JEE Mains · Maths · STD 12 - 6. Application of derivatives
If the volume of a spherical ball is increasing at the rate of \(4 \pi \, cc/sec\), then the rate of increase of its radius (in \(cm/sec\)), when the volume is \(288 \pi \, cc\)
- A \(\frac{1}{6}\)
- B \(\frac{1}{9}\)
- C \(\frac{1}{36}\)
- D \(\frac{1}{24}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{36}\)
Step-by-step Solution
Detailed explanation
Volume of sphere \(V = \frac{4}{3}\pi {r^3}\,\,\,\,\,\,\,\,\,\,......\left( 1 \right)\) \(\frac{{dv}}{{dt}} = \frac{4}{3}.3\pi {r^2}\frac{{dr}}{{dt}}\) \(4\pi = 4\pi {r^2}.\frac{{dr}}{{dt}}\) \(\frac{1}{{{r^2}}} = \frac{{dr}}{{dt}}\) Since, \(V = 288\pi \), therefore from…
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