Question

the radius of a bubble increases by 4%. Calculate the percentage increase in its (a) surface area (b) volume (to three significant figures )​

Answer 1

Explanation:

Given that the radius of a bubble increases by 4%.

Let r be the radius of the spherical bubble, so the new radius of the sphere

`R= r+0.04r = 1.04r\cdots(i)`

(a) Surface area of the bobble,

`s=4\pi r^2`

So, the rate of increase of surface area =
`4\pi R^2 -4\pi r^2= 4\pi(R^2-r^2)`

The percentage change in the surface area
`= \frac {4\pi(R^2-r^2)}{4\pi r^2}* 100`

By using equation (i)

The percentage change in the surface area =
`\frac {(1.04r)^2-r^2}{ r^2}* 100`

`=(1.04^2-1)* 100`

=8.160%

Therefore, the percentage change in the surface area is 8.160%

(b) The volume of the sphere =
`4/3 \pi r^3`

So, the change in the volume =
`4/3 \pi R^3-4/3 \pi r^3`

Percentage change in the volume =

`\frac {4/3 \pi R^3-4/3 \pi r^3}{4/3 \pi r^3}* 100`

`=(1.04^3-1)100` [by using equation (i)]

=12.486%

Therefore, the percentage change in the volume is 12.486%.