JEE PYQ: Units & Measurements - Question ID 1fa38849badf (JEE Main 2026)

Core Formula & Concept:

The density $\rho$ of a uniform cylinder is given by the ratio of its mass to its volume. The volume $V$ of a cylinder is calculated using the formula: $V = \pi r^2 l = \pi \left(\frac{d}{2}\right)^2 l = \frac{\pi d^2 l}{4}$ Thus, the density formula becomes: $\rho = \frac{m}{V} = \frac{4m}{\pi d^2 l}$

The question involves calculating the percentage fractional error in $\rho$. This is derived using the concept of error propagation in multiplication and division. For a function $f = k \cdot x^a y^b z^c$, where $k$ is a constant, the relative (fractional) error in $f$ is given by: $\frac{\Delta f}{f} = |a| \frac{\Delta x}{x} + |b| \frac{\Delta y}{y} + |c| \frac{\Delta z}{z}$ The percentage fractional error is simply $\frac{\Delta f}{f} \times 100\%$.

Step-by-Step Derivation:

Step 1: Express density in terms of measured quantities
From the formula: $\rho = \frac{4m}{\pi d^2 l}$ We treat $\pi$ as a constant with no error. Thus, the relative error in $\rho$ depends on the relative errors in $m$, $d$, and $l$.

Step 2: Compute relative errors in each measured quantity
Given:

• Mass: $m = 97.42 \pm 0.02 \text{ g}$
• Length: $l = 8.35 \pm 0.05 \text{ mm}$
• Diameter: $d = 20.20 \pm 0.02 \text{ mm}$

The relative (fractional) errors are: $\frac{\Delta m}{m} = \frac{0.02}{97.42}, \quad \frac{\Delta l}{l} = \frac{0.05}{8.35}, \quad \frac{\Delta d}{d} = \frac{0.02}{20.20}$

Step 3: Apply error propagation formula
The density formula can be rewritten in terms of exponents: $\rho = \frac{4}{\pi} \cdot m^1 \cdot d^{-2} \cdot l^{-1}$ Using the error propagation rule: $\frac{\Delta \rho}{\rho} = \left|1\right| \frac{\Delta m}{m} + \left|-2\right| \frac{\Delta d}{d} + \left|-1\right| \frac{\Delta l}{l}$ $\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2 \frac{\Delta d}{d} + \frac{\Delta l}{l}$

Step 4: Substitute numerical values and compute
Compute each term: $\frac{\Delta m}{m} = \frac{0.02}{97.42} \approx 0.0002053 \quad (0.02053\%)$ $\frac{\Delta d}{d} = \frac{0.02}{20.20} \approx 0.0009901 \quad (0.09901\%)$ $\frac{\Delta l}{l} = \frac{0.05}{8.35} \approx 0.0059880 \quad (0.59880\%)$ Now, apply the formula: $\frac{\Delta \rho}{\rho} = 0.0002053 + 2 \times 0.0009901 + 0.0059880$ $\frac{\Delta \rho}{\rho} = 0.0002053 + 0.0019802 + 0.0059880 = 0.0081735$

Step 5: Convert to percentage
$\text{Percentage fractional error} = \frac{\Delta \rho}{\rho} \times 100\% = 0.0081735 \times 100\% \approx 0.817\%$ Rounding to two decimal places, we get $0.82\%$.

Common Traps & Exam Tip:

• Incorrect exponent handling: Students often forget that the diameter $d$ is squared in the formula, leading to a factor of 2 in its error contribution. This is the most common mistake.
• Unit consistency: While not an issue here, students sometimes mix units (e.g., mm and cm). Always ensure all quantities are in consistent units before computing errors.
• Rounding errors: Premature rounding of intermediate values (e.g., $\frac{0.02}{20.20}$) can lead to inaccuracies. Keep at least 4-5 decimal places during intermediate steps.
• Sign confusion: The error propagation formula uses absolute values, so negative exponents do not affect the sign of the error contribution.

Final Answer: The percentage fractional error in $\rho$ is 0.82%, which corresponds to option B.