Question : The diameter of the 120 cm long roller is 84 cm. It takes 500 complete revolutions of the roller to level the ground. The cost of levelling the ground at Rs. 1.50 per sq. m. is:
Option 1: Rs. 6,000
Option 2: Rs. 3,762
Option 3: Rs. 2,376
Option 4: Rs. 5,750
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Correct Answer: Rs. 2,376
Solution :
Given: Length, $h$ = 120 cm
Diameter = 84 cm
Radius, $r$ = 42 cm
The curved surface area of the cylinder = $2\pi rh$ where $r$ is the radius and $h$ is the height of the cylinder.
Area covered by roller in 1 revolutions = curved surface area of roller = $2\pi rh$
= $2×\frac{22}{7}×42×120$
= 31680 cm$^2$
Area covered by 500 revolutions = 31860 × 500 cm$^2$
= $\frac{31680×500}{100×100}$ m
2
= 1584 m
2
Cost of levelling at the rate of Rs. 1.50 per sq. m. = 1584 × 1.5
= Rs. 2,376
Hence, the correct answer is Rs. 2,376.
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