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

First, recall the definition of normal stress. We have the basic formula

$$\sigma=\frac{F}{A}$$

where $$\sigma$$ is the normal stress, $$F$$ is the axial load, and $$A$$ is the cross-sectional area on which the load acts.

The rod is circular, so its cross-sectional area is given by

$$A=\frac{\pi d^{2}}{4},$$

where $$d$$ is the diameter.

The material is brass with an elastic (safe) stress limit of $$\sigma_{\text{allow}}=379\;\text{MPa}$$. Since $$1\;\text{MPa}=1\;\text{N/mm}^2$$, we can write

$$\sigma_{\text{allow}} = 379\;\text{N/mm}^2.$$

The allowable stress must not be exceeded, so we set

$$\sigma \le \sigma_{\text{allow}}.$$

For the minimum diameter, we make the working stress exactly equal to the allowable value:

$$\frac{F}{A}= \sigma_{\text{allow}}.$$

The load is $$F = 400\;\text{N}$$, hence

$$\frac{400}{A}=379 \quad\Longrightarrow\quad A=\frac{400}{379}\;\text{mm}^2.$$

Carrying out the division gives

$$A\approx 1.055\;\text{mm}^2.$$

Next, substitute this area into the circular area formula to get the diameter:

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

Replacing $$A$$ by $$1.055\;\text{mm}^2$$ gives

$$d^{2}= \frac{4\times 1.055}{\pi}.$$

First compute the numerator:

$$4\times 1.055 = 4.220.$$

Now divide by $$\pi$$:

$$\frac{4.220}{\pi}\approx \frac{4.220}{3.1416}\approx 1.343.$$

Taking the square root yields

$$d = \sqrt{1.343}\;\text{mm}\approx 1.159\;\text{mm}.$$

Rounding to three significant figures, we obtain

$$d\approx 1.16\;\text{mm}.$$

Hence, the correct answer is Option C.