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 .​

Answer 1

Answer:

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

you'll get ur answer...

Answer 2

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

Step-by-step explanation:

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

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

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

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

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

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

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

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

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

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

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