Question

( A piece of ductile metal is in the form of a cylinder of diameter 1 cm and length 11 cm. It is drawn out into a wire of diameter 1 mm. What will be the length of the wire so obtained?
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Answer 1

$\huge{\underline{\underline{\mathtt{\green{Given}}}}}$

A piece of ductile metal is in the form of a cylinder of diameter 1 cm and length 11 cm. It is drawn out into a wire of diameter 1 mm.

$\huge{\underline{\underline{\mathtt{\green{To\:find}}}}}$

What will be the length of the wire so obtained?

$\huge{\underline{\underline{\mathtt{\green{Solution}}}}}$

Radius of metal ductile in the form of cylinder

= 1/2 = 0.5cm

Length of metal ductile = height of metal ductile

= 11cm

Now ,

Volume of metal ductile in the form of cylinder

πr²h

$\implies\sf π×(0.5)^2×11$

$\implies\sf π×0.25×11$

$\implies\sf 2.75πcm^3$

After recasting in the form of metal wire

Diameter of wire = 1mm = 0.1cm

Radius of wire = 0.05cm

★ Volume of metal ductile in the form of cylinder is equal to the volume of new metal wire

★ Wire is also just like cylinder so we apply volume of cylinder Formula

$\implies\sf 2.75π=πr^2h$

$\implies\sf 2.75=(0.05)^2×h$

$\implies\sf 2.75=0.0025h$

$\implies\sf h=\frac{2.75}{0.0025}=1100cm=11m$

$\huge\underline\frak\red{Important\:points}$

• 1km = 1000m
• 1m = 100cm
• 1mm = 10cm
• 1liter = 1000ml
• 1m³ = 1000liter
• 1cm³ = 1000ml
• 1hour = 60min
• 60min = 3600sec
• 1mm = 1000m

Answer 2

AnswEr :

1100cm.

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

A piece of ductile metal is in the form of a cylinder of diameter 1 cm and length 11 cm. It's drawn out into a wire of diameter 1 mm.

$\setlength{\unitlength}{1.05 cm}}\begin{picture}(12,4)\thicklines\put(3,6){ . }\put(6,6){\circle{4}}\put(6,6){\circle{0.8}}\put(5.34,6){\line(0,1){2.5}}\put(6.66,6){\line(0,1){2.5}}\put(6,8.5){\circle{4}}\put(6,8.5){\circle{0.8}}\put(5.6,6){\line(0,1){2.5}}\put(6.4,6){\line(0,1){2.5}}\put(7,7){\vector(0,1){1.5}}\put(7,7){\vector(0,-1){1}}\put(7.3,7.2){ \sf{11cm }}\put(5.6,4.6){ \sf{1\:cm }}\put(6,5){\vector(1,0){0.5}}\put(6,5){\vector(-1,0){0.5}}\end{picture}$

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The length of the wire so obtained.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

Let the the length of the wire be R

$\bf{We\:have}\begin{cases}\sf{Diameter\:of\:Cylinder\:(d)=1\:cm}\\ \sf{Height\:of\:Cylinder\:(h)=11\:cm}\\ \sf{Diameter\:of\:wire\:(d)=1\:mm}\end{cases}}$

Then;

→ Radius of cylinder = $\sf{\dfrac{d}{2} =\dfrac{1}{2} cm}}$

→ Radius of Wire = $\sf{\dfrac{d}{2} =\dfrac{1}{2\times 10} mm = \dfrac{1}{20} cm}$

$\bf{\red{\underline{\underline{\tt{A.T.Q.\::}}}}}$

Formula use :

$\bf{\boxed{\bf{Volume\:of\:Cylinder=\pi r^{2} h}}}}$

$\mapsto\tt{\orange{Volume\:of\:Wire}\:=\:\orange{Volume\:of\:Cylinder}}\\\\\\\mapsto\tt{\pi r_{1}^{2} R_{1}=\pi r_{2}^{2} R_{2}}\\\\\\\mapsto\tt{R_{1}=\dfrac{\cancel{\pi }r_{2}^{2}R_{2} }{\cancel{\pi }r_{1}^{2}} } \\\\\\\mapsto\tt{R_{1}=\dfrac{(\frac{1}{2})^{2} \times 11 }{(\frac{1}{20}) ^{2} } }\\\\\\\mapsto\tt{R_{1}=\dfrac{\frac{1}{4} \times 11}{\frac{1}{400} } }\\\\\\\mapsto\tt{R_{1}=\dfrac{\frac{11}{4} }{\frac{1}{400} } }\\\\\\\mapsto\tt{R_{1}=\dfrac{11}{4} \times \dfrac{400}{1} }}$

$\mapsto\tt{R_{1}=\cancel{\dfrac{4400}{4}} \:cm}\\\\\\\mapsto\tt{\red{R_{1}=1100\:cm}}$

Thus,

The length of the wire so obtained is 1100 cm.