If the radius of a sphere is increased to four times of its original radius, then find the percentage increase in its total surface area.

Option 2 : 1500%

**Given:**

R = 4r

where, r = original radius

R = new radius

**Formula used:**

The total surface area of a sphere = 4πr^{2 }

where, r = radius of sphere

**Calculation:**

Let the original and new radius of the sphere be r and R respectively.

Original total surface area = 4πr^{2} ----(1)

According to the question,

New radius (R) = 4r

⇒ New total surface area = 4πR2

⇒ New total surface area = 4π × (4 × r)^{2}

⇒ 64πr^{2} ----(2)

From equation (1) and (2), we get

Percentage increase in total surface area = {(64πr^{2} - 4πr^{2})/4πr^{2}} × 100%

⇒ Percentage increase in total surface area = {(60πr^{2})/4πr^{2}} × 100%

⇒ Percentage increase in total surface area = 1500%

**∴**** The percentage increase in its total surface area is 1500%.**

Let the radius be r and R respectively

and R = 4r

and original and new surface area be A and A' respectively

The total surface area of a sphere = 4πr^{2}

According to the formula,

Surface area is directly proportional to square of radius.

A'/A = (R/r)^{2}

⇒ Percentage increase in total surface area = {(R^{2 }- r^{2})/r^{2}} × 100%

⇒ Percentage increase in total surface area = {(16r^{2 }- r^{2})/r^{2}} × 100%

⇒ Percentage increase in total surface area = 1500%

**∴**** The percentage increase in its total surface area is 1500%.**