A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is `tan^(-1 (0.5).` Water is poured into it at a constant rate of 5 cubic metre per hour. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is 4 m.

Options

A. `(2)/(pi)`
B. ` (1)/(5pi)`
C. `(1)/(15pi)`
D. `(1)/(10pi)`

Answer

B

Explanation

Given, semi-vertical angle of right circular cone
`tan ^(-1) ((1)/(2))`
Let `alpha = tan ^(-1) ((1)/(2))`
`rArr tan alpha = (1)/(2)`
`rArr (r ) /( h) = (1)/(2)" " ` [ from fig. `tan alpha = (r ) /(h)`]
`rArr " " r = (1)/(2) h" " ` ... (i)

`because ` Volume of cone is `(V) = (1)/(3) pi r ^(2) h`
` therefore V = (1) /(3) pi ((1)/(2) h)^(2) (h) = (1) /(1 2) pi h ^(3)" " ` [ from Eq. (i)]
On differentiating both sides w.r.t. 't', we get
`(dV)/(dt) = (1)/(12) pi ( 3h ^(2) ) (dh) /(dt)`
`rArr (dh ) /(dt) = ( 4 ) /( pi h ^(2)) (dV) /(d t)`
`rArr (dh ) /(dt) = ( 4) /( pi h^(2)) xx 5 " " ` [ `because ` given `(dV)/(dt) = 5 m^(3) //min`]
Now, at `h = 10 ` m, the rate at which height of water level is rising `= (dh)/(dt):| _(h= 10 )`
`" " ( 4 ) /( pi( 10 )^(2) ) xx 5 = (1) /( 5pi ) m//min`