A solid sphere made of a material of bulk modulus $1.5 \times 10^9$ N/m² is lowered into a lake. The depth below the surface at which its radius decreases by $0.02\%$ is ______ m. (Density of water $= 10^3$ kg/m³, $g = 10\ \text{m/s}^2$)
Numerical value type. Enter your answer.
Answer: 90
Concept. Bulk modulus $B = \dfrac{\Delta P}{|\Delta V/V|}$. For a sphere $V \propto R^3$, so $\dfrac{\Delta V}{V} = 3\dfrac{\Delta R}{R}$. The extra pressure at depth $h$ is $\rho gh$.
Step 1. Volume strain.
$$\frac{\Delta V}{V} = 3 \times 0.02\% = 6 \times 10^{-4}$$
Step 2. Pressure needed.
$$\Delta P = B\frac{\Delta V}{V} = 1.5 \times 10^9 \times 6 \times 10^{-4} = 9 \times 10^5\ \text{Pa}$$
Step 3. Depth.
$$h = \frac{\Delta P}{\rho g} = \frac{9 \times 10^5}{10^3 \times 10} = 90\ \text{m}$$
Answer. $90$.
Solution by Sreeraj P, M.Sc Physics