The diameter of a roller, 1 m long, is 84 cm. If it takes 500 complete revolutions to level a playground,

(b)  1320  m2

Area covered by the roller in 1 revolution $=2 \pi r h$

$=2 \times \frac{22}{7} \times 42 \times 100$

$=44 \times 600$

$=26400 \mathrm{~cm}^{2}$

$\therefore$ Area covered in 50 revolutions $=26400 \times 500 \mathrm{~cm}^{2}$

$=1320 \mathrm{~m}^{2}$