Question

The ratio of the volumes of two spheres is 8:27. If the surface area of the smaller sphere is 64pi what is the diameter of the larger sphere ?

Answer 1

Final answer: Diameter of larger sphere = 12 units    

Given that: We are given the ratio of the volumes of two spheres is 8:27 and the surface area of the smaller sphere is 64$\pi$.

To find: We have to find the diameter of the larger sphere.

Explanation:

• Let r₁ and r₂ be the radii of the given two spheres.

• V₁ be the volume and S₁ be the surface area of the sphere of radius r₁.

• V₂ be the volume of the sphere of radius r₂.

        $V_{1} = \frac{4}{3} \pi r_{1} ^{3}\\\\V_{2} = \frac{4}{3} \pi r_{2} ^{3}$

• Given that the ratio of the volumes of two spheres = 8 : 27

        $\frac{V_{1} }{V_{2}} = \frac{8}{27} \\\\\frac{V_{1} }{V_{2}} = \frac{ \frac{4}{3} \pi r_{1} ^{3}}{\frac{4}{3} \pi r_{2} ^{3}} = \frac{8}{27}$

        $\frac{r_{1}^{3}}{ r_{2} ^{3}} = \frac{8}{27}$

        $\frac{r_{1}}{r_{2}} = (\frac{8}{27})^{\frac{1}{3}}$

        $\frac{r_{1} }{r_{2}}=\frac{2}{3}$

• Surface area of the smaller sphere = 64$\pi$

        $S_{1} = 64\pi \\\\S_{1} = 4\pi r_{1} ^{2} = 64\pi \\\\r_{1} ^{2} = \frac{64\pi }{4\pi } = 16\\\\r_{1} = \sqrt{16} =4$

• r₁ = 4

• We find that the ratio of r₁ and r₂,

         $\frac{r_{1} }{r_{2}} = \frac{2}{3}$

Substitute r₁ = 4:

       $\frac{4}{r_{2}} = \frac{2}{3} \\\\r_{2} = \frac{4}{(\frac{2}{3})} = \frac{4 * 3}{2} = \frac{12}{2} \\\\r_{2} = 6$

Here not given the unit of radius, so we take the unit of radius as units.

• Radius of larger sphere = 6 unit

• Hence,

$Diameter of larger sphere = 2 * Radius of the sphere = 2*6 =12 units$  

• Diameter of larger sphere = 12 units

To know more about the concept please go through the links

https://brainly.in/question/1116370

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