A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is $11.2$ cm and its length is $21$ cm. The metal everywhere is $0.4$ cm thick. The volume of the metal is

### Concept & Volume of Hollow Cylinder The volume of the metal used in a hollow tube is the volume of the outer cylinder minus the volume of the inner cylinder. $$ V = \pi \times h \times (R^2 - r^2) $$ ### Step-by-Step Solution - **Given:** - Length $h = 21$ cm. - Internal diameter $= 11.2$ cm $\implies$ inner radius $r = 5.6$ cm. - Thickness $= 0.4$ cm $\implies$ outer radius $R = 5.6 + 0.4 = 6.0$ cm. - **Calculation:** - Calculate $(R^2 - r^2) = 6^2 - 5.6^2 = 36 - 31.36 = 4.64$. - Volume $= \frac{22}{7} \times 21 \times 4.64$. - Volume $= 22 \times 3 \times 4.64 = 66 \times 4.64 = 306.24$ cm$^3$. ### Exam Strategy & Shortcut Instead of $R^2 - r^2$, use $(R-r)(R+r)$. Here $(6-5.6)(6+5.6) = (0.4)(11.6) = 4.64$. This is much faster to compute mentally. ### Common Pitfall Adding the thickness to the diameter instead of the radius to find the outer dimensions. ### Final Answer Therefore, the correct answer is **306.24 cm³**.