In a Searle's experiment, the diameter of the wire as measured by a screw gauge of least count 0.001

Q: Why do we use the least count as the absolute error?

The least count is the smallest value an instrument can measure. Any measurement taken could be off by at most this amount, making it the maximum possible absolute error.

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

In the formula for Young's modulus, the area is proportional to the square of the diameter ($d^2$). When calculating fractional errors, the power of a variable becomes a multiplier for its fractional error.

Q: Do I need to convert the units to meters when calculating the fractional error?

No! Fractional error is a ratio (e.g., $\Delta d / d$). As long as the absolute error and the measurement are in the same units, the units will cancel out.

Q: Why don't we consider the error in the force (50 N)?

The problem states the weight is exactly 50 N and doesn't provide a least count or error margin for it. Therefore, we treat it as a constant with zero error.

Q: Why do we add the fractional errors instead of subtracting them?

We want to find the *maximum* possible error. Errors can occur in either direction, so to find the worst-case scenario, we always add the absolute values of the fractional errors.

Question

In a Searle's experiment, the diameter of the wire as measured by a screw gauge of least count $0.001	ext{ cm}$ is $0.050	ext{ cm}$. The length, measured by a scale of least count $0.1	ext{ cm}$, is $110.0	ext{ cm}$. When a weight of $50	ext{ N}$ is suspended from the wire, the extension is measured to be $0.125	ext{ cm}$ by a micrometer of least count $0.001	ext{ cm}$. Find the maximum error in the measurement of Young's modulus of the material of the wire from these data.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are in a physics laboratory, standing in front of Searle's apparatus. This classic setup is designed to measure the Young's modulus of a material by stretching a long wire. In our problem, we have a wire of length $L = 110.0 	ext{ cm}$ and a diameter $d = 0.050 	ext{ cm}$. We apply a stretching force by suspending a weight of $F = 50 	ext{ N}$, which produces a tiny extension $l = 0.125 	ext{ cm}$.

But here is the catch—no measurement is perfect! Every instrument has a limitation, known as its least count. The least count represents the maximum possible absolute error in that measurement. For our length, the scale's least count is $\Delta L = 0.1 	ext{ cm}$. For the diameter, the screw gauge gives us $\Delta d = 0.001 	ext{ cm}$. And for the extension, the micrometer provides $\Delta l = 0.001 	ext{ cm}$. Our goal is to find how these tiny uncertainties propagate and affect our final calculation of Young's modulus.

### The Master Equation

Before we can find the error, we need to know the actual value of Young's modulus ($Y$). The fundamental definition of Young's modulus is the ratio of tensile stress to tensile strain:

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

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

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

Now, let's plug in our measured values. **Crucially**, we must convert all lengths from centimeters to meters to ensure our final answer is in standard SI units ($	ext{N/m}^2$).

$$Y = \frac{4 	imes 50 	imes 1.1}{\pi 	imes (5.0 	imes 10^{-4})^2 	imes (1.25 	imes 10^{-3})}$$

After crunching the numbers, we find the value of Young's modulus:

$$Y = 2.24 	imes 10^{11} 	ext{ N/m}^2$$

### Propagating the Errors

Now comes the beautiful part—error analysis. How do the errors in $L$, $d$, and $l$ combine to create an error in $Y$? According to the rules of error propagation for products and quotients, the maximum fractional error in the result is the sum of the fractional errors of the individual variables.

Furthermore, if a variable is raised to a power, that power becomes a multiplier for its fractional error. Since our formula has $d^2$ in the denominator, the fractional error for the diameter will be multiplied by 2.

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

Let's substitute our least counts as the absolute errors. Notice a neat trick here: because fractional error is a dimensionless ratio, we don't need to convert the units to meters! As long as the numerator and denominator are both in centimeters, the units will perfectly cancel out.

$$\frac{\Delta Y}{Y} = \left(\frac{0.1}{110.0}ight) + 2\left(\frac{0.001}{0.050}ight) + \left(\frac{0.001}{0.125}ight)$$

Calculating each term gives us:

$$\frac{\Delta Y}{Y} = 0.000909 + 0.04 + 0.008$$

$$\frac{\Delta Y}{Y} = 0.0489$$

This means our measurement has a maximum fractional error of $0.0489$, or about $4.89\%$.

### Final Calculation

We are almost there! We have the fractional error, but the question asks for the maximum absolute error ($\Delta Y$). To find this, we simply multiply the fractional error by the original value of Young's modulus we calculated earlier.

$$\Delta Y = Y 	imes 0.0489$$

$$\Delta Y = (2.24 	imes 10^{11}) 	imes 0.0489$$

$$\Delta Y = 1.09 	imes 10^{10} 	ext{ N/m}^2$$

And there we have it! The maximum error in the measurement of Young's modulus is **$1.09 	imes 10^{10} 	ext{ N/m}^2$**.

As a final thought, look back at the individual fractional errors. The error from the diameter ($0.04$) was significantly larger than the errors from the length ($0.000909$) and the extension ($0.008$). This is a profound insight: in Searle's experiment, the precision of the screw gauge used to measure the diameter is the most critical factor in determining the overall accuracy of your result!

Visualized Solution

$Searle’s Experiment Setup$

$Young’s Modulus Formula$

$Calculating Y$

$Value of Y$

$Maximum Fractional Error$

$Substituting Errors$

$Computing Fractional Error$

$Maximum Absolute Error$

$Conclusion$

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

Analyzing the Setup

The Master Equation

Propagating the Errors

Final Calculation

Similar Questions

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The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Searle’s Experiment Setup

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

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

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

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

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

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

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

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