JEE PYQ: Units & Measurements - Question ID 84462400579f (JEE Main 2025)

Core Formula & Concept:

In experimental physics, when a quantity \( Q \) depends on multiple measured variables \( x_1, x_2, \dots \) through a functional relationship \[ Q = f(x_1, x_2, \dots), \] the relative error (percentage error) in \( Q \) is determined by the propagation of errors. The key formula for the maximum percentage error in \( Q \) is: \[ \left(\frac{\Delta Q}{Q}\right)_{\text{max}} \approx \left|\frac{\partial \ln Q}{\partial x_1}\right|\Delta x_1 + \left|\frac{\partial \ln Q}{\partial x_2}\right|\Delta x_2 + \dots \] where \( \Delta x_i \) is the absolute error in \( x_i \), and \( \frac{\Delta x_i}{x_i} \) is the relative (percentage) error in \( x_i \).

In this problem, the energy \( E(t) \) is given by \[ E(t) = \alpha^3 e^{-\beta t}, \] and we are asked to find the maximum percentage error in \( E \) at \( t = 5 \) s, given the percentage errors in \( \alpha \) and \( t \).

Step-by-Step Derivation:

Step 1: Express the relative error formula for \( E \).

We start by taking the natural logarithm of \( E \): \[ \ln E = \ln(\alpha^3 e^{-\beta t}) = 3 \ln \alpha - \beta t. \] Now, differentiate \( \ln E \) with respect to each variable: \[ \frac{\partial \ln E}{\partial \alpha} = \frac{3}{\alpha}, \quad \frac{\partial \ln E}{\partial t} = -\beta. \] The maximum relative error in \( E \) is: \[ \left(\frac{\Delta E}{E}\right)_{\text{max}} = \left|\frac{\partial \ln E}{\partial \alpha}\right|\Delta \alpha + \left|\frac{\partial \ln E}{\partial t}\right|\Delta t. \]

Step 2: Convert absolute errors to percentage errors.

Given: \[ \frac{\Delta \alpha}{\alpha} = 1.2\% \quad \text{and} \quad \frac{\Delta t}{t} = 1.6\%. \] Multiply the partial derivatives by the percentage errors: \[ \left|\frac{\partial \ln E}{\partial \alpha}\right| \Delta \alpha = \left|\frac{3}{\alpha}\right| \cdot \Delta \alpha = 3 \cdot \frac{\Delta \alpha}{\alpha} = 3 \times 1.2\% = 3.6\%, \] \[ \left|\frac{\partial \ln E}{\partial t}\right| \Delta t = \left|-\beta\right| \cdot \Delta t = \beta \cdot \Delta t = \beta t \cdot \frac{\Delta t}{t} = 0.3 \times 5 \times 1.6\% = 2.4\%. \]

Step 3: Sum the contributions.

The maximum percentage error in \( E \) is the sum of the two contributions: \[ \left(\frac{\Delta E}{E}\right)_{\text{max}} = 3.6\% + 2.4\% = 6.0\%. \] Common Traps & Exam Tip:

1. Forgetting the chain rule for exponentials: Students often misapply the error propagation to the exponential term. Remember that \( e^{-\beta t} \) contributes an error proportional to \( \beta t \cdot \frac{\Delta t}{t} \). 2. Sign confusion: The derivative \( \frac{\partial \ln E}{\partial t} = -\beta \) is negative, but we take the absolute value for the maximum error. 3. Unit consistency: Ensure \( \beta \) and \( t \) are in compatible units (here both are in seconds). 4. Percentage vs. absolute errors: Always convert percentage errors to absolute errors (or vice versa) consistently.

By carefully following the logarithmic differentiation and summing the absolute contributions, the correct answer is 6%, which corresponds to option A.