JEE MAIN - Mathematics (2005 - No. 14)
A spherical iron ball $$10$$ cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of $$50$$ cm$$^3$$ /min. When the thickness of ice is $$5$$ cm, then the rate at which the thickness of ice decreases is
$${1 \over {36\pi }}$$ cm/min
$${1 \over {18\pi }}$$ cm/min
$${1 \over {54\pi }}$$ cm/min
$${5 \over {6\pi }}$$ cm/min
Explanation
Given that
$${{dv} \over {dt}} = 50\,c{m^3}/\min $$
$$ \Rightarrow {d \over {dt}}\left( {{4 \over 3}\pi {r^3}} \right) = 50$$
$$ \Rightarrow 4\pi {r^2}{{dr} \over {dt}} = 50$$
$$ \Rightarrow {{dr} \over {dt}} = {{50} \over {4\pi {{\left( {15} \right)}^2}}} = {1 \over {18\pi }}\,\,cm/\min \,\,$$
(here $$r=10+5)$$
$${{dv} \over {dt}} = 50\,c{m^3}/\min $$
$$ \Rightarrow {d \over {dt}}\left( {{4 \over 3}\pi {r^3}} \right) = 50$$
$$ \Rightarrow 4\pi {r^2}{{dr} \over {dt}} = 50$$
$$ \Rightarrow {{dr} \over {dt}} = {{50} \over {4\pi {{\left( {15} \right)}^2}}} = {1 \over {18\pi }}\,\,cm/\min \,\,$$
(here $$r=10+5)$$
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