The density and breaking stress of a wire are | vedclass

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(A) Let $\rho$ be the density,$\sigma$ be the breaking stress,$A$ be the cross-sectional area,and $L$ be the length of the wire.At the point of breaki

Vedclass · Accepted Answer

$600$

Vedclass · Answer

(A) Let $ho$ be the density,$\sigma$ be the breaking stress,$A$ be the cross-sectional area,and $L$ be the length of the wire.At the point of breaking,the stress at the top of the wire due to its own weight is equal to the breaking stress.The weight of the wire is $W = mg = (ho A L) g'$.Here,$g'$ is the acceleration due to gravity on the planet,given by $g' = \frac{g}{3} = \frac{10}{3} \ m/s^2$.The breaking stress is $\sigma = \frac{W}{A} = \frac{ho A L g'}{A} = ho L g'$.Rearranging for $L$,we get $L = \frac{\sigma}{ho g'}$.Substituting the given values: $\sigma = 1.2 	imes 10^8 \ N/m^2$,$ho = 6 	imes 10^4 \ kg/m^3$,and $g' = \frac{10}{3} \ m/s^2$.$L = \frac{1.2 	imes 10^8}{6 	imes 10^4 	imes (10/3)} = \frac{1.2 	imes 10^8 	imes 3}{6 	imes 10^4 	imes 10} = \frac{3.6 	imes 10^8}{6 	imes 10^5} = 0.6 	imes 10^3 = 600 \ m$.

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