The increase in the pressure required to decrease the volume $$(\Delta V)$$ of water is $$6.3 \times 10^7$$ N/m². The percentage decrease in the volume is _____. (Bulk modulus of water = $$2.1 \times 10^9$$ N/m².)

The bulk modulus (incompressibility) of a fluid is defined as

$$K = -\frac{\Delta P}{\Delta V/V}$$ $$-(1)$$

where
  $$\Delta P$$ = rise in pressure,
  $$\Delta V$$ = change in volume (negative for compression),
  $$\Delta V/V$$ = fractional change in volume.

We are interested only in the magnitude of the fractional change, so from $$-(1)$$

$$\left|\frac{\Delta V}{V}\right| = \frac{\Delta P}{K}$$ $$-(2)$$

Given data:
  $$\Delta P = 6.3 \times 10^{7}\,\text{N/m}^2$$
  $$K = 2.1 \times 10^{9}\,\text{N/m}^2$$

Substituting in $$-(2)$$:

$$\left|\frac{\Delta V}{V}\right| = \frac{6.3 \times 10^{7}}{2.1 \times 10^{9}}$$

Divide the numbers and powers of 10 separately:

$$\frac{6.3}{2.1} = 3 \quad\text{and}\quad 10^{7}/10^{9} = 10^{-2} = 0.01$$

Hence

$$\left|\frac{\Delta V}{V}\right| = 3 \times 0.01 = 0.03$$

The fractional decrease in volume is $$0.03$$, i.e. a $$3\%$$ decrease.

Therefore, the percentage decrease in the volume of water is $$3\%$$.

Option B which is: $$3\%$$