Question

The diameter of a road roller is 90cm and is 2 m 10 m in length. if it takes 400 revolutions level a playground, find the cost of levelling the ground at rate 2.25 per Square metre.

Answer 1

given :

Diameter of roller = 90 cm

Radius of roller = \frac{90}{2}=45 cm

Height of roller = 2 m 10 cm = 210 cm

Curved surface area of roller = 2 π r h =59400 cm²

We are given that the it takes 400 revolutions to level a playground

Area of playground =

$59400 \: \times 500 = 29700000cm {}^{2}$

Cost of 1 m² = Rs.2.25

Cost of 10000 cm² . = Rs.2.25

Cost of 29700000 cm² . =

$\frac{2.25}{10000} \times 29700000 = 6682.5$

Hence The cost of leveling is ₹ 6682.5

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