Question

a roller is 3.5 m long and has a radius of 90cm .find the number of revolutions it should make to cover an area of 1187 m square.

step by step explanation​

Answer 1

CSA of Cylinder

= (2)(pi)(r)(h)

= (2)(22/7)(90)(350)

(height is taken as 350 as 3.5m = 350cm)

= (44/7)(90)(350)

= (44/7)(31500)

= (1386000/7)

= 198000cm²

cm² to m² = x/1000

= 198000/1000

= 198m²

Now, area to cover = 1187m²

CSA of roller = 198m²

Therefore, 1187/198

= 5.99494949 revolutions

Answer : 5.99 revolutions (approximately) are required by the roller to cover the area of 1187m²

Answer 2

Answer:

Hence the number of revolutions it should make to cover an area of $1187\;m^{2}$ is 1.66 revolutions.

Step-by-step explanation:

As per the data provided in the given question.

The given data is as follows,

A roller is 3.5 m and the radius of a roller (r) is 90 cm.

The total area to cover is $1187\;m^{2} .$

Here the length of a roller represents the height of a roller i.e. (h).

We have to find the number of revolutions it should make to cover an area of $1187\;m^{2} .$

At first, we will convert 3.5 m to cm as below.

[tex]=3.5\times100\\ =350\;cm[/tex]

Now we will calculate the curved surface area of the roller.

Here we will be using the formula of the curved surface area of the cylinder.

$Curved\;surface\;of\;cylinder=2\pi rh$

Now we will substitute the given values in the above formula.

$=2\times\frac{22}{7}\times90\times350$

Now we will 350 by 7 we get,

$=2\times22\times90\times50$

Now we will multiply the above values.

[tex]=44\times90\times50\\ =3960\times50\\ =198000\;cm^{2} [/tex]

Now we will convert 198000 cm to m as below.

[tex]=\frac{198000}{100} \\ =\frac{1980}{1}\\ =1980\;m^{2} [/tex]  

Now we will find the number of revolutions.

$No.\;of\;revolution=\frac{curved\;surface\;area\;of\;cylinder}{total\;curved\;area}$

Now we will substitute the given value in the above formula.

[tex]=\frac{1980}{1187}\\ =1.66\;revolutions[/tex]

Hence the number of revolutions it should make to cover an area of $1187\;m^{2}$ is 1.66 revolutions.