Question

Radius of a sphere in increased by 15% find the percentage change in its volume

Answer 1

Let radius of sphere is R.

so, volume of sphere, V = 4/3πR³

a/c to question,

radius of a sphere is increased by 15% .

so, radius of new sphere, R' = R + 15% of R

= R + 15R/100 = 1.15R

now, volume of new sphere, V' = 4/3πR'³

= 4/3π(1.15R)³

now, % change in volume of sphere = ∆V/V × 100

= [4/3π{(1.15R)³ - R³}]/{4/3πR³} × 100

= {(1.15)³ - 1³} × 100

= (1.520875 - 1) × 100

= 0.520875 × 100

= 52.0875 % ≈ 52%

hence, volume of sphere is increased by 52 %

Answer 2

$\bf{\underline{\underline{Answer:}}}$

$\approx{52}$ %

$\bf{\underline{Given:}}$

Radius of a sphere in increased by 15%

$\bf{\underline{To\:Find:}}$

The percentage change in its volume =?

$\bf{\underline{\underline{Step \:by\:step \:explanation:}}}$

Let radius of sphere is R.

so, volume of sphere, V = 4/3πR³

$\bf{\underline{According \:to\: Question:}}$

Radius of a sphere is increased by 15% .

so, radius of new sphere, R' = R + 15% of R

$⟹ R + \dfrac{15R}{100} = 1.15R$

now, volume of new sphere, V' = 4/3πR'³

= 4/3π(1.15R)³

now, % change in volume of sphere = ∆V/V × 100

= [4/3π{(1.15R)³ - R³}]/{4/3πR³} × 100

= {(1.15)³ - 1³} × 100

= (1.520875 - 1) × 100

= 0.520875 × 100

= 52.0875 % ≈ 52%