Answer :

meerkat18
If the ratio of the radii of the two spheres is the square root of the given ratio of their surface area. That is, the ratio of the radii is sqrt 3: sqrt 2. The ratio of their volumes, on the other hand, is the cube of the ratio of their radii. That is 3^3/2 : 2^3/2. Letting x be the volume of the smaller sphere, the proportion becomes,
                           3^3/2 / 2^3/2 = 2,916 / x 
The volume of the smaller sphere is approximately 1,587.27.

Answer:

The volume of the smaller sphere is 864 in³.

Step-by-step explanation:

Let [tex]r_1[/tex] are [tex]r_2[/tex] are the radius of the spheres,

Thus, the surface area of the first sphere,

[tex]A_1=4\pi (r_1)^2[/tex]

And, the surface area of the second sphere,

[tex]A_2=4\pi (r_2)^2[/tex]

According to the question,

[tex]\frac{A_1}{A_2}=\frac{3}{2}[/tex]

[tex]\frac{4\pi (r_1)^2}{4\pi (r_2)^2}=\frac{3}{2}[/tex]

[tex]\implies \frac{r_1}{r_2}=\frac{3}{2}-------(1)[/tex]

Now,

The volume of first sphere,

[tex]V_1=\frac{4}{3}\pi (r_1)^3[/tex]

And, the volume of second sphere,

[tex]V_2=\frac{4}{3}\pi (r_2)^3[/tex]

[tex]\implies \frac{V_1}{V_2}=\frac{\frac{4}{3}\pi (r_1)^3}{\frac{4}{3}\pi (r_2)^3}

[tex]=(\frac{r_1}{r_2})^3[/tex]

From equation (1),

[tex]\frac{V_1}{V_2}=\frac{27}{8}[/tex]

Given,

[tex]V_1=2,916\text{ cube inches}[/tex]

[tex]\implies \frac{2,916}{V_2}=\frac{27}{8}[/tex]

[tex]8\times 2,916=27V_2/tex]

[tex]23328=27V_2\implies V_2=864\text{ cube in}/tex]

Hence, the volume of the smaller sphere is 864 in³.

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