Question

can anyone give answer​

Answer 1

Given

• Radius of Three Metallic Spheres are 3 cm , 4 cm and 5 cm

To Find

• Radius of New Shape who makes by melting those there sphare

we know that

$\pink{\boxed{\bf{Volume\ of \ Sphare = \dfrac{4}{3}πr³}}}$

Valume of Sphare whose radius is 3

$\small\sf{~~~~~:\implies.Volume= \dfrac{4}{3} \times \dfrac{22}{7} \times 3³}$

$\small\sf{~~~~~:\implies Volume= \dfrac{88}{21} \times 27}$

$\small\sf{~~~~~:\implies Volume= 4.2 \times27}$

$\small\sf{~~~~~:\implies Volume= 112.4}$

• Valume of Sphare whose radius is 3= 112.4cm³

Valume of Sphare whose radius is 4

$\small \sf{~~~~~:\implies Volume= \dfrac{4}{3} \times \dfrac{22}{7} \times 4³}$

$\small\sf{~~~~~:\implies Volume= \dfrac{88}{21} \times 64}$

$\small\sf{~~~~~:\implies Volume= 4.2\times 64}$

$\small\sf{~~~~~:\implies Volume=268.8}$

• Valume of Sphare whose radius is 4=268.8cm³

Valume of Sphare whose radius is 5

$\small\sf{~~~~~:\implies Volume= \dfrac{4}{3} \times \dfrac{22}{7} \times 5³}$

$\small\sf{~~~~~:\implies Volume= \dfrac{88}{21} \times 125}$

$\small\sf{~~~~~:\implies Volume=4.2 \times 125}$

$\small\sf{~~~~~:\implies Volume=525}$

• Valume of Sphare whose radius is 5=525cm³

• Now it is Given that All 3 Sphare are mealted and convert into a new Sphare

Volume of new sphare = 112.4+268.8+225

Volume of new sphare = 906.8

Radius of New Sphare = ?

$\boxed{\bf{Volume\ of \: new \: Sphare = \dfrac{4}{3}πr³}}$

$\sf{~~~~~:\implies 906.8 = \dfrac{4}{3} \times \dfrac{ 22}{7} \times r³}$

$\sf{ ~~~~~:\implies r³ = \dfrac{906.8 \times 21}{88} }$

$\sf{ ~~~~~:\implies r³ = \dfrac{19030.2}{88} }$

$\sf{~~~~~:\implies r³ = 216 }$

$\sf{ ~~~~~:\implies r = \sqrt[3]{216} }$

$\sf{~~~~~:\implies r = \sqrt[3]{6 \times 6 \times 6} }$

$\sf{ ~~~~~:\implies r = \sqrt[3]{ {6}^{3}} }$

$\sf{ ~~~~~:\implies r = 6 }$

★Hance The radius of New Sphare= 6cm.

• Options C is Correct.

Learn More:

$\small{\begin{gathered}\begin{gathered}\small\begin{gathered}\bigstar \: \underline{\mathfrak{More \: Useful \: Formula} } \: \bigstar \\ \begin{gathered}{\boxed{\begin{array} {cccc}{\sf{{\leadsto TSA \: of \: cube \: = \: 6(side)^{2}}}} \\ \\ {\sf{{\leadsto LSA \: of \: cube \:= \: 4(side)^{2}}}} \\ \\{\sf{{\leadsto Volume \: of \: cube \: = \: (side)^{3}}}} \\ \\ {\sf{{\leadsto Diagonal \: of \: cube \: = \: \sqrt(l^{2} + b^{2} + h^{2})}}} \\ \\ {\sf{{\leadsto Perimeter \: of \: cube \: = \: 4(l+b+h)}}} \\ \\ {\sf{{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}} \\ \\ {\sf{{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}} \\ \\{\sf{{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}} \\ \\ {\sf{{\leadsto Diameter \: of \: circle \: = \: 2r}}} \\ \\ {\sf{{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}} \\ \\ {\sf{{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}} \\ \\ {\sf{{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}} \\ \\ {\sf{{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}} \\ \\{\sf{{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}\end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}}$