Question

find the area covered by a road roller of width 80 CM and diameter 140 cm in 40 revolutions

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Answer 1

Answer:

Diagram :

☼ Figure of cylindrical roller with radius and height :

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{70\ cm}}\put(9,17.5){\sf{80\ cm}}\end{picture}$

$\begin{gathered}\end{gathered}$

Given :

• → The width of a roller = 80 cm.
• → The diameter of roller = 140 cm.

$\begin{gathered}\end{gathered}$

To Find :

• → Radius of roller
• → Area covered by roller in 40 revolutions.

$\begin{gathered}\end{gathered}$

Using Formulas :

$\longrightarrow\small{\underline{\boxed{\sf{Radius = \dfrac{Diameter}{2}}}}}$

$\longrightarrow\small{\underline{\boxed{\sf{CSA \: of \: cylinder = 2 \pi rh}}}}$

${\longrightarrow{\small{\underline{\boxed{\sf{Area \: of \: 40 \: revolutions = 40 × CSA \: of \: roller}}}}}}$

$\begin{gathered}\end{gathered}$

Solution :

☼ Firstly, we have to find the find the radius of roller and here we have given that the diameter of roller is 140 cm. So, substituting the values in the formula :

$\longrightarrow \: \: {\sf{Radius = \dfrac{Diameter}{2}}}$

$\longrightarrow \: \: {\sf{Radius = \dfrac{140}{2}}}$

$\longrightarrow \: \: {\sf{Radius = \cancel{\dfrac{140}{2}}}}$

$\longrightarrow \: \: {\sf{Radius = 70 \: cm}}$

$\longrightarrow \: \: {\sf{\underline{\underline{\purple{Radius = 70 \: cm}}}}}$

∴ The radius of roller is 70 cm.

━┅━┅━┅━┅━┅━┅━┅━┅━┅━┅━

☼ Now, we know the radius then now finding the csa of roller by substituting the values in the formula :

$\longrightarrow \: \: {\sf{CSA \: of \: cylinder = 2 \pi rh}}$

${\longrightarrow \: \: {\sf{CSA \: of \: cylinder = 2 \times \dfrac{22}{7} \times 70 \times 80}}}$

${\longrightarrow \: \: {\sf{CSA \: of \: cylinder = 2 \times \dfrac{22}{\cancel{7}} \times \cancel{70} \times 80}}}$

${\longrightarrow \: \: {\sf{CSA \: of \: cylinder = 2 \times 22 \times 10 \times 80}}}$

${\longrightarrow \: \: {\sf{CSA \: of \: cylinder = 44 \times 800}}}$

${\longrightarrow \: \: {\sf{CSA \: of \: cylinder = 35200 \: {cm}^{2}}}}$

${\longrightarrow\:\:{\sf{\underline{\underline{\purple{CSA \: of \: cylinder = 35200 \: {cm}^{2}}}}}}}$

∴ The csa of cylindrical roller is 35200 cm².

☼ Now, let's find out the area covered by the roller in 40 revolutions :

${\longrightarrow \: \: {\sf{Area \: of \: 40 \: revolutions = 40 × CSA \: of \: roller}}}$

${\longrightarrow \: \: {\sf{Area \: of \: 40 \: revolutions = 40 × 35200}}}$

${\longrightarrow \: \: {\sf{Area \: of \: 40 \: revolutions = 1408000 \: {cm}^{2}}}}$

${\longrightarrow \: \: {\sf{\underline{\underline{\purple{Area \: of \: 40 \: revolutions = 1408000 \: {cm}^{2}}}}}}}$

∴ The area covered by roller in 40 revolutions is 1408000 cm².

$\begin{gathered}\end{gathered}$

Learn More :

$\boxed{\begin{minipage}{6.2 cm}\bigstar \:\underline{\textbf{Formulae Related to Cylinder :}}\\\\\sf {\textcircled{\footnotesize\textsf{1}}} \:Area\:of\:Base\:and\:top =\pi r^2 \\\\ \sf {\textcircled{\footnotesize\textsf{2}}} \:\:Curved \: Surface \: Area =2 \pi rh\\\\\sf{\textcircled{\footnotesize\textsf{3}}} \:\:Total \: Surface \: Area = 2 \pi r(h + r)\\ \\{\textcircled{\footnotesize\textsf{4}}} \: \:Volume=\pi r^2h\end{minipage}}$

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