AP EAMCET · Maths · Application of Derivatives
The volume of sphere is increasing at the rate of 1200 \(\mathrm{cu} \mathrm{cm} / \mathrm{s}\). The rate of increase in its surface area when the radius is \(10 \mathrm{~cm}\) is
- A \(120 \mathrm{sq} \mathrm{cm} / \mathrm{s}\)
- B \(240 \mathrm{sq} \mathrm{cm} / \mathrm{s}\)
- C \(200 \mathrm{sq} \mathrm{cm} / \mathrm{s}\)
- D \(100 \mathrm{sq} \mathrm{cm} / \mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(240 \mathrm{sq} \mathrm{cm} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Let \(V\) be the volume, \(S\) be the surface area and \(r\) be the radius of the sphere. It is given that, \(\frac{d V}{d t}=1200 \mathrm{cu} \mathrm{cm} / \mathrm{s}\) and \(r=10 \mathrm{~cm}\) Now, Volume of sphere \(=V=\frac{4}{3} \pi r^3\)…
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