A mass of 1kg is suspended by a thread. It is
1. lifted up with an accleration 4.9`m//s^(2)`.
2. lowered with an acceleration 4.9 `m//s^(2)`.
The ratio of the tensions is

Options

A. `3 : 1`
B. `1 : 3`
C. `1 : 2`
D. `2 : 1`

Answer

To solve the problem, we need to find the tensions in the thread when the mass is lifted and lowered with an acceleration of 4.9 m/s². We will denote the tensions when lifting and lowering as \( T_1 \) and \( T_2 \) respectively.

### Step 1: Analyze the first case (lifting the mass)

1. **Identify Forces**: When the mass is lifted, the forces acting on it are:
- Weight of the mass (\( mg \)) acting downwards.
- Tension (\( T_1 \)) acting upwards.

2. **Write the equation of motion**: According to Newton's second law, the net force is equal to mass times acceleration:
\[ T_1 - mg = ma \]
Here, \( m = 1 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( a = 4.9 \, \text{m/s}^2 \).

3. **Substitute values**:
\[ T_1 - (1 \times 9.8) = 1 \times 4.9 \]
\[ T_1 - 9.8 = 4.9 \]
\[ T_1 = 4.9 + 9.8 = 14.7 \, \text{N} \]

### Step 2: Analyze the second case (lowering the mass)

1. **Identify Forces**: When the mass is lowered, the forces acting on it are:
- Weight of the mass (\( mg \)) acting downwards.
- Tension (\( T_2 \)) acting upwards.

2. **Write the equation of motion**: The net force is given by:
\[ mg - T_2 = ma \]
Here, \( a = 4.9 \, \text{m/s}^2 \) is still the acceleration but now it acts in the direction of gravity.

3. **Substitute values**:
\[ (1 \times 9.8) - T_2 = 1 \times 4.9 \]
\[ 9.8 - T_2 = 4.9 \]
\[ T_2 = 9.8 - 4.9 = 4.9 \, \text{N} \]

### Step 3: Calculate the ratio of tensions

Now that we have both tensions:
- \( T_1 = 14.7 \, \text{N} \)
- \( T_2 = 4.9 \, \text{N} \)

The ratio of the tensions is:
\[ \frac{T_1}{T_2} = \frac{14.7}{4.9} = 3 \]

### Final Answer
The ratio of the tensions \( T_1 : T_2 \) is \( 3 : 1 \).

---