The acceleration due to gravity changes from `9.8 m//s^(2)` to `9.5 m//s^(2)`. To keep the period of pendulum constant, its length must changes by

Options

A. 3 m
B. 0.3 m
C. 0.003 m
D. 3 cm

Explanation

To solve the problem, we need to determine how the length of a pendulum must change in order to keep its period constant when the acceleration due to gravity changes from \( g_1 = 9.8 \, \text{m/s}^2 \) to \( g_2 = 9.5 \, \text{m/s}^2 \).

### Step-by-Step Solution:

1. **Understand the formula for the period of a pendulum:**
The period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Set up the equations for the initial and final conditions:**
Let:
- \( T_1 \) be the initial period with \( g_1 = 9.8 \, \text{m/s}^2 \) and \( L_1 = 1 \, \text{m} \).
- \( T_2 \) be the final period with \( g_2 = 9.5 \, \text{m/s}^2 \) and \( L_2 \) being the new length we need to find.

Since we want the periods to be equal, we have:
\[ T_1 = T_2 \]
This leads to:
\[ 2\pi \sqrt{\frac{L_1}{g_1}} = 2\pi \sqrt{\frac{L_2}{g_2}} \]

3. **Cancel \( 2\pi \) from both sides:**
\[ \sqrt{\frac{L_1}{g_1}} = \sqrt{\frac{L_2}{g_2}} \]

4. **Square both sides to eliminate the square root:**
\[ \frac{L_1}{g_1} = \frac{L_2}{g_2} \]

5. **Rearrange to find \( L_2 \):**
\[ L_2 = L_1 \cdot \frac{g_2}{g_1} \]

6. **Substitute the known values:**
Given \( L_1 = 1 \, \text{m} \), \( g_1 = 9.8 \, \text{m/s}^2 \), and \( g_2 = 9.5 \, \text{m/s}^2 \):
\[ L_2 = 1 \cdot \frac{9.5}{9.8} \]

7. **Calculate \( L_2 \):**
\[ L_2 = \frac{9.5}{9.8} \approx 0.969 \, \text{m} \approx 0.97 \, \text{m} \]

8. **Find the change in length:**
The change in length \( \Delta L \) is given by:
\[ \Delta L = L_1 - L_2 \]
\[ \Delta L = 1 - 0.97 = 0.03 \, \text{m} \]

9. **Convert the change in length to centimeters:**
\[ \Delta L = 0.03 \, \text{m} = 3 \, \text{cm} \]

### Final Answer:
The length of the pendulum must change by **3 cm**.