The elastic limit of brass is 3.5 times 10^{1 | vedclass

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(D) The stress is defined as the force per unit area,$	ext{Stress} = \frac{F}{A}$.For the elastic limit,the stress is given as $3.5 	imes 10^{10} \,

Vedclass · Accepted Answer

$1.55$

Vedclass · Answer

(D) The stress is defined as the force per unit area,$	ext{Stress} = \frac{F}{A}$.For the elastic limit,the stress is given as $3.5 	imes 10^{10} \, N/m^2$.The area of the cross-section of the wire is $A = \pi r^2$,where $r$ is the radius.The diameter is $0.75 \, mm$,so the radius $r = \frac{0.75}{2} 	imes 10^{-3} \, m$.Substituting these values into the formula $F = A 	imes 	ext{Stress}$:$F = \pi 	imes \left( \frac{0.75}{2} 	imes 10^{-3} ight)^2 	imes 3.5 	imes 10^{10}$$F = 3.14 	imes \frac{0.5625}{4} 	imes 10^{-6} 	imes 3.5 	imes 10^{10}$$F = 3.14 	imes 0.140625 	imes 3.5 	imes 10^4$$F \approx 1.545 	imes 10^4 \, N \approx 1.55 	imes 10^4 \, N$.

The elastic limit of brass is $3.5 \times 10^{10} \, N/m^2$. Find the maximum load that can be applied to a brass wire of $0.75 \, mm$ diameter without exceeding the elastic limit $....... \times 10^4 \, N$.

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