If the radius of cylinder is doubled, but height is reduced by 50% . What is the percentage change in volume ?

Answer

To solve the problem of finding the percentage change in volume when the radius of a cylinder is doubled and the height is reduced by 50%, we can follow these steps:

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

### Step 2: Determine the new dimensions of the cylinder.
- The radius is doubled:
\[ \text{New radius} = 2r \]
- The height is reduced by 50%:
\[ \text{New height} = \frac{h}{2} \]

### Step 3: Calculate the new volume of the cylinder.
Using the new dimensions, the new volume \( V' \) is:
\[ V' = \pi (2r)^2 \left(\frac{h}{2}\right) \]
Calculating this gives:
\[ V' = \pi (4r^2) \left(\frac{h}{2}\right) = \pi \cdot 4r^2 \cdot \frac{h}{2} = \pi \cdot 2r^2 h \]

### Step 4: Compare the new volume with the original volume.
The original volume was:
\[ V = \pi r^2 h \]
The new volume is:
\[ V' = 2\pi r^2 h \]

### Step 5: Calculate the change in volume.
The change in volume \( \Delta V \) is:
\[ \Delta V = V' - V = 2\pi r^2 h - \pi r^2 h = \pi r^2 h \]

### Step 6: Calculate the percentage change in volume.
The percentage change in volume can be calculated using the formula:
\[ \text{Percentage Change} = \left(\frac{\Delta V}{V}\right) \times 100\% \]
Substituting the values we have:
\[ \text{Percentage Change} = \left(\frac{\pi r^2 h}{\pi r^2 h}\right) \times 100\% = 100\% \]

### Final Answer:
The percentage change in volume is **100%**.

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