TS EAMCET · Maths · Application of Derivatives
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 \(-\frac{1}{8}\)
- D \(-\frac{1}{2}\)
Answer & Solution
Correct Answer
(D) \(-\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Given, \(p V^{1 / 4}=a\), where \(a\) is constant. \[ p=\frac{a}{V^{1 / 4}} \] \(\therefore\) Decreased volume…
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