Percentage Decrease in Volume using Bulk Modulus

To find the percentage decrease in the volume of water, we use the definition of Bulk Modulus ($B$).

The Bulk Modulus is defined as the ratio of volumetric stress (increase in pressure, $\Delta P$) to volumetric strain ($\frac{-\Delta V}{V}$):

$B = \frac{\Delta P}{\left( \frac{-\Delta V}{V} \right)}$

where:

• $\Delta P$ is the increase in pressure = $6.3 \times 10^7 \text{ N/m}^2$
• $B$ is the Bulk modulus of water = $2.1 \times 10^9 \text{ N/m}^2$
• $\frac{-\Delta V}{V}$ represents the fractional decrease in volume.

Rearranging the formula to find the fractional decrease in volume:

$\frac{-\Delta V}{V} = \frac{\Delta P}{B}$

Substitute the given values into the equation:

$\frac{-\Delta V}{V} = \frac{6.3 \times 10^7 \text{ N/m}^2}{2.1 \times 10^9 \text{ N/m}^2}$

$\frac{-\Delta V}{V} = 3 \times 10^{-2} = 0.03$

The percentage decrease in volume is given by:

$\text{Percentage decrease} = \left( \frac{-\Delta V}{V} \right) \times 100\%$

$\text{Percentage decrease} = 0.03 \times 100\% = 3\%$

Thus, the correct option is B.