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A large tank is filled with water to a height $h$ . A small hole is made at the base of the tank. It takes $T_{1}$ time to decrease the height of water to $\frac{H}{η} (η > 1)$ ; and it takes $T_{2}$ time to take out the rest of water. If $T_{1} = T_{2}$ , then the value of $η$ is:

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detailed solution

Correct option is C

t=Aa2g[H1−H2] Now, T1=Aa2g[H−Hη] andT2=Aa2g[Hη−0] According to problem,T1=T2 ∴ H−Hη=Hη or H=2Hη or η=4

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