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Step-by-Step Solution
Step 1: Understand the problem
Two wires—one made of steel and the other of copper—are connected end to end and stretched by the same load $W$. We know their lengths, Young’s moduli, and radius. The total elongation of the two-wire system is given. We need to find the load $W$ causing this net elongation.
Step 2: Express the elongations for each wire
The elongation $\Delta l$ of a wire under tension is given by
$ \Delta l = \dfrac{F \, l}{Y \, A}, $
where $F$ is the force (or load) on the wire, $l$ is its original length, $Y$ is its Young’s modulus, and $A$ is the cross-sectional area.
For the steel wire (subscript s) of length $l_s$ and Young’s modulus $Y_s$:
$ \Delta l_s = \dfrac{W \, l_s}{Y_s \, A}. $
For the copper wire (subscript c) of length $l_c$ and Young’s modulus $Y_c$:
$ \Delta l_c = \dfrac{W \, l_c}{Y_c \, A}. $
Step 3: Relate the total elongation to the individual elongations
The total elongation given is $1.4 \,\text{mm} = 1.4 \times 10^{-3} \,\text{m}$. Hence,
$ \Delta l_s + \Delta l_c = 1.4 \times 10^{-3}. $
Substitute the expressions for $\Delta l_s$ and $\Delta l_c$:
$ \dfrac{W \, l_s}{Y_s \, A} + \dfrac{W \, l_c}{Y_c \, A} = 1.4 \times 10^{-3}. $
Step 4: Factor out the load $W$
Taking $W$ common:
$ W \Bigl(\dfrac{l_s}{Y_s \, A} + \dfrac{l_c}{Y_c \, A}\Bigr) = 1.4 \times 10^{-3}. $
Solving for $W$:
$ W = \dfrac{1.4 \times 10^{-3}}{\dfrac{l_s}{Y_s \, A} + \dfrac{l_c}{Y_c \, A}}. $
Step 5: Compute the cross-sectional area $A$
Both wires have the same radius $r = 1.4 \,\text{mm} = 1.4 \times 10^{-3} \,\text{m}$. So,
$ A = \pi r^2 = \pi \times (1.4 \times 10^{-3})^2
= 1.96 \times 10^{-6} \,\pi \,\text{m}^2. $
Given $\pi = \tfrac{22}{7}$.
Step 6: Substitute numerical values and simplify
Steel wire length: $l_s = 3.2 \,\text{m}$, Young’s modulus: $Y_s = 2.0 \times 10^{11} \,\text{N/m}^2.$
Copper wire length: $l_c = 4.4 \,\text{m}$, Young’s modulus: $Y_c = 1.1 \times 10^{11} \,\text{N/m}^2.$
Net elongation: $1.4 \times 10^{-3} \,\text{m}.$
Cross-sectional area: $A = 1.96 \times 10^{-6} \,\pi \,\text{m}^2.$
Thus,
$ W = \dfrac{1.4 \times 10^{-3}}{\dfrac{3.2}{(2.0 \times 10^{11}) \times (1.96 \times 10^{-6} \,\pi)} + \dfrac{4.4}{(1.1 \times 10^{11}) \times (1.96 \times 10^{-6} \,\pi)}}. $
Step 7: Final result
Carrying out the calculation (using $\pi \approx \tfrac{22}{7}$), we get:
$ W \approx 154 \,\text{N}. $
Answer: 154 N
