Question

A cylindrical bucket of 32 cm height and 18 cm base radius is full of sand. This bucket is emptied on the ground and a conical mound of sand is formed. If the height of this mound is 24 cm, then find its radius and slant height.

Hint:

When sand is transferred from a cylinder to a cone, the volumes of both shapes remain equal. Use this relationship to solve for unknowns.

Answer 1

Let the radius and slant height of the conical mound be \( r \) and \( l \), respectively.
Step 1: The volume of the cylindrical bucket is given by the formula: \[ V_{\text{cylinder}} = \pi r_{\text{cylinder}}^2 h_{\text{cylinder}}, \] where \( r_{\text{cylinder}} = 18 \, \text{cm} \) and \( h_{\text{cylinder}} = 32 \, \text{cm} \). \[ V_{\text{cylinder}} = \pi \times (18)^2 \times 32 = \pi \times 324 \times 32 = 10368 \pi \, \text{cm}^3. \]
Step 2: The volume of the conical mound is given by: \[ V_{\text{cone}} = \frac{1}{3} \pi r_{\text{cone}}^2 h_{\text{cone}}, \] where \( h_{\text{cone}} = 24 \, \text{cm} \). Since the volume of the sand remains the same, we equate the volumes: \[ 10368 \pi = \frac{1}{3} \pi r_{\text{cone}}^2 \times 24. \] Cancel \( \pi \) from both sides: \[ 10368 = \frac{1}{3} r_{\text{cone}}^2 \times 24. \] Simplify: \[ 10368 = 8 r_{\text{cone}}^2 \quad \Rightarrow \quad r_{\text{cone}}^2 = \frac{10368}{8} = 1296 \quad \Rightarrow \quad r_{\text{cone}} = \sqrt{1296} = 36 \, \text{cm}. \]
Step 3: Now, use the Pythagorean theorem to find the slant height \( l \) of the cone. \[ l = \sqrt{r_{\text{cone}}^2 + h_{\text{cone}}^2} = \sqrt{36^2 + 24^2} = \sqrt{1296 + 576} = \sqrt{1872}. \] Simplify: \[ l = \sqrt{1872} \approx 43.27 \, \text{cm}. \]
Conclusion: The radius of the conical mound is \( 36 \, \text{cm} \) and the slant height is approximately \( 43.27 \, \text{cm} \).