To Determine Young's Modules Of Elasticity Of Material Of Given Wire

Young's Modulus of Elasticity is a fundamental property of materials that quantifies their ability to withstand changes in length when subjected to tensile or compressive forces. Essentially, it measures the stiffness of a material. The higher the Young's Modulus, the stiffer the material is, and the less it will deform under a given load. This property is crucial in engineering and construction, where materials are selected based on their ability to maintain shape and integrity under stress. For example, steel, with a high Young's Modulus, is commonly used in the construction of buildings and bridges due to its strength and rigidity. Conversely, materials like rubber, with a low Young's Modulus, are used in applications requiring flexibility and elasticity, such as tyres and shock absorbers. Understanding Young's Modulus helps engineers design structures and products that are both safe and functional, balancing strength and flexibility as needed.

Aim

To determine Young's modulus of elasticity of the material of a given wire

Apparatus

Searle's apparatus, two long steel wires of the same length and diameter, a meter scale, a screw gauge, eight 0.5 kg slotted weights and a 1 kg hanger.

Theory

If a wire of length L and radius r is loaded by a weight Mg and if $l$ be the increase in length.

$\begin{aligned} & \text { Then, Normal stress }=\frac{M g}{\pi r^2} \\ & \text { and Longitudinal strain }=\frac{l}{L} \\ & \text { Hence, Young's modulus }=\frac{\text { Normal stress }}{\text { Longitudinal strain }} \\ & \text { or } \quad Y=\frac{M g / \pi r^2}{l / L} \\ & \text { or } \quad Y=\frac{M g L}{\pi r^2 l}\end{aligned}$

where

Y= Young's modulus

M= Mass
$
\begin{aligned}
& \mathrm{L}=\text { length of wire } \\
& \mathrm{r}=\text { radius of the wire } \\
& l=\text { Extention }
\end{aligned}
$
Knowing $L$ and $r$, and finding $l$ for known $M g, Y$ can be calculated.

Diagram

Procedure

1. Take two steel wires of the same length and diameter and tight their ends in torsion screws A, B and C, D as shown in the diagram. Wire AB becomes an experimental wire and wire CD becomes auxiliary wire.
2. Suspend a 1 kg dead load from the hook of frame F₂.

3. Suspend a 1 kg hanger and eight 0.5 kg slotted weights from the hook of frame F₁, The experimental wire becomes taut.
4. Remove kinks from the experimental wire by pressing the wire between the nails of the right-hand thumb and first finger (through a handkerchief) and moving them along the length of the wire.

5. Remove all slotted weights from the hanger. Now each wire is equally loaded with 1 kg weight.
6. Measure the length of the experimental wire from tip A to tip B using a meter scale.
7. Find the pitch and the least count of the screw gauge.
8. Measure the diameter of the experimental wire at five different places, equally spaced along the length (near two ends, two-quarter distance from ends and middle). At each place, measure the diameter along with two mutually perpendicular directions. Record the observations in the table.

9. Note the breaking stress for the steel fable of constants. Multiply that by the cross-section area of the wire to find the breaking load of the wire. The maximum load is not to exceed one-third of the breaking load.
10. Find the pitch and the least count of the spherometer screw attached to frame F₁.
11. Adjust the spherometer screw such that the bubble in the spirit level is exactly in the centre. Note the reading of the spherometer disc. This reading is recorded against zero load.

12. Gently slip a 0.5 kg slotted weight into the hanger and wait for two minutes to allow the wire to extend fully. The bubble moves up from the centre.
13. Rotate the spherometer screw to bring the bubble back to the centre. Note the reading of the spherometer disc. This reading is recorded against a 1 kg load in the load-increasing column.
14. Repeat steps 12 and 13 till all eight 0.5 kg slotted weights have been used (now the total load on the experimental wire is 5 kg which must be one-third of the breaking load).

15. Now remove one slotted weight (load decreasing), and wait for two minutes to allow the wire to contract fully. The bubble moves down from the centre.
16. Repeat step 13, The reading is recorded against load in the hanger in load decreasing column.

17. Repeat steps 15 and 16 till all the eight slotted weights are removed (now the load on the experimental wire is 1 kg of the initial load). (Observations for the same load in load increasing and in load decreasing columns must not differ much. Their mean is taken to eliminate the error.)
18. Record your observations.

Result
1. The Young's modulus for steel as determined by Searle's apparatus $=\ldots ... N m^{-2}$
2. Straight-line graph between load and extension shows that stress $\propto$ strain. This verifies Hooke's Law.

Percentage Error
Actual value of $Y$ for steel (from tables) $=\ldots \ldots \mathrm{Nm}^{-2}$
Difference in values $\quad=\ldots \ldots . \mathrm{Nm}^{-2}$
Percentage error $=\frac{\text { Difference in values }}{\text { Actual value }}=\ldots \ldots \%$
It is within the limits of experimental error.

Solved Examples Based on Determine Young's Modules of Elasticity of Material of Given Wire

Example 1: The Young's modulus of a wire of radius R and length L is Y N/M². If the radius and length are changed to 2R and 4 L respectively, then its Young's modulus will be?

1) 2 Y

2) Y

3) O

4) 8 Y

Solution:

$Y=\frac{M g L}{\pi r^2 l}$

Young's modulus is the property of the material and hence its value remains constant.

Hence, the answer is the option (2).

Example 2: In an experiment to determine Young's modulus of a wire of a length of exactly 1 m, the extension in the length of the wire is measured as 0.4 mm with an uncertainty of $\pm 0.02 \mathrm{~mm}$ when a load of 1 kg is applied. The diameter of the wire is measured as 0.4 mm with an uncertainty of $\pm 0.01 \mathrm{~mm}$.The error in the measurement of Young's modulus $(\Delta \mathrm{Y})$ is found to be $\mathrm{x} \times 10^{10} \mathrm{Nm}{ }^{-2}$. The value of x is $\qquad$
$\left(\right.$ take $\left.g=10 \mathrm{~ms}^{-2}\right)$

1) 2

2) 3

3) 4

4) 5

Solution:
$
\begin{aligned}
& \mathrm{L}=1 \mathrm{~m} \\
& \text { Extension }=1=0.4 \mathrm{~mm} \\
& =4 \times 10^{-4} \mathrm{~m} \\
& \Delta \mathrm{L}=0.02 \mathrm{~mm} \\
& =2 \times 10^{-5} \mathrm{~m} \\
& \mathrm{~m}=1 \mathrm{~kg} \\
& \mathrm{~d}=0.4 \mathrm{~mm}=4 \times 10^{-4} \mathrm{~m} \\
& \Delta d=1 \times 10^{-5} \mathrm{~m} \\
& Y=\frac{m g}{A} \times \frac{L}{l}
\end{aligned}
$
$
\begin{aligned}
& \frac{\Delta \mathrm{Y}}{\mathrm{Y}}=\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{\Delta \mathrm{L}}{\mathrm{L}}+\frac{2 \Delta \mathrm{d}}{\mathrm{d}}+\frac{\Delta \mathrm{l}}{\mathrm{l}} \\
& =0+0+2\left(\frac{0.01}{0.4}\right)+\left(\frac{0.02}{0.4}\right) \\
& =\frac{1}{20}+\frac{1}{20}=\frac{2}{20} \\
& \Delta \mathrm{Y}=\frac{\mathrm{Y}}{10}=\frac{\mathrm{mg} \times \mathrm{L}}{(\mathrm{Al}) 10} \\
& \Delta \mathrm{Y}=\frac{1 \times 10 \times 1}{3.14 \times \frac{16 \times 10^{-8}}{4} \times 0.4 \times 10^{-3} \times 10} \\
& \Delta \mathrm{Y}=2 \times 10^{10}
\end{aligned}
$
The Value of $x$ is 2

Hence, the answer is the option (1).

Example 3: A metallic wire of length 2.5 meters and diameter 0.8 mm is subjected to a tensile force of 800 N, causing it to elongate by 2.2 mm. The density of the material of the wire is 7.8 g/cm³. Calculate the Young’s modulus of elasticity of the material of the wire.

1) $9.5 \times 10^{10} \mathrm{~N} / \mathrm{m}^2$
2) $1.81 \times 10^{12} \mathrm{~N} / \mathrm{m}^2$
3) $2.4 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
4) $3.6 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

Solution:

To calculate the Young's modulus of elasticity (E) of the material of the wire, we use the formula:

$\left[E=\frac{F \cdot L}{A \cdot \Delta L}\right]$

Where:
F is the tensile force $(800 \mathrm{~N})$,
L is the original length of the wire $(2.5 \mathrm{~m})$,
A is the cross-sectional area of the wire, $(\Delta L)$ is the elongation of the wire $(2.2 \mathrm{~mm})$.

First, convert the diameter of the wire to meters and find the cross-sectional area:
$
\begin{aligned}
& {\left[\text { Diameter }=0.8 \mathrm{~mm}=0.8 \times 10^{-3} \mathrm{~m}\right]} \\
& {\left[\text { Radius }=\frac{0.8 \times 10^{-3}}{2}=0.4 \times 10^{-3} \mathrm{~m}\right]} \\
& {\left[A=\pi r^2=\pi\left(0.4 \times 10^{-3}\right)^2=\pi \times 0.16 \times 10^{-6}=0.5024 \times 10^{-6} \mathrm{~m}^2\right]}
\end{aligned}
$
Convert the elongation to meters:

$
\left[\Delta L=2.2 \mathrm{~mm}=2.2 \times 10^{-3} \mathrm{~m}\right]
$
Substitute the values into the formula:

$
\begin{aligned}
& {\left[E=\frac{800 \times 2.5}{0.5024 \times 10^{-6} \times 2.2 \times 10^{-3}}\right]} \\
& {\left[E=\frac{2000}{1.10528 \times 10^{-9}}\right]} \\
& {\left[E=1.809 \times 10^{12} \mathrm{~N} / \mathrm{m}^2\right]}
\end{aligned}
$
Thus, Young's modulus of elasticity of the material of the wire is approximately $\left(1.81 \times 10^{12} \mathrm{~N} / \mathrm{m}^2\right)$.

Hence, the answer is the option (2).

Example 4: A copper wire of length 2.0 m and diameter 1.2 mm is suspended vertically from a ceiling. A mass of 4.5 kg is attached to the lower end of the wire. Given that the density of copper is 8.92 g/cm³ , calculate the elongation of the wire due to the weight of the attached mass.

1) 0.000631m

2) 170 m

3) 200 m

4) 20 m

Solution:

The elongation of the wire can be calculated using Hooke’s law, which states that the elongation of a material is directly proportional to the applied force (stress) and inversely proportional to the material’s Young’s modulus.

The formula for elongation (∆L) is given by:

$\Delta L=\frac{F \cdot L}{A \cdot Y}$

Where:

F is the force applied (weight of the mass)

L is the original length of the wire

A is the cross-sectional area of the wire

Y is the Young’s modulus of the material

Given data:

Original length (L) = 2.0 m

Diameter (d) = 1.2 mm

Mass (m) = 4.5 kg

Density of copper (ρ) = 8.92 g/cm³

Young’s modulus of copper (Y ) = 1.25 × 1011 N/m²

First, calculate the cross-sectional area (A) using the diameter:
$
A=\frac{\pi d^2}{4}=\frac{\pi\left(1.2 \times 10^{-3}\right)^2}{4} \approx 1.1309 \times 10^{-6} \mathrm{~m}^2
$
Next, calculate the force (F) due to the weight:

$
\mathrm{F}=\mathrm{m} \cdot \mathrm{g}
$
Where g is the acceleration due to gravity $\left(9.81 \mathrm{~m} / \mathrm{s}^2\right)$.

$
F=4.5 \cdot 9.81 \approx 44.145 \mathrm{~N}
$
Now, plug in the values into the elongation formula:

$
\Delta L=\frac{44.145 \cdot 2.0}{1.1309 \times 10^{-6} \cdot 1.25 \times 10^{11}} \approx 0.000631 \mathrm{~m}
$
The elongation of the wire due to the weight of the attached mass is approximately 0.000631 m.

Hence, the answer is the option (1).

Example 5: A metallic wire of length 2.5 meters and diameter 0.8 mm is subjected to a tensile force of 800 N, causing it to elongate by 2.2 mm. The density of the material of the wire is 7.8 g/cm³ . Calculate the Young’s modulus of elasticity of the material of the wire.

1) $9.5 \times 10^{10} \mathrm{~N} / \mathrm{m}^2$
2) $1.2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
3) $2.4 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
4) $3.6 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

Solution:

Young’s modulus (Y ) is a measure of the stiffness of a material. It’s defined as the ratio of stress (σ) to strain (ϵ) and is given by the formula:
$
Y=\frac{\text { Stress }}{\text { Strain }}
$
Where: Stress $(\sigma)=$ Force / Area Strain $(\epsilon)=$ Change in length / Original length Given data:
Original length $(L)=2.5 \mathrm{~m}$ Diameter $(\mathrm{d})=0.8 \mathrm{~mm}$ Tensile force $(F)=800 \mathrm{~N}$ Elongation $(\Delta \mathrm{L})=2.2 \mathrm{~mm}$ Density $(\rho)=7.8 \mathrm{~g} / \mathrm{cm}^3$ First, let's calculate the cross-sectional area $(A)$ of the wire using its diameter:

$
A=\frac{\pi d^2}{4}
$
Now, convert the diameter to meters and calculate the area:

$
A=\frac{\pi \times\left(0.8 \times 10^{-3}\right)^2}{4} \mathrm{~m}^2
$
Next, calculate the strain ( $\boldsymbol{\varepsilon}$ ) using the elongation and original length:

$
\epsilon=\frac{\Delta L}{L}
$
Convert elongation to meters and calculate strain:

$
\epsilon=\frac{2.2 \times 10^{-3}}{2.5} \mathrm{~m}
$
Now, calculate stress $(\sigma)$ using the formula $\sigma=F / A$ :

$
\sigma=\frac{F}{A}
$

Substitute the given values:
$
\sigma=\frac{800}{\frac{\pi \times\left(0.8 \times 10^{-3}\right)^2}{4}} \mathrm{~N} / \mathrm{m}^2
$
Finally, calculate Young's modulus ( $Y$ ) using the formula $Y=\sigma / \varepsilon$ :
Substitute the calculated values for stress and strain:

$
Y=\frac{\frac{800}{\frac{\pi \times\left(0.8 \times 10^{-3}\right)^2}{4}}}{\frac{2.2 \times 10^{-3}}{2.5}} \mathrm{~N} / \mathrm{m}^2
$
Solve for Y, and you'll find that the value is approximately $1.2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$,

Hence, the answer is the option (2).

Summary

An example of how to calculate Young's modulus of a wire is by measuring its length change when a particular weight is suspended on it. It is done by attaching masses to the suspended wire before using calibrated devices to determine their increase. This would be achieved by plotting out stress or force per unit area against strain or proportional deformation before finding out the gradient in order to see where this falls into line with equilibrium.