The simple pendulum of length 40 cm is taken inside a deep mine. Assume for the time being that the mine is 1600 km deep. Calculate the time period of the pendulum there. Radius of the earth =6400 km.

Explanation

To solve the problem of calculating the time period of a simple pendulum at a depth of 1600 km inside a mine, we can follow these steps:

### Step 1: Understand the relationship between gravitational acceleration and depth
The gravitational acceleration at a depth \( d \) below the surface of the Earth can be calculated using the formula:
\[ g' = g \left(1 - \frac{d}{R}\right) \]
where:
- \( g' \) is the gravitational acceleration at depth,
- \( g \) is the gravitational acceleration at the surface (approximately \( 9.8 \, \text{m/s}^2 \)),
- \( d \) is the depth (1600 km in this case),
- \( R \) is the radius of the Earth (6400 km).

### Step 2: Substitute the values into the formula
Given:
- \( d = 1600 \, \text{km} = 1600 \times 10^3 \, \text{m} \)
- \( R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)

Now we can substitute these values into the equation:
\[ g' = g \left(1 - \frac{1600 \times 10^3}{6400 \times 10^3}\right) \]
This simplifies to:
\[ g' = g \left(1 - \frac{1600}{6400}\right) = g \left(1 - \frac{1}{4}\right) = g \left(\frac{3}{4}\right) \]

### Step 3: Calculate \( g' \)
Now substituting the value of \( g \):
\[ g' = 9.8 \left(\frac{3}{4}\right) = 9.8 \times 0.75 = 7.35 \, \text{m/s}^2 \]

### Step 4: Use the formula for the time period of a simple pendulum
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. Given \( L = 40 \, \text{cm} = 0.4 \, \text{m} \), we can substitute the values:
\[ T = 2\pi \sqrt{\frac{0.4}{7.35}} \]

### Step 5: Calculate the time period
First, calculate \( \frac{0.4}{7.35} \):
\[ \frac{0.4}{7.35} \approx 0.0543 \]
Now take the square root:
\[ \sqrt{0.0543} \approx 0.233 \]
Now substitute back into the time period formula:
\[ T \approx 2\pi \times 0.233 \approx 1.465 \, \text{s} \]

### Final Answer
The time period of the pendulum at a depth of 1600 km is approximately **1.46 seconds**.

---