Question

A solid lead sphere of volume 0.5 m³ is lowered to a depth in the ocean where the water pressure is equal to 2*10^7 N/M² . The bulk modulus of lead is equal to 7.7*10^9 N/M² . What is the change in volume of the sphere??

Answer 1

Question #71873, Physics / Mechanics | Relativity |

A solid lead sphere of volume $0.5\,\mathrm{m}^3$ is lowered to a depth in the ocean where the water pressure is equal to $2*10^7\,\mathrm{N/m}^2$. The bulk modulus of lead is equal to $7.7*10^9\,\mathrm{N/m}^2$. What is the change in volume of the sphere??

**Need to find:** dV - ?

$$V = 0.5\,\mathrm{m}^3$$ $$p_0 = 2*10^7\,\mathrm{N/m}^2$$

The atmospheric pressure is $p = 1.013 \times 10^5\,\mathrm{Pa}$

The bulk modulus of lead is equal to $K = 7.7*10^9\,\mathrm{N/m}^2$

**Solution**

The bulk modulus (K) of a substance measures the substance's resistance to uniform compression. It is defined as the pressure increase needed to cause a given relative decrease in volume.

$$K = -V \left(\frac{dp}{dV}\right)$$

Here, $dp = 2*10^7 - 1.013*10^5 = 1.39987*10^7\,\mathrm{N/m}^2$

Hence, $dV = \left(-V \frac{dp}{K}\right) = -0.5 \cdot \left(\frac{1.98 \cdot 10^7}{7.7 \cdot 10^9}\right) = 1.3 \cdot 10^{-3}\,\mathrm{m}^3$

**Answer** $-V = 1.3 \cdot 10^{-3}\,\mathrm{m}^3$. The negative sign implies that there is a reduction in Volume

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