- Home
- Standard 11
- Physics
A brass rod of length $2\,m$ and cross-sectional area $2.0\,cm^2$ is attached end to end to a steel rod of length $L$ and cross-sectional area $1.0\,cm^2$ . The compound rod is subjected to equal and opposite pulls of magnitude $5 \times 10^4\,N$ at its ends. If the elongations of the two rods are equal, then length of the steel rod $(L)$ is ........... $m$ $(Y_{Brass}=1.0\times 10^{11}\,N/m^2$ and $Y_{Steel} = 2.0 \times 10^{11}\,N/m^2)$
$1.5$
$1.8$
$1$
$2$
Solution
$\ell_{\mathrm{B}}=2 \mathrm{m} \quad \ell_{\mathrm{S}}=\mathrm{L}$
$A_{B}=2 c m^{2} A_{s}=1 \mathrm{cm}^{2}$
$\Delta \ell_{\mathrm{B}}=\Delta \ell_{\mathrm{s}}$
$\frac{\mathrm{F}}{\mathrm{A}_{\mathrm{B}}} \frac{\ell_{\mathrm{B}}}{\mathrm{Y}_{\mathrm{B}}}=\frac{\mathrm{F}}{\mathrm{A}_{\mathrm{S}}} \frac{\ell_{\mathrm{S}}}{\mathrm{Y}_{\mathrm{S}}}$
$\mathrm{L}=\frac{\mathrm{A}_{\mathrm{s}} \mathrm{Y}_{\mathrm{s}}}{\mathrm{A}_{\mathrm{B}} \mathrm{Y}_{\mathrm{B}}} \ell_{\mathrm{B}}=\frac{1}{2} \times \frac{2 \times 10^{11}}{10^{11}} \times 2=2$
