Question

A solid brass sphere of volume 0.305 m³ is dropped in ocean, where water pressure is 2X 10⁷ N / m². The bulk modulus of water is 6.1 x 10¹⁰ N / m². What is change in volume of sphere? (Ans: 10⁻⁴ m³ )

Answer 1

given, water pressure , P = 2 × 10^7 N/m²

Bulk modulus of water , $\beta$ = 6.1 × 10^10 N/m² .

Volume of solid brass sphere , V = 0.305 m³

Now use formula,

$\beta=\frac{P}{\frac{\Delta{V}}{V}}$ [ didn't include sign ]

6.1 × 10^10 = 2 × 10^7/(∆V/0.305)

∆V = 2 × 10^7 × 0.305/6.1 × 10^10

= 0.610 × 10^7/6.1 × 10^10

= 10^-4 m³

Answer 2

Answer:

given, water pressure , P = 2 × 10^7 N/m²

Bulk modulus of water ,

= 6.1 × 10^10 N/m² .

\beta

Volume of solid brass sphere , V = 0.305 m³

Now use formula,

\beta=\frac{P}{\frac{\Delta{V}}{V}}

[ didn't include sign ]

6.1 × 10^10 = 2 × 10^7/(∆V/0.305)

∆V = 2 × 10^7 × 0.305/6.1 × 10^10

= 0.610 × 10^7/6.1 × 10^10

= 10^-4 m³

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