A helicopter of mass `M` is lowering a truck of mass `m` onto the deck of a ship. In first case the helicopter and the truck move downward together (the length of the cable remaine constant). Tension in the cable is `T_(1)` when their downward speed is decreasing at a rate of (g)/(10). In second case when the truck gets close to the deck, the helicopter stops moving downward. While it hovers stationary, it lets out the cabel so that the truck is still moving downward. If the truck is moving downward with a speed decreasing at rate of `(g)/(10)`, tension in string is now `T_(2)`, What is ratio `T_(1)//T_(2)`

Options

A. `(10)/(11)`
B. `(9)/(11)`
C. `1`
D. None

Explanation

To solve the problem, we need to analyze the forces acting on the truck in both cases and find the tensions \( T_1 \) and \( T_2 \).

### Step 1: Analyze the first case
In the first case, both the helicopter and the truck are moving downward together, and the downward speed is decreasing at a rate of \( \frac{g}{10} \).

1. **Identify the forces**:
- Weight of the truck: \( W = mg \)
- Tension in the cable: \( T_1 \)
- The net downward acceleration is \( a = g - \frac{g}{10} = \frac{9g}{10} \).

2. **Apply Newton's second law**:
\[ m \cdot a = mg - T_1 \]
Substituting \( a = \frac{9g}{10} \):
\[ m \cdot \frac{9g}{10} = mg - T_1 \]

3. **Rearranging the equation**:
\[ T_1 = mg - m \cdot \frac{9g}{10} \]
\[ T_1 = mg - \frac{9mg}{10} = mg \cdot \frac{1}{10} = \frac{mg}{10} \]

### Step 2: Analyze the second case
In the second case, the helicopter is stationary, and it lets out the cable while the truck is moving downward with a speed decreasing at a rate of \( \frac{g}{10} \).

1. **Identify the forces**:
- Weight of the truck: \( W = mg \)
- Tension in the cable: \( T_2 \)
- The net downward acceleration is still \( a = g - \frac{g}{10} = \frac{9g}{10} \).

2. **Apply Newton's second law**:
\[ m \cdot a = mg - T_2 \]
Substituting \( a = \frac{9g}{10} \):
\[ m \cdot \frac{9g}{10} = mg - T_2 \]

3. **Rearranging the equation**:
\[ T_2 = mg - m \cdot \frac{9g}{10} \]
\[ T_2 = mg - \frac{9mg}{10} = mg \cdot \frac{1}{10} = \frac{mg}{10} \]

### Step 3: Find the ratio \( \frac{T_1}{T_2} \)
Now we can find the ratio of the tensions:
\[ \frac{T_1}{T_2} = \frac{\frac{mg}{10}}{\frac{mg}{10}} = 1 \]

### Final Answer
The ratio \( \frac{T_1}{T_2} \) is \( 1:1 \).