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Two blocks of masses $1 \mathrm{~kg}$ and $2 \mathrm{~kg}$ are connected by a metal wire going over a smooth pulley. The breaking stress of metal is $\frac{40}{3 \pi} \times 10^6 \mathrm{Nm}^{-2}$. What should be the minimum radius of wire used if it should not break? $\left(g=10 \mathrm{~ms}^{-2}\right)$
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The correct answer is:
1.5 mm
$\begin{aligned} & \text { Breaking stress }=\frac{F}{A} \\ & \text { Here, breaking stress }=\frac{40}{3 \pi} \times 10^6 \mathrm{Nm}^{-2} \\ & \Rightarrow \quad \frac{40}{3 \pi} \times 10^6=\frac{3 \times 10}{\pi r^2} \\ & \Rightarrow \quad r^2=\frac{9}{4} \times 10^{-6}\end{aligned}$
or $\quad \begin{aligned} r & =1.5 \times 10^{-3} \mathrm{~m} \\ & =1.5 \mathrm{~mm}\end{aligned}$
or $\quad \begin{aligned} r & =1.5 \times 10^{-3} \mathrm{~m} \\ & =1.5 \mathrm{~mm}\end{aligned}$
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