A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness t…

Question

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 :

Accepted Answer

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

Solutions · 1