JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Elasticity & SHM

We want the thinnest rod that still keeps the stress at or below the elastic limit. Stress is force over area, and for a circular cross-section of diameter $d$ the area is $\frac{\pi d^2}{4}$, so the limiting condition is $\sigma =\frac{F}{\pi d^2/4}=\frac{4F}{\pi d^2}$. Rearranging for the diameter, $d=\sqrt{\frac{4F}{\pi \sigma }}=\sqrt{\frac{4\times 400}{\pi \times 379\times 10^6}}$. Working the arithmetic gives $d=\sqrt{1.34\times 10^{-6}}\approx 1.16\times 10^{-3}\text{m}$, that is about $1.16\text{mm}$.