A rope lies on a table such that a part of it hangs down the table, when the length of hanging part is 1/3 of entire length the rope just begins to slide. The coefficient of friction between the rope and the table is:-

Options

A. `2//3`
B. `1//2`
C. `1//3`
D. `1//6`

Explanation

To solve the problem, we need to analyze the forces acting on the rope when it just begins to slide off the table. Here's a step-by-step solution:

### Step 1: Define the problem
Let the total length of the rope be \( L \). According to the problem, when the length of the hanging part is \( \frac{1}{3}L \), the rope just begins to slide.

### Step 2: Determine the lengths
- Length of the hanging part: \( L_h = \frac{1}{3}L \)
- Length of the part on the table: \( L_t = L - L_h = L - \frac{1}{3}L = \frac{2}{3}L \)

### Step 3: Calculate the mass of the rope
Assuming the mass per unit length of the rope is \( m \), the total mass of the rope is \( M = mL \).
- Mass of the hanging part:
\[ M_h = \frac{1}{3}M = \frac{1}{3}(mL) = \frac{mL}{3} \]
- Mass of the part on the table:
\[ M_t = \frac{2}{3}M = \frac{2}{3}(mL) = \frac{2mL}{3} \]

### Step 4: Calculate the forces acting on the rope
When the rope just begins to slide, the weight of the hanging part provides the force trying to pull the rope down, while the frictional force opposes this motion.

- Weight of the hanging part:
\[ W_h = M_h g = \frac{mL}{3} g \]

- Normal force acting on the part of the rope on the table:
\[ N = M_t g = \frac{2mL}{3} g \]

### Step 5: Calculate the frictional force
The limiting frictional force can be expressed as:
\[ F_f = \mu N = \mu \left(\frac{2mL}{3} g\right) \]

### Step 6: Set up the equation for equilibrium
At the point of sliding, the weight of the hanging part equals the frictional force:
\[ W_h = F_f \]
Substituting the expressions we found:
\[ \frac{mL}{3} g = \mu \left(\frac{2mL}{3} g\right) \]

### Step 7: Simplify the equation
We can cancel \( \frac{mL}{3} g \) from both sides (assuming \( m, L, g \neq 0 \)):
\[ 1 = 2\mu \]

### Step 8: Solve for the coefficient of friction
\[ \mu = \frac{1}{2} \]

### Conclusion
The coefficient of friction between the rope and the table is \( \mu = \frac{1}{2} \).

---