Question

A sphere of diameter 12 cm is dropped into a right circular cylindrical vessel partly filled with water. If the sphere is emersed completely in the water, the water level in the cylindrical vessel rises by \( \frac{35}{9} \) cm. Find the diameter of the cylindrical vessel.

Hint:

To find the diameter of the cylindrical vessel when a sphere is dropped into it, use the volume of the sphere and equate it to the volume of the displaced water in the cylinder.

Answer 1

Let the radius of the sphere be \( r_{\text{sphere}} \). Since the diameter of the sphere is 12 cm:
\[ r_{\text{sphere}} = \frac{12}{2} = 6 \, \text{cm}. \] The volume of the sphere is given by the formula:
\[ V_{\text{sphere}} = \frac{4}{3} \pi r_{\text{sphere}}^3 = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi \times 216 = 288 \pi \, \text{cm}^3. \] Let the radius of the cylindrical vessel be \( r_{\text{cylinder}} \), and the height rise in the cylindrical vessel be \( h_{\text{rise}} = \frac{35}{9} \, \text{cm} \). The volume of the displaced water, which is the volume of the sphere, is also the volume of the cylindrical part that rises. The volume of a cylinder is given by:
\[ V_{\text{cylinder}} = \pi r_{\text{cylinder}}^2 h_{\text{rise}}. \] Setting this equal to the volume of the sphere:
\[ 288 \pi = \pi r_{\text{cylinder}}^2 \times \frac{35}{9}. \] Canceling \( \pi \) from both sides:
\[ 288 = r_{\text{cylinder}}^2 \times \frac{35}{9}. \] Multiply both sides by 9:
\[ 288 \times 9 = 35 r_{\text{cylinder}}^2 \quad \Rightarrow \quad 2592 = 35 r_{\text{cylinder}}^2. \] Solving for \( r_{\text{cylinder}}^2 \):
\[ r_{\text{cylinder}}^2 = \frac{2592}{35} = 74.06. \] Taking the square root of both sides:
\[ r_{\text{cylinder}} \approx \sqrt{74.06} \approx 8.61 \, \text{cm}. \] Thus, the diameter of the cylindrical vessel is:
\[ \text{Diameter of cylindrical vessel} = 2r_{\text{cylinder}} \approx 2 \times 8.61 = 17.22 \, \text{cm}. \]
Conclusion: The diameter of the cylindrical vessel is approximately \( 17.22 \, \text{cm} \).