COMEDK · Maths · 27. Application of Derivatives
A cone whose height is always equal to its diameter is increasing in volume at the rate of \(40 \mathrm{~cm}^3 / \mathrm{sec}\). The rate at which the radius is increasing when its circular base is \(1 \mathrm{~m}^2\) is
- A \(0.001 \mathrm{~cm} / \mathrm{sec}\)
- B \(0.002 \mathrm{~cm} / \mathrm{sec}\)
- C \(1 \mathrm{~mm} / \mathrm{sec}\)
- D \(2 \mathrm{~mm} / \mathrm{sec}\)
Answer & Solution
Correct Answer
(B) \(0.002 \mathrm{~cm} / \mathrm{sec}\)
Step-by-step Solution
Detailed explanation
Given the height \(h\) of the cone is equal to its diameter \(d = 2r\), so \(h = 2r\). The volume of the cone is \(V = \dfrac{1}{3} \pi r^2 h = \dfrac{1}{3} \pi r^2 (2r) = \dfrac{2}{3} \pi r^3\). Differentiating with respect to time \(t\), we get…
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