A cylindrical roller made of iron is 2 m long. Its inner diameter is 35 cm and the thickness is 7 cm all round. Find the weight of the roller in kg, if 1 cm³ of iron weights 8 g.

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

= $\dfrac{35}{2}$ = 17.5 cm

External radius = Thickness + Internal radius = 7 + 17.5 = 24.5 cm

Height = 2 m = 2 × 100 cm = 200 cm.

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

Putting values we get,

$\therefore \text{Volume of roller} = \dfrac{22}{7} \times ((24.5)^2 - (17.5)^2) \times 200 \\[1em] = \dfrac{22}{7} \times (600.25 - 306.25) \times 200 \\[1em] = \dfrac{22 \times 294 \times 200}{7} \\[1em] = 22 \times 42 \times 200 \\[1em] = 184800 \text{ cm}^3.$

Since, 1 cm³ of iron weights 8 g.

∴ 184800 cm³ weights = 184800 × 8 = 1478400 g.

Since 1 kg = 1000g or 1g = $\dfrac{1}{1000}kg$.

∴ 1478400 g = $1478400 \times \dfrac{1}{1000} = 1478.4$ kg.

Hence, the weight of the roller = 1478.4 kg.