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(B) The tension $T$ in the wire when two masses $m_1 = 1 \ kg$ and $m_2 = 2 \ kg$ are connected over a pulley is given by $T = \frac{2 m_1 m_2 g}{m_1

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(B) The tension $T$ in the wire when two masses $m_1 = 1 \ kg$ and $m_2 = 2 \ kg$ are connected over a pulley is given by $T = \frac{2 m_1 m_2 g}{m_1 + m_2}$.Substituting the values: $T = \frac{2 	imes 1 	imes 2 	imes 10}{1 + 2} = \frac{40}{3} \ N$.Breaking stress is defined as $\sigma = \frac{T}{A}$,where $A = \pi r^2$ is the cross-sectional area.Given $\sigma = \frac{40}{3 \pi} 	imes 10^6 \ N m^{-2}$.Equating the two: $\frac{40}{3 \pi} 	imes 10^6 = \frac{40/3}{\pi r^2}$.Simplifying: $10^6 = \frac{1}{r^2} \Rightarrow r^2 = 10^{-6} \ m^2$.Therefore,$r = 10^{-3} \ m = 1 \ mm$.

Two blocks of masses $1 \ kg$ and $2 \ kg$ are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is $\frac{40}{3 \pi} \times 10^6 \ N m^{-2}$. What should be the minimum radius of the wire used if it should not break (in $mm$)? $\left(g = 10 \ m s^{-2}\right)$

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