Question

The diameter of a roller 120 cm long is 84 cm. If it takes 500 complete revolutions to level a playground, determine the cost of levelling at the rate of Rs 50 per square metre.

[Use π =$\frac{22}{7}$]

Answer 1

2r = 84 cm ∴ r =$\frac{84}{2}$ cm = 42 cm , h = 120 cm.
Area of the playground levelled in 1 complete revolution
= 2πrh = 2×$\frac{22}{7}$ ×42×120
= 31680 cm²
Area of the playground levelled in 500 revolutions
= 31680×500 cm² = $\frac{31680\times 500}{100\times 100}{m}^2=1584m^2.$
∴ Cost of levelling Rs. 50 per square metre
= Rs. 1584×50
= Rs. 79200.