Question

A spherical iron ball $10cm$ in radius is coated with a layer of ice of uniform thickness that metals at a rate of $50c{m}^3/min$. when the thickness of ice is $15cm$, then the rate of which the thickness of ice decreases, is

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

Answer 1

The correct option is D

$\frac{1}{18\pi }cm/min$

Explanation for correct option:

Step-1: Simplify the given data.

Given, $\frac{dv}{dt}=50c{m}^3/min$

$\Rightarrow$$\frac{d\left({\displaystyle \frac{4}{3}}\pi r^3\right)}{dt}=50c{m}^3/min\left(\mathbf{\because }\mathit{v}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}\mathit{\pi }{\mathit{r}}^{\mathbf{3}}\right)$

$\Rightarrow$ $3r^2\frac{dr}{dt}=\frac{50\times 3}{4\pi }$

$\Rightarrow$ $\frac{dr}{dt}=\frac{50\times 3}{4\pi \times 3r^2}$

$\Rightarrow$ $\frac{dr}{dt}=\frac{50}{4\pi r^2}$

Step-2: Substitute $r=15$.

At, $r=15$

$\Rightarrow$ ${\left(\frac{dr}{dt}\right)}_{r=15}=\frac{50}{4\pi 225}$

$\Rightarrow$ ${\left(\frac{dr}{dt}\right)}_{r=15}=\frac{1}{18\pi }cm/min$

Hence, correct answer is option $\left(D\right)$.