Question

four cylindrical rods with various cross - sectional areas and initial lengths are stretched by an applied force, as shown in figure below. the resulting change in length of each rod is given in the following table. task: rank these rods in order of increasing youngs modulus. indicate ties where appropriate.

rod	applied force	cross - sectional area	initial length	change in length
2	f	2a	2l	δl
3	2f	2a	l	2δl
4	3f	a	l/2	δl

a. y1 = y3 = y2 = y4
b. y4 = y1 < y2 < y3
c. y1 < y4 < y2 < y3
d. y1 = y3 < y2 < y4
e. y3 < y1 = y2 < y4

Explanation:

Step1: Recall Young's modulus formula

The formula for Young's modulus $Y$ is $Y=\frac{F/A}{\Delta L/L}=\frac{FL}{A\Delta L}$.

Step2: Calculate Young's modulus for rod 1

For rod 1, with $F = F$, $A = A$, $L = L$ and $\Delta L=\Delta L$, $Y_1=\frac{F\times L}{A\times\Delta L}=\frac{FL}{A\Delta L}$.

Step3: Calculate Young's modulus for rod 2

For rod 2, $F = F$, $A = 2A$, $L = 2L$ and $\Delta L=\Delta L$. Then $Y_2=\frac{F\times2L}{2A\times\Delta L}=\frac{FL}{A\Delta L}$.

Step4: Calculate Young's modulus for rod 3

For rod 3, $F = 2F$, $A = 2A$, $L = L$ and $\Delta L = 2\Delta L$. So $Y_3=\frac{2F\times L}{2A\times2\Delta L}=\frac{FL}{2A\Delta L}$.

Step5: Calculate Young's modulus for rod 4

For rod 4, $F = 3F$, $A = A$, $L=\frac{L}{2}$ and $\Delta L=\Delta L$. Then $Y_4=\frac{3F\times\frac{L}{2}}{A\times\Delta L}=\frac{3FL}{2A\Delta L}$.

Step6: Compare the values of Young's modulus

Comparing $Y_1=\frac{FL}{A\Delta L}$, $Y_2=\frac{FL}{A\Delta L}$, $Y_3=\frac{FL}{2A\Delta L}$ and $Y_4=\frac{3FL}{2A\Delta L}$, we have $Y_3

Answer:

e. $Y_3