The increase in the pressure required to decrease the volume ( $\Delta V$ ) of water is $6.3 imes

Question

The increase in the pressure required to decrease the volume ( $\Delta V$ ) of water is $6.3 \times 10^{7}\text{ }N/m^{2}$. The percentage decrease in the volume is $\_\_\_\_$ .

(Bulk modulus of water $= 2.1 \times 10^{9}\text{ }N/m^{2}$.)

Accepted Answer

Sol. [Key Concept: Bulk modulus $\beta = - \frac{\Delta\ P}{\Delta\ V/V}$; percentage decrease $= \frac{\Delta\ P}{\beta} 	imes 100$.]Bulk modulus $\beta = \frac{- \Delta P}{\left( \frac{\Delta V}{V} ight)}$$$100 	imes \frac{\Delta V}{V} = \frac{- \Delta P}{\beta} 	imes 100 = - \frac{6.3 	imes 10^{7}}{2.1 	imes 10^{9}} 	imes 100 = - 3\% $$Percentage decrease $= 3\%$

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