A cylinderical tank of height 0.4 m is open at the top and has a diameter 0.16m. Water is filled in it up to height of 0.16 m. Find the time taken to empty the tank through a hole of radius `5 xx 10^(-3)m` in its bottom.

Options

A. `46.26 sec`.
B. `4.6 sec`.
C. `462.6 sec`.
D. `0.46 sec`.

Answer

A

Explanation

`A_(1)v_(1) = A_(2)v_(2)`
`piR^(2)dh//dt = pir^(2)v ….(i)`
`v = sqrt(2gh) …….(ii)`
from equation `(ii)` put the value of `v` in equation `(i)`
`piR^(2) dh//dt = pir^(2) sqrt(2gh)`
`rArr int(R^(2)dh)/(r^(2)sqrt(2gh)) = intdt`

`(R^(2))/(r^(2)sqrt(2g))overset(0)underset(h)(int)(dh)/(sqrt(h)) = overset(T)underset(0)(int)dt`
`T = (R^(2))/(r^(2)) sqrt((2h)/(g))`
on solving
`t = 46.26` second.