An accurate pendulum clock is mounted on ground floor of a high building. How much time will it lose or gain in one day if its is transferred to top storey of a building which is h = 200m higher than the ground floor? Radius of earth is `6.4 xx 10^(6)`

Options

A. it will lose 6.2 s
B. it will lose 2.7 s
C. it will gain 5.2s
D. it will gain 1.6 s

Answer

To solve the problem of how much time a pendulum clock will lose or gain when moved from the ground floor to the top storey of a building 200 m high, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Effect of Height on Gravity**:
The acceleration due to gravity \( g' \) at a height \( h \) above the Earth's surface can be approximated using the formula:
\[ g' = g \left(1 - \frac{2h}{R_e}\right) \]
where \( g \) is the acceleration due to gravity at the surface, \( R_e \) is the radius of the Earth, and \( h \) is the height above the surface.

2. **Determine the Time Period of the Pendulum**:
The time period \( T \) of a pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum.

3. **Calculate the Time Period at Ground Level (\( T_1 \))**:
At the ground floor, the time period \( T_1 \) is:
\[ T_1 = 2\pi \sqrt{\frac{L}{g}} \]

4. **Calculate the Time Period at the Top Storey (\( T_2 \))**:
At the height \( h \), the time period \( T_2 \) becomes:
\[ T_2 = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{g \left(1 - \frac{2h}{R_e}\right)}} \]

5. **Relate \( T_2 \) to \( T_1 \)**:
Using the approximation for \( g' \):
\[ T_2 = T_1 \sqrt{\frac{1}{1 - \frac{2h}{R_e}}} \]
Applying the binomial approximation \( \sqrt{1 - x} \approx 1 - \frac{x}{2} \) for small \( x \):
\[ T_2 \approx T_1 \left(1 + \frac{h}{R_e}\right) \]

6. **Calculate the Change in Time Period**:
The change in time period per cycle is:
\[ T_2 - T_1 = T_1 \left(\frac{h}{R_e}\right) \]

7. **Determine the Time Lost in One Day**:
To find out how much time is lost in one day (86400 seconds), we can calculate:
\[ \text{Time lost} = \frac{T_2 - T_1}{T_1} \times \text{Total seconds in a day} \]
Substituting the values:
\[ \text{Time lost} = \left(\frac{h}{R_e}\right) \times 86400 \]
Where \( h = 200 \, \text{m} \) and \( R_e = 6.4 \times 10^6 \, \text{m} \):
\[ \text{Time lost} = \left(\frac{200}{6.4 \times 10^6}\right) \times 86400 \]

8. **Calculate the Final Value**:
\[ \text{Time lost} = \frac{200 \times 86400}{6.4 \times 10^6} \approx 2.7 \, \text{seconds} \]

### Conclusion:
The pendulum clock will lose approximately 2.7 seconds in one day when moved to the top storey of the building.