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A large tank is filled with water to a height \mathrm{H}. A small hole is made at the base of the tank. It takes \mathrm{T_1} time to decrease the height of water to \mathrm{\mathrm{H} / \eta,(\eta>1)} and it takes \mathrm{T_2} time to take out the rest of water. If \mathrm{T_1=T_2}, then the value of \mathrm{\eta} is:

Option: 1

2


Option: 2

3


Option: 3

4


Option: 4

2 \sqrt{2}


Answers (1)

Volume decrease = Volume outlet

\mathrm{Adh =\mathrm{a} \sqrt{2 \mathrm{gh}} \mathrm{dt}}

\mathrm{ \begin{aligned} & \frac{-d h}{d t}=\frac{a \sqrt{2 g h}}{A} \Rightarrow \int_H^{H / \eta} \frac{d h}{\sqrt{2 g h}}=-\int_0^t \frac{a}{A} d t \\ & t_1=\left[-\sqrt{\frac{H}{\eta}}+\sqrt{H}\right] \end{aligned} }

\mathrm{ \begin{aligned} & \text { Similarly } t_2=\sqrt{\frac{H}{\eta}} \\ & \Rightarrow t_1=t_2 \Rightarrow 2 \sqrt{\frac{H}{\eta}}=\sqrt{H} \\ & \eta=4 \end{aligned} }

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Kshitij

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