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 the ratio of the pressure increase \( \Delta P \) to the relative volume decrease \( \Delta V / V \), expressed as:\[B = - \frac{\Delta P}{\Delta V / V}.\]To find \( \Delta P \), the equation can be rearranged to:\[\Delta P = - B \times \frac{\Delta V}{V}.\]The negative sign signifies a volume reduction; however, only the magnitude is relevant here.With \( B = 2.15 \times 10^9 \, \text{Nm}^{-2} \) and a volume change of \( 0.2\% \) or \( 0.002 \), \( P \) is calculated as:\[P = B \times 0.002 = 2.15 \times 10^9 \times 0.002 = 4.3 \times 10^6 \, \text{Nm}^{-2}.\]Therefore, \( P \) equals \( 4.6 \, \text{Nm}^{-2} \). Final Answer: 4.6.