Question

Metallic sphere of radius 7 CM is kept in a metallic cubical tin of length equal to diameter of sphere how much air space is left in the cube
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Answer 1

${\pmb{\underline{\sf{ Required \ Solution ... }}}}$

As We know that Metallic Sphere kept in a metallic cubical tin of length equal to diameter of sphere.

$\colon\longrightarrow{\sf{Radius \ of \ Sphere = 7 \ CM }}$

Now, It's time to find the diameter of the Sphere.

$\colon\longrightarrow{\sf{ D = 2r }} \\ \\ \colon\longrightarrow{\sf{ D = 2 \times 7 }} \\ \\ \colon\longrightarrow{\sf{ D = 14 _{(CM)} }}$

So, We also Know that Length of cubical tin is Equal to the diameter of the Sphere.

>> Length (cube) = 14 CM

${\pmb{\underline{\sf\gray{Calculation... }}}}$

Firstly, We can Find the Volume of the sphere as by using the Radius of it.

$\circ \ {\underline{\boxed{\sf\orange{ Volume_{(Sphere)} = \dfrac{4}{3} πr^3 }}}}$

Now, It's time to Find the volume of the Sphere using Radius of the Sphere as.

$\colon\implies{\sf{ \dfrac{4}{3} π (7)^3 }} \\ \\ \colon\implies{\sf{ \dfrac{4}{3} \times \dfrac{22}{ \cancel{7} } \times 7 \times 7 \times \cancel{7} }} \\ \\ \colon\implies{\sf{ 1401.4 \ CM^3 _{(Volume)} }}$

At this time but not Last, we have to find the Volume of the Metallic Cubical tin using length of it.

$\circ \ {\underline{\boxed{\sf\green{ Volume_{(Cube)} = a^3 }}}}$

As by Applying above mentioned formula to find the volume of the cube as.

$\colon\implies{\sf{ a^3 }} \\ \\ \colon\implies{\sf{ (14)^3 }} \\ \\ \colon\implies{\sf{ 14 \times 14 \times 14 }} \\ \\ \colon\implies{\sf{ 2744 \ CM^3 _{(volume)} }}$

So Finally, Using got Data's we can Subtract Volume of sphere from the volume of cube to find the left space in Metallic Cubical tin.

$\colon\longrightarrow{\sf{ Space_{(Left)} = 2744-1401.4 }} \\ \\ \colon\longrightarrow{\sf{ 1342.6 \ CM^3 }}$

Hence,

We can conclude that Metallic Cubical tin have 1342.6 cm³ space left inside it.

Answer 2

Answer:

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