The diameter of roller 1.5 long is 84 cm.if it takes 100 revolutions to level a playground, find the cost of leveling this ground at the rate of 50 paise per square metre.
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For roller
r= 1.5
______
2 m = 0.75 m
h =84 cm =0.84 m
therefore ,curved surface area = 2πrh
=2 × 22/7 ×0.75 ×0.84
= 3.96 m ^2
therefore area of the ground levelled in 1 revolution =3. 96 m^2
therefore area of the ground levelled in 100 revolution
= 3.96 × 100 m2 = 396m ^2
therefore cost of levelling
= rupees 396 × 50 /100
= rupees 198
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