JEE PYQ: Units & Measurements - Question ID f1401ef49862 (JEE Main 2023)

Core Formula & Concept:

The density $\rho$ of a cylindrical wire is defined as its mass per unit volume. For a cylinder of mass $m$, length $L$, and radius $r$, the volume $V$ is given by: $V = \pi r^2 L$ Therefore, the density is: $\rho = \frac{m}{V} = \frac{m}{\pi r^2 L}$

When quantities are measured with uncertainties (errors), the maximum possible error in a derived quantity like density is found using the rule of error propagation for multiplication/division. If a quantity $Q$ depends on variables $x, y, z$ as: $Q = k \cdot x^a y^b z^c$ where $k$ is a constant, then the relative (percentage) error in $Q$ is: $\frac{\Delta Q}{Q} = |a| \frac{\Delta x}{x} + |b| \frac{\Delta y}{y} + |c| \frac{\Delta z}{z}$ This rule assumes errors are small and independent, and gives the maximum possible error when errors add up in the worst-case scenario.

In this problem, we are to find the maximum percentage error in the density of the wire, given the errors in mass, length, and radius.

--- Step-by-Step Derivation:

Step 1: Write the expression for density

$\rho = \frac{m}{\pi r^2 L}$

Step 2: Identify the powers of each variable

- $m$: power = $+1$ - $r$: power = $-2$ (since $r^2$ is in denominator) - $L$: power = $-1$ (in denominator)

Step 3: Apply the error propagation formula

The relative error in $\rho$ is: $$\frac{\Delta \rho}{\rho} = \left| \frac{\partial \rho}{\partial m} \cdot \frac{\Delta m}{\rho} \right| + \left| \frac{\partial \rho}{\partial r} \cdot \frac{\Delta r}{\rho} \right| + \left| \frac{\partial \rho}{\partial L} \cdot \frac{\Delta L}{\rho} \right|$$ But using the power rule: $\frac{\Delta \rho}{\rho} = \left|1\right| \frac{\Delta m}{m} + \left|-2\right| \frac{\Delta r}{r} + \left|-1\right| \frac{\Delta L}{L} = \frac{\Delta m}{m} + 2 \frac{\Delta r}{r} + \frac{\Delta L}{L}$

Step 4: Convert all quantities to consistent units (optional but good practice)

We can compute errors in any consistent unit system. Here, we'll use cm and grams. - Mass: $m = 0.4 \text{ g}$, $\Delta m = 0.01 \text{ g}$ - Length: $L = 8 \text{ cm}$, $\Delta L = 0.04 \text{ cm}$ - Radius: $r = 6 \text{ mm} = 0.6 \text{ cm}$, $\Delta r = 0.03 \text{ mm} = 0.003 \text{ cm}$

Step 5: Compute relative errors

- $\frac{\Delta m}{m} = \frac{0.01}{0.4} = 0.025 = 2.5\%$ - $\frac{\Delta r}{r} = \frac{0.003}{0.6} = 0.005 = 0.5\%$ - $\frac{\Delta L}{L} = \frac{0.04}{8} = 0.005 = 0.5\%$

Step 6: Apply the error propagation formula

$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2 \frac{\Delta r}{r} + \frac{\Delta L}{L} = 2.5\% + 2 \times 0.5\% + 0.5\% = 2.5\% + 1\% + 0.5\% = 4\%$

Step 7: Interpret the result

The maximum percentage error in the density is $4\%$.

Therefore, the correct answer is Option C: 4%

--- Common Traps & Exam Tip:

1. Forgetting to square the radius in error propagation:

Many students incorrectly use $\frac{\Delta r}{r}$ only once, forgetting that since $r$ is squared in the volume, its error contribution is doubled. This leads to underestimating the total error.

2. Unit inconsistency:

Mixing mm and cm without conversion can lead to incorrect relative error calculations. Always ensure all quantities are in consistent units when computing ratios.

3. Confusing absolute and relative error:

The question asks for percentage (relative) error, not absolute error. Students sometimes compute $\Delta \rho$ in $\text{g/cm}^3$, which is unnecessary and time-consuming.

4. Ignoring the sign of powers:

Even though density has $r^2$ in the denominator, the error propagation formula uses the absolute value of the power. So negative exponents don't reduce the error — they increase it.

Exam Tip:

In error propagation questions, always:

• Write down the formula for the derived quantity.
• Identify the power of each variable.
• Apply the rule: relative error = sum of (absolute value of power × relative error of variable).
• Convert all errors to percentages for final answer.

This method ensures accuracy and speed in exams.