Question

$\large\underline\mathfrak\red{Question \: :- }$
Some plastic balls of radius 1 cm were melted and cast into a tube. The thickness, length and outer radius of the tube were 2 cm, 90 cm and 30 cm respectively. How many balls were melted to make the tube ?

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Answer 1

Given that

• Radius of plastic balls= 1cm
• Thickness of tube=2cm
• Length of tube(h)=90
• Outer Radius of tube (R)=30cm

To Find

• How many balls were melted to make the tube?

Formula used

${\underline{\boxed{ { \bf{ \: Volume~ of~ plastic \: balls= \green{ \cfrac{4}{3} \pi {r}^{3} \: \: }}}}}}$

${\underline{\boxed{ { \bf{ \: Volume~ of~ tube(hollow \: cylinder)= \green{ πh{ ({R}^{2} - {r}^{2} ) \: \: }}}}}}}$

Solution

${\sf Volume~ of~ plastic \: balls= { \cfrac{4}{3}\pi {r}^{3} \: \: }}$

${\sf Volume~ of~ plastic \: balls= { \cfrac{4}{3}\pi \times {1}^{3} \: \: }}$

${\sf Volume~ of~ plastic \: balls= { \cfrac{4}{3}\pi \: \: }}$

Now

• Thickness of tube=2cm
• Length of tube(h)=90
• Outer Radius of tube (R)=30cm

$\\\text{\sf Inner Radius= Outer Radius - Thickness of tube}$

$\sf{Inner Radius= 30-2}$

$\sf{Inner Radius= 28cm}$

${ \sf \: Volume~ of~ tube(hollow \: cylinder)= πh{ ({R}^{2} - {r}^{2} ) \: \: }}$

${ \sf \: Volume~ of~ tube(hollow \: cylinder)= π \times 90 \times { ({30}^{2} - {28}^{2} ) \: \: }}$

${ \sf \: Volume~ of~ tube(hollow \: cylinder)= π \times 90 \times { (900 - 784) \: \: }}$

${ \sf \: Volume~ of~ tube(hollow \: cylinder)= 90 \pi\times116\: \: }$

${ \sf \: Volume~ of~ tube(hollow \: cylinder)= 10440\pi\: \: }$

${\sf Number \: of \: balls \: were \: melted \: to \: make \: the \: tube= \dfrac{Volume~ of~ tube(hollow \: cylinder)}{Volume~ of~ plastic \: balls}}$

${\sf Number \: of \: balls \: were \: melted \: to \: make \: the \: tube= \dfrac{10440\pi}{ \dfrac{4}{3} \pi}}$

${\sf Number \: of \: balls \: were \: melted \: to \: make \: the \: tube= \dfrac{10440 \times 3}{ 4}}$

${\sf Number \: of \: balls \: were \: melted \: to \: make \: the \: tube= \dfrac{31320}{ 4}}$

${\sf Number \: of \: balls \: were \: melted \: to \: make \: the \: tube= 7830}$

$\\ \\ \\ \text{ \bf{Number of balls were melted to make the tube }}\\\sf= \pink{ 7830}$

Figure of the Tube

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){6}}\multiput(1,0)(2,0){2}{\line(0,1){6}}\multiput(0,0)(0,6){2}{\qbezier(0,0)(2,-0.6)(4,0)}\multiput(0,0)(0,6){2}{\qbezier(0,0)(2,0.6)(4,0)}\multiput(1,0)(0,6){2}{\qbezier(0,0)(1,-0.2)(2,0)}\multiput(1,0)(0,6){2}{\qbezier(0,0)(1,0.2)(2,0)}\multiput(2,0.07)(0,0.3){20}{\line(0,1){0.2}}\multiput(2,4)(0.3,0){7}{\line(1,0){0.2}}\multiput(2,2)(-0.27,0){4}{\line(-1,0){0.2}}\put(1.4,1.5){\bf\large r}\put(3.35,3.45){\bf R}\put(1.4,3){\bf H}\end{picture}$

More Important Formula

$\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}$