The mass and diameter of a planet are twice those of earth. The period of oscillation of pendulum on this planet will be (if it is a second's pendulum on earth )

Options

A. `1//sqrt(2)` second
B. `2//sqrt(2)` second
C. 2 second
D. (1/2) second

Explanation

To solve the problem, we need to determine the period of oscillation of a pendulum on a new planet, given that the mass and diameter of this planet are twice those of Earth. We know that the period of a pendulum is influenced by the acceleration due to gravity, which is affected by both the mass and radius of the planet.

### Step-by-Step Solution:

1. **Understanding the Period of a Pendulum**:
The 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 period of a seconds pendulum on Earth is \( T_E = 2 \) seconds.
- The mass of the new planet \( M_P = 2M_E \) (twice the mass of Earth).
- The diameter of the new planet \( D_P = 2D_E \) (twice the diameter of Earth), which means the radius \( R_P = 2R_E \).

3. **Finding the Acceleration Due to Gravity**:
The acceleration due to gravity \( g \) is given by:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant.

For Earth:
\[ g_E = \frac{GM_E}{R_E^2} \]

For the new planet:
\[ g_P = \frac{G(2M_E)}{(2R_E)^2} = \frac{G(2M_E)}{4R_E^2} = \frac{1}{2} \cdot \frac{GM_E}{R_E^2} = \frac{g_E}{2} \]

4. **Finding the Ratio of Gravity**:
The ratio of the gravitational acceleration on Earth to that on the new planet is:
\[ \frac{g_E}{g_P} = \frac{g_E}{\frac{g_E}{2}} = 2 \]

5. **Relating the Periods**:
Since the period of the pendulum is inversely proportional to the square root of \( g \):
\[ \frac{T_E}{T_P} = \sqrt{\frac{g_P}{g_E}} = \sqrt{\frac{1}{2}} \]
Therefore:
\[ T_P = T_E \cdot \sqrt{2} \]

6. **Calculating the Period on the New Planet**:
Substituting \( T_E = 2 \) seconds:
\[ T_P = 2 \cdot \sqrt{2} \text{ seconds} \]

7. **Final Answer**:
Hence, the period of oscillation of the pendulum on the new planet is:
\[ T_P = 2\sqrt{2} \text{ seconds} \]

### Conclusion:
The correct answer is \( \frac{2}{\sqrt{2}} \) seconds.