Q109 / 160 · AP EAMCET 2013 · Mathematics
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
- A$-\frac{1}{8}$
- B$\frac{1}{16}$
- C$-\frac{1}{8}$
- Correct answerD$-\frac{1}{2}$
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
$$
\begin{aligned}
& =\frac{\frac{200 a}{199 V^{1 / 4}}-\frac{a}{V^{1 / 4}}}{\frac{a}{V^{1 / 4}}} \times 100 \\
& =\left(\frac{200}{199}-1\right) \times 100 \\
& =\frac{100}{199} \approx \frac{1}{2}(\text { approximate })
\end{aligned}
$$
Hence, the correct answer is option: 'D'.