Question

A cylinder pillar is 50cm in diameter and 3.5m high find the cost of white washing its curved surface at the rate of rs 1.25per square metre.​

Answer 1

${\large{\pmb{\sf{\underline{RequirEd \; Solution...}}}}}$

${\bigstar \: {\pmb{\sf{\red{Understanding \; the \; question...}}}}}$

⋆ This question says that we have to find out the cost of white washing pillar's curved surface at the rate of rupees 1.25 per square metre. The pillar is in the shape of cylinder. It is 50 centimetres in diameter. And 3.5 metres in height. Let us solve this question!

${\bigstar \: {\pmb{\sf{\red{Given \: that...}}}}}$

⋆ Pillar is given

⋆ Pillar is in the shape of cylinder

⋆ Diameter = 50 centimetres

⋆ Height = 3.5 metres

${\bigstar \: {\pmb{\sf{\red{To \: find...}}}}}$

⋆ The cost of white washing pillar's curved surface at the rate of rupees 1.25 per square metre.

${\bigstar \: {\pmb{\sf{\red{Solution...}}}}}$

⋆ The cost of white washing pillar's curved surface at the rate of rupees 1.25 per square metre = 6.875 Rupees

${\bigstar \: {\pmb{\sf{\red{Using \: concepts...}}}}}$

⋆ Formula to find out the curved surface area of the cylinder

⋆ Formula to convert centimetres into metres.

⋆ Formula to convert diameter into radius.

⋆ Formula to find rate(according to the question)

${\bigstar \: {\pmb{\sf{\red{Using \: formulas...}}}}}$

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

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

${\small{\underline{\boxed{\sf{\star \: 1 \: m \: = 100 \: cm}}}}}$

$\quad \quad \quad{\pmb{\sf{\dag \: Henceforth,}}}$

${\small{\underline{\boxed{\sf{\star \: 1 \: cm \: = \dfrac{1}{100} \: m}}}}}$

${\small{\underline{\boxed{\sf{\star \: Rate \: = CSA \times Given \: rate}}}}}$

${\bigstar \: {\pmb{\sf{\red{Where...}}}}}$

• CSA denotes Curved Surface Area

• π is pronounced as pi

• The value of π is 22/7 or 3.14

• h denotes height

• r denotes radius

${\bigstar \: {\pmb{\sf{\red{Full \; Solution...}}}}}$

~ Firstly let us convert diameter into radius by using the formula!

${\small{\underline{\boxed{\sf{Radius \: = \dfrac{Diameter}{2}}}}}} \\ \\ :\implies \sf Radius \: = \dfrac{Diameter}{2} \\ \\ :\implies \sf Radius \: = \dfrac{50}{2} \\ \\ :\implies \sf Radius \: = \cancel{\dfrac{50}{2}} \\ \\ :\implies \sf Radius \: = 25 \: cm$

${\pmb{\sf{Henceforth, \: 25 \: cm \: is \: the \: radius \: we \: get!}}}$

~ Now let's convert centimetres into metres by using formula!

${\small{\underline{\boxed{\sf{1 \: cm \: = \dfrac{1}{100} \: m}}}}} \\ \\ :\implies \sf 1 \: cm \: = \dfrac{1}{100} \: m \\ \\ :\implies \sf \dfrac{25}{100} \\ \\ :\implies \sf 0.25 \: metres \\ \\ \sf Henceforth, \: converted!$

${\pmb{\sf{Means \: radius \: is \: 0.25 \: m}}}$

~ Now let's find out the curved surface area of the cylinder by using the formula!

${\small{\underline{\boxed{\sf{CSA \: of \: cylinder \: = 2 \pi rh}}}}} \\ \\ :\implies \sf CSA \: of \: cylinder \: = 2 \pi rh \\ \\ :\implies \sf CSA \: of \: cylinder \: pillar \: = 2 \pi rh \\ \\ :\implies \sf CSA \: of \: cylinder \: pillar \: = 2 \times 3.14 \times 0.25 \times 3.5 \\ \\ :\implies \sf CSA \: of \: cylinder \: pillar \: = 6.28 \times 0.25 \times 3.5 \\ \\ :\implies \sf CSA \: of \: cylinder \: pillar \: \approx 5.5 \: metres \: sq.$

${\pmb{\sf{Henceforth, \: 5.5 \: metres \: sq. \: is \: the \: CSA}}}$

~ Now let's find out the rate!

${\small{\underline{\boxed{\sf{Rate \: = CSA \times Given \: rate}}}}} \\ \\ :\implies \sf Rate \: = CSA \times Given \: rate \\ \\ :\implies \sf Rate \: = 5.5 \times 1.25 \\ \\ :\implies \sf Rate \: = 6.875 \: Rs$

${\pmb{\sf{Henceforth, \: rate \: is \: 6.875 \: Rupees}}}$

${\bigstar \: {\pmb{\sf{\red{Diagram \: regards \: Question...}}}}}$

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{0.25 \: m}}\put(9,17.5){\sf{3.5 \: m}}\end{picture}$

${\bigstar \: {\pmb{\sf{\red{Some \: formulas \: regards \: cylinder...}}}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto Volume \: of \: cylinder \: = \: \pi r^{2}h}}$

$\; \; \; \; \; \; \;{\sf{\leadsto Surface \: area \: of \: cylinder \: = \: 2 \pi rh + 2 \pi r^{2}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto Lateral \: area \: of \: cylinder \: = \: 2 \pi rh}}$

$\; \; \; \; \; \; \;{\sf{\leadsto Base \: area \: of \: cylinder \: = \: \pi r^{2}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto Height \: of \: cylinder \: = \: \dfrac{v}{\pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto Radius \: of \: cylinder \: = \:\sqrt \dfrac{v}{\pi h}}}$