Question

The ratio of the surface area of two spheres is 3:2. The volume of the larger sphere is 2,916 in3. What is the volume of the smaller sphere?

Answer 1

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 2

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.