Math, asked by yadavabhishek4409, 1 year ago

The diameter of a road-roller is 90 cm and is 2 m 10 cm in length. If it takes 400 revolutions to level a playground,
22
find the cost of levelling the ground at 2.25 per square metre​

Answers

Answered by Anonymous
49

Given :-

Diameter of the roller = 90cm

length= 2m 10 cm = 2.1 m

So , radius would be 45 cm or 0.45 m

To find :-

Find the cost of levelling the ground at ₹2.25 per square meter.

Solution :-

The part which will touch the ground will be the C.S.A. of roller

now , C.S.A. of roller = 2πrh

➞ 2 × (22/7) × 0.45 × 2.1

➞ 5.94 m²

It takes 400 revolutions , area of ground

➞ 5.94 × 400

➞ 2376 m²

Now cost to level = rate × area

➞ 2376 × 2.25

➞ 5346

➞ 5346 Rs.

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