Question

A brass rod of length 2 m and cross-sectional area $2{\text{cm}}^2$ is attached end to end to a steel rod of length L and cross-sectional area $1{\text{cm}}^2$. The compound rod is subjected to equal and opposite pulls of magnitude $5\times {10}^4\text{N}$ at its ends. If the elongations of the two rods are equal, then the length (in m) the steel rod(L) is___

$[Y_B=1\times {10}^{11}\text{N}/{\text{m}}^2$ and $Y_S=2\times {10}^{11}\text{N}/{\text{m}}^2]$

Answer 1

$\Delta L_{\text{S}}=\Delta L_{\text{B}}$

$\frac{{\text{FL}}_{\text{S}}}{{\text{Y}}_{\text{S}}{\text{A}}_{\text{S}}}=\frac{{\text{FL}}_{\text{B}}}{{\text{Y}}_{\text{B}}{\text{A}}_{\text{B}}}$

$\text{L}=\frac{2}{{10}^{11}\times 2}\times 2\times {10}^{11}\times 1=2\text{m}$