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JEE Mains · Maths · STD 12 - 6. Application of derivatives
A spherical iron ball of \(10 \;\mathrm{cm}\) radius is coated with a layer of ice of uniform thickness the melts at a rate of \(50\; \mathrm{cm}^{3} / \mathrm{min}\). When the thickness of ice is \(5 \;\mathrm{cm},\) then the rate (in \(\mathrm{cm} / \mathrm{min.}\) ) at which of the thickness of ice decreases, is
- A \(\frac{1}{36 \pi}\)
- B \(\frac{5}{6 \pi}\)
- C \(\frac{1}{18 \pi}\)
- D \(\frac{1}{54 \pi}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{18 \pi}\)
Step-by-step Solution
Detailed explanation
Let thickness of ice be 'h'. \({\text { Vol. of ice }=\mathrm{v}=\frac{4 \pi}{3}\left((10+\mathrm{h})^{3}-10^{3}\right)}\) \({\qquad \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{4 \pi}{3}\left(3(10+\mathrm{h})^{2}\right) \cdot \frac{\mathrm{dh}}{\mathrm{dt}}}\) Given…
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