MHT CET · Maths · Application of Derivatives
A spherical iron ball of \(10 \mathrm{~cm}\) radius is coated with a layer of ice of uniform thickness that melts at the rate of \(50 \mathrm{~cm}^3 / \mathrm{min}\). If the thickness of ice is \(5 \mathrm{~cm}\), then the rate at which the thickness of ice decrease is
- A \(\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}\)
- B \(\frac{2}{9 \pi} \mathrm{cm} / \mathrm{min}\)
- C \(\frac{-1}{18 \pi} \mathrm{cm} / \mathrm{min}\)
- D \(\frac{1}{3 \pi} \mathrm{cm} / \mathrm{min}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & v=\frac{4}{3} \pi r^3 \\ & \frac{d v}{d t}=50 \quad r=(10+5)=15 \mathrm{~cm} \\ & \Rightarrow \frac{d v}{d t}=4 \pi r^2 \frac{d r}{d t} \\ & \Rightarrow 50=4 \pi \times 15 \times 15 \times \frac{d r}{d t} \\ & \Rightarrow \frac{d r}{d t}=\frac{50}{4 \pi \times 15 \times 15}=\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}\end{aligned}\)
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