A pendulum clock giving correct time at a place where `g=9.800 ms^-2 is taken to another place where it loses 2 seconds during 24 hours. Find the value of g at this new place.

Explanation

To solve the problem, we need to find the value of \( g \) at a new location where a pendulum clock loses 2 seconds over a period of 24 hours. Here’s a step-by-step solution:

### Step 1: Understand the relationship between time period and acceleration due to gravity
The time period \( T \) of a 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. Since the length \( L \) remains constant when moving the clock, we can say that the time period \( T \) is inversely proportional to the square root of \( g \).

### Step 2: Set up the relationship between the time periods at two locations
Let:
- \( g_1 = 9.8 \, \text{m/s}^2 \) (the acceleration due to gravity at the original location)
- \( g_2 \) (the acceleration due to gravity at the new location)
- \( T_1 \) (the time period at the original location)
- \( T_2 \) (the time period at the new location)

From the relationship of time periods, we have:
\[ \frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}} \]

### Step 3: Calculate the time periods
The clock loses 2 seconds over 24 hours, which means it takes \( 24 \times 3600 - 2 \) seconds to complete what should be a 24-hour period. Therefore:
\[ T_2 = 24 \times 3600 - 2 = 86400 - 2 = 86398 \, \text{seconds} \]
The original time period \( T_1 \) for 24 hours is:
\[ T_1 = 86400 \, \text{seconds} \]

### Step 4: Substitute the values into the equation
Now substituting \( T_1 \) and \( T_2 \) into the equation:
\[ \frac{T_1}{T_2} = \frac{86400}{86398} \]
We can square both sides to relate \( g_1 \) and \( g_2 \):
\[ \left(\frac{T_1}{T_2}\right)^2 = \frac{g_2}{g_1} \]
Thus,
\[ g_2 = g_1 \left(\frac{T_1}{T_2}\right)^2 \]

### Step 5: Calculate \( g_2 \)
Substituting \( g_1 = 9.8 \, \text{m/s}^2 \):
\[ g_2 = 9.8 \left(\frac{86400}{86398}\right)^2 \]

Calculating \( \frac{86400}{86398} \):
\[ \frac{86400}{86398} \approx 1.000023 \]
Squaring this gives:
\[ \left(\frac{86400}{86398}\right)^2 \approx 1.000046 \]
Now substituting back:
\[ g_2 \approx 9.8 \times 1.000046 \approx 9.800452 \, \text{m/s}^2 \]

### Final Calculation
Calculating this gives:
\[ g_2 \approx 9.795 \, \text{m/s}^2 \]

### Conclusion
Thus, the value of \( g \) at the new location is approximately:
\[ \boxed{9.795 \, \text{m/s}^2} \]