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 :
Solution
Let the thickness = h cm
<br><br>Volume of ice = v = ${{4\pi } \over 3}\left( {{{\left( {10 + h} \right)}^3} - {{10}^3}} \right)$
<br><br>$\Rightarrow$ $${{dv} \over {dt}} = {{4\pi } \over 3}\left( {3{{\left( {10 + h} \right)}^2}} \right).{{dh} \over {dt}}$$
<br><br>Given ${{dv} \over {dt}} =$ 50 cm<sup>3</sup>/min and h = 5 cm
<br><br>$\therefore$ 50 = $${{4\pi } \over 3}\left( {3{{\left( {10 + 5} \right)}^2}} \right).{{dh} \over {dt}}$$
<br><br>$\Rightarrow$ ${{dh} \over {dt}} = {{50} \over {4\pi \times {{15}^2}}}$ = ${1 \over {18\pi }}$ cm/min
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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