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
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☈ 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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