The relation between pressure $p$ and volume $V$ is given by $p V^{1 / 4}=$ constant. If the percentage…

The relation between pressure $p$ and volume $V$ is given by $p V^{1 / 4}=$ constant. If the percentage decrease in volume is $\frac{1}{2}$, then the percentage increase in pressure is
  1. $-\frac{1}{8}$
  2. $\frac{1}{16}$
  3. $-\frac{1}{8}$
  4. $-\frac{1}{2}$

Solution

Given, $p V^{1 / 4}=a$, where $a$ is constant. $ p=\frac{a}{V^{1 / 4}} $ $\therefore$ Decreased volume $ \begin{aligned} & =V^{1 / 4}-\frac{V^{1 / 4}}{200}=\frac{199}{200} V^{1 / 4} \\ & \text { Increased pressure }=\frac{a}{\frac{199}{200} V^{1 / 4}}=\frac{200 a}{199 V^{1 / 4}} \end{aligned} $ $\therefore$ Percentage increase in pressure $ \begin{aligned} & =\frac{\frac{200 a}{199 V^{1 / 4}}-\frac{a}{V^{1 / 4}}}{\frac{a}{V^{1 / 4}}} \times 100 \\ & =\left(\frac{200}{199}-1\right) \times 100 \\ & =\frac{100}{199} \approx \frac{1}{2}(\text { approximate }) \end{aligned} $

Asked in: AP EAMCET 2013

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