A fully filled hemispherical tank of radius R has an orifice of small area a at its bottom. Time required to completely empty the tank, will be (assume that the top surface area of liquid is always much greater than the orifice area)

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$\frac{5 π}{\sqrt{2 g}} \frac{R^{5 / 2}}{a}$

$\frac{9}{\sqrt{2 g}} \frac{π R^{5 / 2}}{a}$

$\frac{7 \sqrt{2}}{15} \frac{π R^{5 / 2}}{a \sqrt{g}}$

$\frac{5 π R^{5 / 2}}{7 a \sqrt{g}}$

answer is C.

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Detailed Solution

$\begin{array}{l} A_{1} v_{1} = A_{2} v_{2} \\ \Rightarrow π x^{2} (- \frac{d h}{d t}) = a \sqrt{2 g h} \\ {\Rightarrow \int}_{0}^{t} a d t = -_{R}^{0} \frac{π [R^{2} - {(R - h)}^{2}] d h}{\sqrt{2 g h}} \\ \Rightarrow t = \frac{7 \sqrt{2} π R^{5 / 2}}{15 a \sqrt{g}} \end{array}$

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