If the volume of spherical ball is increasing at the rate of `4pi" cc/s"`, then the rate of change of its surface area when the volume is 288 `pi` cc is

Options

A. `4 pi cm^(2)//s`
B. `2/3 pi cm^(2)//s`
C. `4 pi cm^(2)//s`
D. `2 pi cm^(2)//s`

Answer

A

Explanation

Let V and r be the volume and radius of spherical ball, respectively.
Volume of spherical ball `= 4/3 pir^(3)`
`rArrV=4/3pir^(3)"..."(i)`
`rArr288pi=4/3pir^(3)`[given, `V = 288 cm^(3)`]
`rArr 288 =4/3 r^(3)`
`rArr r^(3)=72xx3 = 8xx27`
`rArr r^(3)=2^(2)xx3^(3)`
[taking cube roots both sides]
`rArr r = 6`
On differentiating Eq. (i) w.r.t. 't', we get
`(dV)/(dt) = 4pir^(2)'(dr)/(dt)`.
`:. 4pi = 4pir^(2) '(dr)/(dt)`
[Given `(dV)/(dt) = 4pi` cubic cm/s]
`rArr 1 = (6)^(2) '(dr)/(dt) " " [ :' r = 6]`
`rArr (dr)/(dT) = 1/36`
Now, surface area of spherical ball, `(s) = 4pir^(2)`
`rArr s = 4pir^(2)`
On differentiating both sides, w.r.t. 't' we get
`(ds)/(dt) = 4 xx 2 pir '(dr)/(dt)`
`= 8 xx pi xx 6 xx 1/36`
`[ :' r " and" (dr)/(dt) "and" (dr)/(dT) = 1/(36)]`
`rArr (ds)/(dT) = (4pi)/(3) cm^(2)//s`