JEE PYQ: Units & Measurements - Question ID 0ffe031c5053 (JEE Main 2022)

Core Formula & Concept:

In electrical circuits, the heat dissipated (often called Joule heating) by a resistor is given by the formula:

$H = I^2 R t$

where:

• $H$ = Heat dissipated (in joules)
• $I$ = Current flowing through the resistor (in amperes)
• $R$ = Resistance of the resistor (in ohms)
• $t$ = Time for which the current flows (in seconds)

When dealing with errors in measurements, especially percentage errors, we use the concept of relative error propagation. The relative (percentage) error in a product or power of measured quantities is determined by the sum of the relative errors of each quantity, weighted by their exponents.

For a general function $f = x^a y^b z^c$, the maximum percentage error in $f$ is given by:

$\left( \frac{\Delta f}{f} \right)_{\text{max}} = |a| \left( \frac{\Delta x}{x} \right) + |b| \left( \frac{\Delta y}{y} \right) + |c| \left( \frac{\Delta z}{z} \right)$

This formula is derived from logarithmic differentiation and is crucial in error analysis.

Step-by-Step Derivation:

We are given:

• Percentage error in resistance, $\frac{\Delta R}{R} = 1\%$
• Percentage error in current, $\frac{\Delta I}{I} = 2\%$
• Percentage error in time, $\frac{\Delta t}{t} = 3\%$

We need to find the maximum percentage error in $H = I^2 R t$.

Let’s express the relative error in $H$ using the error propagation rule:

$\frac{\Delta H}{H} = \frac{\Delta (I^2 R t)}{I^2 R t}$

Using the formula for relative error in a product of powers:

$\frac{\Delta H}{H} = 2 \cdot \frac{\Delta I}{I} + 1 \cdot \frac{\Delta R}{R} + 1 \cdot \frac{\Delta t}{t}$

Note:

• The exponent of $I$ is 2, so its relative error is multiplied by 2.
• The exponents of $R$ and $t$ are 1, so their relative errors are multiplied by 1.

Now, substitute the given percentage errors (expressed as decimals for calculation):

$\frac{\Delta H}{H} = 2 \times 0.02 + 1 \times 0.01 + 1 \times 0.03$ $\frac{\Delta H}{H} = 0.04 + 0.01 + 0.03 = 0.08$

Convert back to percentage:

$\frac{\Delta H}{H} \times 100\% = 8\%$

Thus, the maximum percentage error in the detection of the dissipated heat is 8%.

Common Traps & Exam Tip:

Common Mistakes:

• Ignoring the exponent on current: Many students forget that $I$ is squared in the formula, so they multiply the error in $I$ by 1 instead of 2. This leads to an incorrect total error of $1\% + 2\% + 3\% = 6\%$, which is option C — a trap.
• Adding errors incorrectly: Some students add the percentage errors directly without considering the exponents, leading to $1 + 2 + 3 = 6\%$ error.
• Confusing absolute and relative errors: Using absolute errors instead of relative (percentage) errors can lead to dimensional inconsistencies and wrong results.

Exam Tip: Always remember: When a quantity is raised to a power, its relative error is multiplied by that power. So for $I^2$, the error contribution is $2 \times \text{error in } I$. This rule applies to all physical formulas involving powers, products, or quotients.

In this question, recognizing that $H = I^2 R t$ and applying the correct error propagation rule is the key to selecting the correct option — D: 8.