Question

% Question A pendulum clock gives correct time at $20^\circ$C. If the coefficient of linear expansion of the pendulum's material is \alpha = 1.2 \times 10^{-5} /^\circC, the clock will lose time per day at $40^\circ$C by approximately:

• A. 10.4 s
• B. 5.2 s
• C. 20.8 s
• D. 2.6 s

Hint:

Key Exam Tip:
Time lost per day by a pendulum clock is approximately $86400 \times (\frac{1}{2} \alpha \Delta \theta)$, where $\Delta T / T_0 \approx \frac{1}{2} \alpha \Delta \theta$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The time period of a simple pendulum is given by \(T = 2\pi \sqrt{\frac{L}{g}}\).
This indicates that the time period is dependent on the length of the pendulum.
When temperature increases, most materials expand. The length \(L\) of the pendulum increases, which in turn increases the time period \(T\).
A larger time period means the clock takes more time to complete one "tick", so it lags behind the actual time, resulting in a "loss" of time.
Step 2: Key Formula or Approach:
The fractional change in the time period due to a change in temperature \(\Delta \theta\) is:
\[ \frac{\Delta T}{T} = \frac{1}{2} \alpha \Delta \theta \]
The total time lost in a duration \(t\) is:
\[ \Delta t = \left( \frac{1}{2} \alpha \Delta \theta \right) \times t \]
Where:
\(\alpha\) = coefficient of linear expansion.
\(\Delta \theta\) = change in temperature.
\(t\) = total time interval (usually one day = 86400 seconds).
Step 3: Detailed Explanation:
Given values:
\(\alpha = 1.2 \times 10^{-5} / ^{\circ}\text{C}\)
Initial Temperature \(T_{initial} = 20^{\circ}\text{C}\)
Final Temperature \(T_{final} = 40^{\circ}\text{C}\)
Change in temperature \(\Delta \theta = 40 - 20 = 20^{\circ}\text{C}\)
Total time in one day \(t = 24 \times 60 \times 60 = 86400 \, \text{s}\)
Substitute into the formula:
\[ \Delta t = \frac{1}{2} \times (1.2 \times 10^{-5}) \times 20 \times 86400 \]
Simplify the constants:
\[ \Delta t = (0.6 \times 10^{-5}) \times 20 \times 86400 \]
\[ \Delta t = (1.2 \times 10^{-4}) \times 86400 \]
Convert to standard decimal notation for easier multiplication:
\[ \Delta t = 0.00012 \times 86400 \]
Calculation:
\[ 12 \times 864 = 10368 \]
Adjusting for the decimal places:
\[ \Delta t = 10.368 \, \text{s} \]
Rounding this value to the nearest tenth as provided in the options gives 10.4 s.
Step 4: Final Answer:
The clock will lose approximately 10.4 seconds per day.