptying the remaining \[\frac{3}{4}\] of the tank is given by \[{{t}_{2}}=\frac{A}{{{A}_{0}}}\sqrt{\frac{2}{g}\left( \frac{3h}{4} \right)}\] ?(iii) Dividing Eq. (i) by Eq. (ii), we have \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{\frac{A}{{{A}_{0}}}\sqrt{\frac{2}{g}\times \frac{h}{4}}}{\frac{A}{{{A}_{0}}}\sqrt{\frac{2}{g}\left( \frac{3h}{4} \right)}}=\sqrt{\frac{2}{g}\times \frac{h}{4}\times \frac{4g}{6h}}=\sqrt{\frac{1}{3}}\] \[=\frac{1}{\sqrt{3}}\]
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