Volume and Radius Decrease in Lake – Rankers Physics
Topic: Solid and Fluids
Subtopic: Solids

Volume and Radius Decrease in Lake

A rubber sphere is taken in a lake to a depth \(1800\text{ m}\). If bulk modulus of rubber is \(6 \times 10^8\text{ N/m}^2\), then radius of this rubber sphere will decrease by:
1%
2%
3%
4%

Solution:

The pressure change is \(dP = \rho g h = 10^3 \times 10 \times 1800 = 1.8 \times 10^7\text{ N/m}^2\). The fractional volume change is \(\frac{dV}{V} = \frac{dP}{B} = \frac{1.8 \times 10^7}{6 \times 10^8} = 3\%\). Since \(\frac{dV}{V} = 3\frac{dr}{r}\), the radius decreases by \(\frac{3\%}{3} = 1\%\).

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