If the pressure on a body is increased from 2 | vedclass

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(C) The change in pressure is $\Delta P = 250 	ext{ kPa} - 200 	ext{ kPa} = 50 	ext{ kPa} = 50 	imes 10^3 	ext{ Pa} = 5 	imes 10^4 	ext{ Pa}$.T

Vedclass · Accepted Answer

$5 	imes 10^{-8}$

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(C) The change in pressure is $\Delta P = 250 	ext{ kPa} - 200 	ext{ kPa} = 50 	ext{ kPa} = 50 	imes 10^3 	ext{ Pa} = 5 	imes 10^4 	ext{ Pa}$.The fractional change in volume is $\frac{\Delta V}{V} = 0.25 \% = \frac{0.25}{100} = 2.5 	imes 10^{-3}$.Bulk modulus $B$ is given by $B = -\frac{\Delta P}{\Delta V / V}$.Compressibility $K$ is the reciprocal of the bulk modulus,$K = \frac{1}{B} = \frac{\Delta V / V}{\Delta P}$.Substituting the values: $K = \frac{2.5 	imes 10^{-3}}{5 	imes 10^4} = 0.5 	imes 10^{-7} 	ext{ m}^2 	ext{ N}^{-1} = 5 	imes 10^{-8} 	ext{ m}^2 	ext{ N}^{-1}$.

If the pressure on a body is increased from $200 \text{ kPa}$ to $250 \text{ kPa}$,the volume of the body decreases by $0.25 \%$. The compressibility of the material of the body is (in $m^2 \text{ N}^{-1}$)

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