A student performs an experiment to determine the Young's modulus of a wire, exactly 2 m long, by Se

Q: Why do we ignore the errors in mass, gravity, and length?

The problem explicitly states that the length is 'exactly 2 m', the load is 'exactly 1.0 kg', and $g$ is 'exact'. Exact values have zero uncertainty, so their fractional errors are zero.

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

In the formula for Young's modulus, the area of the cross-section is $\frac{\pi d^2}{4}$. When we take the natural logarithm and differentiate to find the fractional error, the exponent 2 on the diameter comes down as a multiplier.

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

When calculating the maximum permissible error, we always add the absolute values of the fractional errors. This ensures we account for the worst-case scenario where all errors compound in the same direction.

Q: Why did we round off the final error to $0.2 \times 10^{11}$ instead of keeping $0.225 \times 10^{11}$?

The uncertainties in the given measurements ($\pm 0.05$ and $\pm 0.01$) have only one significant figure. According to the rules of significant figures, the final calculated error should not be more precise than the least precise measurement, so we round it to one significant digit.

Q: What is Searle's method?

Searle's method is a classic laboratory experiment used to determine the Young's modulus of a material. It involves hanging a known weight from a wire and measuring the microscopic extension using a micrometer or vernier scale attached to a reference wire.

Question

A student performs an experiment to determine the Young's modulus of a wire, exactly $2	ext{ m}$ long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be $0.8	ext{ mm}$ with an uncertainty of $\pm 0.05	ext{ mm}$ at a load of exactly $1.0	ext{ kg}$. The student also measures the diameter of the wire to be $0.4	ext{ mm}$ with an uncertainty of $\pm 0.01	ext{ mm}$. Take $g = 9.8	ext{ m/s}^2$ (exact). The Young's modulus obtained from the reading is close to

Options:

(A) $(2.0 \pm 0.3) 	imes 10^{11}	ext{ N/m}^2$

(B) $(2.0 \pm 0.2) 	imes 10^{11}	ext{ N/m}^2$

(C) $(2.0 \pm 0.1) 	imes 10^{11}	ext{ N/m}^2$

(D) $(2.0 \pm 0.05) 	imes 10^{11}	ext{ N/m}^2$

Sigma Percentile · Accepted Answer

Imagine you are in a physics laboratory, standing in front of a classic setup known as Searle's apparatus. You have a long, thin wire suspended from a rigid support, and you are about to determine one of its most fundamental properties: the Young's modulus. This problem is a beautiful journey that combines pure mechanical theory with the rigorous reality of experimental uncertainties.

### The Master Equation

We begin by recalling the definition of Young's modulus, $Y$, which is the ratio of tensile stress to tensile strain.

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

Here, the force $F$ is provided by the suspended mass, so $F = Mg$. The cross-sectional area $A$ of the wire can be expressed in terms of its diameter $d$ as $A = \frac{\pi d^2}{4}$. Substituting these into our equation gives us the master formula:

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

### Calculating the Base Value

Before we worry about the errors, let's calculate the base value of $Y$ using the exact measurements provided. The problem is generous enough to give us exact values for the mass ($1.0	ext{ kg}$), the acceleration due to gravity ($9.8	ext{ m/s}^2$), and the length of the wire ($2	ext{ m}$).

We must be careful to convert the diameter and extension from millimeters to meters to maintain SI units:

$$Y = \frac{4(1.0)(9.8)(2)}{\pi (0.4 	imes 10^{-3})^2 (0.8 	imes 10^{-3})}$$

Evaluating this expression yields:

$$Y = 2.0 	imes 10^{11} 	ext{ N/m}^2$$

### The Reality of Experimental Error

In the real world, no measurement is perfect. The diameter and the extension were measured with instruments that have inherent uncertainties. To find out how these small errors propagate into our final calculated value of $Y$, we use the method of fractional errors.

By taking the natural logarithm of our master equation and differentiating, we find the maximum permissible fractional error:

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

Since $M$, $g$, and $L$ are given as exact values, their uncertainties are zero. This simplifies our equation significantly:

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

**Notice the factor of 2!** Because the diameter is squared in the area formula, its fractional error contributes twice as much to the final error. This is a classic trap, so always watch out for exponents.

### Final Calculation

Now, we substitute the given uncertainties into our simplified error equation:

$$\Delta Y = Y \left( 2\frac{\Delta d}{d} + \frac{\Delta l}{l} ight)$$

$$\Delta Y = (2.0 	imes 10^{11}) \left( 2 	imes \frac{0.01}{0.4} + \frac{0.05}{0.8} ight)$$

$$\Delta Y = (2.0 	imes 10^{11}) (0.05 + 0.0625) = 0.225 	imes 10^{11} 	ext{ N/m}^2$$

Following the rules of significant figures, since our input uncertainties (like $0.01$ and $0.05$) have only one significant digit, we must round our final error to one significant digit as well:

$$\Delta Y \approx 0.2 	imes 10^{11} 	ext{ N/m}^2$$

Combining our base value with this uncertainty, we arrive at the final, scientifically rigorous result:

$$Y = (2.0 \pm 0.2) 	imes 10^{11} 	ext{ N/m}^2$$

This perfectly matches option (b). The beauty of this problem lies in how it forces us to respect both the mathematical theory and the physical limitations of our measuring instruments.

Select Answer:

Visualized Solution

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

$Calculating Y$

$Y = 2.0 \times 1 0^{11} N/m^{2}$

$Error Analysis$

$\frac{Δ Y}{Y} = 2 \frac{Δ d}{d} + \frac{Δ l}{l}$

$Substituting Errors$

$Δ Y = 0.2 \times 1 0^{11} N/m^{2}$

$Final Result$

The Sigma Insight: Errors in Measurement

The Master Equation

Calculating the Base Value

The Reality of Experimental Error

Final Calculation

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Select Answer:

Visualized Solution

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

.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; } Y=2.0×1011 N/m2

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } YΔY​=2dΔd​+lΔl​

.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; } ΔY=0.2×1011 N/m2

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