If the radius of a cylinder is doubled and the beight is reduced by `50%`, then by how much percent does the volume increase/decrease?

Options

A. `66.67%` decrease
B. `100%` increase
C. `50%` decrease
D. `75%` increase

Answer

To solve the problem, we need to determine how the volume of a cylinder changes when its radius is doubled and its height is reduced by 50%.

### Step-by-Step Solution:

1. **Understand the formula for the volume of a cylinder**:
The volume \( V \) of a cylinder is given by the formula:
\[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height of the cylinder.

2. **Define the initial dimensions**:
Let the initial radius be \( r \) and the initial height be \( h \). Therefore, the initial volume \( V_1 \) is:
\[ V_1 = \pi r^2 h \]

3. **Determine the new dimensions**:
- The radius is doubled, so the new radius \( r' \) is:
\[ r' = 2r \]
- The height is reduced by 50%, so the new height \( h' \) is:
\[ h' = \frac{h}{2} \]

4. **Calculate the new volume**:
The new volume \( V_2 \) with the new dimensions is:
\[ V_2 = \pi (r')^2 (h') = \pi (2r)^2 \left(\frac{h}{2}\right) \]
Simplifying this:
\[ V_2 = \pi (4r^2) \left(\frac{h}{2}\right) = \pi \cdot 4r^2 \cdot \frac{h}{2} = \pi \cdot 2r^2 h \]

5. **Compare the volumes**:
Now we have:
- Initial volume \( V_1 = \pi r^2 h \)
- New volume \( V_2 = 2 \pi r^2 h \)

6. **Calculate the change in volume**:
The change in volume is:
\[ \text{Change} = V_2 - V_1 = 2\pi r^2 h - \pi r^2 h = \pi r^2 h \]

7. **Calculate the percentage change**:
To find the percentage increase in volume, we use the formula:
\[ \text{Percentage Change} = \left(\frac{\text{Change}}{V_1}\right) \times 100 \]
Substituting the values:
\[ \text{Percentage Change} = \left(\frac{\pi r^2 h}{\pi r^2 h}\right) \times 100 = 100\% \]

### Conclusion:
The volume of the cylinder increases by **100%**.