A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of $$10^5$$ N at the top. Considering infinitesimally small displacement in the upper edge, the angular displacement $$\theta$$ of the rod axis from its original position would be : (shear moduli, $$G = 10^{10}$$ N/m$$^2$$)

The cylindrical rod is subjected to a horizontal (shear) force at its upper end while the lower end is fixed. Such a loading produces a uniform shear stress over the circular cross-section.

Step 1 : Shear stress on the cross-section
Shear stress $$\tau$$ is force per unit area: $$\tau = \dfrac{F}{A}$$

The cross-sectional area is $$A = \pi r^{2}$$.
Given radius $$r = 4\text{ cm} = 0.04\text{ m}$$, so
$$A = \pi (0.04)^{2} = \pi \times 0.0016 = 0.0016\,\pi\;\text{m}^{2}$$.

With the applied force $$F = 10^{5}\text{ N}$$,
$$\tau = \dfrac{10^{5}}{0.0016\,\pi} = \dfrac{10^{5}\times625}{\pi} = \dfrac{6.25\times10^{7}}{\pi}\;\text{N m}^{-2}$$ $$-(1)$$

Step 2 : Shear strain produced
For an elastic material, shear strain $$\gamma$$ is related to shear stress by the shear modulus $$G$$:
$$\gamma = \dfrac{\tau}{G}$$.

Here $$G = 10^{10}\;\text{N m}^{-2}$$. Substitute $$\tau$$ from $$(1)$$:
$$\gamma = \dfrac{6.25\times10^{7}/\pi}{10^{10}} = \dfrac{6.25}{\pi\times10^{3}} = \dfrac{0.00625}{\pi}$$.

Step 3 : Relation between shear strain and angular displacement
For a small angular displacement of the top relative to the fixed bottom, shear strain equals the angle in radians:
$$\theta \approx \gamma$$.

Therefore,
$$\theta = \dfrac{0.00625}{\pi} = \dfrac{1}{160\,\pi}\;\text{radian}$$.

Result
Angular displacement $$\theta = \dfrac{1}{160\pi}\;\text{rad}$$.

Hence, the correct option is Option A.