Volumes Practice Test 2

Here, 2π (R - r) h = 44
⇒ $2\times \frac{22}{7}$ (R - r) × 14 = 44
∴ (R - r) = $\frac{1}{2}$
Again, $\mathit{\pi }\left({\mathit{R}}^2-{\mathit{r}}^2\right)\mathit{h}$ = 99
⇒ $\frac{22}{7}$ (R + r) (R - r) × 14 = 99
⇒ $\frac{22]}{7}\left(\mathit{R}+\mathit{r}\right)$ × $\frac{1}{2}\times 14$ = 99
∴ (R + r) = $\frac{99}{22}$ = $\frac{9}{2}$
Since, R = $\frac{1}{2}\left(\frac{9}{2}+\frac{1}{2}\right)$ = $\frac{5}{2}$ = 2.5 cm
r = $\frac{1}{2}\left(\frac{9}{2}-\frac{1}{2}\right)$ = 2 cm
Hence, outer and inner radii of the pipe are 2.5 cm and 2 cm.