A cylindrical bucket of height 36 cm and radius 21 cm is filled with sand. The bucket is emptied on the ground and a conical heap of sand is formed, the height of the heap being 12 cm. The radius of the heap at the base is :

Options

A. 63 cm
B. 53 cm
C. 56 cm
D. 66 cm

Answer

To solve the problem step by step, we will first calculate the volume of the cylindrical bucket and then equate it to the volume of the conical heap formed from the sand.

### Step 1: Calculate the volume of the cylindrical bucket
The formula for the volume of a cylinder is given by:
\[ V_{\text{cylinder}} = \pi r^2 h \]
Where:
- \( r \) = radius of the cylinder = 21 cm
- \( h \) = height of the cylinder = 36 cm

Substituting the values:
\[ V_{\text{cylinder}} = \pi (21)^2 (36) \]
\[ = \pi \times 441 \times 36 \]
\[ = 15876\pi \, \text{cm}^3 \]

### Step 2: Set up the volume of the conical heap
The formula for the volume of a cone is given by:
\[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h \]
Where:
- \( r \) = radius of the cone (which we need to find)
- \( h \) = height of the cone = 12 cm

### Step 3: Equate the volumes
Since the volume of the sand in the cylindrical bucket is equal to the volume of the conical heap, we have:
\[ V_{\text{cylinder}} = V_{\text{cone}} \]
\[ 15876\pi = \frac{1}{3} \pi r^2 (12) \]

### Step 4: Simplify the equation
We can cancel \( \pi \) from both sides:
\[ 15876 = \frac{1}{3} r^2 (12) \]
\[ 15876 = 4 r^2 \]

### Step 5: Solve for \( r^2 \)
Now, we can solve for \( r^2 \):
\[ r^2 = \frac{15876}{4} \]
\[ r^2 = 3969 \]

### Step 6: Find \( r \)
Taking the square root of both sides gives us:
\[ r = \sqrt{3969} \]
\[ r = 63 \, \text{cm} \]

### Conclusion
The radius of the heap at the base is \( 63 \, \text{cm} \).

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