the diameter of a roller is 42 cm and its length is 100 cm. it takes 400 complete revolutions moving once over the level of a playground . determine the area of the playground. also,find the cost of leveling the playground a rs50 per 105sqm?
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Answered by
73
Area of ground = 400 × CSA of cylindrical roller
= 2 × pi× r×h ×400
= 2×22/7 × 21 × 100 × 400
=5280000 cm^2
=5280000/10000 m^2
= 528 m^2
Cost of levelling 105 m^2 = rs.50
so cost of leveling 1 m^2 = 50/105
So cost of leveling 528 m^2 = 50/105 × 528 = RS.25.14
= 2 × pi× r×h ×400
= 2×22/7 × 21 × 100 × 400
=5280000 cm^2
=5280000/10000 m^2
= 528 m^2
Cost of levelling 105 m^2 = rs.50
so cost of leveling 1 m^2 = 50/105
So cost of leveling 528 m^2 = 50/105 × 528 = RS.25.14
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21
Answer:
upper answer is right
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