The elastic limit of brass is $379\,MPa.$ .......... $mm$ should be the minimum diameter of a brass rod if it is to support a $400\,N$ load without exceeding its elastic limit .
JEE MAIN 2019, Medium
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$\frac{{400}}{{\frac{\pi }{4}{d^2}}} = 379 \times {10^6}$
${d^2} = \frac{{4 \times 400 \times {{10}^{ - 6}}}}{{\pi \times 379}} = 0.336 \times {10^{ - 6}} \times 4$
$d = 2\sqrt {0.336} \times {10^{ - 3}}M \simeq 1.16\,mm$
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