Question

the diameter of a roller 120 cm long is 84 CM if it take 500 complete revolution of a playground determine the cost of levelling it at the rate of 30 per square metre​

Answer 1

Answer:

Clearly the roller is a right circular cylinder of height h=120 cm

and radius of its base =r=42 cm

∴ area covered by the roller in one revolution = curved surface area of the roller 

           =2×722×42×120cm2=31680cm2

so, area covered by the roller is 500 revolution = (31680×500)cm2

                    =100×10031680×500m2=1584m2

hence , cost of levelling the playground = 1584×10030=Rs475.20

Answer 2

$\large\underline{ \underline{ \sf \maltese{ \: Question:- }}}$

• the diameter of a roller 120 cm long is 84 CM if it take 500 complete revolution of a playground determine the cost of levelling it at the rate of 30 per square metre

$\large\underline{ \underline{ \sf \maltese{ \: Given:- }}}$

Clearly, The roller is a right is a right circular cylinder of

• $\sf\green{ Height\: = \: \fbox \red{120cm} \: and \: }$

• $\sf\green{Radius \:of \: it's \: base \: = \: 82 \div 2 \: = \: \fbox \red{42cm }}$

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$\large\underline{ \underline{ \sf \maltese{ \: Solution:- }}}$

Area Covered by the roller in 1 revolution = Curved Surface Of The Roller

$\begin{gathered}\begin{gathered}\begin{gathered}: \implies \underline\blue{ \boxed{\displaystyle \sf \bold\orange{\: Area \: Covered \: in \: 1 \: Revolution \: = 2\pi \: × \: R\: × \: H }} }\\ \\\end{gathered}\end{gathered}\end{gathered}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ 2 \: \times \: \frac{22}{ \cancel7} \: \times \cancel{42} \: \times \: 120 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ 2 \: \times \: {22} \: \times 6 \: \times \: 120 }}$

$\quad {:} \longrightarrow \underline \red{\boxed{\sf{Area \: Covered \: in \: 1 \: Revolution \: = \: 31680 \: {cm}^{2} }}}$

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So, Area Covered by the roller in 500 revolution = $\blue{\fbox \green{ \: 31680 \: × \: 500 }} \: {cm}^{2}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{31680 \: \times \: 500}{100 \: \times \: 100} \: {m}^{2} }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{3168 \cancel0 \: \times \: 5 \cancel{00}}{10 \cancel0 \: \times \: 1 \cancel{00}} \: {m}^{2} }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{3168 \: \times \: \cancel5 }{ \cancel{10}} \: {m}^{2} }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ \frac{ \cancel{3168} \: }{ \cancel 2} \: {m}^{2} }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{ \:Area \: Covered \: in \: 500 \: Revolution \: = \:1584{m}^{2} }}}$

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Cost of levelling the playground :-

$\qquad \quad {:} \longrightarrow \sf{\sf{ 1584 \: \times \: \frac{30}{100} }}$

$\bf \therefore \;Cost \; of \; Levelling= Rs. 475.20$

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$\begin{gathered}\begin{gathered}\qquad \therefore\: \sf{ Cost \: of \: Levelling \: the \: Playground\: = \underline {\underline{ Rs. 475.20}}}\\\end{gathered}\end{gathered}$

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