Let's first identify the relationships between the cone's dimensions and the sphere's dimensions.
The formula for the volume of a cone is: \[ V_{cone} = \frac{1}{3} \pi r^2 h \] where \( r \) is the radius and \( h \) is the height.
Given that the volume of the cone is \( \frac{25}{3} \pi , \text{cm}^3 \), we can set the expression for the cone's volume equal to this value: \[ \frac{1}{3} \pi r^2 h = \frac{25}{3} \pi \]
Since we can eliminate \( \pi \) from both sides (assuming \( \pi \neq 0 \)), we have: \[ \frac{1}{3} r^2 h = \frac{25}{3} \]
Multiplying both sides by 3 gives us: \[ r^2 h = 25 \]
Next, we need to determine the volume of the sphere using the formula: \[ V_{sphere} = \frac{4}{3} \pi r^3 \]
From the problem, we know that the height \( h \) of the cone is equal to the diameter of the sphere. Since the diameter of the sphere is \( 2r \), we have: \[ h = 2r \]
Substituting \( h \) into our earlier equation \( r^2 h = 25 \): \[ r^2 (2r) = 25 \] \[ 2r^3 = 25 \]
Now, we can solve for \( r^3 \): \[ r^3 = \frac{25}{2} \]
We can now substitute \( r^3 \) into the volume formula for the sphere: \[ V_{sphere} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left(\frac{25}{2}\right) \] \[ = \frac{4 \times 25 \pi}{6} = \frac{100 \pi}{6} = \frac{50 \pi}{3} \]
Thus, the volume of the sphere is: \[ \frac{50}{3} \pi , \text{cm}^3 \]
The correct answer is: Start Fraction 50 over 3 End Fraction pi cm cubed