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 `T_1` time to decrease the height of water to H/n(ngt1) and it takes `T_(2)` time to take out the remaining water. If `T_(1)=T_(2)`, then the value of n is

Options

A. 2
B. 3
C. 4
D. `2sqrt(2)`

Explanation

`-A(dy)/(dt)=asqrt(2gy)`
`(2A)/(asqrt(2)g)(sqrt(H)-sqrt(H/n))=T_(1)`
`(2A)/(asqrt(2)g)(sqrt(H/n)-0)=T_(2)`
`T_(1)=T_(2)`
n=4.