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AP EAMCET · Maths · Application of Derivatives

The volume of a spherical ball is increasing at a rate of \(4 \pi \mathrm{cm}^3 \mathrm{~s}^{-1}\). The rate at which its radius increases, when its volume is \(288 \pi \mathrm{cm}^3\), is.......\(\mathrm{cm} \mathrm{s}^{-1}\)

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

Answer & Solution

Correct Answer

(B) \(\frac{1}{36}\)

Step-by-step Solution

Detailed explanation

Given, \(\frac{d v}{d t}=4 \pi \mathrm{cm}^3 / \mathrm{s}\) ...(i) To find, \(r / d t=\) ? When, \(V=288 \pi \mathrm{cm}^3\) Let \(r\) be the radius of spherical ball \(\therefore\) Volume \((V)=\frac{4}{3} \pi r^3\) On differentiating w.r.t. \(t\),…