Question

the diameter of a road roller is 90cm and is 2m 10cm length. If it takes 400 revolutions to level a playground, find the cost of levelling the ground at rupees 2.25 per square meter (take π =22/7)

Answer 1

The cost of leveling playground is Rs.6682.5

Step-by-step explanation:

Roller is in cylindrical shape

Diameter of roller = 90 cm

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

1 m = 100 cm

Height of roller = 2 m 10 cm = 210 cm

Curved surface area of roller = $2 \pi r h = 2 \times \frac{22}{7} \times 45 \times 210 =59400 cm^2$

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

Area of playground = $59400 \times 500 = 29700000 cm^2$

Cost of 1 sq.m. = Rs.2.25

Cost of 10000 sq.cm. = Rs.2.25

Cost of 29700000 sq.cm. = $\frac{2.25}{10000} \times 29700000 = 6682.5$

Hence The cost of leveling playground is Rs.6682.5

#Learn more:

The diameter of a road roller, 130 cm long is 91 cm if it takes 450 complete revolutions to level a playground, then find the cost of levelling the ground at 80 paise per square metre.​

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