Question

When surrounded by air at a pressure of 1.0×105N/m2, a basketball has a radius of 0.11 m. Find the volume that the ball would have if you take it 30 m below the water surface where the pressure is 3.0×105N/m2.

Answer 1

Solution:

Given:

P₁ = 1 x 10⁵ N/m²

P₂ = 3 x 10⁵ N/m²

radius, r₁ = 0.11 m

let the volume of the ball have at location 2 be V₂.

Then: P₁ V₁ = P₂ V₂

${\color{Golden} \therefore }$ (1 x 10⁵ N/m²) { (4/3) ($\pi$) (0.11 m)³ } = (3 x 10⁵ N/m²) V₂

${\color{Golden} \therefore }$ V₂ = 1.858 x 10^-3 m³

the volume of the ball under the water is V₂ = 1.858 x 10^-3 m³.