Question

the volume of the sphere is 4851cm cube, find its diameter​

Answer 1

Answer :

• Diameter of sphere is 21cm

Given :

• Volume of the sphere is 4851cm³

To find :

• Diameter

Solution :

As we know that :

• Volume of sphere = 4/3 πr³

Given Volume of sphere is 4851cm

• As we know that sphere is 3 dimensional object which is perfectly round in shape

⟶ 4/3 πr² = 4851

⟶ r³ = 4851 × 3 /4× π

⟶ r³ = 14553/12. 564

⟶ r³ = 1158.309

⟶ r = √1158.309

⟶ r = 10.5

Now we have to find the diameter

• Diameter of sphere = 2 × radius = 2 × 10.5 = 21cm

Hence , Diameter of sphere is 21cm

Answer 2

AnswEr-:

• $\underline {\boxed{\dag{\mathrm {\blue{Diameter \:of\:Sphere \:-: 21\:cm}}}}}$

Explanation-:

• $\mathrm { Given-:}$

• The volume of sphere -: 4851cm³

• $\mathrm{ To\:Find-:}$

• The Diameter of sphere .

$\dag{\mathcal{Solution \:of\:Question -:}}$

• $\underbrace {\mathrm { Understanding \:The\:Concept \:-:}}$

• We have to find the Diameter of Sphere when Volume of sphere is given .

• For this we have to put the given volume of sphere in the Formula for Volume of Sphere .By doing this

• We can get the Diameter of Circle.

As , We know that ,

• $\underline {\boxed{\dag{\mathrm {\red{Volume \:of\:Sphere \:-:\dfrac{4}{3} \times \pi \times Radius^{3}\: cubic.units }}}}}$

• $\mathrm { Here-:}$

• $\longrightarrow {\mathrm {Volume =4851 cm^{3}}}$

• $\pi=3.14\:or\:\dfrac{22}{7}$

Now , By Putting Known Values-:

• $\longrightarrow {\mathrm { 4851cm^{3} = \dfrac{4}{3} \times \dfrac{22}{7} \times Radius^{3}}}$

• $\longrightarrow {\mathrm { 4851cm^{3} = \dfrac{4}{3} \times 3.14 \times Radius^{3}}}$

• $\longrightarrow {\mathrm { \dfrac{4851}{3.14} = \dfrac{4}{3} \times Radius^{3}}}$

• $\longrightarrow {\mathrm { \dfrac{4851 \times 3 }{3.14\times 4 } = Radius^{3}}}$

Or ,

• $\longrightarrow {\mathrm { Radius ^{3} = \dfrac{4851 \times 3 }{3.14\times 4 } = }}$

• $\longrightarrow {\mathrm {Radius ^{3} = \dfrac{14553 }{12.56 } }}$

• $\longrightarrow {\mathrm {Radius ^{3} = \dfrac{\cancel {14553} }{\cancel{12.56} } }}$

• $\longrightarrow {\mathrm { Radius ^{3} = 1158.678 }}$

• $\longrightarrow {\mathrm {Radius = \sqrt[3]{1158.678} }}$

• $\longrightarrow {\mathrm { Radius = 10.5 cm }}$

Therefore ,

• $\underline {\boxed{\dag{\mathrm {\blue{Radius \:of\:Sphere \:-: 10.5 \:cm}}}}}$

As , We know that ,

• $\underline {\boxed{\dag{\mathrm {\red{Diameter \:of\:Circle \:-: Radius \times 2 }}}}}$

$\mathrm { Here-:}$

• $\longrightarrow {\mathrm { Radius = 10.5 cm }}$

Now , By Putting known Values-:

• $\longrightarrow {\mathrm { Diameter =10.5 \times 2 }}$

• $\longrightarrow {\mathrm { Diameter =21cm }}$

Hence,

• $\underline {\boxed{\dag{\mathrm {\blue{Diameter \:of\:Sphere \:-: 21 \:cm}}}}}$

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