Question

The Diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolution to move once over to level playground.Find the cost of levelling it's at the rate of RS 1.20 paise per sq. m.

Answer 1

The curved surface area of roller x 500 = area of level ground

csa= 2*3.14*42*120 *500
= 31651.2 cm sq. *500
=3.16 m sq. *500
= 1580 m sq.

cost of levelling = csa*1.20
=1580*1.2
= Rs 1896

Answer 2

Given,
Diameter of a roller = 84 cm
Length of roller = 120 cm
No. of revolutions = 500
Cost of levelling the ground = 1.20 paise/ sq. m

Here,
Radius of roller = Diameter of roller / 2
= 84 / 2
= 42 cm
Also,
Length of roller = Height of the cylinder
Therefore,
Height of the cylinder = 120 cm

Now,
Area of the cylindrical roller = 2πrh
= 2• 314/ 100 • 120
= 628/ 10 • 12
= 753.6 sq. cm
Now,
Area of Ground Levelled = n • Area of roller
where,
n = No. of revolutions of roller
= 500
Area of roller = 753.6 sq. cm

Therefore,
Area of Ground levelled = 500 • 753.6
= 356800.0
= 356800 sq. cm
As we know,
10000 sq. cm = 1 sq. m
356800 sq. cm = ( ? )
Therefore,
356800 • 1 / 10000
35.68 sq. m

Cost of levelling the Ground = 1.20 paise / sq.m
35.68 sq. m = ( ? )
Therefore,
35.68 • 1.20
= 42.816 paise.

Hence Solved........