What is the time period of the seconds pendulum, on a planet similar to earth, but mass and radius three times as that of the earth?

Options

A. 2 s
B. 6 s
C. `(2)/(sqrt3)s`
D. `2sqrt3s`

Answer

To find the time period of a seconds pendulum on a planet that has a mass and radius three times that 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: Identify the values for the new planet
We know that the new planet has:
- Mass \( M' = 3M \) (three times the mass of Earth)
- Radius \( R' = 3R \) (three times the radius of Earth)

### Step 3: Calculate the acceleration due to gravity on the new planet
The acceleration due to gravity \( g' \) on the surface of the new planet can be calculated using the formula:
\[ g' = \frac{GM'}{(R')^2} \]
Substituting the values for \( M' \) and \( R' \):
\[ g' = \frac{G(3M)}{(3R)^2} = \frac{3GM}{9R^2} = \frac{1}{3} \cdot \frac{GM}{R^2} = \frac{1}{3} g \]
where \( g \) is the acceleration due to gravity on Earth.

### Step 4: Relate the time periods of the pendulum on Earth and the new planet
Using the relationship derived from the time period formula:
\[ \frac{T'}{T} = \sqrt{\frac{g}{g'}} \]
Substituting \( g' = \frac{1}{3} g \):
\[ \frac{T'}{T} = \sqrt{\frac{g}{\frac{1}{3}g}} = \sqrt{3} \]
This means:
\[ T' = T \sqrt{3} \]

### Step 5: Calculate the time period of the seconds pendulum on Earth
The time period of a seconds pendulum on Earth is \( T = 2 \) seconds.

### Step 6: Calculate the time period on the new planet
Substituting \( T = 2 \) seconds into the equation:
\[ T' = 2 \sqrt{3} \text{ seconds} \]

### Final Answer
The time period of the seconds pendulum on the new planet is:
\[ T' = 2\sqrt{3} \text{ seconds} \]