The diameter of a roller 120 cm long is 84 cm. if it takes 500 complete revolutions to level a playground, determine the cost of levelling it at the rate of 30 paise per square metre
Answers
Answered by
7
Step-by-step explanation:
radius of the roller=d\2=120\2=60cm
length of roller is 84cm
the roller is of cylindrical shape
area of roller=
=2*3.14*60(60+84)
=6.28*60(144)
=54259.2cm^2
=542.592m^2
=542.59m^2
cost of levelling =30 paise per m^2
the roller takes 500 revolution
cost of levelling the playground=500*542.59
=271295 paise
1rupee=100 Paisa
=2712.95 rupees
the cost of levelling the playground is 2712.95 rupees
hope it's correct
Answered by
6
diameter=84cm
therefore radius=42cm
height=120cm
area rolled in 1 revolution=c.s.a
=2*22/7*r*h=2*22/7*42*120
316800 cm^2 = 3.168 m^2
area of field=500*3.168
=1584 m^2
cost=30 p/m^2
therefore, total cost=1584*30
47520 p/m^2
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