A) \[\frac{T}{\sqrt{2}}\]
B) \[\frac{T}{\sqrt{3}}\]
C) \[\frac{T}{2}\]
D) \[\frac{T}{2\sqrt{2}}\]
Correct Answer: A
Solution :
The velocity of efflux through a hole is given by \[v=\sqrt{2gh}.\]If the hole is at the top of the tank, v is obviously zero. Therefore, the average velocity of efflux \[{{V}_{a}}=\frac{1}{2}\sqrt{2gh}=\sqrt{\frac{gh}{2}}.\] If the volume of water in the tank when it is full, is V and A the cross sectional area of the hole, then the time taken by the taken to the emptied is \[T=\frac{V}{A{{v}_{a}}}=\frac{\sqrt{2}V}{A\sqrt{gh}}\] when the tank is half-full, V is replaced by \[\frac{v}{2}\And \,h\]is replaced by\[h/2.\] Therefore, the time taken is mow \[T'=\frac{\sqrt{2}V/2}{A\sqrt{gh}/2}=\frac{V}{A\sqrt{gh}}=\frac{T}{\sqrt{2}}\] Hence, the correction option is (a).
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