Question

The volume of a spherical balloon is increased by 700% what is the percentage increase in its surface area

Answer 1

Answer:

Let the initial volume be “V₁” and surface area be “S₁”.

After volume is increased by 700%, let the new volume be denoted as “V₂” i.e.,

V₂ = V₁+(700% of V₁) = 8 V₁ ….. (i)

And let the new surface area is denoted as “S₂”.

The formula for volume and surface area of a sphere is given by,

V = (4πr³)/3

And,

S = 4πr²

 

Now, from equation (i), we have

V₂/V₁= 8

Or, [(4πr₂³)/3] / [(4πr₁³)/3] = 8

Or, r₂³ / r₁³ = 8  

Multiplying the power on both sides by 2/3, we get

Or, r₂² / r₁² = (8)^(2/3)

Or, r₂² / r₁² = 4 ….. (ii)

We can also write eq. (ii) as,

4πr₂² / 4πr₁² = 4

Or, S₂ / S₁ = 4

Or, S₂= 4 S₁

Thus,  

The percentage increase in the surface area is,

= [(S2 – S1) / S1] * 100

= [(4S1 – S1) / S1] * 100

= [3S1/ S1] * 100

= 300 %