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Q. A spherical, iron ball of radius 10 cm, coated with a layer of ice of uniform thickness, melts at the rate of $ T=3mg $ . The rate at which the thickness of decreases when the thickness of ice is 5 .cm, is

JamiaJamia 2015

Solution:

Given, $ \frac{dv}{dt}=100\pi c{{m}^{3}}/\min $ where, V is the volume of spherical ball. $ \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\,\,cm/\min $