Young's moduli of two wires A and B are in the ratio 7:4. Wire A is 2 m long and has radius R. Wire B is 1.5 m long and has radius 2 mm. If the two wires stretch by the same length for a given load, the value of R is close to:

For a uniform wire that is stretched within the elastic limit, the extension produced by a load is given by the standard elastic‐string formula

$$\Delta L=\frac{F\,L}{A\,Y},$$

where $$F$$ is the applied force, $$L$$ is the original length, $$A$$ is the cross-sectional area and $$Y$$ is Young’s modulus of the material.

Because the two wires are subjected to the same load and are said to “stretch by the same length”, we have for wires A and B

$$\frac{F\,L_A}{A_A\,Y_A}=\frac{F\,L_B}{A_B\,Y_B}.$$

The force $$F$$ and the factor $$\pi$$ hidden inside each area will cancel, so we can write

$$\frac{L_A}{r_A^{\,2}\,Y_A}=\frac{L_B}{r_B^{\,2}\,Y_B},$$

where $$r_A=R$$ (unknown) and $$r_B=2\;\text{mm}$$ (given).

The ratio of Young’s moduli is given as $$Y_A:Y_B=7:4$$, so we may set $$Y_B=\dfrac{4}{7}\,Y_A$$.

Substituting this value together with the given lengths $$L_A=2\;\text{m}$$ and $$L_B=1.5\;\text{m}$$ gives

$$\frac{2}{R^{2}\,Y_A}=\frac{1.5}{(2\;\text{mm})^{2}\,\left(\dfrac{4}{7}Y_A\right)}.$$

The factor $$Y_A$$ now cancels out completely, leading to

$$\frac{2}{R^{2}}=\frac{1.5\times7}{(2\;\text{mm})^{2}\times4}.$$

Taking the reciprocal of both sides so that $$R^{2}$$ stands alone, we obtain

$$R^{2}=\frac{2\times4}{1.5\times7}\,(2\;\text{mm})^{2}.$$

Simplifying each numerical factor step by step:

$$\frac{2}{1.5}=\frac{4}{3}, \quad \text{so} \quad R^{2}=\frac{4}{3}\times\frac{4}{7}\times(2\;\text{mm})^{2}.$$

Next, multiply the first two fractions:

$$\frac{4}{3}\times\frac{4}{7}=\frac{16}{21}.$$

Because $$(2\;\text{mm})^{2}=4\;\text{mm}^{2},$$ we have

$$R^{2}=\frac{16}{21}\times4\;\text{mm}^{2}=\frac{64}{21}\;\text{mm}^{2}.$$

Evaluating the fraction numerically,

$$\frac{64}{21}\approx3.0476\;\text{mm}^{2}.$$

Taking the square root finally yields

$$R=\sqrt{3.0476}\;\text{mm}\approx1.7457\;\text{mm}.$$

The value rounds to about $$1.7\;\text{mm}$$, which matches the fourth option in the list.

Hence, the correct answer is Option D.