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:

Radius of cylinderical roller = 42 cm and height = 120 cm

=> Distance covered in 1 revolution by the roller = Curved surface area of the roller = $$2\pi rh$$

= $$2\times\frac{22}{7}\times42\times120$$

= $$44\times6\times120=31680$$ $$cm^2=3.168$$ $$m^2$$

=> Total distance covered in 500 revolutions = $$500\times3.168=1584$$ $$m^2$$

Now, cost of levelling the $$1$$ $$m^2$$ ground = Rs. $$1.50$$

$$\therefore$$ Total cost required = $$1584\times1.50=Rs.$$ $$2376$$

=> Ans - (A)