Two blocks with masses `m_(1) = 0.2` kg and `m_(2) = 0.3` kg hang one under other as shown in figure. Find the tensions in the strings (massless) in the following situations :
`(g = 10 m//s^(2))`
if maximum allowable tension is 10 N. What is maximum possible upward acceleration ?

Explanation

To solve the problem of finding the tensions in the strings and the maximum possible upward acceleration, we will follow these steps:

### Step 1: Identify the Forces Acting on Each Block
We have two blocks:
- Block 1 (mass \( m_1 = 0.2 \) kg)
- Block 2 (mass \( m_2 = 0.3 \) kg)

The forces acting on each block are:
- For Block 1:
- Weight \( W_1 = m_1 \cdot g = 0.2 \cdot 10 = 2 \) N (acting downwards)
- Tension \( T_1 \) (acting upwards)

- For Block 2:
- Weight \( W_2 = m_2 \cdot g = 0.3 \cdot 10 = 3 \) N (acting downwards)
- Tension \( T_2 \) (acting upwards)

### Step 2: Write the Equations of Motion
Assuming both blocks are accelerating upwards with acceleration \( a \):
1. For Block 1:
\[ T_1 - W_1 = m_1 \cdot a \implies T_1 - 2 = 0.2a \tag{1} \]

2. For Block 2:
\[ T_2 - W_2 = m_2 \cdot a \implies T_2 - 3 = 0.3a \tag{2} \]

### Step 3: Relate the Tensions
Since Block 1 is above Block 2, the tension in the string supporting Block 1 must be greater than the tension in the string supporting Block 2:
\[ T_1 = T_2 + W_1 \tag{3} \]

### Step 4: Substitute \( T_1 \) in Terms of \( T_2 \)
From equation (3), we can express \( T_1 \) in terms of \( T_2 \):
\[ T_1 = T_2 + 2 \]

### Step 5: Substitute \( T_1 \) into Equation (1)
Substituting \( T_1 \) from equation (3) into equation (1):
\[ (T_2 + 2) - 2 = 0.2a \]
This simplifies to:
\[ T_2 = 0.2a \tag{4} \]

### Step 6: Substitute \( T_2 \) into Equation (2)
Now substituting \( T_2 \) from equation (4) into equation (2):
\[ 0.2a - 3 = 0.3a \]
Rearranging gives:
\[ -3 = 0.3a - 0.2a \implies -3 = 0.1a \implies a = -30 \text{ m/s}^2 \]
This negative acceleration indicates that our assumption of upward acceleration is incorrect.

### Step 7: Find the Maximum Allowable Tension
Given that the maximum allowable tension is \( 10 \) N, we can set \( T_1 = 10 \) N:
\[ 10 - 2 = 0.2a \implies 8 = 0.2a \implies a = 40 \text{ m/s}^2 \]

### Step 8: Conclusion
The maximum possible upward acceleration is \( 40 \text{ m/s}^2 \).