Bulk modulus of water is 2 times 10^9 N/m^2. | vedclass

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Question

(C) The formula for Bulk modulus $(K)$ is given by $K = -\frac{p}{\Delta V / V}$,where $p$ is the change in pressure and $\Delta V / V$ is the volumet

Vedclass · Accepted Answer

$2 	imes 10^6$

Vedclass · Answer

(C) The formula for Bulk modulus $(K)$ is given by $K = -\frac{p}{\Delta V / V}$,where $p$ is the change in pressure and $\Delta V / V$ is the volumetric strain.Given,Bulk modulus $K = 2 	imes 10^9 \ N/m^2$.The fractional change in volume is $\frac{\Delta V}{V} = 0.1 \% = \frac{0.1}{100} = 10^{-3}$.Since we are increasing the volume,the pressure change $p$ is negative (tensile stress),but we are looking for the magnitude of pressure required.Using the magnitude: $p = K 	imes \left( \frac{\Delta V}{V} ight)$.Substituting the values: $p = (2 	imes 10^9) 	imes (10^{-3})$.$p = 2 	imes 10^6 \ N/m^2$.

Bulk modulus of water is $2 \times 10^9 \ N/m^2$. The pressure required to increase the volume of water by $0.1 \%$ in $N/m^2$ is:

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