MHT CET · Maths · Application of Derivatives

A spherical iron ball \(10 \mathrm{~cm}\) in radius is coated with a layer of ice of uniform thickness that melts at a rate of \(50 \mathrm{~cm}^3 / \mathrm{min}\). When the thickness of ice is \(5 \mathrm{~cm}\), then the rate at which the thickness of ice decreases, is

A \(\frac{1}{36 \pi} \mathrm{cm} / \mathrm{min}\)
B \(\frac{5}{6 \pi} \mathrm{cm} / \mathrm{min}\)
C \(\frac{1}{54 \pi} \mathrm{cm} / \mathrm{min}\)
D \(\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & V=\frac{4}{3} \pi r^3 \\ & \Rightarrow \frac{\mathrm{d} v}{\mathrm{~d} t}=4 \pi r^2 \frac{\mathrm{d} r}{\mathrm{~d} t} \\ & \Rightarrow 50=4 \pi \times 15^2 \times \frac{\mathrm{d} r}{\mathrm{~d} t} \\ & \Rightarrow \frac{\mathrm{d} r}{\mathrm{~d} t}=\frac{50}{4 \pi \times 15^2}=\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}\end{aligned}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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