9.18 A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

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Vishal Baghel | Contributor-Level 10

8 months ago

Cross-sectional area of wire A, $a_{1}$ = 1 ${m m}^{2}$ = 1 $* 10^{- 6} m^{2}$

Cross-sectional area of wire B, $a_{2}$ = 2 ${m m}^{2}$ = 2 $* 10^{- 6} m^{2}$

Young's modulus for steel, $Y_{1}$ = 2 $* 10^{11}$ N/ $m^{2}$

Young's modulus for aluminium, $Y_{2}$ = 7 $* 10^{10}$ N/ $m^{2}$

Stress in the wire = $\frac{F o r c e}{a r e a}$ = $\frac{F}{a}$

If the two wires have equal stresses, then

$\frac{F_{1}}{a_{1}}$ = $\frac{F_{2}}{a_{2}}$ or $\frac{F_{1}}{F_{2}}$ = $\frac{a_{1}}{a_{2}}$ = $\frac{1}{2}$ ………(i)

Where $F_{1}$ is the force exerted on steel wire and $F_{2}$ is the

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