A tank having cross-section 'A' is filled with water upto height `h_(1)`. What will be the time taken by water to come out of a hole of cross-section 'a' at the bottom to decrease the level from `h_(1)` to `h_(2)`·

Options

A. A. a . `sqrt(2g).(h_(1) -h_(2))`
B. `A/asqrt(2/g)(sqrt(h_(1))-sqrt(h_(2)))`
C. `A/asqrt(2g)(sqrt(h_(1))-sqrt(h_(2)))`
D. `A/asqrt(2/g)(sqrt(h_(1))-sqrt(h_(2)))`

Answer

B

Explanation

Let .h. be the height at any instant and set the rate of decrease of level be `-(dh)/(dt)` Therefore,
`=pirh^(2)dg,`
`A(-(du)/(dt))=av=asqrt(2gh)`
or `(dh)/(dt)=-a/A(2gh)^(1//2)`
`(h_(2))/(h_(1))(dh)/((h)^(1//2))=a/Asqrt(2g)_(0)^(t)dt`
or `-2[sqrth]_(h_(1))^(h_(2))=a/Asqrt(2gt)`
`thereforet=2[sqrt(h_(1))-sqrt(h_(2))]A/axx1/(sqrt(2g))`
`=A/asqrt(2/g)[sqrt(h_(1))-sqrt(h_(2))]`
`therefore` Correct choice is (b).