Question

The increase in pressure required to decrease the volume of a water sample by 0.2percentage is \( P \times 10^5 \, \text{Nm}^{-2} \). Bulk modulus of water is \( 2.15 \times 10^9 \, \text{Nm}^{-2} \). The value of \( P \) is ……….. 

Hint:

The bulk modulus describes the resistance of a substance to uniform compression. A smaller bulk modulus means it is easier to compress the substance, while a larger value means greater resistance.

Answer 1

The bulk modulus \( B \) is defined as the ratio of the increase in pressure \( \Delta P \) to the relative decrease in volume \( \Delta V / V \): \[ B = - \frac{\Delta P}{\Delta V / V}. \] Rearranging the equation to solve for \( \Delta P \), we get: \[ \Delta P = - B \times \frac{\Delta V}{V}. \] The negative sign indicates a decrease in volume, but we are concerned only with the magnitude. Given that \( B = 2.15 \times 10^9 \, \text{Nm}^{-2} \) and the volume change is \( 0.2\% = 0.002 \), we can calculate \( P \) as: \[ P = B \times 0.002 = 2.15 \times 10^9 \times 0.002 = 4.3 \times 10^6 \, \text{Nm}^{-2}. \] Thus, the value of \( P \) is \( 4.6 \, \text{Nm}^{-2} \). Final Answer: 4.6.