Question

a roller of cylindrical shape has radius 2 metre and length 10 metre find the area covered by the roller in 100 rotations
answer it fast..​

Answer 1

$\Large {\underline { \sf \orange{Clarification :}}}$

Here, as per the provided question we are given that the radius of the cylindrical shaped roller is 2 m and its length is 10 m. We have to find the area covered by the roller in 100 rotations.

Using concept, its curved surface area is the area covered by the roller in 1 rotation. So, at first we'll calculate the curved surface area of the roller. After that, we'll multiply it with 100 in order to find the area covered by the roller in 100 rotations.

$\Large {\underline { \sf \orange{Explication \: of \: steps :}}}$

Given :

• Radius of the cylindrical roller (r) = 2 m

• Length (h) = 10 m

To calculate :

• Area covered by the roller in 100 rotations.

Calculations :

Let's calculate area covered by the roller in 1 rotation first.

$\bigstar \: \boxed{\sf { Area_{(1 \: Rotation)} = L.S.A_{(Roller)} }}$

$\longrightarrow \sf {Area_{(1 \: Rotation)} = 2\pi r h }$

$\longrightarrow \sf {Area_{(1 \: Rotation)} = 2 \times 3.14 \times 2 \times 10 \: {m}^{2} }$

• Value of π is 3.14 or 22/7

$\longrightarrow \sf {Area_{(1 \: Rotation)} = 2 \times \dfrac{314}{100} \times 2 \times 10 \: {m}^{2}}$

$\longrightarrow \sf {Area_{(1 \: Rotation)} = \dfrac{12560}{100} \: {m}^{2}}$

$\longrightarrow \sf {Area_{(1 \: Rotation)} = 125.6 \: {m}^{2}}$

So, in one rotation it covers 125.6 m². We can say that,

$\bigstar \: \boxed{\sf {Area_{(100 \: Rotations)} = Area_{(1 \: Rotation)} \times 100 }}$

$\longrightarrow \sf {Area_{(100 \: Rotations)} = 125.6 \times 100 \: {m}^{2}}$

$\longrightarrow \sf {Area_{(100 \: Rotations)} = \dfrac{1256}{\cancel{10}} \times \cancel{100} \: {m}^{2}}$

$\longrightarrow \sf {Area_{(100 \: Rotations)} = 1256 \times 10 \: {m}^{2}}$

$\longrightarrow$ ❝ $\boxed{ \sf \orange {Area_{(100 \: Rotations)} = 12560 \: {m}^{2} }}$ ❞

❝ Therefore, area covered by the roller in 100 rotations is 12560 m² ❞

Answer 2

Answer:

Here, as per the provided question we are given that the radius of the cylindrical shaped roller is 2 m and its length is 10 m. We have to find the area covered by the roller in 100 rotations.

Using concept, its curved surface area is the area covered by the roller in 1 rotation. So, at first we'll calculate the curved surface area of the roller. After that, we'll multiply it with 100 in order to find the area covered by the roller in 100 rotations.

\Large {\underline { \sf \orange{Explication \: of \: steps :}}}Explicationofsteps:

Given :

• Radius of the cylindrical roller (r) = 2 m

• Length (h) = 10 m

To calculate :

• Area covered by the roller in 100 rotations.

Calculations :

Let's calculate area covered by the roller in 1 rotation first.

\begin{gathered}\bigstar \: \boxed{\sf { Area_{(1 \: Rotation)} = L.S.A_{(Roller)} }} \\ \end{gathered}★Area(1Rotation)=L.S.A(Roller)

\longrightarrow \sf {Area_{(1 \: Rotation)} = 2\pi r h }⟶Area(1Rotation)=2πrh

\begin{gathered} \longrightarrow \sf {Area_{(1 \: Rotation)} = 2 \times 3.14 \times 2 \times 10 \: {m}^{2} } \\ \end{gathered}⟶Area(1Rotation)=2×3.14×2×10m2

Value of π is 3.14 or 22/7

\begin{gathered} \longrightarrow \sf {Area_{(1 \: Rotation)} = 2 \times \dfrac{314}{100} \times 2 \times 10 \: {m}^{2}} \\ \end{gathered}⟶Area(1Rotation)=2×100314×2×10m2

\begin{gathered} \longrightarrow \sf {Area_{(1 \: Rotation)} = \dfrac{12560}{100} \: {m}^{2}}\\ \end{gathered}⟶Area(1Rotation)=10012560m2

\begin{gathered} \longrightarrow \sf {Area_{(1 \: Rotation)} = 125.6 \: {m}^{2}}\\\end{gathered}⟶Area(1Rotation)=125.6m2

So, in one rotation it covers 125.6 m². We can say that,

\begin{gathered}\bigstar \: \boxed{\sf {Area_{(100 \: Rotations)} = Area_{(1 \: Rotation)} \times 100 }} \\ \end{gathered}★Area(100Rotations)=Area(1Rotation)×100

\begin{gathered} \longrightarrow \sf {Area_{(100 \: Rotations)} = 125.6 \times 100 \: {m}^{2}}\\\end{gathered}⟶Area(100Rotations)=125.6×100m2

\begin{gathered} \longrightarrow \sf {Area_{(100 \: Rotations)} = \dfrac{1256}{\cancel{10}} \times \cancel{100} \: {m}^{2}}\\\end{gathered}⟶Area(100Rotations)=101256×100m2

\begin{gathered} \longrightarrow \sf {Area_{(100 \: Rotations)} = 1256 \times 10 \: {m}^{2}}\\\end{gathered}⟶Area(100Rotations)=1256×10m2

\begin{gathered} \longrightarrow \\ \end{gathered}⟶ ❝ \begin{gathered} \boxed{ \sf \orange {Area_{(100 \: Rotations)} = 12560 \: {m}^{2} }} \\\end{gathered}Area(100Rotations)=12560m2 ❞

❝ Therefore, area covered by the roller in 100 rotations is 12560 m² ❞