Question

A spherical iron ball $10cm$ in radius is coated with a layer of ice of uniform thickness that melts at a rate of $50{cm}^3/min$. When the thickness of ice is $5cm$, then the rate at 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 $\frac{1}{18\pi }cm/min$

Radius of spherical ball $=10cm$.

Rate of mealy $=\frac{dV}{dt}=\frac{50c{m}^5}{mm}.$

Thickness of ice $=5cm\left(d^{\prime }\right)$

Total radius $=5+10=15cm$

Volume $=\frac{4}{3}\pi (10+d^{\prime }{)}^3$

$\frac{dv}{dt}=\frac{4}{5}\pi \times 3(10+d^{\prime }{)}^2\times \left(\frac{d{d}^{\prime }}{dt}\right)$.

$\Rightarrow 50=4\pi (10+d^{\prime }{)}^2\times \frac{d{d}^{\prime }}{dt}$

$\Rightarrow 50=4\pi (15{)}^2\times \frac{dv}{dt}.$

$\frac{dv}{dt}=\frac{5010}{4\pi \times 15\times 15}=\frac{10}{4\pi \times 3}=\frac{5}{2\pi \times 3}$

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