AP EAMCET · Maths · Application of Derivatives
The radius of a sphere increases at the rate of \(0.04 \mathrm{~cm} / \mathrm{sec}\). The rate of increase in the volume of that sphere with respect to its surface area, when its radius is \(10 \mathrm{~cm}\) is
- A \(16 \pi\)
- B 25
- C 20
- D 5
Answer & Solution
Correct Answer
(D) 5
Step-by-step Solution
Detailed explanation
Let \(r\) be the radius of the sphere. Given, rate of change in radius \(\frac{d r}{d t}=0.04 \mathrm{~cm} / \mathrm{sec}\) Volume of sphere \((V)=\frac{4}{3} \pi r^3\) Differentiating w.r.t, \(t\), we get…
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