Answer:
Step-by-step 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
[tex]R= r+0.04r = 1.04r\cdots(i)[/tex]
(a) Surface area of the bobble,
[tex]s=4\pi r^2[/tex]
So, the rate of increase of surface area = [tex]4\pi R^2 -4\pi r^2= 4\pi(R^2-r^2)[/tex]
The percentage change in the surface area [tex]= \frac {4\pi(R^2-r^2)}{4\pi r^2}\times 100[/tex]
By using equation (i)
The percentage change in the surface area = [tex]\frac {(1.04r)^2-r^2}{ r^2}\times 100[/tex]
[tex]=(1.04^2-1)\times 100[/tex]
=8.160%
Therefore, the percentage change in the surface area is 8.160%
(b) The volume of the sphere = [tex]4/3 \pi r^3[/tex]
So, the change in the volume = [tex]4/3 \pi R^3-4/3 \pi r^3[/tex]
Percentage change in the volume =
[tex]\frac {4/3 \pi R^3-4/3 \pi r^3}{4/3 \pi r^3}\times 100[/tex]
[tex]=(1.04^3-1)100[/tex] [by using equation (i)]
=12.486%
Therefore, the percentage change in the volume is 12.486%.
