Question

a radius of a road roller is 35cm and it is 1m long. if it takes 200 revelutions to level a play ground. find the cost of levelling a ground at the rate of rs 3 per m^2​

Answer 1

${\large{\sf{\pmb{\underline{\green{Given:-}}}}}}$

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• Radius of roller (r) = 35 cm

• Height of roller (h) = 1 m

• Cost of levelling = Rs. 3 per m²

• Number of revelutions = 200

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${\large{\sf{\pmb{\underline{\green{To\:find:-}}}}}}$

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• Cost of levelling of the ground

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${\large{\sf{\pmb{\underline{\green{Solution:-}}}}}}$

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• Radius of roller (r) = 35 cm = 0.35 m

• Height (h) = 1 m

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We know that

$\boxed{\pink{\sf Curved~surface~area~of~cylinder=2πr}}$

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${:\implies{\sf{CSA\:of\:the\:roller=2×\dfrac{22}{7}×0.35×1}}}$

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${:\implies{\sf{CSA\:of\:the\:roller=2×\dfrac{22}{7}×\dfrac{35}{100}×1}}}$

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${:\implies{\sf{CSA\:of\:the\:roller=\dfrac{220}{100}}}}$

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${:\implies{\sf{CSA\:of\:the\:roller=2.2}}}$

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Now ,

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${:\implies{\sf{Area\:of\:road=2.2×200=440\:m^{2}}}}$

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${:\implies{\sf{Cost\:of\:levelling=440×3}}}$

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${:\implies{\sf{Cost\:of\:levelling=1320}}}$

${\underline{\sf{\purple{Hence\:cost\:of\:levelling\:the\:ground=\:Rs.1320}}}}$

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⠀⠀ ________________

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⠀⠀⠀${\large{\sf{\pmb{\underline{\green{More\:Formulas\:of\:Cylinder}}}}}}$

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

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Note:- Please see these formulas from website Brainly.in

Answer 2

Answer:

Given:−

Given:−

⠀

Radius of roller (r) = 35 cm

Height of roller (h) = 1 m

Cost of levelling = Rs. 3 per m²

Number of revelutions = 200

⠀

{\large{\sf{\pmb{\underline{\green{To\:find:-}}}}}}

Tofind:−

Tofind:−

⠀

Cost of levelling of the ground

⠀

{\large{\sf{\pmb{\underline{\green{Solution:-}}}}}}

Solution:−

Solution:−

⠀

Radius of roller (r) = 35 cm = 0.35 m

Height (h) = 1 m

⠀

We know that

\boxed{\pink{\sf Curved~surface~area~of~cylinder=2πr}}

Curved surface area of cylinder=2πr

⠀

{:\implies{\sf{CSA\:of\:the\:roller=2×\dfrac{22}{7}×0.35×1}}}:⟹CSAoftheroller=2×

7

22

×0.35×1

⠀

{:\implies{\sf{CSA\:of\:the\:roller=2×\dfrac{22}{7}×\dfrac{35}{100}×1}}}:⟹CSAoftheroller=2×

7

22

×

100

35

×1

⠀

{:\implies{\sf{CSA\:of\:the\:roller=\dfrac{220}{100}}}}:⟹CSAoftheroller=

100

220

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{:\implies{\sf{CSA\:of\:the\:roller=2.2}}}:⟹CSAoftheroller=2.2

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Now ,

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{:\implies{\sf{Area\:of\:road=2.2×200=440\:m^{2}}}}:⟹Areaofroad=2.2×200=440m

2

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{:\implies{\sf{Cost\:of\:levelling=440×3}}}:⟹Costoflevelling=440×3

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{:\implies{\sf{Cost\:of\:levelling=1320}}}:⟹Costoflevelling=1320

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

{\underline{\sf{\purple{Hence\:cost\:of\:levelling\:the\:ground=\:Rs.1320}}}}

Hencecostoflevellingtheground=Rs.1320

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⠀⠀ ________________