Question

A brass rod of length 2 m and cross-sectional area 2.0 $c{m}^2$ is attached end to end to a steel rod of length L and cross-sectional area 1.0 $c{m}^2.$ 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 the length of the steel rod (L) is
($Y_{Brass}=1.0\times {10}^{11}N/m^{2}and{Y}_{Steel}=2.0\times {10}^{11}N/m^2$)

• A. 1.5 m
• B. 1.8 m
• C. 1 m
• D. 2 m

Answer 1

The correct option is D 2 m

$k=\frac{YA}{L}$

$k_S=\frac{Y_S{A}_S}{L},k_B=\frac{Y_B{A}_B}{L_B}$

$F=5\times {10}^{4}N$

$\Delta l_S=\Delta l_b=\Delta l$

$F=k_B\Delta l$

$\frac{Y_S{A}_S}{L}=\frac{Y_B{A}_B}{L_B}$

$\left(\frac{Y_S}{Y_B}\right).\left(\frac{L_B}{L}\right)=\frac{A_B}{A_S}$

$2\times \frac{2}{L}=2$

$L=2m$