A wire can be broken by applying a load of 20 | vedclass

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(C) Breaking stress is a material property and remains constant for a given material.Breaking stress $= \frac{F}{A} = 	ext{constant}$.Let $F_1 = 20 \

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$80$

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(C) Breaking stress is a material property and remains constant for a given material.Breaking stress $= \frac{F}{A} = 	ext{constant}$.Let $F_1 = 20 \, kg \, wt$ be the force required to break a wire of radius $r_1$.Let $F_2$ be the force required to break a wire of radius $r_2 = 2r_1$ (since the diameter is doubled).Since the breaking stress is constant: $\frac{F_1}{A_1} = \frac{F_2}{A_2}$.$F_2 = F_1 	imes \frac{A_2}{A_1} = F_1 	imes \frac{\pi r_2^2}{\pi r_1^2}$.Substituting $r_2 = 2r_1$: $F_2 = F_1 	imes \frac{(2r_1)^2}{r_1^2} = F_1 	imes 4$.$F_2 = 20 	imes 4 = 80 \, kg \, wt$.

$A$ wire can be broken by applying a load of $20 \, kg$ weight. The force required to break the wire of twice the diameter is .......... $kg \, wt$.

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