Assuming the earth as an spherical body, for seconds pendulum
`{:("Column - I","Column - II"),("(a) at pole","(p) T"gt 2s),("(b) on a satellite","(q) T" lt 2s),("(c) at mountain","(r) T = 2s"),("(d) at centre of earth","(s) T = 0"),(,"(t) T"gt oo):}`

Options

A. a - q, b - t, c - p, d - t
B. a - q, b - s, c - t, d - t
C. a - t, b - p, c - s, d - t
D. a - t, b - s, c - q, d - t

Answer

To solve the problem, we need to analyze how the time period of a seconds pendulum (which has a time period of 2 seconds) changes under different conditions: at the poles, on a satellite, at a mountain, and at the center of the Earth. The time period \( T \) of a simple pendulum is given by the formula:

\[ T = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}} \]

where \( g_{\text{effective}} \) is the effective acceleration due to gravity.

### Step 1: At the Pole
At the poles, the effective gravity \( g_{\text{effective}} \) is equal to \( g \) (the acceleration due to gravity at the surface of the Earth) because the effect of Earth's rotation is negligible. Therefore, the time period remains:

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

**Hint:** Remember that at the poles, the gravitational pull is maximum and unaffected by the Earth's rotation.

### Step 2: On a Satellite
In a satellite, the effective gravity \( g_{\text{effective}} \) is zero because the satellite is in free fall. Thus, the time period becomes:

\[ T = 2\pi \sqrt{\frac{L}{0}} \rightarrow T \to \infty \]

**Hint:** Consider the concept of weightlessness in a satellite; the pendulum does not experience any gravitational pull.

### Step 3: At a Mountain
As we ascend a mountain, the effective gravity \( g_{\text{effective}} \) decreases. The formula for effective gravity at a height \( h \) above the Earth's surface is:

\[ g_{\text{effective}} = g \left(1 - \frac{2h}{R}\right) \]

where \( R \) is the radius of the Earth. Since \( g_{\text{effective}} \) is less than \( g \), the time period will be greater than 2 seconds:

\[ T > 2 \text{ seconds} \]

**Hint:** Higher altitude means less gravitational pull, which increases the time period.

### Step 4: At the Center of the Earth
At the center of the Earth, the effective gravity \( g_{\text{effective}} \) is zero. Thus, similar to the satellite case, the time period becomes:

\[ T = 2\pi \sqrt{\frac{L}{0}} \rightarrow T \to \infty \]

**Hint:** The center of the Earth has no gravitational pull acting on the pendulum.

### Summary of Results
- (a) At the pole: \( T = 2 \text{ seconds} \) (matches with \( r \))
- (b) On a satellite: \( T \to \infty \) (matches with \( t \))
- (c) At a mountain: \( T > 2 \text{ seconds} \) (matches with \( p \))
- (d) At the center of the Earth: \( T \to \infty \) (matches with \( t \))

### Final Matching
- (a) at pole → (r) \( T = 2s \)
- (b) on a satellite → (t) \( T \to \infty \)
- (c) at mountain → (q) \( T > 2s \)
- (d) at center of earth → (s) \( T = 0 \)

### Conclusion
The correct matches are:
- (a) → (r)
- (b) → (t)
- (c) → (q)
- (d) → (s)