A cylindrical road roller made of metal is one meter long. Its inner radius is 27 cm and the thickness of the metal sheet rolled into it is 9 cm. What is the weight of the roller, if 1 cm$$^3$$ of the metal weighs 8 g ?

Given, length(height) of the cylindrical road roller(h) = 1 meter = 100 cm

Inner radius of the cylindrical road roller($$r_1$$) = 27 cm

Thickness of the metal sheet = 9 cm

Outer radius of the cylindrical road roller($$r_2$$) = 27 + 9 = 36 cm

Volume of the cylindrical road roller = $$\pi\ r_2^2h-\pi\ r_1^2h$$

$$=\pi\ \left(36\right)^2\times100-\pi\ \left(27\right)^2\times100$$

$$=100\pi\ \left[\left(36\right)^2-\left(27\right)^2\right]$$

$$=100\pi\ \left(36+27\right)\left(36-27\right)$$

$$=100\pi\ \left(63\right)\left(9\right)$$

$$=100\pi\ \times567$$ cm$$^3$$

$$\therefore\ $$Weight of the roller $$=\frac{100\pi\ \times567}{8}$$ grams

$$=\frac{100\pi\ \times567}{8\times1000}$$ kg

$$=$$ 453.6 $$\pi$$ kg

Hence, the correct answer is Option A