Answer:
The volume of the smaller sphere is 864 in³.
Step-by-step explanation:
Let \(r_1\) are \(r_2\) are the radius of the spheres,
Thus, the surface area of the first sphere,
\(A_1=4\pi (r_1)^2\)
And, the surface area of the second sphere,
\(A_2=4\pi (r_2)^2\)
According to the question,
\(\frac{A_1}{A_2}=\frac{3}{2}\)
\(\frac{4\pi (r_1)^2}{4\pi (r_2)^2}=\frac{3}{2}\)
\(\implies \frac{r_1}{r_2}=\frac{3}{2}-------(1)\)
Now,
The volume of first sphere,
\(V_1=\frac{4}{3}\pi (r_1)^3\)
And, the volume of second sphere,
\(V_2=\frac{4}{3}\pi (r_2)^3\)
\(\implies \frac{V_1}{V_2}=\frac{\frac{4}{3}\pi (r_1)^3}{\frac{4}{3}\pi (r_2)^3}
[tex]=(\frac{r_1}{r_2})^3\)
From equation (1),
\(\frac{V_1}{V_2}=\frac{27}{8}\)
Given,
\(V_1=2,916\text{ cube inches}\)
\(\implies \frac{2,916}{V_2}=\frac{27}{8}\)
[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.
