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

Explanation

To solve the problem step-by-step, we will follow these steps:

### Step 1: Understand the Problem
A pendulum clock that gives the correct time at a place with \( g = 9.8 \, \text{m/s}^2 \) loses 24 seconds over 24 hours. We need to find the value of \( g \) at the new location.

### Step 2: Calculate the Loss of Time per Hour
The clock loses 24 seconds in 24 hours. Therefore, the loss of time per hour is:
\[ \text{Loss per hour} = \frac{24 \, \text{seconds}}{24 \, \text{hours}} = 1 \, \text{second/hour} \]

### Step 3: Determine the Total Time in Seconds for One Day
The total time in seconds for one day is:
\[ \text{Total seconds in a day} = 24 \times 60 \times 60 = 86400 \, \text{seconds} \]

### Step 4: Calculate the Effective Time for the Clock
Since the clock loses 1 second every hour, the effective time for the clock in one day is:
\[ \text{Effective time} = 86400 + 24 = 86424 \, \text{seconds} \]

### Step 5: Calculate the Number of Oscillations in One Day
The time period of the pendulum clock at the original location is given as \( T = 2 \, \text{seconds} \). The number of oscillations in one day is:
\[ \text{Number of oscillations} = \frac{86400 \, \text{seconds}}{2 \, \text{seconds}} = 43200 \]

### Step 6: Calculate the New Time Period
At the new location, the effective time period \( T' \) can be calculated using the effective time:
\[ T' = \frac{86424 \, \text{seconds}}{43200} = 2.000555 \, \text{seconds} \]

### Step 7: Relate Time Period with Gravity
The time period \( T \) of a pendulum is related to the acceleration due to gravity \( g \) by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
From this, we can derive that:
\[ \frac{T'}{T} = \sqrt{\frac{g}{g'}} \]
Where \( g' \) is the new acceleration due to gravity.

### Step 8: Substitute Known Values
Substituting the values we have:
\[ \frac{2.000555}{2} = \sqrt{\frac{9.8}{g'}} \]

### Step 9: Square Both Sides
Squaring both sides gives:
\[ \left(\frac{2.000555}{2}\right)^2 = \frac{9.8}{g'} \]

### Step 10: Solve for \( g' \)
Rearranging gives:
\[ g' = \frac{9.8 \times 4}{(2.000555)^2} \]
Calculating this will yield:
\[ g' \approx 9.79 \, \text{m/s}^2 \]

### Final Answer
The value of \( g \) at the new place is approximately \( 9.79 \, \text{m/s}^2 \).

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