Question:
The radius of a sphere shrinks from 10 to 9.8 cm. Find approximately the decrease in its volume.
Solution:
Let $r$ be the radius of the sphere.
$r=10 \mathrm{~cm}$
$r+\Delta r=9.8 \mathrm{~cm}$
$\Rightarrow \Delta r=10.0-9.8=0.2 \mathrm{~cm}$
Volume of the sphere, $V=\frac{4}{3} \pi r^{3}$
$\Rightarrow \frac{d V}{d r}=\frac{4}{3} \pi \times 3 r^{2}=4 \pi r^{2}$
$\Rightarrow\left(\frac{d V}{d r}\right)_{r=10 \mathrm{~cm}}=4 \pi(10)^{2}=400 \pi \mathrm{cm}^{3} / \mathrm{cm}$
Change in the volume of the sphere, $\Delta V=\frac{d V}{d r} \times d r=400 \pi \times 0.2=80 \pi \mathrm{cm}^{3}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.