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Question

# A large open tank is filled with water upto a height H. A small hole is made at the base of the tank. It takes T1 time to decrease the height of water to Hn(n>1) and it takes T2 time to take out the remaining water. If T1=T2, then the value of n is :

A
2
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B
3
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C
4
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D
22
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Solution

## The correct option is C 4If A0 is the area of the orifice at a depth y below the free surface and A that of the container, the volume of the liquid coming out of the orifice(hole) per second isdVdt=vA0=A0√2gy ....(1)Since dV=−Ady, substituting in (1) we get,−Adydt=A0√2gy∴∫T10dt=−AA0(2g)−12∫HnHy−12dy⇒T1=AA0√2g(√H−√Hn) ...(4)For the liquid to go from height Hn to 0 time T2 ia taken as given.∴T2=AA0√2g(√Hn−0) .....(5)Given T1=T2⇒AA0√2g(√H−√Hn)=AA0√2g(√Hn−0)∴2√Hn=√H⇒4n=1⇒n=4

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