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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 \(288 \pi \, cc\)

  1. A \(\frac{1}{6}\)
  2. B \(\frac{1}{9}\)
  3. C \(\frac{1}{36}\)
  4. D \(\frac{1}{24}\)
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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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