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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(d)
Stress $=\frac{ F }{ A } ;$ For elastic limit, stress $=3.5 \times 10^{10} N / m ^2$ (given).
Thus, $F=\left(\pi r^2\right) \times$ stress
$=3.14 \times\left(\frac{0.75}{2} \times 10^{-3}\right)^2 \times 3.5 \times 10^{10}$
$=3.14 \times \frac{(0.75)^2}{4} \times 3.5 \times 10^4=1.55 \times 10^4\,N$
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