Two blocks of masses 3 , kg and 5 , kg are co | vedclass

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(C) The tension $T$ in the wire connecting two masses $m_1$ and $m_2$ over a smooth pulley is given by:$T = \frac{2 m_1 m_2 g}{m_1 + m_2}$Substituting

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$12.5$

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(C) The tension $T$ in the wire connecting two masses $m_1$ and $m_2$ over a smooth pulley is given by:$T = \frac{2 m_1 m_2 g}{m_1 + m_2}$Substituting the given values $m_1 = 3 \, kg$,$m_2 = 5 \, kg$,and $g = 10 \, ms^{-2}$:$T = \frac{2 	imes 3 	imes 5 	imes 10}{3 + 5} = \frac{300}{8} = 37.5 \, N$Stress is defined as force per unit area,so $	ext{Stress} = \frac{T}{A} = \frac{T}{\pi R^2}$.Given the breaking stress is $\frac{24}{\pi} 	imes 10^2 \, Nm^{-2}$,we set:$\frac{24}{\pi} 	imes 10^2 = \frac{37.5}{\pi R^2}$$2400 = \frac{37.5}{R^2}$$R^2 = \frac{37.5}{2400} = \frac{375}{24000} = \frac{1}{64} \, m^2$$R = \sqrt{\frac{1}{64}} = \frac{1}{8} \, m = 0.125 \, m$Converting to centimeters: $R = 0.125 	imes 100 \, cm = 12.5 \, cm$.

Two blocks of masses $3 \, kg$ and $5 \, kg$ are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is $\frac{24}{\pi} \times 10^{2} \, Nm^{-2}$. What is the minimum radius of the wire? (Take $g = 10 \, ms^{-2}$) (in $cm$)

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