The length of a second's pendulum on the surface of the earth, where `g=9.8 m//s^(2)`, is approximately equal to

Options

A. 0.8 m
B. 1 m
C. 1.2m
D. 1.1 m

Answer

To find the length of a second's pendulum on the surface of the Earth where \( g = 9.8 \, \text{m/s}^2 \), we can use the formula for the period of a simple pendulum:

\[ T = 2\pi \sqrt{\frac{L}{g}} \]

Where:
- \( T \) is the period of the pendulum,
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.

Since we are dealing with a second's pendulum, its period \( T \) is 2 seconds.

### Step 1: Substitute the known values into the formula
We know:
- \( T = 2 \, \text{s} \)
- \( g = 9.8 \, \text{m/s}^2 \)

Substituting these values into the formula gives:

\[ 2 = 2\pi \sqrt{\frac{L}{9.8}} \]

### Step 2: Simplify the equation
First, we can divide both sides by \( 2 \):

\[ 1 = \pi \sqrt{\frac{L}{9.8}} \]

### Step 3: Isolate the square root
Next, we isolate the square root by dividing both sides by \( \pi \):

\[ \frac{1}{\pi} = \sqrt{\frac{L}{9.8}} \]

### Step 4: Square both sides
To eliminate the square root, we square both sides:

\[ \left(\frac{1}{\pi}\right)^2 = \frac{L}{9.8} \]

### Step 5: Solve for \( L \)
Now, multiply both sides by \( 9.8 \) to solve for \( L \):

\[ L = 9.8 \left(\frac{1}{\pi}\right)^2 \]

### Step 6: Calculate \( L \)
Using \( \pi \approx 3.14 \):

\[ L = 9.8 \left(\frac{1}{3.14}\right)^2 \]

Calculating \( \left(\frac{1}{3.14}\right)^2 \):

\[ \left(\frac{1}{3.14}\right)^2 \approx \frac{1}{9.8596} \approx 0.1013 \]

Now substituting this back into the equation for \( L \):

\[ L \approx 9.8 \times 0.1013 \approx 0.993 \, \text{m} \]

This rounds to approximately 1 meter.

### Final Answer
The length of a second's pendulum on the surface of the Earth is approximately \( 1 \, \text{meter} \).

---