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(C) The breaking force of a wire is determined by its breaking stress,which is a material property. The breaking force $F$ is given by the product of

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$4F$

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(C) The breaking force of a wire is determined by its breaking stress,which is a material property. The breaking force $F$ is given by the product of breaking stress $\sigma$ and the cross-sectional area $A$ of the wire.$F = \sigma 	imes A = \sigma 	imes (\pi R^2)$Since the material is the same (copper),the breaking stress $\sigma$ remains constant.Therefore,$F \propto R^2$.If $F_1 = F$ for radius $R_1 = R$,and $F_2$ is the force for radius $R_2 = 2R$,then:$\frac{F_2}{F_1} = \left(\frac{R_2}{R_1}ight)^2 = \left(\frac{2R}{R}ight)^2 = 4$$F_2 = 4F_1 = 4F$.

$A$ force $F$ is needed to break a copper wire having radius $R$. The force needed to break a copper wire of radius $2R$ will be

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