If the breaking stress of a wire is 3.18 time | vedclass

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(D) $1$. Calculate the tension $T$ in the string using the formula for a pulley system with masses $m_1 = 1 \, kg$ and $m_2 = 4 \, kg$:$T = \frac{2 m_

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$4 	imes 10^{-5} \, m$

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(D) $1$. Calculate the tension $T$ in the string using the formula for a pulley system with masses $m_1 = 1 \, kg$ and $m_2 = 4 \, kg$:$T = \frac{2 m_1 m_2}{m_1 + m_2} g = \frac{2 	imes 1 	imes 4}{1 + 4} 	imes 10 = \frac{8}{5} 	imes 10 = 16 \, N$.$2$. The breaking force is given by the product of breaking stress and the cross-sectional area $A = \pi r^2$:$F_{breaking} = 	ext{Stress} 	imes A = 3.18 	imes 10^{10} 	imes \pi r^2$.$3$. For the wire not to break,the tension $T$ must be less than or equal to the breaking force:$16 = 3.18 	imes 10^{10} 	imes \pi r^2$.$4$. Solve for the radius $r$:$r^2 = \frac{16}{3.18 	imes 10^{10} 	imes 3.14} \approx \frac{16}{9.9852 	imes 10^{10}} \approx 1.602 	imes 10^{-9} \approx 16 	imes 10^{-10} \, m^2$.$r = \sqrt{16 	imes 10^{-10}} = 4 	imes 10^{-5} \, m$.

Column $I$	Column $II$
$A$. Hooke's law	$1$. Tangential strain
$B$. Shearing strain	$2$. Temporary loss of elastic property
$C$. Bulk strain	$3$. Elastic limit
$D$. Elastic fatigue	$4$. $3$ times the linear strain

If the breaking stress of a wire is $3.18 \times 10^{10} \, N/m^2$,what should be its minimum radius so that the wire does not break? (Refer to the figure for the masses attached to the pulley system,assume $g = 10 \, m/s^2$)

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Match the following:
Column $I$ Column $II$
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