JEE PYQ: Units & Measurements - Question ID 605eaee8ed80 (JEE Main 2026)

Core Formula & Concept:

The question involves the determination of the percentage error in the spring constant \( k \) of a simple harmonic oscillator (SHO). The time period \( T \) of a SHO is given by:

$T = 2\pi \sqrt{\frac{m}{k}}$

Rearranging this formula to solve for the spring constant \( k \), we get:

$k = \frac{4 \pi^2 m}{T^2}$

The problem requires us to compute the percentage error in \( k \) based on the errors in the measurements of mass \( m \) and time period \( T \).

Key concepts involved:

• Error Propagation: When a quantity depends on multiple measured variables, the relative (percentage) error in the quantity is determined by the relative errors in the individual variables. For a function \( k = f(m, T) \), the relative error in \( k \) is given by:
$\frac{\Delta k}{k} = \left| \frac{\partial k}{\partial m} \frac{\Delta m}{k} \right| + \left| \frac{\partial k}{\partial T} \frac{\Delta T}{k} \right|$
• Percentage Error: The percentage error is simply the relative error multiplied by 100.
• Resolution and Accuracy: The resolution of the measuring instrument (e.g., watch) and the accuracy of the measurement (e.g., mass) contribute to the absolute error in the respective quantities.

Step-by-Step Derivation:

Given data:

• Measured mass, \( m = 10 \, \text{g} = 0.01 \, \text{kg} \) (converted to SI units for consistency, though grams can also be used as long as units are consistent).
• Accuracy in mass measurement, \( \Delta m = 10 \, \text{mg} = 0.01 \, \text{g} = 0.00001 \, \text{kg} \).
• Time for 50 oscillations, \( t = 60 \, \text{s} \).
• Resolution of the watch, \( \Delta t_{\text{resolution}} = 2 \, \text{s} \).

Step 1: Compute the time period \( T \) and its error \( \Delta T \).

The time period \( T \) is the time for one oscillation, so:

$T = \frac{t}{50} = \frac{60}{50} = 1.2 \, \text{s}$

The error in measuring 50 oscillations is due to the resolution of the watch. Since the watch has a resolution of 2 s, the absolute error in measuring 60 s is \( \Delta t = 2 \, \text{s} \).

The error in the time period \( T \) is:

$\Delta T = \frac{\Delta t}{50} = \frac{2}{50} = 0.04 \, \text{s}$

Step 2: Express \( k \) in terms of \( m \) and \( T \).

From the formula for \( k \):

$k = \frac{4 \pi^2 m}{T^2}$

Step 3: Compute the relative errors in \( m \) and \( T \).

The relative error in mass \( m \) is:

$\frac{\Delta m}{m} = \frac{0.01 \, \text{g}}{10 \, \text{g}} = 0.001$

The relative error in time period \( T \) is:

$\frac{\Delta T}{T} = \frac{0.04}{1.2} \approx 0.0333$

Step 4: Propagate the errors to find the relative error in \( k \).

Taking the natural logarithm of both sides of the equation \( k = \frac{4 \pi^2 m}{T^2} \):

$\ln k = \ln(4 \pi^2) + \ln m - 2 \ln T$

Differentiating both sides with respect to each variable:

$\frac{dk}{k} = \frac{dm}{m} - 2 \frac{dT}{T}$

The relative error in \( k \) is the sum of the absolute values of the relative errors (since errors add in quadrature for independent measurements, but here we take the worst-case scenario by adding them directly):

$\frac{\Delta k}{k} = \left| \frac{\Delta m}{m} \right| + 2 \left| \frac{\Delta T}{T} \right|$

Substituting the values:

$\frac{\Delta k}{k} = 0.001 + 2 \times 0.0333 = 0.001 + 0.0666 = 0.0676$

Step 5: Compute the percentage error in \( k \).

$\text{Percentage error in } k = \frac{\Delta k}{k} \times 100 = 0.0676 \times 100 = 6.76\%$ Common Traps & Exam Tip:

Students often make the following mistakes in this question:

1. Incorrect Error Propagation: Forgetting that the error in \( T \) is multiplied by 2 in the relative error formula for \( k \) because \( T \) is squared in the denominator. This is a common oversight when dealing with exponents in error propagation.
2. Unit Inconsistency: Not converting all quantities to consistent units (e.g., using grams for mass and kilograms in the formula). While the relative error calculation is unit-independent, it's good practice to ensure consistency.
3. Resolution vs. Absolute Error: Confusing the resolution of the watch with the absolute error in the time measurement. The resolution directly gives the absolute error in the time measurement for 50 oscillations.
4. Number of Oscillations: Forgetting to divide the total time by the number of oscillations to get the time period \( T \) and its error. The error in \( T \) is the error in the total time divided by the number of oscillations.
5. Rounding Errors: Rounding intermediate values too early can lead to inaccuracies in the final percentage error. It's best to keep more decimal places during calculations and round only the final answer.

Exam Tip: Always double-check the formula for error propagation when dealing with exponents. For a quantity \( Q = x^n \), the relative error in \( Q \) is \( n \) times the relative error in \( x \). This is a recurring theme in error propagation problems.