The height of a circular cylinder is increased six times and the base area is decreased to one-ninth of its value. The factor by which of lateral surface of the cylinder increases is

Options

A. 2
B. `(1)/(2)`
C. `(2)/(3)`
D. `(3)/(2)`

Answer

To solve the problem, we need to analyze how the lateral surface area of a circular cylinder changes when the height is increased and the base area is decreased.

### Step-by-Step Solution:

1. **Understand the formulas**:
- The lateral surface area (LSA) of a cylinder is given by the formula:
\[ \text{LSA} = 2\pi rh \]
where \( r \) is the radius and \( h \) is the height of the cylinder.

2. **Initial dimensions**:
- Let the initial radius of the cylinder be \( r \) and the initial height be \( h \).
- Therefore, the initial lateral surface area (LSA) is:
\[ \text{LSA}_{\text{initial}} = 2\pi rh \]

3. **Changes in dimensions**:
- The height of the cylinder is increased six times:
\[ h_{\text{new}} = 6h \]
- The base area is decreased to one-ninth of its original value. The base area \( A \) is given by:
\[ A = \pi r^2 \]
- If the base area is decreased to one-ninth, we have:
\[ \pi r_{\text{new}}^2 = \frac{1}{9} \pi r^2 \]
- From this, we can find the new radius:
\[ r_{\text{new}}^2 = \frac{1}{9} r^2 \implies r_{\text{new}} = \frac{r}{3} \]

4. **Calculate the new lateral surface area**:
- Now, substituting the new values of \( r \) and \( h \) into the LSA formula:
\[ \text{LSA}_{\text{new}} = 2\pi r_{\text{new}} h_{\text{new}} = 2\pi \left(\frac{r}{3}\right)(6h) \]
- Simplifying this:
\[ \text{LSA}_{\text{new}} = 2\pi \left(\frac{r}{3}\right)(6h) = 2\pi r h \cdot 2 = \frac{2\pi rh}{3} \cdot 6 = 4\pi rh \]

5. **Finding the factor of increase**:
- Now, we compare the new lateral surface area with the initial lateral surface area:
\[ \text{Factor of increase} = \frac{\text{LSA}_{\text{new}}}{\text{LSA}_{\text{initial}}} = \frac{4\pi rh}{2\pi rh} = 2 \]

### Conclusion:
The factor by which the lateral surface area of the cylinder increases is **2**.