Diameter and mass of a planet is double that earth. Then time period of a pendulum at surface of planet is how much times of time period at earth surface :-

Options

A. `1/sqrt(2)` times
B. `2sqrt(2)` times
C. Equal
D. None of these

Answer

To solve the problem, we need to determine how the time period of a pendulum on a new planet compares to that on Earth, given that the new planet has double the diameter and mass of Earth.

### Step-by-Step Solution:

1. **Understanding the Time Period Formula**:
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. **Acceleration due to Gravity on Earth**:
The acceleration due to gravity \( g \) on the surface of Earth is given by:
\[ g_e = \frac{G M_e}{R_e^2} \]
where \( G \) is the universal gravitational constant, \( M_e \) is the mass of Earth, and \( R_e \) is the radius of Earth.

3. **Parameters for the New Planet**:
Given that the diameter and mass of the new planet are double that of Earth:
- The radius of the new planet \( R_n = 2 R_e \)
- The mass of the new planet \( M_n = 2 M_e \)

4. **Calculating Acceleration due to Gravity on the New Planet**:
The acceleration due to gravity \( g_n \) on the new planet can be calculated as:
\[ g_n = \frac{G M_n}{R_n^2} = \frac{G (2 M_e)}{(2 R_e)^2} \]
Simplifying this gives:
\[ g_n = \frac{G (2 M_e)}{4 R_e^2} = \frac{1}{2} \cdot \frac{G M_e}{R_e^2} = \frac{1}{2} g_e \]

5. **Finding the Time Period on the New Planet**:
Using the time period formula for the new planet:
\[ T_n = 2\pi \sqrt{\frac{L}{g_n}} = 2\pi \sqrt{\frac{L}{\frac{1}{2} g_e}} = 2\pi \sqrt{\frac{2L}{g_e}} = \sqrt{2} \cdot 2\pi \sqrt{\frac{L}{g_e}} = \sqrt{2} \cdot T_e \]

6. **Conclusion**:
Therefore, the time period of the pendulum on the new planet is:
\[ T_n = \sqrt{2} \cdot T_e \]
This means the time period of the pendulum at the surface of the new planet is \( \sqrt{2} \) times that of the time period at the surface of Earth.

### Final Answer:
The time period of a pendulum at the surface of the new planet is \( \sqrt{2} \) times the time period at the surface of Earth.

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