Question

the diameter of a cylindrical roller 120 cm long is 84 cm it takes 500 complete resolutions to level a playground find the cost of leveling it at the rate of rupees 2 per square meter

Answer 1

SOLUTION :

The diameter of a cylindrical roller = 120 cm     (given)

The height if the cylindrical roller = 84 cm   (given)

Here

Radius of cylindrical roller = diameter ÷ 2

                                            = 120 ÷ 2

                                            = 60 cm

So,

The roller is in cylindrical shape

According to the question,

The roller takes 500 complete revolutions to level the playground field

Only the curved surface area is needed

So,

The curved surface area of a cylinder = 2πrh  unit sq.

[ ∴ In which r is radius and h is height of the cylinder ]

So,

The curved surface area of the cylindrical roller will be

= $2 \times \frac{22}{7} \times 60 \times 84$ cm sq.

= 31680 cm sq.

Converting 31680 cm sq. into m sq.

⇒ $\bf{ \frac{31680}{10000} }$ m sq.     [ As 1 m sq. = 10000 cm sq. ]

⇒ 3.168 m sq.

Now,

1 revolution = Curved surface area

So,

500 revolution = 500 × 3.168

                         = 1584 m sq.

So,

The area of the playground is 1584 m sq.

Now,

If the cost of leveling 1 m sq. is Rs 2

Than the cost of leveling the playground will be

= 2 × 1584

= Rs 3168

Hence,

The cost of leveling the playground is Rs 3168

Answer 2

Answer:

Step-by-step explanation:

given diameter of roadroller = 140cm = 1.4m

radius will be 0.7

length of roller = 2.2 m

therefore, CSA = 2×22/7×0.7×2.2

= 9.68 cm^2

It takes 400 revolutions to level a stretch 2.2m wide

therefore the surface area of road levelled =

400 × 9.68 m^2

= 3872.00 m^2