To solve the problem, we first need to establish the relationship between the volumes of a cylinder, sphere, and cone. The formulas for their volumes are:
- Volume of a cylinder: \( V_{cylinder} = \pi r^2 h \)
- Volume of a sphere: \( V_{sphere} = \frac{4}{3} \pi r^3 \)
- Volume of a cone: \( V_{cone} = \frac{1}{3} \pi r^2 h \)
Given that the volume of the cylinder is \( 36 , \text{cm}^3 \) and that the height of the cylinder is equal to the diameter of the sphere, we can set up the relationships as follows:
Let \( r \) be the radius of the cylinder (and sphere) and \( h \) be the height of the cylinder. Since the diameter of the sphere is twice the radius:
- The diameter of the sphere \( = 2r \)
- The height of the cylinder \( h = 2r \)
Substituting \( h = 2r \) into the volume formula for the cylinder gives:
\[ V_{cylinder} = \pi r^2 (2r) = 2\pi r^3 \]
We know the volume of the cylinder is \( 36 , \text{cm}^3 \):
\[ 2\pi r^3 = 36 \]
Now, we can solve for \( r^3 \):
\[ \pi r^3 = 18 \quad \Rightarrow \quad r^3 = \frac{18}{\pi} \]
Now we can use this value of \( r \) to find the volume of the sphere:
\[ V_{sphere} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left(\frac{18}{\pi}\right) \]
Simplifying this gives:
\[ V_{sphere} = \frac{4}{3} \cdot 18 = \frac{72}{3} = 24 , \text{cm}^3 \]
Therefore, the volume of the sphere is \( 24 , \text{cm}^3 \).
The correct answer is \( \boxed{24 , \text{cm}^3} \).