Math, asked by Anonymous, 25 days ago

504 cones, each of diameter 3.5 cm and height 3 cm are melted and recast into a metallic sphere.find the diameter of the sphere and hence find its surface area. [ use π = 22/7]​

Answers

Answered by mathdude500
36

\large\underline{\sf{Solution-}}

Given that,

Diameter of cone = 3.5 cm = 7/2 cm

So, radius of cone, r = 7/4 cm

Height of cone, h = 3 cm

We know, Volume of cone of radius r and height h is given by

\boxed{\rm{  \:Volume_{(cone)} \:  =  \:  \frac{\pi}{3} \:  {r}^{2} \: h \:  \: }} \\

So, using this result, we have

\rm \:  \:Volume_{(1 \: cone)} \:  =  \:  \frac{\pi}{3} \:  {r}^{2} \: h \:  \: \\

So,

\rm \:  \:Volume_{(504 \: cones)} \:  =  \:  504 \times \frac{\pi}{3} \:  {r}^{2} \: h \:  \: \\

\rm \:  \:Volume_{(504 \: cones)} \:  =  \:  168 \:\pi \:   {r}^{2} \: h \: -  -  - (1)  \: \\

Now, it is given that,

504 cones, each of diameter 3.5 cm and height 3 cm are melted and recast into a metallic sphere.

Let assume that radius of sphere be R cm.

So, as 504 cones melted and recast in to a metallic sphere.

\rm \: Volume_{(504 \: cones)} = Volume_{(sphere)} \\

\rm \: 168\pi \:  {r}^{2}h \:  =  \:  \dfrac{4}{3}\pi \:  {R}^{3}  \\

\rm \: 42\pi \:  {r}^{2}h \:  =  \:  \dfrac{1}{3}\pi \:  {R}^{3}  \\

\rm \:3 \times  42\:  \times  \frac{7}{4} \times  \frac{7}{4} \times 3 \:  =  \:   {R}^{3}  \\

\rm \:3 \times  21\:  \times  \frac{7}{2} \times  \frac{7}{4} \times 3 \:  =  \:   {R}^{3}  \\

\rm \:3 \times  7 \times 3\:  \times  \frac{7}{2} \times  \frac{7}{2 \times 2} \times 3 \:  =  \:   {R}^{3}  \\

\rm \: {\bigg(\dfrac{21}{2} \bigg) }^{3}  \:  =  \:   {R}^{3}  \\

\rm\implies \:R =  \dfrac{21}{2} \: cm  \\

Hence,

\rm\implies \:Diameter \: of \: sphere \:  =  \: 21 \: cm \\

Now,

\rm \: Surface \: Area_{(sphere)} \\

\rm \: =  \:4 \: \pi \:  {r}^{2}  \\

\rm \: =  \:4 \times \dfrac{22}{7}  \times \dfrac{21}{2}  \times \dfrac{21}{2}  \\

\rm \: =  \:22 \times 3 \times 21 \\

\rm \: =  \:1386 \:  {cm}^{2}  \\

\rm\implies \:\rm \: Surface \: Area_{(sphere)}  \:  =  \: 1386 \:  {cm}^{2} \\

\rule{190pt}{2pt}

Additional Information

\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r  \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} =  \dfrac{4}{3}\pi {r}^{3}  }\\ \\ \bigstar \: \bf{Volume_{(cube)} =  {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}

Answered by PopularStar
101

Solution:-

Volume of 504 cone=Volume of Sphere

⇢504× \sf \dfrac{1}{3}× \sf \dfrac{22}{7}× \sf \dfrac{3.5}{2}× \sf \dfrac{3.5}{2}×3= \sf \dfrac{4}{3}π \sf{r^2}

 \sf{r^3} =1157.625

 \sf \pink{r=10.5 \ cm}

Diameter of Sphere=2×r= \sf{4π \ r^2}

=4× \sf \dfrac{22}{7}×10.5

 \sf \pink{1386 \ cm^2}

More formulas...

Surface area of cuboid=2(lb+bh+hl)

Surface area of cube= \sf \pink{6 \ a^2}

Curved Surface Area of a cylinder= \sf \pink{2π \ rh}

Total surface area of cylinder= \sf \pink{2πr(r+h)}

Curved surface area of cone= \sf \dfrac{1}{2}×l× \sf \pink{2πr}= \sf \pink{πrl}

Total surface area of cone= \sf \pink{πrl+π \ r^2}=πr(l+r)

Surface area of sphere= \sf \pink{4π \ r^2}

Volume of cube= \sf \pink{a^3}

Volume of Hemisphere= \sf \dfrac{2}{3} \sf \pink{π \ r^3}

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