In an experiment to find Young's modulus, three different wires W₁, W₂ and W₃ of the same material have been used and obtained three curves G₁, G₂ and G₃ respectively for the graph of applied tensile force F with extension Δ L as shown in the figure. Which of the following statements, made for the reason of obtaining different graphs, is true?

(3) The wire W₃ may have the same area of cross-section as W₁ but a length larger than W₁.

(1) The wire W₁ may have a larger length and a smaller area of cross-section than W₂.

(2) The wire W₁ may have the same length as W₂ but a smaller area of cross-section than W₂.

(4) The wire W₂ may have a smaller area of cross-section, but a larger length than W₃.

(5) The wire W₃ may have a larger value for the ratio, (Area of cross-section)/Length than that of W₁.

2018 A/L Physics Past Paper · Question 30 (English Medium)

Young's modulus — F–ΔL graphs of three wires

In an experiment to find Young's modulus, three different wires $W_1$ , $W_2$ and $W_3$ of the same material have been used and obtained three curves $G_1$ , $G_2$ and $G_3$ respectively for the graph of applied tensile force $F$ with extension $\Delta L$ as shown in the figure. Which of the following statements, made for the reason of obtaining different graphs, is true?

(1) The wire $W_1$ may have a larger length and a smaller area of cross-section than $W_2$ .
(2) The wire $W_1$ may have the same length as $W_2$ but a smaller area of cross-section than $W_2$ .
(3) The wire $W_3$ may have the same area of cross-section as $W_1$ but a length larger than $W_1$ .
(4) The wire $W_2$ may have a smaller area of cross-section, but a larger length than $W_3$ .
(5) The wire $W_3$ may have a larger value for the ratio, $\dfrac{\text{Area of cross-section}}{\text{Length}}$ than that of $W_1$ .

Answer and explanation

Answer: (3) The wire $W_3$ may have the same area of cross-section as $W_1$ but a length larger than $W_1$ .

What the question asks

We want: why the graphs of the three wires have different slopes.

Relevant theory

In the distortions under tensile and compressive forces, E is referred to as Young's modulus (Y)

Young's modulus:

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

So the slope of the graph is:

$\frac{F}{\Delta L} = \frac{YA}{L}$

All three are the same material. So $Y$ is the same.

So the slope goes with $A/L$ .

Reading the options

In the figure $G_1$ is the steepest. Then $G_2$ . Then $G_3$ .

So we need:

$\left(\frac{A}{L}\right)_1 > \left(\frac{A}{L}\right)_2 > \left(\frac{A}{L}\right)_3$

Test each statement against this.

Reasoning and calculation

Statement (1): $W_1$ is longer and thinner. So $(A/L)_1 < (A/L)_2$ . That does not fit the graph.

Statement (2): the same length, $W_1$ thinner. So $(A/L)_1 < (A/L)_2$ . It does not fit.

Statement (3): the same $A$ , $W_3$ longer. So $(A/L)_3 < (A/L)_1$ . It fits the graph.

Statement (4): $W_2$ thinner and longer. So $(A/L)_2 < (A/L)_3$ . It does not fit.

Statement (5): $(A/L)_3 > (A/L)_1$ . Then $G_3$ would be the steepest. It does not fit.

The four rejected statements make the same mistake.

They give a smaller $A/L$ to the graph with the larger slope.

Why the other options fail

This option says: $W_1$ is longer and thinner than $W_2$ .

Then $W_1$ has the smaller $A/L$ . But $G_1$ is the steepest.

So it is wrong.

This option says: $W_1$ has the same length as $W_2$ and a smaller area.

At the same length, a smaller area gives a smaller slope. But $G_1$ is steeper than $G_2$ .

This option says: $W_2$ is thinner and longer than $W_3$ .

Then $W_2$ has a smaller $A/L$ than $W_3$ . But $G_2$ is steeper than $G_3$ .

This option says: $W_3$ has a larger $A/L$ than $W_1$ .

Then $G_3$ would be the steepest. But it is the least steep.

Conclusion

Only statement (3) is left.

With the same area, a longer wire stretches more for the same force.

So its graph $G_3$ is the least steep.

A quick check.

A very long wire stretches easily. Its slope gets close to zero.

A very thick wire hardly stretches. Its slope is very large.

Both fit "the slope goes with $A/L$ ".

$\frac{F}{\Delta L} = \frac{YA}{L}$

The slope is:

$\frac{F}{\Delta L} = \frac{YA}{L}$

$G_3$ is the least steep. So $W_3$ has the smallest $A/L$ .

The same $A$ with a larger $L$ does that.

The answer is option (3).

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