In order to determine the Young's modulus of a wire of radius 0.2 cm (measured using a scale of leas

Q: Why do we add the fractional errors even though some quantities are in the denominator?

When calculating the maximum possible error, we always assume the worst-case scenario where all individual errors compound together. Therefore, whether a quantity is multiplied or divided, its relative error is always added.

Q: Why is the fractional error of the radius multiplied by 2?

In the formula for Young's modulus, the cross-sectional area is $\pi R^2$. According to the rules of error propagation, if a quantity is raised to a power $n$, its relative error is multiplied by $n$.

Q: Do I need to convert all the measurements into standard SI units (meters, kilograms) before calculating the error?

Not necessarily! The beauty of fractional error ($\frac{\Delta x}{x}$) is that it is a dimensionless ratio. As long as the absolute error ($\Delta x$) and the measured value ($x$) are in the *same* unit, they will cancel out.

Q: What about the acceleration due to gravity ($g$) and the constant $\pi$? Don't they have errors?

In the context of this experiment, $g$ is considered a known constant, and $\pi$ is a mathematical constant. Constants have an absolute error of zero, so they do not contribute to the fractional error of the final result.

Q: The question asks for 'fractional error', but the options have percentage signs. Is there a difference?

Strictly speaking, fractional error is just the ratio (e.g., $0.014$). Percentage error is the fractional error multiplied by $100$ (e.g., $1.4\%$). In JEE problems, if the options have a $\%$ sign, you are expected to find the percentage error.

Question

In order to determine the Young's modulus of a wire of radius $0.2	ext{ cm}$ (measured using a scale of least count $= 0.001	ext{ cm}$) and length $1	ext{ m}$ (measured using a scale of least count $= 1	ext{ mm}$), a weight of mass $1	ext{ kg}$ (measured using a scale of least count $= 1	ext{ g}$) was hanged to get the elongation of $0.5	ext{ cm}$ (measured using a scale of least count $0.001	ext{ cm}$). What will be the fractional error in the value of Young's modulus determined by this experiment?

Options:

(A) $0.14\%$

(B) $0.9\%$

(C) $9\%$

(D) $1.4\%$

Sigma Percentile · Accepted Answer

### The Physical Setup: Stretching a Wire

Imagine you are in a physics laboratory, tasked with determining the Young's modulus of a metallic wire. You suspend the wire from a rigid support and attach a heavy mass to its free end. As gravity pulls the mass downwards, the wire experiences a stretching force, causing it to elongate slightly.

This simple setup is the foundation of our problem. We are given four key measurements: the original length of the wire ($L$), its radius ($R$), the mass attached ($m$), and the resulting elongation ($\Delta l$). However, no measurement is perfect. Each instrument used—whether it's a meter scale, a vernier caliper, or a screw gauge—has a least count, which represents the maximum possible absolute error in that measurement.

### The Mathematical Engine: Young's Modulus

To understand how these individual errors affect our final result, we first need the mathematical relationship connecting them. Young's modulus ($Y$) is defined as the ratio of longitudinal stress to longitudinal strain.

$$Y = \frac{	ext{Stress}}{	ext{Strain}}$$

The stress is the restoring force per unit area. Here, the force is the weight of the mass ($mg$), and the cross-sectional area of the wire is $\pi R^2$. The strain is the fractional change in length, given by $\frac{\Delta l}{L}$. Substituting these into our definition, we get the master equation:

$$Y = \frac{mgL}{\pi R^2 \Delta l}$$

### The Art of Error Propagation

Now comes the crucial step: translating our physical formula into an error propagation equation. When physical quantities are multiplied or divided, their maximum fractional (or relative) errors add up.

We take the relative error of each variable in our master equation. Constants like the acceleration due to gravity ($g$) and $\pi$ are exact values in this context, so their error contribution is zero.

There is a vital trap to avoid here: the radius ($R$) is squared in the formula. According to the rules of error analysis, the power of a quantity becomes a multiplier for its relative error. Therefore, the fractional error equation becomes:

$$\frac{\Delta Y}{Y} = \frac{\Delta m}{m} + \frac{\Delta L}{L} + 2\frac{\Delta R}{R} + \frac{\Delta (\Delta l)}{\Delta l}$$

### Taming the Units: Substitution

Before we plug in the numbers, we must carefully extract the measured values and their absolute errors (least counts) from the problem statement.

- Radius: $R = 0.2	ext{ cm}$, $\Delta R = 0.001	ext{ cm}$
- Length: $L = 1	ext{ m} = 1000	ext{ mm}$, $\Delta L = 1	ext{ mm}$
- Mass: $m = 1	ext{ kg} = 1000	ext{ g}$, $\Delta m = 1	ext{ g}$
- Elongation: $\Delta l = 0.5	ext{ cm}$, $\Delta (\Delta l) = 0.001	ext{ cm}$

Notice that we don't necessarily need to convert everything to standard SI units (like meters and kilograms). Because fractional error is a ratio (e.g., $\frac{\Delta m}{m}$), the units will perfectly cancel out as long as the numerator and denominator of each individual term share the same unit.

Substituting these values into our error equation:

$$\frac{\Delta Y}{Y} = \frac{1	ext{ g}}{1000	ext{ g}} + \frac{1	ext{ mm}}{1000	ext{ mm}} + 2\left(\frac{0.001	ext{ cm}}{0.2	ext{ cm}}ight) + \frac{0.001	ext{ cm}}{0.5	ext{ cm}}$$

### The Final Tally: Calculating the Error

Now, we execute the arithmetic. Let's simplify each fraction step-by-step to avoid any silly mistakes.

$$\frac{\Delta Y}{Y} = \frac{1}{1000} + \frac{1}{1000} + 2\left(\frac{1}{200}ight) + \frac{1}{500}$$

Converting these fractions into decimals makes the final addition straightforward:

$$\frac{\Delta Y}{Y} = 0.001 + 0.001 + 0.010 + 0.002$$

Adding them all together yields the total fractional error:

$$\frac{\Delta Y}{Y} = 0.014$$

The question asks for the fractional error, but a quick glance at the options reveals they are all in percentages. To convert a fractional error to a percentage error, we simply multiply by $100\%$.

$$	ext{Percentage Error} = 0.014 	imes 100\% = 1.4\%$$

### A Crucial Insight: The Dominant Error

Before we move on, let's look at the numbers we just added. The term for the radius ($0.010$) contributed the vast majority of the total error ($0.014$).

Why? Because the radius is a very small value, making its relative error large, and it is squared in the formula, which doubles its impact! If you were an engineer trying to improve the accuracy of this experiment, your primary focus should be on measuring the radius of the wire with a much more precise instrument.

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Visualized Solution

$Y o u n g^{'} s M o d u l u s S e t u p$

$F or m u l a f or Y o u n g^{'} s M o d u l u s$

$F r a c t i o na l E r r or E q u a t i o n$

$E x t r a c t in g G i v e n V a l u es$

$S u b s t i t u t in g V a l u es$

$S im pl i f y in g F r a c t i o n s$

$C a l c u l a t in g T o t a l F r a c t i o na l E r r or$

$P er ce n t a g e E r r or$

$R e f l ec t i o n$

The Sigma Insight: Errors in Measurement

The Physical Setup

The Mathematical Engine

The Art of Error Propagation

Taming the Units

The Final Tally

A Crucial Insight

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Young′s Modulus Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Formula for Young′s Modulus

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fractional Error Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Extracting Given Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying Fractions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Total Fractional Error

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Percentage Error

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reflection