To solve the problem, we are given the volume of a cone and need to find the volume of a sphere with specific relationships between their dimensions.
-
The volume \( V \) of a cone is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \] where \( r \) is the radius and \( h \) is the height of the cone.
-
We know the volume of the cone is \( \frac{25}{3} \pi \) cm³. Therefore, we can set up the equation: \[ \frac{1}{3} \pi r^2 h = \frac{25}{3} \pi \] Cancelling \( \pi \) from both sides: \[ \frac{1}{3} r^2 h = \frac{25}{3} \] Multiplying both sides by 3 gives: \[ r^2 h = 25 \]
-
We also know that the height of the cone \( h \) is equal to the diameter of the sphere. The diameter of the sphere is \( 2r \), so: \[ h = 2r \]
-
Now, substitute \( h = 2r \) into the equation \( r^2 h = 25 \): \[ r^2 (2r) = 25 \] Simplifying gives: \[ 2r^3 = 25 \] Therefore: \[ r^3 = \frac{25}{2} \]
-
Now we can find the volume \( V_s \) of the sphere using the formula: \[ V_s = \frac{4}{3} \pi r^3 \] Substituting \( r^3 = \frac{25}{2} \): \[ V_s = \frac{4}{3} \pi \left( \frac{25}{2} \right) \] Simplifying: \[ V_s = \frac{4 \cdot 25}{3 \cdot 2} \pi = \frac{100}{6} \pi = \frac{50}{3} \pi \]
Thus, the volume of the sphere is: \[ \boxed{\frac{50}{3} \pi \text{ cm}^3} \]