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The diameter of roller 1.5 m long is 84 cm. If it takes 100 revolutions to level a playground, find the cost of levelling this ground at the rate of 50 paise per square meter. |
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Answer» Given, Diameter of roller = 84 cm So radius of roller =\(\cfrac{84}2\) = 42 cm = .42 m Length of roller = 1.5 m Area covered by roller in one revolution = 2πrh = 2 \(\times\cfrac{22}7\times \)4.2 \(\times\) 1.5 = 3.96 m2 Area covered by it in 100 revolution = 100 × 3.96 = 396 m2 Cost of levelling 1m2 area = Rs. .50 ∴cost of levelling 396m2 = .50 × 396 = Rs. 198 |
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