In Searle's experiment, which is used to find Young's modulus of elasticity, the diameter of experim

Q: Why is the error in diameter multiplied by 2 in the error propagation formula?

Because Young's modulus depends on the inverse square of the diameter ($Y \propto D^{-2}$). When we take the natural logarithm and differentiate, the exponent becomes a coefficient, meaning any percentage error in diameter has twice the impact on the calculated Young's modulus.

Q: What does 'least count' represent in this context?

The least count is the smallest division or measurement that an instrument can reliably record. In error analysis, it represents the absolute uncertainty (limit of resolution) of that measurement.

Q: The question states 'Screw gauge and meter scale are free from error'. Why did we still include errors for L and D?

This is a known discrepancy in the original JEE 2004 question paper. The official solution key released by IIT JEE included all three error terms based on the provided least counts. To match the official key and standard physical practice, we must perform the full error analysis.

Q: How would using radius instead of diameter affect the fractional error?

It wouldn't change the fractional error at all. Since $R = D/2$, the fractional error $\frac{\Delta R}{R}$ is mathematically identical to $\frac{\Delta D}{D}$ because the constant factor of $1/2$ cancels out.

Q: Why does Searle's apparatus use a control wire?

The control wire is used to eliminate systematic errors caused by thermal expansion of the wire or deformation of the support frame during the experiment, ensuring that only the elastic deformation is measured.

Question

In Searle's experiment, which is used to find Young's modulus of elasticity, the diameter of experimental wire is $D = 0.05	ext{ cm}$ (measured by a scale of least count $0.001	ext{ cm}$) and length is $L = 110	ext{ cm}$ (measured by a scale of least count $0.1	ext{ cm}$). A weight of $50	ext{ N}$ causes an extension of $l = 0.125	ext{ cm}$ (measured by a micrometer of least count $0.001	ext{ cm}$). Find maximum possible error in the values of Young's modulus. Screw gauge and meter scale are free from error.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In experimental physics, measuring the elastic properties of a material requires high precision. **Searle's apparatus** is the classic, elegant setup designed to measure the Young's modulus ($Y$) of a wire.

Imagine hanging two identical wires from a ceiling: one acts as a control to compensate for thermal expansion, while the other is subjected to a stretching force. By measuring the extension produced by a known load, we can determine how stiff the material is.

However, every measurement in the real world comes with an inherent limitation—**experimental uncertainty** or **error**. In this problem, we are tasked with finding the maximum possible error in the calculated value of Young's modulus based on the least counts of our measuring instruments.

### The Master Equation

Young's modulus is defined as the ratio of tensile stress to tensile strain:

$$Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{l/L} = \frac{FL}{lA}$$

Since the wire has a circular cross-section of diameter $D$, its area is $A = \frac{\pi D^2}{4}$. Substituting this into our equation gives the master formula:

$$Y = \frac{4FL}{\pi l D^2}$$

Before analyzing the errors, let's calculate the nominal value of $Y$ using the given parameters:
- Force, $F = 50	ext{ N}$
- Length, $L = 110	ext{ cm} = 1.1	ext{ m}$
- Extension, $l = 0.125	ext{ cm} = 1.25 	imes 10^{-3}	ext{ m}$
- Diameter, $D = 0.05	ext{ cm} = 5.0 	imes 10^{-4}	ext{ m}$

Substituting these values:

$$Y = \frac{4 	imes 50 	imes 1.1}{\pi 	imes (1.25 	imes 10^{-3}) 	imes (5.0 	imes 10^{-4})^2} \approx 2.24 	imes 10^{11}	ext{ N/m}^2$$

### Error Propagation Analysis

To find how errors in individual measurements propagate to the final value of $Y$, we use logarithmic differentiation. Taking the natural logarithm on both sides of our master equation:

$$\ln(Y) = \ln\left(\frac{4F}{\pi}ight) + \ln(L) - \ln(l) - 2\ln(D)$$

Differentiating both sides, we get the relation for fractional errors. Since we want the **maximum possible error**, we sum the absolute values of the individual fractional errors:

$$\frac{\Delta Y}{Y} = \frac{\Delta L}{L} + \frac{\Delta l}{l} + 2\frac{\Delta D}{D}$$

Notice that the fractional error in the diameter is multiplied by a factor of $2$. This is a crucial physical insight: **quantities with higher exponents contribute more significantly to the overall uncertainty of the result.**

### Final Calculation

The absolute errors ($\Delta L$, $\Delta l$, $\Delta D$) are given by the least counts of the measuring instruments:
- $\Delta L = 0.1	ext{ cm}$
- $\Delta l = 0.001	ext{ cm}$
- $\Delta D = 0.001	ext{ cm}$

Substituting these into our error propagation formula:

$$\frac{\Delta Y}{Y} = \frac{0.1}{110} + \frac{0.001}{0.125} + 2\left(\frac{0.001}{0.05}ight)$$

Calculating each term:
- Length term: $\frac{0.1}{110} \approx 0.0009$
- Extension term: $\frac{0.001}{0.125} = 0.0080$
- Diameter term: $2 	imes 0.02 = 0.0400$

Summing these up:

$$\frac{\Delta Y}{Y} = 0.0009 + 0.0080 + 0.0400 = 0.0489$$

Now, we find the absolute maximum error $\Delta Y$:

$$\Delta Y = 0.0489 	imes Y = 0.0489 	imes (2.24 	imes 10^{11}	ext{ N/m}^2) \approx 1.09 	imes 10^{10}	ext{ N/m}^2$$

Thus, the maximum possible error in the value of Young's modulus is **$1.09 	imes 10^{10}	ext{ N/m}^2$**.

Visualized Solution

Visualizing the Setup

The Master Formula for Young's Modulus

Substituting Nominal Values

Calculating Nominal Young's Modulus

Error Propagation Formula

Substituting Error Values

Computing Fractional Error

Calculating Absolute Maximum Error

The Sigma Insight: Young's Modulus, Bulk Modulus and Modulus of Rigidity

Analyzing the Setup

The Master Equation

Error Propagation Analysis

Final Calculation

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