TS EAMCET · Maths · Application of Derivatives
The volume of a sphere is increasing at the rate of \(4 \pi \mathrm{cc} / \mathrm{sec}\). When its volume is \(288 \pi \mathrm{cc}\), the rate of increase (in \(\mathrm{cm} / \mathrm{sec}\) ) in its radius is
- A \(\frac{1}{36}\)
- B \(\frac{1}{6}\)
- C \(\frac{1}{7}\)
- D \(\frac{1}{49}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{36}\)
Step-by-step Solution
Detailed explanation
Given that, Rate of increase in, \(\frac{d V}{d t}=4 \pi \mathrm{cc} / \mathrm{sec}\) Volume, \(V=288 \pi \mathrm{cc}\) As we know, volume of a sphere of radius \((r)\), \(V=\frac{4}{3} \pi r^3\) \(\ldots(\mathrm{i})\) Differentiating above expression w.r.t. ' \(t\) ' on both…
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