Question

9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

Diameter of road roller, d = 84 cm

So,

Radius of road roller, r = 42 cm

Length of road roller, h = 1 m = 100 cm

Road roller is in the shape of cylinder.

We know,

Area covered by road roller in 1 revolution = Curved Surface Area of cylindrical road roller.

So,

$\red{\rm :\longmapsto\:Area_{(covered \: in \: 1 \: revolution) } = 2\pi \: rh}$

So,

$\red{\rm :\longmapsto\:Area_{(covered \: in \: 750 \: revolution) } = 750 \times 2\pi \: rh}$

$\red{\rm :\longmapsto\:Area_{(covered \: in \: 750 \: revolution) } = 1500\pi \: rh}$

On substituting the values of r and h, we get

$\red{\rm :\longmapsto\:Area_{(covered \: in \: 750 \: revolution) } = 1500 \times \dfrac{22}{7} \times 42 \times 100}$

$\red{\rm :\longmapsto\:Area_{(covered \: in \: 750 \: revolution) } = 19800000 \: {cm}^{2} }$

$\red{\rm :\longmapsto\:Area_{(covered \: in \: 750 \: revolution) } = 1980 \: {m}^{2} }$

Additional Information :-

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²