A mass of 1 kg is suspended by means of a thread. The system is (i) lifted up with an acceleration of `4.9ms^(2)` (ii) lowered with an acceleration of `4.9ms^(-2)`. The ratio of tension in the first and second case is

Options

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

Explanation

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 \( T_1 \) and when it is lowered as \( T_2 \).

### Step 1: Calculate Tension when the mass is lifted (T1)
When the mass is lifted with an acceleration \( a = 4.9 \, \text{m/s}^2 \), the forces acting on the mass are:
- The upward tension \( T_1 \)
- The downward gravitational force \( mg \)

The net force acting on the mass can be expressed as:
\[ T_1 - mg = ma \]
Rearranging this gives:
\[ T_1 = mg + ma \]
Substituting \( m = 1 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( a = 4.9 \, \text{m/s}^2 \):
\[ T_1 = 1 \times 9.8 + 1 \times 4.9 = 9.8 + 4.9 = 14.7 \, \text{N} \]

### Step 2: Calculate Tension when the mass is lowered (T2)
When the mass is lowered with the same acceleration \( a = 4.9 \, \text{m/s}^2 \), the forces acting on the mass are:
- The upward tension \( T_2 \)
- The downward gravitational force \( mg \)

The net force in this case can be expressed as:
\[ mg - T_2 = ma \]
Rearranging this gives:
\[ T_2 = mg - ma \]
Substituting \( m = 1 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( a = 4.9 \, \text{m/s}^2 \):
\[ T_2 = 1 \times 9.8 - 1 \times 4.9 = 9.8 - 4.9 = 4.9 \, \text{N} \]

### Step 3: Calculate the ratio of tensions (T1/T2)
Now we can find the ratio of the tensions:
\[ \frac{T_1}{T_2} = \frac{14.7}{4.9} \]
Calculating this gives:
\[ \frac{T_1}{T_2} = 3 \]

### Final Answer
The ratio of tension in the first case (lifted) to the second case (lowered) is:
\[ \boxed{3:1} \]