A spherical iron ball of radius
Question:

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

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

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

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

  4. (4) $\frac{1}{9 \pi}$


Correct Option: 1

Solution:

(1) Given that ice melts at a rate of $50 \mathrm{~cm}^{3} / \mathrm{min}$.

$\therefore \frac{d V_{\text {ice }}}{d t}=50$

$V_{\text {ice }}=\frac{4}{3} \pi(10+r)^{3}-\frac{4}{3} \pi(10)^{3}$

$\Rightarrow \frac{d V}{d t}=\frac{4}{3} \pi 3(10+r)^{2} \frac{d r}{d t}=4 \pi(10+r)^{2} \frac{d r}{d t}$

Substitute $r=5$,

$50=4 \pi(225) \frac{d r}{d t} \Rightarrow \frac{d r}{d t}=\frac{50}{4 \pi(225)}=\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}$

 

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.