A sphere of radius 100 mm shrinks to radius 98 mm, then the approximate decrease in its volume is

(c) $80000 \pi \mathrm{mm}^{3}$

Let x be the radius of the sphere and y be its volume.

$x=100, x+\Delta x=98$

$\Rightarrow \Delta x=-2$

$y=\frac{4}{3} \pi x^{3}$

$\Rightarrow \frac{d y}{d x}=4 \pi x^{2}$

$\Rightarrow\left(\frac{d y}{d x}\right)_{x=100}=40000 \pi$

$\therefore \Delta y=d y=\frac{d y}{d x} d x=40000 \pi \times(-2)=-80000 \pi$

Hence, the decrease in the volume of the sphere is $80000 \pi \mathrm{mm}^{3}$.