Question

Two wires of equal cross-section but one made of steel and the other of copper, are joined end to end. When the combination is kept under tension, the elongations in the two wires are found to be equal. The ratio of the lengths of the two wires is found to be $\frac{10n}{11}$. Then, the value of $n$ is

Young’s modulus of steel $=2.0\times {10}^{11}{\text{N/m}}^2$ and that of copper$=1.1\times {10}^{11}{\text{N/m}}^2$

Answer 1

$\Delta l_{st}=\Delta l_{cu}$

Using Hooke's law,
$\frac{\Delta l_{st}}{l_{st}}=\frac{\text{Stress}}{Y_{st}}$...(1)

$\frac{\Delta l_{cu}}{l_{cu}}=\frac{\text{Stress}}{Y_{cu}}$.......(2)

Since stress is same and strain is also same,
Divide (1) by (2)
$\frac{l_{st}}{l_{cu}}=\frac{Y_{st}}{Y_{cu}}$ ...(3)
$Y_{st}=2\times {10}^{11}N{m}^{-2}$
$Y_{cu}=1.1\times {10}^{11}N{m}^{-2}$
Substituting values in (3)
$\frac{l_{st}}{l_{cu}}=\frac{20}{11}$...(4)
By comparing (4) with the given expression
Value of $n=2$