Question

A spherical medicine ball when dropped in water dissolves in such a way that the rate of decrease of volume at any instant is proportional to its surface area. Calculate the rate of decrease of its radius.

Answer 1

Let r = Radius of the sphere

V = Volume of the sphere

S = Surface area of the sphere

Given, `(dV)/(dt) α  s`

`(dV)/(dt) = -kS`

Where k is a proportionality constant and (−) indicates that the volume is decreasing with time.

`d/(dt)(4/3pir^3) = -k xx 4pir^2`

Here, V = `4/3pir^3 and S = 4pir^2`

`4/3pi3r^2 (dr)/(dt) = -k xx 4pir^2`

`r^2(dr)/(dt) = -kr^2`

Since, `(dr)/(dt) = -k`

Where k is a constant.

The rate of change in radius is constant.