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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

  1. A \(\frac{\pi}{2} \mathrm{cc} / \mathrm{sec}\)
  2. B \(\pi \mathrm{cc} / \mathrm{sec}\)
  3. C \(\frac{\pi}{5} \mathrm{cc} / \mathrm{sec}\)
  4. D \(\frac{\pi}{10} \mathrm{cc} / \mathrm{sec}\)
Verified Solution

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,…