The time period of a seconds pendulum on a planet, whose mass is twice that of the earth and radius is equal to the diameter of the earth, will be

Options

A. 2.828 s
B. 3.742 s
C. 1.414 s
D. 2 s

Explanation

To solve the problem of finding the time period of a seconds pendulum on a planet with twice the mass of Earth and a radius equal to the diameter of Earth, we can follow these steps:

### Step 1: Understand the formula for the time period of a 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.

### Step 2: Determine the gravitational acceleration on the new planet
The acceleration due to gravity \( g \) on a planet is given by:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

In this case, the mass of the new planet is \( 2M_e \) (twice the mass of Earth) and the radius is \( R = 2R_e \) (equal to the diameter of Earth).

### Step 3: Substitute the values into the formula for gravitational acceleration
Substituting the values into the formula for \( g \):
\[ g' = \frac{G(2M_e)}{(2R_e)^2} \]
This simplifies to:
\[ g' = \frac{2GM_e}{4R_e^2} = \frac{GM_e}{2R_e^2} \]
Since \( g_e = \frac{GM_e}{R_e^2} \) (the acceleration due to gravity on Earth), we can express \( g' \) as:
\[ g' = \frac{g_e}{2} \]

### Step 4: Substitute \( g' \) into the time period formula
Now, we can substitute \( g' \) back into the time period formula:
\[ T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{\frac{g_e}{2}}} = 2\pi \sqrt{\frac{2L}{g_e}} \]
This can be rewritten as:
\[ T' = \sqrt{2} \cdot 2\pi \sqrt{\frac{L}{g_e}} = \sqrt{2} \cdot T \]
where \( T \) is the time period of the pendulum on Earth (which is 2 seconds for a seconds pendulum).

### Step 5: Calculate the new time period
Now we can calculate \( T' \):
\[ T' = \sqrt{2} \cdot 2 \text{ seconds} \approx 1.414 \cdot 2 \approx 2.828 \text{ seconds} \]

### Final Answer
Thus, the time period of the seconds pendulum on the new planet is approximately **2.828 seconds**.