A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. if the lift accelerates upwards with an acceleration g/4, then the period of the pendulum will be

Options

A. `T`
B. `T/4`
C. `(2T)/(sqrt(5))`
D. `2Tsqrt(5)`

Explanation

To solve the problem, we need to determine the new period of a simple pendulum when the lift it is in accelerates upwards with an acceleration of \( g/4 \).

### Step-by-Step Solution:

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

2. **Considering the Lift's Acceleration**:
When the lift accelerates upwards with an acceleration of \( g/4 \), the effective acceleration due to gravity that the pendulum experiences changes. The effective gravitational acceleration \( g' \) can be calculated as:
\[ g' = g + \frac{g}{4} = g + 0.25g = \frac{5g}{4} \]

3. **Calculating the New Period**:
The new period \( T' \) of the pendulum in the accelerating lift can be expressed using the modified effective gravitational acceleration \( g' \):
\[ T' = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( g' = \frac{5g}{4} \):
\[ T' = 2\pi \sqrt{\frac{L}{\frac{5g}{4}}} = 2\pi \sqrt{\frac{4L}{5g}} \]

4. **Relating to the Original Period**:
We can rewrite the expression for \( T' \) in terms of the original period \( T \):
\[ T' = 2\pi \sqrt{\frac{4L}{5g}} = 2\pi \sqrt{\frac{4}{5}} \sqrt{\frac{L}{g}} = \sqrt{\frac{4}{5}} \cdot T \]
Since \( \sqrt{4} = 2 \), we have:
\[ T' = \frac{2}{\sqrt{5}} T \]

5. **Final Result**:
Therefore, the new period of the pendulum when the lift accelerates upwards with an acceleration of \( g/4 \) is:
\[ T' = \frac{2}{\sqrt{5}} T \]

### Summary:
The new period of the pendulum is \( \frac{2}{\sqrt{5}} T \).