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 thickness is 0.4 cm. Calculate the volume of the metal.

Internal diameter = 11.2 cm

Internal radius = $\dfrac{\text{Internal diameter}}{2}$

= $\dfrac{11.2}{2}$ = 5.6 cm.

   Thickness = External radius - Internal radius
⇒ 0.4 = External radius - Internal radius
⇒ External radius = 0.4 + Internal radius
⇒ External radius = 0.4 + 5.6 = 6.0 cm

Volume of hollow cylinder = π(R² - r²)h, where R = External radius and r = Internal radius.

Putting values we get,

$\therefore \text{Volume of metal} = \dfrac{22}{7} \times (6^2 - (5.6)^2) \times 21 \\[1em] = \dfrac{22}{7} \times (36 - 31.36) \times 21 \\[1em] = \dfrac{22 \times 4.64 \times 21}{7} \\[1em] = 22 \times 4.64 \times 3 \\[1em] = 306.24 \text{ cm}^3.$

Hence, volume of metal = 306.24 cm³.