If the volume of a sphere increases at constant rate `((dv)/(dt)=4)`. If radius of the sphere is denoted by `r` then surface area of the sphere inceases at the rate`:-`

Options

A. `(4)/(r )`
B. `(8)/(r )`
C. `(12)/(r )`
D. `(16)/(r )`

Answer

B

Explanation

`V=(4)/(3)pir^(2)rArr(dV)/(dt)=(dV)/(dt)xx(dr)/(dt)=[(4)/(3)pi(2r)](dr)/(dt)=4`
`rArr(dr)/(dt)=(4)/(4pir^(2))`
`5=4pir^(2)rArr(dS)/(dt)=(dS)/(dr)xx(dr)/(dt)=[4pi(2r)](dr)/(dt)`
`(dS)/(dt)=(8pir)((4)/(4pir^(2)))=(8)/(r)`