JEE PYQ: Units & Measurements - Question ID 25bc43c9be74 (JEE Main 2019)

In a simple pendulum, the time period \( T \) of small oscillations is related to the acceleration due to gravity \( g \) and the length of the pendulum \( L \) by the formula:

Rearranging this, we get the expression for \( g \):

This formula is the foundation for determining \( g \) experimentally using a pendulum. The time period \( T \) is obtained by measuring the time for multiple oscillations and dividing by the number of oscillations.

The question involves calculating the percentage error in \( g \) due to uncertainties (errors) in the measurement of time and length. The percentage error in a quantity that depends on multiple measured variables is found using the concept of relative error propagation.

For a function \( g = f(L, T) \), the relative error in \( g \) is given by:

This simplifies to:

The percentage error in \( g \) is then:

Step 1: Extract given data and compute time period

• Number of oscillations, \( n = 20 \)
• Mean time for 20 oscillations, \( t = 30 \, \text{s} \)
• Least count of watch (time measurement), \( \Delta t = 1 \, \text{s} \)
• Length of pendulum, \( L = 55.0 \, \text{cm} = 0.550 \, \text{m} \)
• Least count of meter scale (length measurement), \( \Delta L = 1 \, \text{mm} = 0.001 \, \text{m} \)

The time period \( T \) is:

Step 2: Compute absolute errors in \( T \) and \( L \)

The error in measuring time for 20 oscillations is \( \Delta t = 1 \, \text{s} \). Since \( T = t / n \), the error in \( T \) is:

The error in length \( \Delta L = 0.001 \, \text{m} \) (given by the least count of the meter scale).

Step 3: Compute relative errors in \( L \) and \( T \)

Step 4: Apply error propagation formula for \( g \)

From the formula \( g = \frac{4\pi^2 L}{T^2} \), the relative error in \( g \) is:

Substitute the values:

Step 5: Convert to percentage error

This value is close to option D (6.8%).

1. Misinterpreting least count as absolute error: Many students confuse the least count of the measuring instrument with the actual error in measurement. While the least count gives the smallest division, the absolute error in a single measurement is typically taken as the least count. However, when multiple measurements are averaged (as in time for 20 oscillations), the error in the mean value is still considered as the least count unless specified otherwise.

2. Incorrect error propagation formula: Students often forget that the error in \( T \) is multiplied by 2 in the formula for \( g \), because \( T \) appears squared in the denominator. This is a common oversight and leads to underestimating the error in \( g \).

3. Unit inconsistency: Mixing units (e.g., using length in cm and time in seconds without converting to consistent units) can lead to incorrect calculations. Always ensure all quantities are in SI units (meters and seconds) before applying formulas.

4. Rounding errors prematurely: Rounding intermediate values (like \( \Delta T / T \)) too early can lead to inaccuracies in the final percentage error. Keep more decimal places during calculations and round only at the end.

Exam Tip: When dealing with percentage errors in derived quantities, always:

• Identify the formula connecting the derived quantity to measured quantities.
• Apply the relative error propagation rule: for multiplication/division, add relative errors; for powers, multiply the relative error by the exponent.
• Double-check the least count and how it translates to absolute error in the measured quantity.

In this question, the dominant source of error is the time measurement, not the length. This is because the error in \( T \) is amplified by a factor of 2 in the error propagation formula. Always look for such amplifying factors in error analysis.