Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is 1 : 4, the ratio of their diameters is:

We begin with the expression for the elastic (strain) energy stored per unit volume in a stretched wire. For a material that obeys Hooke’s law the strain energy density is

$$u \;=\; \frac12\,\sigma\,\varepsilon$$

where $$\sigma$$ is the normal stress and $$\varepsilon$$ is the linear strain. Because strain for an elastic material is related to stress through Young’s modulus $$E$$ by $$\varepsilon = \dfrac{\sigma}{E}$$, we can substitute and obtain

$$u \;=\; \frac12\,\sigma \left(\frac{\sigma}{E}\right) \;=\; \frac{\sigma^{2}}{2E}.$$

Both wires are made of steel, so the Young’s modulus $$E$$ is the same for each. Hence, for comparison of the two wires, the factor $$\dfrac{1}{2E}$$ is common and the energy density is directly proportional to the square of the stress:

$$u \;\propto\; \sigma^{2}.$$

Each wire carries the same load $$F$$, but because their diameters are different, their cross-sectional areas differ. For a circular cross-section of diameter $$d$$, the area is

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

Stress is force divided by area, so

$$\sigma \;=\; \frac{F}{A} \;=\; \frac{F}{\pi d^{2}/4} \;=\; \frac{4F}{\pi d^{2}}.$$

This shows that

$$\sigma \;\propto\; \frac{1}{d^{2}}.$$

Substituting this proportionality into the proportionality for energy density, we get

$$u \;\propto\; \sigma^{2} \;\propto\; \left(\frac{1}{d^{2}}\right)^{2} \;=\; \frac{1}{d^{4}}.$$

Therefore, for two wires (1) and (2) with diameters $$d_{1}$$ and $$d_{2}$$, the ratio of their strain energies per unit volume is

$$\frac{u_{1}}{u_{2}} \;=\; \frac{1/d_{1}^{4}}{1/d_{2}^{4}} \;=\; \frac{d_{2}^{4}}{d_{1}^{4}}.$$

The problem states that this ratio is $$1:4$$, i.e.

$$\frac{u_{1}}{u_{2}} \;=\; \frac{1}{4}.$$

Equating the two expressions, we have

$$\frac{d_{2}^{4}}{d_{1}^{4}} \;=\; \frac{1}{4}.$$

Cross-multiplying gives

$$4d_{2}^{4} \;=\; d_{1}^{4}.$$

Taking the fourth root of both sides,

$$d_{1} \;=\; \sqrt{2}\,d_{2}.$$

Thus the ratio of the diameters is

$$d_{1}:d_{2} \;=\; \sqrt{2}:1.$$

Hence, the correct answer is Option A.