The diameter of a roller, 1 m 40 cm long, is 80 cm. If it takes 600 complete revolutions to level a play ground, find the cost of levelling the ground at 75 paise per sq metre.
With solution please
Answers
Answer:
Cost of leveling the ground at Rs 15 per square meter = RS. 31651.2
Step-by-step explanation:
It is given that, the diameter of a road roller, 1m 40cm long is 80cm and it takes 600 revolutions to level a playground
Curved surface area CSA of cylinder = 2πrh
r - radius and h height
consider the roller as a cylinder
To find the curved surface area of roller
r = 80/2 = 40cm = 0.4m
h = 1m 40cm = 1.4 m
C.S.A 2πrh = 2 x 3.14 x 0.4 x 1.4 =3.5168 square meter
To find the cost
It is given that,the cost of leveling the ground at Rs 15 per square meter
And it takes 600 revolutions to level a playground
Therefore cost = 3.5168 x 15 x 600 =31651.2
Total cost = Rs. 31651.2
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Answer:
1584
is your answer
Step-by-step explanation:
1. we have to find the surface area of one revolution=2¶rh=2*22/7*80*140=35200cm
as this is in cm we divide it by 100*100.
2. let's keep the equation as 35200/100*100
3. As we want the answer for 600 revolution. Let's multiply it by 600. When we multiply.We must cut the zeroes.After this we must multiply it by 0.75