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Q. A spherical iron ball of radius $10\, cm$, coated with a layer of ice of uniform thickness, melts at a rate of $ 100\,\pi \,c{{m}^{3}}/min $ . The rate at which the thickness of decreases when the thickness of ice is $5\, cm$, is

KEAMKEAM 2008Application of Derivatives

Solution:

Given, $ \frac{dV}{dt}=100\pi \,c{{m}^{3}}/\min $
$ \therefore $ $ \frac{d}{dt}\left( \frac{4}{3}\pi {{r}^{3}} \right)=100\pi $
$ \Rightarrow $ $ 3{{r}^{2}}\frac{dr}{dt}=\frac{300\pi }{4\pi } $
$ \Rightarrow $ $ {{\left( \frac{dr}{dt} \right)}_{(r=5)}}=\frac{300}{4\times 3\times 25} $
$=1\text{ }cm/min $