COMEDK · Maths · 27. Application of Derivatives
The slant height of a cone is fixed at \(7 \mathrm{~cm}\). If the rate of increase of its height is \(0.3 \mathrm{~cm} / \mathrm{sec}\), then the rate of increase of its volume when its height is \(4 \mathrm{~cm}\) is
- A \(\frac{\pi}{2} \mathrm{cc} / \mathrm{sec}\)
- B \(\pi \mathrm{cc} / \mathrm{sec}\)
- C \(\frac{\pi}{5} \mathrm{cc} / \mathrm{sec}\)
- D \(\frac{\pi}{10} \mathrm{cc} / \mathrm{sec}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi}{10} \mathrm{cc} / \mathrm{sec}\)
Step-by-step Solution
Detailed explanation
We have, slant height of cone \((l)=7 \mathrm{~cm}\) \[ \begin{array}{ll} \Rightarrow & l^{2}=h^{2}+r^{2} \quad \text{...(i)} \\ \Rightarrow & r^{2}=7^{2}-4^{2} \quad \text{[when h=4 cm]} \\ \Rightarrow & r^{2}=33 \\ \Rightarrow & r=\sqrt{33} \mathrm{~cm} \end{array} \] Now,…
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