Radius of a right circular cone is decreased by 20%. To maintain the same volume, the height will have to be increased by?

⇒ Let radius be r, height be h and volume be V

⇒ Then initial volume of cone is

⇒ V = (1/3) × (π × r2 × h)

⇒ Now if radius is decreased by 20% then new radius will be = r – 20% of r

⇒ New radius = 0.8r

⇒ Suppose new height increased by x %

⇒ Then new height = h + x% of h

⇒ New height = (100h + x × h)/100

⇒ New height = (h/100) × (100 + x)

⇒ New volume = (1/3) × (π × 0.8r2 × (h/100) × (100 + x)

⇒ As volume remains the same so ratio of volume will be same

⇒ (Initial volume)/(New Volume) = 1 ----- 1

⇒ Putting the value of each in 1 we get

⇒ {(1/3) × (π × r2 × h) }/{(1/3) × (π × 0.8r2 × (h/100) × (100 + x) } = 1

⇒ 100/(0.64) × (100 + x) = 1

∴ x = 56.25%