JEE PYQ: Units & Measurements - Question ID 3b304c88fbe0 (JEE Main 2021)

Core Formula & Concept:

Young's Modulus ($Y$) of a wire is defined as the ratio of longitudinal stress to longitudinal strain within the elastic limit. The formula is:

$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L} = \frac{F \cdot L}{A \cdot \Delta L}$

Where:

• $F$ = Applied force (weight of the mass hung) = $m \cdot g$
• $A$ = Cross-sectional area of the wire = $\pi r^2$
• $L$ = Original length of the wire
• $\Delta L$ = Elongation of the wire
• $r$ = Radius of the wire

The fractional error in $Y$ is calculated using the formula for relative error in a product/quotient of measured quantities. If $Y = k \cdot \frac{x_1^{a} x_2^{b}}{x_3^{c} x_4^{d}}$, then the fractional error in $Y$ is:

$\frac{\Delta Y}{Y} = |a| \frac{\Delta x_1}{x_1} + |b| \frac{\Delta x_2}{x_2} + |c| \frac{\Delta x_3}{x_3} + |d| \frac{\Delta x_4}{x_4}$

Here, $\Delta x_i$ represents the absolute error (least count) in the measurement of $x_i$.

--- Step-by-Step Derivation:

Step 1: Express Young's Modulus in terms of measured quantities

$Y = \frac{F \cdot L}{A \cdot \Delta L} = \frac{m g L}{\pi r^2 \Delta L}$

Step 2: Identify the powers of each measured quantity

Let’s rewrite $Y$ as: $Y = \frac{g}{\pi} \cdot \frac{m L}{r^2 \Delta L}$ Here: - $m$ appears to the power of $+1$ - $L$ appears to the power of $+1$ - $r$ appears to the power of $-2$ (since $r^2$ is in the denominator) - $\Delta L$ appears to the power of $-1$

Step 3: Compute fractional errors in each quantity

We are given: - Radius $r = 0.2 \text{ cm}$, least count $\Delta r = 0.001 \text{ cm}$ - Length $L = 1 \text{ m} = 100 \text{ cm}$, least count $\Delta L_{\text{length}} = 1 \text{ mm} = 0.1 \text{ cm}$ - Mass $m = 1 \text{ kg} = 1000 \text{ g}$, least count $\Delta m = 1 \text{ g}$ - Elongation $\Delta L = 0.5 \text{ cm}$, least count $\Delta (\Delta L) = 0.001 \text{ cm}$ Now compute fractional errors: - $\frac{\Delta m}{m} = \frac{1 \text{ g}}{1000 \text{ g}} = 0.001 = 0.1\%$ - $\frac{\Delta L}{L} = \frac{0.1 \text{ cm}}{100 \text{ cm}} = 0.001 = 0.1\%$ - $\frac{\Delta r}{r} = \frac{0.001 \text{ cm}}{0.2 \text{ cm}} = 0.005 = 0.5\%$ - $\frac{\Delta (\Delta L)}{\Delta L} = \frac{0.001 \text{ cm}}{0.5 \text{ cm}} = 0.002 = 0.2\%$

Step 4: Apply the formula for fractional error in $Y$

Since $Y \propto m^1 L^1 r^{-2} (\Delta L)^{-1}$, the fractional error is: $\frac{\Delta Y}{Y} = \left| \frac{\Delta m}{m} \right| + \left| \frac{\Delta L}{L} \right| + 2 \left| \frac{\Delta r}{r} \right| + \left| \frac{\Delta (\Delta L)}{\Delta L} \right|$ Substitute the values: $\frac{\Delta Y}{Y} = 0.001 + 0.001 + 2(0.005) + 0.002$ $= 0.001 + 0.001 + 0.010 + 0.002$ $= 0.014$ Convert to percentage: $\frac{\Delta Y}{Y} \times 100 = 1.4\%$

Step 5: Match with given options

The fractional error is $1.4\%$, which corresponds to option C. --- Common Traps & Exam Tip:

Trap 1: Confusing absolute and fractional errors. Students often add absolute errors directly instead of converting them to fractional errors first. Always divide the least count by the measured value to get the fractional error.

Trap 2: Misapplying the power rule. The error in $r^2$ is $2 \cdot \frac{\Delta r}{r}$, not $\left(\frac{\Delta r}{r}\right)^2$. Remember: for $x^n$, the fractional error is $|n| \cdot \frac{\Delta x}{x}$.

Trap 3: Ignoring unit consistency. Ensure all quantities are in consistent units (e.g., convert meters to centimeters or vice versa) before computing fractional errors.

Exam Tip: In error propagation, always use the formula for relative error in products/quotients. Memorize: for $Y = \frac{ab}{cd}$, the fractional error is $\frac{\Delta Y}{Y} = \frac{\Delta a}{a} + \frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}$.