Water is dropped at the rate of `2m^(2)//s` into a cone of semivertical angel of `45^(@)`. The rate at which periphery of water surface changes when height of water in the cone is 2 m, is

Options

A. 2 m/s
B. 1m/s
C. 3 m/s
D. 4 m/s

Answer

a

Explanation

Given that, `(dV)/(dt) =2`
` rArr d/(dt)(1/3 pir^(3)) = 2 rArr pi r^(2) (dr)/(dt) = 2`
` rArr (dr)/(dt) = 2/(pir^(2))`
` rArr d/(dt) (2pir) = 4/(r^(2))" "` …(i)
when H = 2m and r = 2m , then
` (dp)/(dt) = 4/4 = 1m//s ` [ where p =` 2 pi r` = perimeter ]