Question

If the radius of the sphere is increased by 25%, then the percentage increased in it's volume is ____________(approximately).

Answer 1

Original question: If the radius of a sphere is increased by 5 percent, then what is the percent increase in its volume with proper steps?

The volume of a sphere can be derived using the equation

v=(4πr3)3)/3

Let us assume that the original radius of the sphere as 'r', and the increased radius as 's'. As described in the question, we can deduce that s=r+(5%)r

=> s=r+(5r/100)

=> s=1.05r

The modified volume of the sphere, denoted by v2, can then be calculated using the above formula:

v2=(4πs^3)/3

=> v2=(4π{1.05r}^3)/3

=> v2=(4πr^3)(1.05)^3/3

=> v2={(1.05)^3}v

=> v2=1.158v

This means that the modified volume is 1.158 times the original volume of the sphere due to 5% increase in its radius.

Deducting the increased volume with original volume and multiplying it 100 parts would hence give us the percentage increase in volume of the sphere.

=> % increase in volume of sphere = (v2 - v)*100

= (1.158v - v )*100

= 0.158v*100

= 15.8v

Hence, increasing the radius of a sphere by 5% would result in an approximate increase of 15.8% percent in its volume.

Answer 2

let original radius be r

then increased radius = r+0.25r

= 1.25r

vol,V= 4/3 pi r^3

vol increased ,V'= 4/3 pi (1.25r)^3

V'= 1.25^3×V

increase in volume is 1.953125