JEE MAIN - Mathematics (2020 - 9th January Morning Slot - No. 5)
A spherical iron ball of 10 cm radius is
coated with a layer of ice of uniform
thickness the melts at a rate of 50 cm3/min.
When the thickness of ice is 5 cm, then the rate
(in cm/min.) at which of the thickness of ice
decreases, is :
$${1 \over {18\pi }}$$
$${1 \over {36\pi }}$$
$${1 \over {54\pi }}$$
$${5 \over {6\pi }}$$
Explanation
Let the thickness = h cm
Volume of ice = v = $${{4\pi } \over 3}\left( {{{\left( {10 + h} \right)}^3} - {{10}^3}} \right)$$
$$ \Rightarrow $$ $${{dv} \over {dt}} = {{4\pi } \over 3}\left( {3{{\left( {10 + h} \right)}^2}} \right).{{dh} \over {dt}}$$
Given $${{dv} \over {dt}} = $$ 50 cm3/min and h = 5 cm
$$ \therefore $$ 50 = $${{4\pi } \over 3}\left( {3{{\left( {10 + 5} \right)}^2}} \right).{{dh} \over {dt}}$$
$$ \Rightarrow $$ $${{dh} \over {dt}} = {{50} \over {4\pi \times {{15}^2}}}$$ = $${1 \over {18\pi }}$$ cm/min
Volume of ice = v = $${{4\pi } \over 3}\left( {{{\left( {10 + h} \right)}^3} - {{10}^3}} \right)$$
$$ \Rightarrow $$ $${{dv} \over {dt}} = {{4\pi } \over 3}\left( {3{{\left( {10 + h} \right)}^2}} \right).{{dh} \over {dt}}$$
Given $${{dv} \over {dt}} = $$ 50 cm3/min and h = 5 cm
$$ \therefore $$ 50 = $${{4\pi } \over 3}\left( {3{{\left( {10 + 5} \right)}^2}} \right).{{dh} \over {dt}}$$
$$ \Rightarrow $$ $${{dh} \over {dt}} = {{50} \over {4\pi \times {{15}^2}}}$$ = $${1 \over {18\pi }}$$ cm/min
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