The relation between pressure p and volume V | vedclass

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(A) Given the relation $p V^{1/4} = C$,where $C$ is a constant. Taking the natural logarithm on both sides: $\ln p + \frac{1}{4} \ln V = \ln C$. Diffe

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$\frac{1}{8} \%$

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(A) Given the relation $p V^{1/4} = C$,where $C$ is a constant. Taking the natural logarithm on both sides: $\ln p + \frac{1}{4} \ln V = \ln C$. Differentiating both sides with respect to $V$: $\frac{1}{p} \frac{dp}{dV} + \frac{1}{4V} = 0$. This gives $\frac{dp}{p} = -\frac{1}{4} \frac{dV}{V}$. We are given that the percentage decrease in volume is $\frac{dV}{V} 	imes 100 = -\frac{1}{2} \%$. Substituting this into the differential relation: $\frac{dp}{p} 	imes 100 = -\frac{1}{4} \left( \frac{dV}{V} 	imes 100 ight)$. $\frac{dp}{p} 	imes 100 = -\frac{1}{4} \left( -\frac{1}{2} \% ight) = \frac{1}{8} \%$. Thus,the percentage increase in pressure is $\frac{1}{8} \%$.

The relation between pressure $p$ and volume $V$ is given by $p V^{1/4} = \text{constant}$. If the percentage decrease in volume is $\frac{1}{2} \%$,then the percentage increase in pressure is:

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