A seconds pendulum at the equator `(g = 9.78 ms^(-2))` is taken to the poles `(g = 9.83 ms^(-2))`. If it is to act as second's pendulum its length should be

Options

A. increased by nearly 0.5 cm
B. decreased by nearly 0.5 cm
C. increased by nearly 0.05 cm
D. increased by nearly 1 cm

Explanation

To solve the problem, we need to determine the new length of a seconds pendulum when it is moved from the equator to the poles, where the acceleration due to gravity (g) changes.

### Step-by-Step Solution:

1. **Understanding the Seconds Pendulum:**
A seconds pendulum is defined as a pendulum that has a time period of 2 seconds. The formula for the time period (T) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity.

2. **Setting Up the Initial Condition:**
At the equator, we have:
- \(g_{eq} = 9.78 \, \text{m/s}^2\)
- For a seconds pendulum, \(T = 2 \, \text{s}\).

Rearranging the formula for the time period to find the length \(L_{eq}\):
\[ 2 = 2\pi \sqrt{\frac{L_{eq}}{9.78}} \]
Dividing both sides by \(2\):
\[ 1 = \pi \sqrt{\frac{L_{eq}}{9.78}} \]
Squaring both sides:
\[ 1 = \pi^2 \frac{L_{eq}}{9.78} \]
Thus,
\[ L_{eq} = \frac{9.78}{\pi^2} \]

3. **Calculating the Length at the Poles:**
At the poles, the value of \(g\) changes to:
- \(g_{pole} = 9.83 \, \text{m/s}^2\)

For the pendulum to still act as a seconds pendulum at the poles, we need to find the new length \(L_{pole}\):
\[ 2 = 2\pi \sqrt{\frac{L_{pole}}{9.83}} \]
Following the same steps as before:
\[ 1 = \pi \sqrt{\frac{L_{pole}}{9.83}} \]
Squaring both sides:
\[ 1 = \pi^2 \frac{L_{pole}}{9.83} \]
Thus,
\[ L_{pole} = \frac{9.83}{\pi^2} \]

4. **Finding the Change in Length:**
Now, we can find the change in length \(\Delta L\):
\[ \Delta L = L_{pole} - L_{eq} = \frac{9.83}{\pi^2} - \frac{9.78}{\pi^2} \]
Simplifying this:
\[ \Delta L = \frac{9.83 - 9.78}{\pi^2} = \frac{0.05}{\pi^2} \]

5. **Calculating the Numerical Value:**
Using \(\pi \approx 3.14\):
\[ \pi^2 \approx 9.86 \]
Therefore,
\[ \Delta L \approx \frac{0.05}{9.86} \approx 0.00507 \, \text{m} \approx 0.5 \, \text{cm} \]

### Final Answer:
The length of the seconds pendulum at the poles should be increased by approximately **0.5 cm**.