A spherical iron ball of radius 10cm is coated with a layer of ice of uniform thickness that melts at a rate of 50cm^3/min. When the thickness of the…

We know that the volume of a sphere of radius $r$r is $\frac{4}{3}{\pi r}^{3}.$34​πr3.

Given the radius of the spherical iron ball is $R=10cm$R=10cm and the thickness of the ice is $xcm.$xcm.

Volume of ice: $V=\frac{4}{3}\pi {(R+x)}^{3}-\frac{4}{3}\pi {R}^{3}$V=34​π(R+x)3−34​πR3

$$\Rightarrow V=\frac{4}{3}\pi {(10+x)}^{3}-\frac{4}{3}\pi {(10)}^{3}$$⇒V=34​π(10+x)3−34​π(10)3

Differentiating with respect to $t,$t, we get

$$\frac{dV}{dt}=4\pi {(10+x)}^{2}\frac{dx}{dt}-0$$dtdV​=4π(10+x)2dtdx​−0

Given $\frac{dV}{dt}=50c{m}^{3}/min$dtdV​=50cm3/min

$$\Rightarrow 50=4\pi {(10+x)}^{2}\frac{dx}{dt}$$⇒50=4π(10+x)2dtdx​

$$\Rightarrow \frac{dx}{dt}=\frac{50}{4\pi {(10+x)}^{2}}$$⇒dtdx​=4π(10+x)250​

$$\Rightarrow {\frac{dx}{dt}|}_{x=5}=\frac{50}{4\pi {(15)}^{2}}=\frac{1}{18\pi }cm/min.$$⇒dtdx​∣x=5​=4π(15)250​=18π1​cm/min.