A pendulum suspended from the ceiling of the train has a time period of two seconds when the train is at rest, then the time period of the pendulum, if the train accelerates 10 `m//s^(2)` will be `(g=10m//s^(2))`

Options

A. (a)increase
B. (b)decrease
C. (c)remain unaffected
D. (d)become infinite

Explanation

To solve the problem, we need to determine how the time period of a pendulum changes when the train it is suspended from accelerates.

### Step-by-Step Solution:

1. **Understanding the Time Period of a Pendulum**:
The time 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. **Given Information**:
- The time period of the pendulum when the train is at rest is \( T_0 = 2 \) seconds.
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).
- The acceleration of the train \( a = 10 \, \text{m/s}^2 \).

3. **Calculating the Length of the Pendulum**:
From the time period formula, we can express the length \( L \) of the pendulum when the train is at rest:
\[ T_0 = 2\pi \sqrt{\frac{L}{g}} \implies 2 = 2\pi \sqrt{\frac{L}{10}} \]
Dividing both sides by \( 2\pi \):
\[ \frac{1}{\pi} = \sqrt{\frac{L}{10}} \implies \left(\frac{1}{\pi}\right)^2 = \frac{L}{10} \implies L = 10 \left(\frac{1}{\pi}\right)^2 \]

4. **Effect of Train's Acceleration**:
When the train accelerates, a pseudo force acts on the pendulum. The effective gravitational acceleration \( g' \) acting on the pendulum becomes:
\[ g' = \sqrt{g^2 + a^2} \]
Substituting the values of \( g \) and \( a \):
\[ g' = \sqrt{10^2 + 10^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2} \]

5. **Calculating the New Time Period**:
Now, we can find the new time period \( T' \) of the pendulum when the train is accelerating:
\[ T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{10 \left(\frac{1}{\pi}\right)^2}{10\sqrt{2}}} \]
Simplifying:
\[ T' = 2\pi \sqrt{\frac{1}{\sqrt{2}}} = 2\pi \cdot \frac{1}{\sqrt[4]{2}} = \frac{2\pi}{\sqrt[4]{2}} = 2\pi \cdot 2^{-1/4} \]

6. **Conclusion**:
Since \( g' > g \), the time period \( T' \) will be less than the original time period \( T_0 \). Therefore, the time period of the pendulum will **decrease** when the train accelerates.

### Final Answer:
The time period of the pendulum will **decrease** when the train accelerates.