Question

a metal sphere  of radius 21cm is melted,andconverted into a wire of uniform cross-section area.if the radius of the wire is 7 cm,then find the length of the wire

Answer 1

r=21cm
volume= 4/3$\pi$r$x^{3}$
therefore
38808 cm$x^{3}$
volume of wire 
$\pi$r2 h
38808 =154h
length =
252cm
2.52m

Answer 2

Solution :

A metal sphere of radius 21 cm is melted and converted into a wire having an uniform cross sectional area.

The radius of the wire is 7 cm.

We have to find the length of the wire.

$\setlength{\unitlength}{1.2cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3)\qbezier(-2.3,0)(0,-1)(2.3,0)\qbezier(-2.3,0)(0,1)(2.3,0)\thinlines\qbezier (0,0)(0,0)(0.2,0.3)\qbezier (0.3,0.4)(0.3,0.4)(0.5,0.7)\qbezier (0.6,0.8)(0.6,0.8)(0.8,1.1)\qbezier (0.9,1.2)(0.9,1.2)(1.1,1.5)\qbezier (1.2,1.6)(1.2,1.6)(1.38,1.9)\put(0.2,1){\bf r \: = \: 21 \: cm }\end{picture}$

Volume of the sphere

>> (4/3) π r³

>> (4/3) × (22/7) × 21 × 21 × 21 cm³

>> 4 × 22 × 21 × 21 cm³

>> 441 × 88 cm³

>> 38,808 cm³

Now, this is the volume of the wire.

The wire, can be considered as a cylinder having an uniform cross sectional area.

Assuming that the wire isn't hollow ;

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

Volume of a cylinder

>> πr²l

>> (22/7) × 7 × 7 × l

>> 154 l

So

154l = 38,808

>> l = 252 cm

Answer : The length of the wire is 252 cm.

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Check these out :

$\boxed{\begin{minipage}{6.2 cm}\bigstar \:\underline{\textbf{Formulae Related to Cylinder :}}\\\\\sf {\textcircled{\footnotesize\textsf{1}}} \:Area\:of\:Base\:and\:top =\pi r^2 \\\\ \sf {\textcircled{\footnotesize\textsf{2}}} \:\:Curved \: Surface \: Area =2 \pi rh\\\\\sf{\textcircled{\footnotesize\textsf{3}}} \:\:Total \: Surface \: Area = 2 \pi r(h + r)\\ \\{\textcircled{\footnotesize\textsf{4}}} \: \:Volume=\pi r^2h\end{minipage}} </p><p></p><p>$ $\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}$

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