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A large tank is filled with water to a h...

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 `(H)/(eta) (eta gt 1)`, and it takes `T_(2)` time to take out the rest of water. If `T_(1) = T_(2)`, then the value of `eta` is

A

2

B

3

C

4

D

`2sqrt2`

Text Solution

Verified by Experts

The correct Answer is:
C
`t=(A)/(A_0)sqrt(2/g)[sqrt(H_1)-sqrt(H_2)]`
`(T_1)/(T_2)=((A)/(A_0)sqrt(2/g)[sqrt(H)-sqrt(H/eta)])/((A)/(A_0)sqrt(2/g)[sqrt(H/eta)-0])`
`implies sqrt(H/eta)=sqrt(H)-sqrt(H/eta)`
`implies 4 H/eta = H implies eta=4`.
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