A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60° with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is: (take g = 10 m/s2)

The correct option is:
$100\sqrt{3}\text{N}$

Step-by-Step Derivation and Explanation:

Let us model the forces acting on the uniform rod.
Let:
- m = 20 kg be the mass of the rod.
- L = 5 m be the length of the rod.
- θ = 60° be the angle the rod makes with the smooth vertical wall. This means the angle the rod makes with the horizontal floor is 90° - 60° = 30°.
- g = 10 m/s² be the acceleration due to gravity.

The forces acting on the rod are:
1. The weight of the rod, W = mg = 20 × 10 = 200 N, acting vertically downwards at its center of gravity (at a distance of L/2 from either end).
2. The normal reaction force from the vertical wall, N₁, acting horizontally away from the wall (since the wall is smooth, there is no vertical friction force at the wall).
3. The normal reaction force from the floor, N₂, acting vertically upwards at the base of the rod.
4. The friction force, f, exerted by the floor, acting horizontally towards the wall to prevent the bottom of the rod from sliding away.

Since the rod is in static equilibrium, the net force in both the horizontal and vertical directions must be zero:
- Horizontal equilibrium:
$f=N_1$
- Vertical equilibrium:
$N_2=W=mg=200\text{N}$

To find N₁, we apply the torque equilibrium condition. Taking the torque about the base of the rod (contact point with the floor):
- The torque due to the weight W acts clockwise and is given by:
${\tau }_W=W\times \frac{L}{2}cos(30°)$
(since the rod makes an angle of 30° with the horizontal floor).
- The torque due to the wall's normal force N₁ acts counter-clockwise and is given by:
${\tau }_{N_1}=N_1\times Lsin(30°)$

Setting the net torque about the base to zero:
$N_{1}Lsin(30°)=W\frac{L}{2}cos(30°)$

Dividing both sides by L and rearranging to solve for N₁:
$N_1=\frac{W}{2}cot(30°)$

Substitute the values of W = 200 N and cot(30°) = $\sqrt{3}$:
$N_1=\frac{200}{2}\sqrt{3}=100\sqrt{3}\text{N}$

Since the horizontal friction force equals the normal reaction of the wall, we have:
$f=100\sqrt{3}\text{N}$