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. `0.5m//s`
B. `2m//s`
C. `3m//s`
D. `1m//s`

Answer

D

Explanation

Given that `(dV)/(dt)=2`
`rArr" "(d)/(dt)((1)/(3)pir^(3))=2`
`rArr" "pir^(2).(dr)/(dt)=2`
`rArr" "(dr)/(dt)=(2)/(pir^(2))`
`rArr" "(d)/(dt)(2pir)=(4)/(r^(2))" (i)"`
When h = 2 m
`rArr" "r=2m`
Hence, `(d(2pir))/(dt)=(4)/(4)=1m//s`