the diameter of a roller is 250cm and length is 140 cm if it takes 500 complete revelutions to level a playground determine the cost of leveling at the rate of 50 paise per sq m
Answers
Answer:
2750₹
Step-by-step explanation:
csa= 2x22/7x 125x140.
= 11000000cm^2 .
area covered by the roller in 500 revolutions.= 110000 x 500= 55000000cm^2
= 5500m^2 .
therefore cost of levelling
5500x0.5
= 2750₹ .
hope it helps you....
Given,
- The diameter of the roller = 250 cm
- The length of the roller = 140 cm
- Number of revolutions = 500
- Cost of leveling per square meter of the playground = 50 paise
To find,
- We have to determine the total cost of leveling at the rate of 50 paise per sq m.
Solution,
The diameter of a roller is 250cm and length is 140 cm if it takes 500 complete revolutions to level a playground then the cost of leveling the playground at the rate of 50 paise per sq m is Rs.2750.
Diameter of the roller = 250 cm
Radius = diameter/2
Radius of the roller = 250/2
= 125 cm
Height of the roller = 140 cm
Area covered by the roller in one revolution = Curved surface area of the roller
Curved surface area = 2πrh
Area covered by the roller in one revolution = 2 * 22/7 * 125 * 140
= 110000 cm²
As we know, 1 cm = 1/100 m, then 1 cm² = 1/10000 m²
Area covered by the roller in one revolution in m² = 110000/10000
= 11 m²
Area covered in 500 revolutions = 11 *500
= 5500 m²
Now, the cost of leveling of 1 m² of the playground = 50 paise
As we know, 1paise = 1/100 rupee
50 paise = 50/100 rupee
Then, the cost of leveling of 1 m² of the playground in rupees = 50/100
cost of leveling of 5500 m² of the playground = 5500 * 50/100
= Rs. 2750
∴ The cost of leveling of 5500 m² of the playground is Rs. 2750.
Hence, the diameter of a roller is 250cm and length is 140 cm if it takes 500 complete revolutions to level a playground then the cost of leveling the playground at the rate of 50 paise per sq m is Rs.2750.