Question

A solid sphere of brass (bulk modulus of 14.0 ✕ 1010 N/m2) with a diameter of 2.20 m is thrown into the ocean. By how much does the diameter of the sphere decrease as it sinks to a depth of 2.40 km? (Take the density of ocean water to be 1,030 kg/m3.)

Answer 1

Diameter decreases by the diameter of 0.0312 m.

Step-by-step explanation:

Given that,

Bulk modulus = 14.0 × 10¹⁰ N/m²

Diameter d = 2.20 m

Depth = 2.40 km

Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000

= 24.25 × 10⁶ N/m²

Volume =
`(4)/(3) \pi r^3`

`(\Delta V)/(V)=((\Delta r)^3)/(r^3)`

Bulk modulus is equal to

`B = -(\Delta P)/((\Delta V)/(V) )`

now

`B = -(24.25 * 10^6)/(((\Delta r)^3)/(r^3) )`

`B = -(24.25 * 10^6)/(((\Delta r)^3)/(1.1^3) )`

`(\Delta r)^3 = (24.25 * 10^6 * 1.1^3)/(14.0 * 10^(10))`

Δ r = -0.0156 m

change in diameter

Δ d = -2 × 0.0156

Δ d = -0.0312 m

Diameter decreases by the diameter of 0.0312 m.