Young's modulus of a wire by Searle's method

Aim

To find Young's modulus of the material of a wire by Searle's method.

Why a long thin wire?

Hang a heavy bag from a steel wire. The wire gets longer, but only by a fraction of a millimetre. We cannot see this with a ruler. So the experiment needs two things: a way to apply known loads, and a way to measure a very small extension. Searle's apparatus does both.

The idea

Stretch a wire of length $L$ and cross-section $A$ with a force $F$. Its length increases by $\Delta L$. Three quantities describe what happens:

• Stress is the force on each unit of area, $\dfrac{F}{A}$, in N/m² (pascal).
• Strain is the fractional change of length, $\dfrac{\Delta L}{L}$. It has no unit.
• Young's modulus $Y$ is the ratio of the two, for a wire that is not stretched too far:

For a round wire of diameter $d$, $A = \dfrac{\pi d^2}{4}$. The force is the weight of the load, $F = Mg$. So

Within the elastic limit, the extension is in proportion to the load (Hooke's law). The graph of $\Delta L$ against $M$ is then a straight line through the origin, and $\dfrac{\Delta L}{M}$ is its slope. So

Steel has $Y = 2.0 \times 10^{11}\ \text{N/m}^2$. Copper has about $1.1 \times 10^{11}\ \text{N/m}^2$ and brass about $0.9 \times 10^{11}\ \text{N/m}^2$.

The apparatus

Searle's apparatus has two wires of the same material, the same length and the same diameter. They hang side by side from one rigid support.

• The reference wire carries a small constant weight, which keeps it straight.
• The experimental wire carries a hanger for slotted weights.
• Each wire ends in a metal frame. A spirit level is hinged to the frame of the reference wire. Its other end rests on the tip of a micrometer screw fixed to the frame of the experimental wire. The screw has a pitch of $1$ mm and $100$ divisions, so its least count is $0.01$ mm.

Why two wires? If the room warms up, or the support sags a little, both wires change together, and the spirit level does not tilt. Only the extra load on the experimental wire tilts the level. So the extension of that wire alone is measured.

The steps

1. Measure the length $L$ of the experimental wire from the clamp to the frame with a metre scale.
2. Measure the diameter $d$ with a screw gauge at five places along the wire, and in two directions at right angles at each place. Correct for the zero error and take the mean.
3. With only the dead weight on the experimental wire, turn the micrometer screw until the bubble of the level is at the centre. Read the screw. This is the zero of the load scale.
4. Add a $1$ kg slotted weight. The wire stretches and the bubble moves. Turn the screw until the bubble is at the centre again, and read the screw. Repeat for $2, 3, \ldots, 6$ kg.
5. Unload in the same steps and read the screw each time.
6. For each load take the mean of the loading and the unloading readings. The extension $\Delta L$ is that mean minus the zero reading.

Why unload as well? A wire does not return exactly to the same length at once after the load is taken off. The two sets of readings differ a little, and their mean cancels this elastic after-effect.

The readings

A steel wire has $L = 1.50$ m and a mean diameter of $0.70$ mm. The extensions are:

One reading. Take $M = 4$ kg, $\Delta L = 0.77$ mm.

1. The force. $F = Mg = 4\ \text{kg} \times 9.8\ \text{m/s}^2$ $= 39.2$ N.
2. The area. $A = \dfrac{\pi d^2}{4}$ $= \dfrac{\pi \times (0.70 \times 10^{-3}\ \text{m})^2}{4}$ $= 3.85 \times 10^{-7}\ \text{m}^2$.
3. The modulus. $Y = \dfrac{FL}{A\,\Delta L}$. The numerator is $FL = 39.2\ \text{N} \times 1.50\ \text{m}$ $= 58.8\ \text{N m}$. The denominator is $A\,\Delta L = 3.85 \times 10^{-7}\ \text{m}^2$ $\times 0.77 \times 10^{-3}\ \text{m}$ $= 2.96 \times 10^{-10}\ \text{m}^3$. So $Y = \dfrac{58.8\ \text{N m}}{2.96 \times 10^{-10}\ \text{m}^3}$ $= 1.98 \times 10^{11}\ \text{N/m}^2$.

From the graph. The least-squares line through the origin has a slope of $0.191$ mm/kg. Then $Y = \dfrac{4gL}{\pi d^2 \times \text{slope}}$ $= 2.00 \times 10^{11}\ \text{N/m}^2$, which is the accepted value for steel. A graph is better than one reading, because it uses all six.

The error. The load is known to $0.01$ kg, $L$ to $0.001$ m, $d$ to $0.01$ mm and $\Delta L$ to $0.01$ mm. For the reading above,

So $Y$ $= (1.98 \pm 0.09) \times 10^{11}\ \text{N/m}^2$. The diameter counts twice, so it gives most of the error. The extension is the second largest part.

Set $M = 4$ kg, then halve the diameter. By what factor does the extension change? Why is the wire of a Searle's apparatus thin?

Precautions and sources of error

• Same wires. The two wires must have the same length, diameter and material, or the reference wire does not cancel the changes.
• A dead weight first. It straightens the wires, and the first reading is not spoiled by kinks.
• The elastic limit. Never load the wire past it. A wire that is stretched too far does not return, and Hooke's law fails. The graph bends away from a straight line.
• Steady readings. Wait a few seconds after each change of load. Bring the bubble to the centre by turning the screw in one direction only, to avoid backlash.
• The diameter. A wire is seldom perfectly round or even. Measure it at many places. A $0.01$ mm error in $d$ matters most.
• Slipping. A wire that slips in its clamp adds a false extension. Tighten the clamps first.

Practise with new numbers

Try it in Python

The program takes the table of the lesson, fits the line through the origin, and finds Young's modulus. It also finds the stress at the greatest load.

Common mistakes

• Taking $d$ as the radius. In $\dfrac{4MgL}{\pi d^2 \Delta L}$, $d$ is the diameter.
• Mixing units. Put the extension in metre when the other lengths are in metre.
• Using $L$ as the full length of the wire. $L$ is the length that is stretched, from the clamp to the frame.
• Loading past the elastic limit. Then the extension is no longer in proportion to the load.
• Forgetting the weight of the hanger. Count it in the first load, or measure the extension from the hanger alone.

Remember these

• $Y = \dfrac{F/A}{\Delta L/L} = \dfrac{4MgL}{\pi d^2\,\Delta L}$, and $Y = \dfrac{4gL}{\pi d^2 \times \text{slope}}$ from the graph of $\Delta L$ against $M$.
• Within the elastic limit, the graph is a straight line through the origin.
• The reference wire removes the effect of temperature and of a sagging support.
• $\dfrac{\Delta Y}{Y}$ has the term $2\dfrac{\Delta d}{d}$, which is usually the largest.
• Load and unload, and take the mean of the two readings.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why does Searle's apparatus use a second wire, the reference wire?

2. A wire of length $2.00$ m and diameter $0.60$ mm stretches by $0.55$ mm under a load of $2.0$ kg. What is its Young's modulus? Take $g = 9.8\ \text{m/s}^2$.

3. A graph of extension against load for a wire starts straight and then bends upwards. What does the bend show?

4. Which of $L$, $d$ and $\Delta L$ gives the largest share of the percentage error in $Y$ for the values $L = 1.50 \pm 0.001$ m, $d = 0.70 \pm 0.01$ mm and $\Delta L = 0.77 \pm 0.01$ mm?

References

• NCERT, Physics Class XI, Chapter 8 (Mechanical Properties of Solids), the section on Young's modulus.
• NCERT, Laboratory Manual: Physics, Class XI, the experiment on Young's modulus of a wire by Searle's method.
• OpenStax University Physics Volume 1, Chapter 12 (Static Equilibrium and Elasticity), the section on stress, strain and elastic moduli.