A uniform rope of mass M=0.1kg and length L=10m hangs from the celling. `[g=10m//s^(2)]` :- (1)speed of transverse wave in the rope increases linearly from top to the bottom. (2)speed of the transverse wave in the rope decreases linearly from bottom to the top. (3)speed of the transverse wave in the rope remain constant along the length of the rope (4)time taken by the transverse wave to travel the full length of the rope is 2 sec.

Explanation

To solve the problem, we will analyze the behavior of a uniform rope hanging from the ceiling and determine the speed of transverse waves in the rope, as well as the time taken for the wave to travel the full length of the rope.

### Step-by-Step Solution:

1. **Understanding the Rope Setup**:
- We have a uniform rope of mass \( M = 0.1 \, \text{kg} \) and length \( L = 10 \, \text{m} \).
- The acceleration due to gravity is given as \( g = 10 \, \text{m/s}^2 \).
- The rope hangs vertically from the ceiling.

2. **Finding Linear Mass Density**:
- The linear mass density \( \mu \) of the rope is given by:
\[ \mu = \frac{M}{L} = \frac{0.1 \, \text{kg}}{10 \, \text{m}} = 0.01 \, \text{kg/m} \]

3. **Calculating Tension at a Point**:
- Consider a small element of the rope at a distance \( x \) from the bottom. The tension \( T(x) \) at this point is due to the weight of the rope below it:
\[ T(x) = \text{mass of rope below} \times g = \left(\frac{M}{L} \cdot (L - x)\right) \cdot g = \mu (L - x) g \]

4. **Finding the Speed of the Wave**:
- The speed \( v \) of the transverse wave in the rope is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
- Substituting for \( T(x) \):
\[ v(x) = \sqrt{\frac{\mu (L - x) g}{\mu}} = \sqrt{(L - x) g} \]
- This shows that the speed of the wave depends on \( x \).

5. **Analyzing the Speed Behavior**:
- As \( x \) increases from 0 to \( L \), \( v(x) \) increases from \( 0 \) to \( \sqrt{Lg} \) (maximum speed at the top).
- Therefore, the speed of the transverse wave increases linearly from the bottom to the top of the rope.

6. **Time Taken to Travel the Length of the Rope**:
- The time \( dt \) taken for the wave to travel a small distance \( dx \) is:
\[ dt = \frac{dx}{v(x)} = \frac{dx}{\sqrt{(L - x)g}} \]
- To find the total time \( t \) to travel the full length of the rope, we integrate:
\[ t = \int_0^L \frac{dx}{\sqrt{(L - x)g}} = \frac{1}{\sqrt{g}} \int_0^L \frac{dx}{\sqrt{(L - x)}} \]
- The integral evaluates to:
\[ \int_0^L \frac{dx}{\sqrt{(L - x)}} = 2\sqrt{L} \]
- Thus, we have:
\[ t = \frac{2\sqrt{L}}{\sqrt{g}} = 2\sqrt{\frac{L}{g}} \]

7. **Substituting Values**:
- Substituting \( L = 10 \, \text{m} \) and \( g = 10 \, \text{m/s}^2 \):
\[ t = 2\sqrt{\frac{10}{10}} = 2 \, \text{s} \]

### Conclusion:
- The correct option is that the time taken by the transverse wave to travel the full length of the rope is \( 2 \, \text{s} \).