Question

A road roller is 140 CM. It's diameter is 84 CM. It takes 1000 complete revolution once over to level the road. Find the cost of levelling the road at random. 50. per 100 square meters. ​

Answer 1

$\bold{ANSWER}$

•Area $\bold{covered}$ in one complete revolution
=surface area of $\bold{road}$ =$2\pi \: rh$

=2 x 22/7 x 42 x 140 cm^2

=12 x 22 x 140 cm^2

= $= 36960 {cm}^{2}$

☞covered $\bold{area}$ in 1000 revolution =1000 x surface area of road roller

=$36960 \times 10^3 cm^3$

=$3696m^2$

then,

$\bold{cost}$ = surface area of roller x rate

=3696 x 50/100

=$\bold{1848\: Rs}$

Answer 2

Answer:

Rs 1848

Step-by-step explanation:

Height of road roller = 140 cm

Radius of road roller = 84/2 = 42 cm.

Area covered in one complete rotation is equal to the surface area of road.

Surface area = 2 π r h cm²

⇒ 2 * 22/7 * 42 * 140 cm²

⇒ 44 * 840 cm²

⇒ 36960 cm²

Area covered in 1000 revolution is equal to the product of surface area if road and 1000.

⇒ 36960 * 10³ cm²

⇒ 3696 m²

Then,

Cost = Surface area * rate

⇒ cost = 3696 * 50 / 100

⇒ cost = Rs 1848

Hence, the answer is Rs 1848.