If the base radius of a cone is doubled and its height is halved , then which of the following is true regarding its volume ?

Options

A. Increases by 200 %
B. Decreases by 200 %
C. Increases by 100 %
D. Decreases by 100 %

Answer

To solve the problem, we need to understand the formula for the volume of a cone and how changes in its dimensions affect its volume.

### Step-by-Step Solution:

1. **Understand the Volume Formula:**
The volume \( V \) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
where \( r \) is the base radius and \( h \) is the height of the cone.

2. **Define the Original Dimensions:**
Let the original radius be \( r \) and the original height be \( h \). The volume of the original cone (Cone 1) can be expressed as:
\[ V_1 = \frac{1}{3} \pi r^2 h \]

3. **Change the Dimensions:**
According to the problem:
- The base radius is doubled: new radius \( r_2 = 2r \)
- The height is halved: new height \( h_2 = \frac{h}{2} \)

4. **Calculate the Volume of the New Cone:**
Now, we can find the volume of the new cone (Cone 2) using the new dimensions:
\[ V_2 = \frac{1}{3} \pi (r_2)^2 h_2 = \frac{1}{3} \pi (2r)^2 \left(\frac{h}{2}\right) \]
Simplifying this:
\[ V_2 = \frac{1}{3} \pi (4r^2) \left(\frac{h}{2}\right) = \frac{1}{3} \pi \cdot 4r^2 \cdot \frac{h}{2} \]
\[ V_2 = \frac{1}{3} \pi \cdot 2r^2 h = 2 \left(\frac{1}{3} \pi r^2 h\right) = 2V_1 \]

5. **Conclusion:**
The volume of the new cone (Cone 2) is twice the volume of the original cone (Cone 1). Therefore, we can conclude that:
\[ V_2 = 2V_1 \]

### Final Answer:
The volume of the cone doubles when the base radius is doubled and the height is halved.