1. The diameter of a 120 cm long roller is 84 cm. It takes 500 complete revolutions of the roller to level a ground. The cost of levelling the ground at ₹1.50 per sq.m. is:
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By: anil on 05 May 2019 02.19 am
Radius of cylinderical roller = 42 cm and height = 120 cm => Distance covered in 1 revolution by the roller = Curved surface area of the roller = $$2pi rh$$ = $$2 imesfrac{22}{7} imes42 imes120$$ = $$44 imes6 imes120=31680$$ $$cm^2=3.168$$ $$m^2$$ => Total distance covered in 500 revolutions = $$500 imes3.168=1584$$ $$m^2$$ Now, cost of levelling the $$1$$ $$m^2$$ ground = Rs. $$1.50$$ $$ herefore$$ Total cost required = $$1584 imes1.50=Rs.$$ $$2376$$ => Ans - (A)
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