Question

A road roller is cylindrical in shape. Its circular end has a diameter 250 cm and its width is
1 m 40cm Find the least number of revolutions that the roller must make in order to level
a playground 110 m * 25 m. prove​

Answer 1

$\huge{\text{\underline{Correct\:Question:-}}}$

A road roller is cylindrical in shape its circular and has a diameter 250 cm and its width is 1 m 40 centimetre Find the least number of revolution that the roller must make in order to level up in a gram 10 metre into 25 metre.

$\huge{\text{\underline{Solution:-}}}$

★ Given:-

$\implies$Diameter of roller = 250 cm

$\implies$ 250 cm = 2.5 m

$\implies$Width of roller = 1.4 m

$\implies$Radius of the roller = 1.25 m

★ We know that:-

Curved surface area = 2 × 22 / 7 × 1.25 × 1.40

$\implies$44 × 1.25 × 0.20

$\implies{\boxed{\tt{11\:m^2}}}$

Area of ground = length × breath

$\implies$ 25 × 10

$\implies{\boxed{\tt{ 250\:m^2}}}$

Number of revolutions to cover the field = Area of field / CSA of roller

$\implies$ 250 / 11

$\implies{\boxed{\tt{22.727\:(approx)}}}$

Therefore, the revolutions made by the roller is 22.727

__________________________________

Answer 2

$\bold{\underline{\underline{\huge{\sf{AnsWer:}}}}}$

Revolution required by the roller in order to level the ground is 250

$\bold{\underline{\underline{\huge{\sf{StEp\:by\:stEp\:explanation:}}}}}$

Road Roller :

GiVeN :

• A road roller is cylindrical in shape.
• Its circular end has a diameter 250 cm
• Width of 1 m 40 cm

Playground :

GiVeN :

• Length = 110 m
• Breadth = 25 m

To FiNd :

• The least number of revolutions that the roller must make in order to level the playing

SoLuTiOn :

We have the diameter of the circular end which is 250 cm.

Radius of the circular end,

$\longrightarrow$$\sf{\dfrac{Diamter}{2}}$

$\longrightarrow$$\sf{\dfrac{250}{2}}$

$\longrightarrow$$\sf{125cm}$

•°• Radius of the circular end of the cylindrical pipe is 125 cm

Width = 1 m 40 cm

1 m = 100 cm

•°• Width = 100 + 40 = 140 cm

We need to calculate the Curved Surface Area of the cylindrical road roller.

Formula :

$\bold{\boxed{\large{\rm{Curved\:surface\:area\:=\:2\:\pi\:r\:h}}}}$

Where,

• r = radius = 125 cm
• h = height = 140 cm

Block in the values,

$\longrightarrow$ $\sf{2\:\times\:{\dfrac{22}{7}\:\times\:125\:\times\:140}}$

$\longrightarrow$ $\sf{\dfrac{44\:\times\:125\:\times\:140}{7}}$

$\longrightarrow$ $\sf{\dfrac{44\:\times\:17500}{7}}$

$\longrightarrow$ $\sf{\dfrac{770000}{7}}$

$\longrightarrow$ $\sf{110000}$

$\sf{\therefore{Curved\:surface\:area\:of\:cylindrical\:road\:roller\:=\:110000\:cm^2}}$

PlAyGrOuNd :

Length = 110 m

Breadth = 25 m

1 m = 100 cm

•°• Length = 110 (100)

Length = 11000 cm

Breadth = 25 (100)

•°• Breadth = 2500 cm

FoRmUlA :

$\bold{\boxed{\large{\rm{Area\:of\:playground\:=\:length\:\times\:breadth}}}}$

Block in the values,

$\longrightarrow$ $\sf{11000\:\times\:2500}$

$\longrightarrow$ $\sf{275000000}$

•°• Area = 27500000 cm²

Number of revolutions :

We can find the number of revolutions by dividing the area of playground by curved surface area of the cylindrical road roller.

$\bold{\boxed{\sf{Number\:of\:revolutios\:=\:\:{\dfrac{Area\:of\:playground\:}{Curved\:surfae\:area}}}}}$

Block in the values,

$\longrightarrow$ $\sf{\dfrac{27500000}{110000}}$

$\longrightarrow$ $\sf{\dfrac{2750}{11}}$

$\longrightarrow$ $\sf{250}$

•°•Revolutions made by roller is 250.