JEE Mains · Maths · STD 12 - 6. Application of derivatives
A spherical iron ball of radius \(10\,cm\) is coated with a layer of ice of uniform thickness that melts at a rate of \(50\,cm^3/min.\) When the thickness of the ice is \(5\,cm,\) then the rate at which the thickness (in \(cm/min\) ) of ice decreases is
- A \(\frac{1}{{36\pi }}\)
- B \(\frac{5}{{6\pi }}\)
- C \(\frac{1}{{9\pi }}\)
- D \(\frac{1}{{18\pi }}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{{18\pi }}\)
Step-by-step Solution
Detailed explanation
\(V = \frac{4}{3}\pi \left( {\left( {10 + {h^3}} \right) - {{10}^3}} \right)\) \(\frac{{dV}}{{dt}} = 4\pi {\left( {10 + h} \right)^2}\frac{{dh}}{{dt}}\) \( - 50 = 4\pi {\left( {10 + 5} \right)^2}\frac{{dh}}{{dt}}\)…
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