$1.05\, m$ लंबाई तथा नगण्य द्रव्यमान की एक छड़ को बराबर लंबाई के दो तारों, एक इस्पात का (तार $A$ ) तथा दूसरा ऐलुमिनियम का तार (तार $B$) द्वारा सिरों से लटका दिया गया है, जैसा कि चित्र में दिखाया गया है। $A$ तथा $B$ के तारों के अनुप्रस्थ परिच्छेद के क्षेत्रफल क्रमशः $1.0\, mm ^{2}$ और $2.0\, mm ^{2}$ हैं। छड़ के किसी बिंदु से एक द्रव्यमान $m$ को लटका दिया जाए ताकि इस्पात तथा ऐलुमिनियम के तारों में $(a)$ समान प्रतिबल तथा $(b)$ समान विकृति उत्पन्न हो।

(a) $0.7 m$ from the steel-wire end $0.432 m$ from the steel-wire end

Cross-sectional area of wire $A , a_{1}=1.0 mm ^{2}=1.0 \times 10^{-6} m ^{2}$

Cross-sectional area of wire $B , a_{2}=2.0 mm ^{2}=2.0 \times 10^{-6} m ^{2}$

Young's modulus for steel, $Y_{1}=2 \times 10^{11} Nm ^{-2}$

Young's modulus for aluminium, $Y_{2}=7.0 \times 10^{10} Nm ^{-2}$

Let a small mass $m$ be suspended to the rod at a distance $y$ from the end where wire $A$ is attached. Stress in the wire $=\frac{\text { Force }}{\text { Area }}=\frac{F}{a}$

If the two wires have equal stresses, then:

$\frac{F_{1}}{a_{1}}=\frac{F_{2}}{a_{2}}$

$F_{1}=$ Force exerted on the steel wire

$F_{2}=$ Force exerted on the aluminum wire

$\frac{F_{1}}{F_{2}}=\frac{a_{1}}{a_{2}}=\frac{1}{2}$

The situation is shown in the following figure.

Taking torque about the point of suspension, we have:

$F_{1} y=F_{2}(1.05-y)$

$\frac{F_{1}}{F_{2}}=\frac{(1.05-y)}{y}$

$\frac{(1.05-y)}{y}=\frac{1}{2}$

$2(1.05-y)=y$

$2.1-2 y=y$

$3 y=2.1$

$\therefore y=0.7 m$

In order to produce an equal stress in the two wires, the mass should be suspended at a distance of $0.7 \;m$ from the end where wire $A$ is attached.

Young's modulus $=\frac{\text { Stress }}{\text { Strain }}$

Strain $=\frac{\text { Stress }}{\text { Young's modulus }}=\frac{\frac{F}{a}}{Y}$

If the strain in the two wires is equal, then:

$\frac{\frac{F_1}{a_{1}}}{Y_{1}}=\frac{\frac{F_2}{a_{2}}}{Y_{2}}$

$\frac{F_{1}}{F_{2}}=\frac{a_{1}}{a_{2}} \frac{Y_{1}}{Y_{2}}=\frac{1}{2} \times \frac{2 \times 10^{11}}{7 \times 10^{10}}=\frac{10}{7}$

Taking torque about the point where mass $m$, is suspended at a distance $y_{1}$ from the side where wire A attached, we get:

$F_{1} y_{1}=F_{2}\left(1.05-y_{1}\right)$

$\frac{F_{1}}{F_{2}}=\frac{\left(1.05-y_{1}\right)}{y_{1}}$

$\frac{\left(1.05-y_{1}\right)}{y_{1}}=\frac{10}{7}$

$7\left(1.05-y_{1}\right)=10 y_{1}$

$17 y_{1}=7.35$

$\therefore y_{1}=0.432 m$

In order to produce an equal strain in the two wires, the mass should be suspended at a distance of $0.432 m$ from the end where wire $A$ is attached.