Question

If the length of the pendulum in pendulum clock increases by 0.1%, then the error in time per day is :

• A. 86.4 s
• B. 8.64 s
• C. 43.2 s
• D. 4.32 s

Hint:

Remember the relationship between fractional errors for power laws. If \(y = kx^n\), then \(\frac{\Delta y}{y} = n \frac{\Delta x}{x}\). Here, \(T \propto L^{1/2}\), so \(n = 1/2\). An increase in length leads to an increase in the time period, making the clock run slow (lose time). A decrease in length makes it run fast (gain time).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The length of a pendulum clock increases, which will change its time period. We need to find the total error in time (time lost or gained) over a period of one day.
Step 2: Key Formula or Approach:
The time period (T) of a simple pendulum is given by \(T = 2\pi\sqrt{\frac{L}{g}}\).
For small changes, the fractional change in the time period can be found using error analysis:
\[ \frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta L}{L} \] The total error in time (\(\Delta t\)) in a day is this fractional change multiplied by the total seconds in a day.
Step 3: Detailed Explanation:
Given that the length (L) increases by 0.1%, the fractional change in length is:
\[ \frac{\Delta L}{L} = 0.1% = \frac{0.1}{100} = 0.001 \] Now, we calculate the fractional change in the time period (T):
\[ \frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta L}{L} = \frac{1}{2} \times 0.001 = 0.0005 \] Since the length increases, the time period also increases (\(\Delta T\) is positive). This means the clock runs slower and loses time.
The total number of seconds in one day is:
\[ t = 24 \, \text{hours} \times 60 \, \text{minutes/hour} \times 60 \, \text{seconds/minute} = 86400 \, \text{s} \] The total error in time per day (\(\Delta t\)) is:
\[ \Delta t = \left(\frac{\Delta T}{T}\right) \times t = 0.0005 \times 86400 \, \text{s} \] \[ \Delta t = \frac{1}{2000} \times 86400 \, \text{s} = \frac{864}{20} \, \text{s} = \frac{86.4}{2} \, \text{s} = 43.2 \, \text{s} \] Step 4: Final Answer:
The error in time per day is 43.2 seconds. The clock will lose 43.2 seconds. This corresponds to option (C).