A spherical iron ball of
Question:

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

  1. (1) $\frac{5}{6 \pi}$

  2. (2) $\frac{1}{54 \pi}$

  3. (3) $\frac{1}{36 \pi}$

  4. (4) $\frac{1}{18 \pi}$


Correct Option: , 4

Solution:

Let the thickness of ice layer be $=x \mathrm{~cm}$

Total volume $V=\frac{4}{3} \pi(10+x)^{3}$

$\frac{d V}{d t}=4 \pi(10+x)^{2} \frac{d x}{d t}$ ……(1)

Since, it is given that

$\frac{d V}{d t}=50 \mathrm{~cm}^{3} / \mathrm{min}$………(2)

From (1) and (2), $50=4 \pi(10+x) \frac{d x}{d t}$

$\Rightarrow 50=4 \pi(10+5)^{2} \frac{d x}{d t}[\because$ thickness of ice $x=5]$

$\Rightarrow \quad \frac{d x}{d t}=\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}$

 

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