Q.
A light rod of length $2m$ suspended from the ceiling horizontally by means of two vertical wires of equal length. A weight $W$ is hung from a light rod as shown in figure. The rod hung by means of a steel wire of cross-sectional area $A_{1} = 0.1\ cm^{2}$ and brass wire of cross-sectional area $A_{2} = 0.2\ cm^{2}$. To have equal stress in both wires, $T_{1}/T_{2} =$
A) 1/3
B) 1/4
C) 4/3
D) 1/2
Explanation
(d) Equal stress
$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} \Rightarrow \frac{F_{1}}{F_{2}} = \frac{0.1}{0.2} = \frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.