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.......\(\mathrm{cm} \mathrm{s}^{-1}\)
- A \(\frac{1}{6}\)
- B \(\frac{1}{36}\)
- C \(\frac{1}{9}\)
- D \(\frac{1}{24}\)
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. \(t\),…
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