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A steel and a brass wire, each of length $50 \mathrm{~cm}$ and cross-sectional area $0.005 \mathrm{~cm}^{2}$ hang from a ceiling and are $15 \mathrm{~cm}$ apart. Lower ends of the wires are attached to a light horizontal bar. A suitable downward load is applied to the bar so that each of the wires extends in length by $0.1 \mathrm{~cm}$. At what distance from the steel wire the load must be applied? [Young's modulus of steel is $2 \times 10^{12}$ dynes $/ \mathrm{cm}^{2}$ and that of brass is $1 \times 10^{12}$ dynes $\left./ \mathrm{cm}^{2}\right]$
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1632 Upvotes
Verified Answer
The correct answer is:
$5 \mathrm{~cm}$
Hint:
At equilibrium, Taking torque about point $\mathrm{O}$
$\begin{aligned}
T_{1} x=T_{2}(15-x) & \Rightarrow Y_{1} x=Y_{2}(15-x) \\
\left[Y=\frac{T}{A} \times \frac{L}{\ell} ; T \propto Y\right] & 1 \times 10^{12} x=2 \times 10^{12}(15-x) ; 3 x=15, \quad \text { So, } x=5 \mathrm{~cm}
\end{aligned}$
At equilibrium, Taking torque about point $\mathrm{O}$
$\begin{aligned}
T_{1} x=T_{2}(15-x) & \Rightarrow Y_{1} x=Y_{2}(15-x) \\
\left[Y=\frac{T}{A} \times \frac{L}{\ell} ; T \propto Y\right] & 1 \times 10^{12} x=2 \times 10^{12}(15-x) ; 3 x=15, \quad \text { So, } x=5 \mathrm{~cm}
\end{aligned}$
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