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 for a wire of radius $r_1$ and area $A_1 = \pi r_1^2$.Let $F_2$ be the force for a wire of radius $r_2 = 2r_1$ and area $A_2 = \pi r_2^2 = \pi (2r_1)^2 = 4\pi r_1^2$.Since the breaking stress is constant,$\frac{F_1}{A_1} = \frac{F_2}{A_2}$.$F_2 = F_1 	imes \frac{A_2}{A_1} = 20 	imes \frac{4\pi r_1^2}{\pi r_1^2} = 20 	imes 4 = 80\,kg\,wt$.

$A$ 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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