Q115 / 225 · JEE Main 2005 · Mathematics

A spherical iron ball $10 \mathrm{~cm}$ in radius is coated with a layer of ice of uniform thickness than melts at a rate of $50 \mathrm{~cm}^3 / \mathrm{min}$. When the thickness of ice is $5 \mathrm{~cm}$, then the rate at which the thickness of ice decreases, is

  • A
    $\frac{1}{36 \pi} \mathrm{cm} / \mathrm{min}$
  • Correct answer
    B
    $\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}$
  • C
    $\frac{1}{54 \pi} \mathrm{cm} / \mathrm{min}$
  • D
    $\frac{5}{6 \pi} \mathrm{cm} / \min$

Solution

$$ \begin{aligned} & \frac{d v}{d t}=50 \\ & 4 \pi r^2 \frac{d r}{d t}=50 \\ & \Rightarrow \frac{d r}{d t}=\frac{50}{4 \pi(15)^2} \quad \text { where } r=15 \\ & =\frac{1}{16 \pi} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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