By melting a metallic solid sphere with radius 9 cm , some cones are recasted . If the radius and height of cones recasted are 3 cm and 6 cm

Question

By melting a metallic solid sphere with radius 9 cm , some cones are recasted . If the radius and height of cones recasted are 3 cm and 6 cm respectively , then find the number of cones recasted .​

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Isabelle 2 weeks 2021-10-01T13:50:50+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-01T13:52:14+00:00

    \huge\sf\mathbb\color{white} \underline{\colorbox{black}{☯SoLuTiOn☯}}

    Step-by-step explanation:

    Radii  \: of  \: three  \: iron \:  balls  \: are  \: given \:   r_{1} = 6cm \: , r_{3} = 8cm,  r_{3} = 10cm

    ∴  \: Volume \: of \: first \: ball \:   V_{1} =  \frac{4}{3} \pi \:  {r_{1}}^{3}  =  \frac{4}{3} \pi. {6}^{3}  \\

    Volume \: of \: second \: ball \:  V_{2} =  \frac{4}{3}  {\pi \:  r_{2} }^{3}  =  \frac{4}{3} \pi. {8}^{3}  \\

    Volume \: of \: third \: ball \:  V_{3}  =  \frac{4}{3} \pi \:  { r_{3} }^{3}  =  \frac{4}{3} \pi. {(10)}^{3}  \\

    If  \: the \:  radius  \: of \:  a \:  new  \: sphere  \: made  \: is  \:  r and  \: volume \:  V,  \: then \:  V =  \frac{4}{3} \pi {r}^{3}  \\

    Since,  \: sphere \:   is  \: made  \: by  \: melting  \: the \:  three \:  balls.

    Hence, \:  Volume \:  of \:  sphere =  \: Sun \:  of \:  the  \: volume  \: of \:  the \:  balls \\

    ⇒ \:  \frac{4}{3} \pi {r}^{3}  =  \frac{4}{3} \pi. {6}^{3}  +  \frac{4}{3} \pi {8}^{3}  +  \frac{4}{3} \pi {(10)}^{3}  \\

    ⇒ \:  \:  {r}^{3}  =  {6}^{3}  +  {8}^{3}  +  {10}^{3}  \\  = 216 + 512 + 1000 = 1728

    ∴ \:  \:  \:  \: r \sqrt[3]{1728}  = 12cm

    So, \:  Radius \:  of  \: new  \: sphere \:  is \:  12 cm.  \\

    0
    2021-10-01T13:52:20+00:00

    Answer:

    bro just find the volume of each cone and divide the volume of sphere by it

    you’ll get ur answer…

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