A mass 1 kg is suspended by a thread. It is
(i) lifted up with an acceleration `4.9m//s^(2)`
(ii) lowered with an acceleration `4.9m//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 tension in the thread when a mass of 1 kg is lifted and lowered with an acceleration of 4.9 m/s². We will denote the tension when the mass is lifted as T1 and when it is lowered as T2.

### Step-by-Step Solution:

1. **Identify the forces acting on the mass when lifted:**
- When the mass is lifted with an acceleration of 4.9 m/s², the tension (T1) in the thread must overcome both the weight of the mass and provide the additional force due to acceleration.
- The weight of the mass (W) is given by \( W = mg \), where \( m = 1 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).
- Thus, \( W = 1 \times 9.8 = 9.8 \, \text{N} \).

2. **Apply Newton's second law for the upward motion:**
- According to Newton's second law, the net force (F_net) acting on the mass is equal to the mass times its acceleration (F_net = ma).
- For upward motion, we have:
\[ T1 - W = ma \]
- Substituting the values:
\[ T1 - 9.8 = 1 \times 4.9 \]
- Rearranging gives:
\[ T1 = 9.8 + 4.9 = 14.7 \, \text{N} \]

3. **Identify the forces acting on the mass when lowered:**
- When the mass is lowered with an acceleration of 4.9 m/s², the tension (T2) in the thread must be less than the weight of the mass since it is moving downward.
- The equation for the downward motion is:
\[ W - T2 = ma \]
- Substituting the values:
\[ 9.8 - T2 = 1 \times 4.9 \]
- Rearranging gives:
\[ T2 = 9.8 - 4.9 = 4.9 \, \text{N} \]

4. **Calculate the ratio of the tensions:**
- Now we have T1 and T2:
\[ T1 = 14.7 \, \text{N}, \quad T2 = 4.9 \, \text{N} \]
- The ratio of the tensions is:
\[ \frac{T1}{T2} = \frac{14.7}{4.9} = 3 \]

### Final Answer:
The ratio of the tensions \( T1 : T2 \) is \( 3 : 1 \).