Q.A steel wire of length 4.7 m and cross-sectional area $3.0 \times 10^{-5}\ \text{m}^{2}$ stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of $4.0 \times 10^{-5}\ \text{m}^{2}$ under a given load. What is the ratio of the Young's modulus of steel to that of copper?

The key idea is Young's Modulus, which relates stress to strain in an elastic material.

1. We are given that both wires experience the same load $F$ and stretch by the same amount $\Delta L$.
2. For steel, $Y_s = \frac{F L_s}{A_s \Delta L}$.
3. For copper, $Y_c = \frac{F L_c}{A_c \Delta L}$.
4. Taking the ratio $\frac{Y_s}{Y_c} = \frac{F L_s / (A_s \Delta L)}{F L_c / (A_c \Delta L)} = \frac{L_s A_c}{L_c A_s}$.
5. Substituting the given values: $\frac{Y_s}{Y_c} = \frac{(4.7\ \text{m}) \times (4.0 \times 10^{-5}\ \text{m}^2)}{(3.5\ \text{m}) \times (3.0 \times 10^{-5}\ \text{m}^2)} = \frac{4.7 \times 4.0}{3.5 \times 3.0} = \frac{18.8}{10.5} \approx 1.79$

When a material is subjected to a deforming force, it undergoes a change in shape or size. For a wire pulled along its length, this change is an elongation. Young's Modulus is a fundamental property of a material that quantifies its stiffness or resistance to elastic deformation under tensile or compressive stress. It tells us how much a material will stretch or compress when a certain force is applied.

The core idea is that stress (force per unit area) causes strain (fractional change in length). Young's Modulus ($Y$) is defined as the ratio of stress to strain:

For a wire of original length $L$, cross-sectional area $A$, subjected to a tensile force $F$ that causes an elongation $\Delta L$:

• Stress $= \frac{F}{A}$
• Strain $= \frac{\Delta L}{L}$

Therefore, the formula for Young's Modulus becomes:

This formula is key because it connects the material property ($Y$) with the physical dimensions ($L, A$) and the observed deformation ($\Delta L$) under a given force ($F$). In this problem, we are told that both wires experience the "same amount" of stretch ($\Delta L$) under a "given load" ($F$). This means $F$ and $\Delta L$ are identical for both wires, allowing us to set up an equation to find the ratio of their Young's moduli.

1. Identify the given information and the goal.

   We are given the following parameters for a steel wire and a copper wire:

   • Steel wire (subscript $s$):
     • Length, $L_s = 4.7\ \text{m}$
     • Cross-sectional area, $A_s = 3.0 \times 10^{-5}\ \text{m}^2$
   • Copper wire (subscript $c$):
     • Length, $L_c = 3.5\ \text{m}$
     • Cross-sectional area, $A_c = 4.0 \times 10^{-5}\ \text{m}^2$

   We are also told two crucial conditions:

   • Both wires stretch by the same amount: $\Delta L_s = \Delta L_c = \Delta L$
   • Both wires are under a given (same) load: $F_s = F_c = F$

   Our goal is to find the ratio of Young's modulus of steel to that of copper, i.e., $\frac{Y_s}{Y_c}$.

2. Express elongation in terms of Young's Modulus.

   From the Young's Modulus formula $Y = \frac{FL}{A\Delta L}$, we can rearrange it to solve for the elongation $\Delta L$:

This equation shows that for a given force $F$, a longer wire ($L$), a smaller cross-sectional area ($A$), or a smaller Young's Modulus ($Y$) will result in a greater elongation.

3. Apply the elongation formula to both wires.

For the steel wire, the elongation is:

For the copper wire, the elongation is:

4. Use the condition of equal elongation and load. Since $\Delta L_s = \Delta L_c$ and $F_s = F_c = F$, we can set the two expressions for elongation equal to each other:

The force $F$ cancels out from both sides, as it is the same for both wires:

5. Solve for the ratio of Young's moduli. We need to find $\frac{Y_s}{Y_c}$. Rearranging the equation from the previous step:

Now, substitute the given numerical values:

Notice that the units of length (m) and area ($\text{m}^2$) cancel out, as do the powers of $10^{-5}$, leaving a dimensionless ratio, which is expected for a ratio of Young's moduli.

Rounding to two significant figures, consistent with the input data:

Step 1: deltaL=FL/(AY). Step 2: equal deltaL for both wires under same F: L_s/(A_sY_s)=L_c/(A_cY_c). Step 3: Y_s/Y_c=(L_sA_c)/(A_sL_c). Step 4: =(4.74.0e-5)/(3.0e-53.5)=18.8/10.5~=1.8.