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The diameter of a roller, 120 cm long, is 84 cm. If it takes 500 complete revolutions to level a playground, find the cost of levelling it at Rs 5 per square metre. |
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Answer» Radius of the roller, r = 42 cm, and its length, h = 120 cm Area covered by the roller in 1 revolution = curved surface area of the roller `= (2pi rh)` sq units `= (2 xx (22)/(7) xx 42 xx 120) cm^(2) = 31680 cm^(2)` Area covered by the roller in 500 revolutions `(31680 xx 500) cm^(2) = ((31680 xx 500)/(100 xx 100)) m^(2) = 1584 m^(2)` `:.` area of the playground `= 1584 m^(2)` Cost of levelling the playground `= Rs (1584 xx 5) = Rs 7920` |
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