A ladder of weight ‘w’ rests against a smooth vertical wall and rests on rough horizontal ground, the coefficient of friction between the ladder and the ground being $$\frac{1}{4}$$. The maximum angle of inclination of the ladder to the vertical, if a man of weight ‘w’ is to walk to the top of it safely, is tan-1x, where x is A. $$\frac{1}{4}$$ B. $$\frac{1}{3}$$ C. 3 D. 4

The correct answer is $\boxed{\frac{1}{3}}$.

The maximum angle of inclination of the ladder to the vertical is the angle at which the ladder is just about to slip. At this angle, the force of friction between the ladder and the ground is equal to the maximum possible force of friction, which is $\mu N$, where $\mu$ is the coefficient of friction and $N$ is the normal force. The normal force is equal to the weight of the ladder, $w$.

The force of friction is also equal to the component of the weight of the ladder parallel to the ground, which is $w \sin \theta$, where $\theta$ is the angle of inclination of the ladder to the vertical. Therefore, we have the equation

$$\mu N = w \sin \theta$$

Substituting $N = w$, we get

$$\mu w = w \sin \theta$$

Solving for $\theta$, we get

$$\theta = \sin^{-1} \mu = \sin^{-1} \frac{1}{4} = \boxed{\frac{1}{3}}$$

Option A is incorrect because the coefficient of friction is $\frac{1}{4}$, not $\frac{1}{3}$. Option B is incorrect because the angle of inclination of the ladder to the vertical is not $\frac{1}{3}$ radians. Option C is incorrect because the maximum angle of inclination of the ladder to the vertical is not 3 degrees. Option D is incorrect because the maximum angle of inclination of the ladder to the vertical is not 4 degrees.