The length of second pendulum is 1 m on earth. If mass and diameter of the planet is doubled than that of earth, then length becomes

Options

A. 1 m
B. 2 m
C. 0.5 m
D. 4 m

Explanation

To solve the problem, we need to find the new length of the pendulum on a planet where both the mass and diameter are doubled compared to Earth. We start with the formula for the time period of a pendulum.

### Step-by-step Solution:

1. **Understanding the Second Pendulum**:
A second pendulum has a time period of 2 seconds. The formula for the time period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Acceleration due to Gravity on Earth**:
For Earth, we have:
\[ T = 2 \text{ seconds} \quad \text{and} \quad L = 1 \text{ meter} \]
Plugging in these values, we can express \( g \) for Earth:
\[ 2 = 2\pi \sqrt{\frac{1}{g}} \implies 1 = \pi \sqrt{\frac{1}{g}} \implies g = \frac{\pi^2}{1} \approx 9.87 \text{ m/s}^2 \]

3. **New Planet's Parameters**:
The mass \( M \) and diameter \( D \) of the new planet are both double that of Earth. Therefore:
- New mass \( M' = 2M \)
- New diameter \( D' = 2D \)
- New radius \( R' = \frac{D'}{2} = 2 \times \frac{D}{2} = 2R \)

4. **Acceleration due to Gravity on the New Planet**:
The formula for gravitational acceleration \( g' \) on the new planet is:
\[ g' = \frac{GM'}{(R')^2} \]
Substituting the new values:
\[ g' = \frac{G(2M)}{(2R)^2} = \frac{2GM}{4R^2} = \frac{1}{2} \cdot \frac{GM}{R^2} = \frac{g}{2} \]

5. **Finding the New Length of the Pendulum**:
Now, we need to find the new length \( L' \) of the pendulum on this planet. The time period remains the same (2 seconds):
\[ 2 = 2\pi \sqrt{\frac{L'}{g'}} \]
Substituting \( g' = \frac{g}{2} \):
\[ 2 = 2\pi \sqrt{\frac{L'}{\frac{g}{2}}} \implies 2 = 2\pi \sqrt{\frac{2L'}{g}} \implies 1 = \pi \sqrt{\frac{2L'}{g}} \]

6. **Solving for \( L' \)**:
Squaring both sides:
\[ 1 = \pi^2 \cdot \frac{2L'}{g} \implies 2L' = \frac{g}{\pi^2} \implies L' = \frac{g}{2\pi^2} \]
Since we know \( g = \pi^2 \), we can substitute:
\[ L' = \frac{\pi^2}{2\pi^2} = \frac{1}{2} \text{ meters} = 0.5 \text{ meters} \]

### Final Answer:
The new length of the pendulum on the new planet is **0.5 meters**.