Question

help me.............. ​

Answer 1

$\bf GivEn \begin{cases} & \sf{Radius\;of\;cylindrical\;oil\;can,\;r = \bf{7.7\;cm}} \\ & \sf{Height\;of\;cylindrical\;oil\;can,\;h = \bf{56\;cm}} \end{cases}$

Need to find: The quantity of oil in litres that can be stored in the can.

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$\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}$

Quantity of oil in that can be stored in can = Volume of oil can

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$\dag\;{\underline{\frak{As\;we\;know\;that,}}}$

$\star\;{\boxed{\sf{\pink{Volume_{\;(cylinder)} = \pi r^2 h}}}}$

Here,

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r = 7.7 cm

h = 56 cm

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$\setlength{\unitlength}{1.3mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(16,2)(0,32){2}{\sf{\footnotesize 7.7 cm}}\put(14,17.5){\sf{\footnotesize 56 cm}}\end{picture}$

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$\dag\;{\underline{\frak{Putting\;values\;in\;formula,}}}$

$:\implies\sf Volume_{\;(oil\;can)} = \dfrac{22}{ \cancel{7}} \times 7.7 \times 7.7 \times \cancel{56}\\ \\ :\implies\sf Volume_{\;(oil\;can)} = 22 \times 7.7 \times 7.7 \times 8\\ \\ :\implies\sf Volume_{\;(oil\;can)} = 176 \times 7.7 \times 7.7\\ \\ :\implies{\underline{\boxed{\frak{\purple{Volume_{\;(oil\;can)} = 10435.04</p><p>\;cm^3 }}}}}\;\bigstar$

$\dag\;{\underline{\frak{Quantity\;of\;oil\;in\;Litre,}}}$

We know that,

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1 Litre = 1000 cm³

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$:\implies\sf Volume_{\;(oil\;can)} = 10.43504\;L$

$\therefore$ Quantity of oil in litres that can be stored in the can is 10.43504 L.

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$\boxed{\underline{\underline{\bigstar \: \bf\:Formulas\:related\:to\: cylinder\:\bigstar}}}$

$\sf (i)\;Curved\;surface\;area\;of\;cylinder\; = \; \red{2 \pi rh}$

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$\sf (ii)\;Total\;surface\;area\;of\;cylinder\; = \; \purple{2 \pi r(h + r)}$

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$\sf (iii)\;Volume\;of\;cylinder\; = \; \pink{ \pi r^2 h}$

Answer 2

$\bf GivEn \begin{cases} & \sf{Radius\;of\;cylindrical\;oil\;can,\;r = \bf{7.7\;cm}} \\ & \sf{Height\;of\;cylindrical\;oil\;can,\;h = \bf{56\;cm}} \end{cases}$

Need to find: The quantity of oil in litres that can be stored in the can.

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━

$\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}$

Quantity of oil in that can be stored in can = Volume of oil can

⠀⠀

$\dag\;{\underline{\frak{As\;we\;know\;that,}}}$

$\star\;{\boxed{\sf{\pink{Volume_{\;(cylinder)} = \pi r^2 h}}}}$

Here,

⠀⠀

r = 7.7 cm

h = 56 cm

⠀⠀

$\setlength{\unitlength}{1.3mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(16,2)(0,32){2}{\sf{\footnotesize 7.7 cm}}\put(14,17.5){\sf{\footnotesize 56 cm}}\end{picture}$

⠀⠀

$\dag\;{\underline{\frak{Putting\;values\;in\;formula,}}}$

$:\implies\sf Volume_{\;(oil\;can)} = \dfrac{22}{ \cancel{7}} \times 7.7 \times 7.7 \times \cancel{56}\\ \\ :\implies\sf Volume_{\;(oil\;can)} = 22 \times 7.7 \times 7.7 \times 8\\ \\ :\implies\sf Volume_{\;(oil\;can)} = 176 \times 7.7 \times 7.7\\ \\ :\implies{\underline{\boxed{\frak{\purple{Volume_{\;(oil\;can)} = 10435.04</p><p>\;cm^3 }}}}}\;\bigstar$

$\dag\;{\underline{\frak{Quantity\;of\;oil\;in\;Litre,}}}$

We know that,

⠀⠀

1 Litre = 1000 cm³

⠀⠀

$:\implies\sf Volume_{\;(oil\;can)} = 10.43504\;L$

$\therefore$ Quantity of oil in litres that can be stored in the can is 10.43504 L.

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\boxed{\underline{\underline{\bigstar \: \bf\:Formulas\:related\:to\: cylinder\:\bigstar}}}$

$\sf (i)\;Curved\;surface\;area\;of\;cylinder\; = \; \red{2 \pi rh}$

⠀⠀⠀⠀⠀⠀⠀

$\sf (ii)\;Total\;surface\;area\;of\;cylinder\; = \; \purple{2 \pi r(h + r)}$

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$\sf (iii)\;Volume\;of\;cylinder\; = \; \pink{ \pi r^2 h}$