The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit?

Properties of Matter PYQ — JEE Main 2019 · 10 Apr · Shift 2 · Q67 with Solution

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Correct answer: A

The rod must not exceed its elastic limit, so the maximum allowed stress is

\sigma = 379\,\text{MPa} = 379 \times 10^6\,\text{N/m}^2.

Also,

\sigma = \frac{F}{A}

so the minimum required cross-sectional area is

A = \frac{F}{\sigma} = \frac{400}{379 \times 10^6}\,\text{m}^2.

A \approx 1.055 \times 10^{-6}\,\text{m}^2.

For a circular rod,

A = \frac{\pi d^2}{4}.

Hence,

d = \sqrt{\frac{4A}{\pi}}.

Substituting,

d = \sqrt{\frac{4(1.055 \times 10^{-6})}{\pi}}.

d \approx \sqrt{1.343 \times 10^{-6}} \approx 1.159 \times 10^{-3}\,\text{m}.

Converting to mm:

d \approx 1.159\,\text{mm} \approx 1.16\,\text{mm}.

The required minimum diameter is:

\boxed{1.16\,\text{mm}}

So the correct option is A.