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Q9.1:
A steel wire of length 4.7 m and cross-sectional area $3.0 \times 10^{-5} m^2$ stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of $4.0 \times 10^{-5} m^2$ under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Solution :

Step 1: Identifying Given Variables

For the steel wire (subscript $s$):

  • Length ($L_s$) = $4.7 \, \text{m}$
  • Area of cross-section ($A_s$) = $3.0 \times 10^{-5} \, \text{m}^2$
  • Extension ($\Delta L_s$) = $\Delta L$
  • Load (Force) ($F_s$) = $F$

For the copper wire (subscript $c$):

  • Length ($L_c$) = $3.5 \, \text{m}$
  • Area of cross-section ($A_c$) = $4.0 \times 10^{-5} \, \text{m}^2$
  • Extension ($\Delta L_c$) = $\Delta L$
  • Load (Force) ($F_c$) = $F$

Step 2: Defining the Young's Modulus Formula

By the definition of Young's Modulus ($Y$), the ratio of longitudinal stress to longitudinal strain is constant for a given material [Per Hooke's Law]:

$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A \Delta L}$

Step 3: Setting up the Ratio Equation

Since the load $F$ and the extension $\Delta L$ are identical for both wires, we express $Y_s$ and $Y_c$ as follows:

$Y_s = \frac{F \cdot L_s}{A_s \cdot \Delta L}$

$Y_c = \frac{F \cdot L_c}{A_c \cdot \Delta L}$

To find the ratio $\frac{Y_s}{Y_c}$, we perform the division:

$\frac{Y_s}{Y_c} = \left( \frac{F \cdot L_s}{A_s \cdot \Delta L} \right) / \left( \frac{F \cdot L_c}{A_c \cdot \Delta L} \right)$

By algebraic simplification [Canceling common terms $F$ and $\Delta L$]:

$\frac{Y_s}{Y_c} = \frac{L_s}{A_s} \times \frac{A_c}{L_c} = \frac{L_s \cdot A_c}{A_s \cdot L_c}$

Step 4: Calculation

Substitute the given numerical values into the derived equation:

$\frac{Y_s}{Y_c} = \frac{4.7 \times 4.0 \times 10^{-5}}{3.0 \times 10^{-5} \times 3.5}$

$\frac{Y_s}{Y_c} = \frac{4.7 \times 4.0}{3.0 \times 3.5}$

$\frac{Y_s}{Y_c} = \frac{18.8}{10.5}$

$\frac{Y_s}{Y_c} \approx 1.79047...$

Visual Representation of Experimental Setup

Steel Load F Copper Load F Both experience extension ΔL

Final Solution: The ratio of the Young’s modulus of steel to that of copper is 1.79 : 1.


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