A simple pendulum has time period (T_1). The point of suspension is now moved upward according to the relation `y = K t^2, (K = 1 m//s^2)` where (y) is the vertical displacement. The time period now becomes (T_2). The ratio of `(T_1^2)/(T_2^2)` is `(g = 10 m//s^2)`.

Options

A. `6//5`
B. `5//6`
C. 1
D. `4//5`

Answer

To solve the problem, we need to analyze the changes in the time period of a simple pendulum when the point of suspension is moved upward according to the relation \( y = K t^2 \), where \( K = 1 \, \text{m/s}^2 \).

### Step-by-Step Solution:

1. **Understanding the Time Period of a Simple 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. **Initial Condition**:
Let the initial time period be \( T_1 \) when the pendulum is at rest and the point of suspension is stationary. Thus, we have:
\[ T_1 = 2\pi \sqrt{\frac{L}{g}} \]
where \( g = 10 \, \text{m/s}^2 \).

3. **Acceleration of the Suspension Point**:
The point of suspension is moved upward according to the relation \( y = K t^2 \). Given \( K = 1 \, \text{m/s}^2 \), we can differentiate to find the velocity and acceleration:
- Velocity \( v = \frac{dy}{dt} = 2Kt = 2t \)
- Acceleration \( a = \frac{d^2y}{dt^2} = 2K = 2 \, \text{m/s}^2 \)

4. **Effective Gravity**:
When the suspension point accelerates upwards, the effective acceleration due to gravity \( g' \) acting on the pendulum changes. The new effective gravity is given by:
\[ g' = g + a = 10 + 2 = 12 \, \text{m/s}^2 \]

5. **New Time Period**:
The new time period \( T_2 \) with the effective gravity \( g' \) is:
\[ T_2 = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{12}} \]

6. **Finding the Ratio of Time Periods**:
We need to find the ratio \( \frac{T_1^2}{T_2^2} \):
\[ \frac{T_1^2}{T_2^2} = \frac{(2\pi \sqrt{\frac{L}{g}})^2}{(2\pi \sqrt{\frac{L}{g'}})^2} = \frac{\frac{L}{g}}{\frac{L}{g'}} = \frac{g'}{g} \]
Substituting the values of \( g' \) and \( g \):
\[ \frac{T_1^2}{T_2^2} = \frac{12}{10} = \frac{6}{5} \]

### Final Answer:
The ratio \( \frac{T_1^2}{T_2^2} \) is \( \frac{6}{5} \).