A simple seconds pendulum is constructed out of a very thin string of thermal coefficient of linear expansion `alpha = 20 xx 10^(-4)//^(@)C` and a heavy particle attached to one end. The free end of the string is suspended from the ceiling of an elevator at rest. the pendulum keeps correct time at `0^(@)C`. when the temperature rises to `50^(@)C`, the elevator operator of mass `60 kg` being a student of Physics accelerates the elevator vertically, to have the pendulum correct time. the apparent weight of the operator when the pendulum keeps correct time at `50^(@)C` is `("Take " g = 10m//s^(2)`)

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Calculate the Change in Length of the Pendulum String

The change in length (\( \Delta L \)) of the pendulum string due to thermal expansion can be calculated using the formula:

\[ \Delta L = \alpha L_0 \Delta T \]

Where:
- \( \alpha = 20 \times 10^{-4} \, \text{°C}^{-1} \) (coefficient of linear expansion)
- \( \Delta T = 50 \, \text{°C} - 0 \, \text{°C} = 50 \, \text{°C} \)
- \( L_0 \) is the original length of the pendulum string.

Thus, we can express the new length \( L \) as:

\[ L = L_0 + \Delta L = L_0 + \alpha L_0 \Delta T = L_0(1 + \alpha \Delta T) \]

### Step 2: Write the Time Period Formula at 0°C

The time period \( T \) of a simple pendulum is given by:

\[ T = 2\pi \sqrt{\frac{L_0}{g}} \]

Where \( g = 10 \, \text{m/s}^2 \).

### Step 3: Write the Time Period Formula at 50°C with Elevator Acceleration

When the temperature rises to 50°C and the elevator accelerates upwards with acceleration \( a \), the time period formula becomes:

\[ T = 2\pi \sqrt{\frac{L}{g + a}} \]

### Step 4: Set the Time Periods Equal

Since the pendulum keeps correct time at both temperatures, we can set the two expressions for \( T \) equal to each other:

\[ 2\pi \sqrt{\frac{L_0}{g}} = 2\pi \sqrt{\frac{L}{g + a}} \]

### Step 5: Square Both Sides and Rearrange

Squaring both sides gives:

\[ \frac{L_0}{g} = \frac{L}{g + a} \]

Rearranging this, we have:

\[ L_0(g + a) = L \]

### Step 6: Substitute for \( L \)

Substituting \( L \) from Step 1 into the equation gives:

\[ L_0(g + a) = L_0(1 + \alpha \Delta T) \]

Dividing both sides by \( L_0 \) (assuming \( L_0 \neq 0 \)):

\[ g + a = 1 + \alpha \Delta T \]

### Step 7: Solve for Acceleration \( a \)

Rearranging gives:

\[ a = \alpha \Delta T g - g \]

Substituting the values:

\[ a = (20 \times 10^{-4})(50)(10) - 10 \]

Calculating this:

\[ a = (10)(10^{-2}) - 10 = 0.1g - g = -0.9g \]

This indicates the elevator is accelerating downwards.

### Step 8: Calculate Apparent Weight of the Operator

The apparent weight \( W' \) of the operator when the pendulum keeps correct time is given by:

\[ W' = m(g + a) \]

Substituting \( m = 60 \, \text{kg} \):

\[ W' = 60(g + a) = 60(g - 0.9g) = 60(0.1g) \]

Substituting \( g = 10 \, \text{m/s}^2 \):

\[ W' = 60(0.1 \times 10) = 60 \times 1 = 60 \, \text{N} \]

### Final Answer

The apparent weight of the operator when the pendulum keeps correct time at \( 50 \, \text{°C} \) is:

\[ W' = 66 \, \text{N} \]