JEE PYQ: Units & Measurements - Question ID 15baec74fd49 (JEE Main 2023)

Core Formula & Concept:

In experimental physics, when a physical quantity \( P \) is expressed as a product or quotient of other measured quantities (say \( a, b, c, d \)), the relative error (or percentage error) in \( P \) is determined using the rules of error propagation. The key concept is:

• For a quantity \( P \) given by: $P = \frac{a^m \cdot b^n}{c^p \cdot d^q}$ the relative error in \( P \) is: $\frac{\Delta P}{P} = m \cdot \frac{\Delta a}{a} + n \cdot \frac{\Delta b}{b} + p \cdot \frac{\Delta c}{c} + q \cdot \frac{\Delta d}{d}$ where \( \frac{\Delta x}{x} \) represents the relative error in quantity \( x \).
• The percentage error is simply the relative error multiplied by 100.

This rule arises from logarithmic differentiation, where we take the natural logarithm of both sides and then differentiate, treating errors as small differentials.

Step-by-Step Derivation:

Given the physical quantity:

$P = \frac{a^2 b^3}{c \sqrt{d}}$

First, express \( \sqrt{d} \) as \( d^{1/2} \), so the expression becomes:

$P = \frac{a^2 b^3}{c^1 d^{1/2}} = a^2 b^3 c^{-1} d^{-1/2}$

Now, take the natural logarithm of both sides:

$\ln P = 2 \ln a + 3 \ln b - \ln c - \frac{1}{2} \ln d$

Differentiate both sides with respect to the variables (treating errors as small changes):

$\frac{dP}{P} = 2 \frac{da}{a} + 3 \frac{db}{b} - \frac{dc}{c} - \frac{1}{2} \frac{dd}{d}$

Since errors are always added in quadrature (or in absolute value for maximum error estimation), we take absolute values of each term:

$\frac{\Delta P}{P} = 2 \frac{\Delta a}{a} + 3 \frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{1}{2} \frac{\Delta d}{d}$

Now, substitute the given percentage errors (converted to relative errors):

• \( \frac{\Delta a}{a} = 1\% = 0.01 \)
• \( \frac{\Delta b}{b} = 2\% = 0.02 \)
• \( \frac{\Delta c}{c} = 3\% = 0.03 \)
• \( \frac{\Delta d}{d} = 4\% = 0.04 \)

Plug these into the error propagation formula:

$\frac{\Delta P}{P} = 2(0.01) + 3(0.02) + 1(0.03) + \frac{1}{2}(0.04)$

Calculate each term:

• \( 2 \times 0.01 = 0.02 \)
• \( 3 \times 0.02 = 0.06 \)
• \( 1 \times 0.03 = 0.03 \)
• \( \frac{1}{2} \times 0.04 = 0.02 \)

Sum them up:

$\frac{\Delta P}{P} = 0.02 + 0.06 + 0.03 + 0.02 = 0.13$

Convert the relative error to percentage error:

$\text{Percentage error in } P = 0.13 \times 100 = 13\%$

Thus, the percentage error in \( P \) is 13%, which corresponds to option B.

Common Traps & Exam Tip:

Students often make the following mistakes in such questions:

1. Ignoring the sign of exponents: Some students forget that negative exponents (like \( c^{-1} \)) still contribute positively to the error. The error is always added in absolute terms.
2. Miscounting powers: Misidentifying the exponent of a variable (e.g., treating \( \sqrt{d} \) as \( d^1 \) instead of \( d^{1/2} \)) leads to incorrect error calculation.
3. Adding percentage errors directly: Simply adding 1% + 2% + 3% + 4% = 10% is incorrect. The exponents must be accounted for.
4. Forgetting to convert relative error to percentage: The final answer must be multiplied by 100 to express it as a percentage.

Exam Tip: Always write down the expression in exponential form (e.g., \( d^{1/2} \)) before applying error propagation. This avoids confusion with roots and fractional powers.