Two blocks of masses 10 kg and 9 kg are connected with light string which is passing over an ideal pulley as shown in figure. If system is released from rest then the acceleration of 9 kg block at the given instant is about (Assume all the surfaces are smooth and `g = 10 m/s^2`)

Options

A. `1.8 m/s^2`
B. `4.73 m/s^2`
C. `3.5 m/s^2`
D. `3 m/s^2`

Answer

To find the acceleration of the 9 kg block in the given system, we can follow these steps:

### Step 1: Identify the forces acting on the blocks
- For the 10 kg block (Block A), the forces acting on it are:
- Weight (downward) = \( W_A = m_A \cdot g = 10 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 100 \, \text{N} \)
- Tension (upward) = \( T \)

- For the 9 kg block (Block B), the forces acting on it are:
- Weight (downward) = \( W_B = m_B \cdot g = 9 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 90 \, \text{N} \)
- Tension (upward) = \( T \)

### Step 2: Write the equations of motion for both blocks
- For Block A (10 kg):
\[ W_A - T = m_A \cdot a_A \quad \text{(1)} \]
Substituting the values:
\[ 100 - T = 10 \cdot a_A \quad \text{(1)} \]

- For Block B (9 kg):
\[ W_B - T = m_B \cdot a_B \quad \text{(2)} \]
Substituting the values:
\[ 90 - T = 9 \cdot a_B \quad \text{(2)} \]

### Step 3: Relate the accelerations of the blocks
Since the blocks are connected by a string over a pulley, the acceleration of the two blocks will be related. If we assume the downward acceleration of the 9 kg block is \( a \), then the upward acceleration of the 10 kg block will be \( a_A = a \).

### Step 4: Solve the equations
From equation (1):
\[ 100 - T = 10a \quad \Rightarrow \quad T = 100 - 10a \quad \text{(3)} \]

From equation (2):
\[ 90 - T = 9a \quad \Rightarrow \quad T = 90 - 9a \quad \text{(4)} \]

### Step 5: Set equations (3) and (4) equal to each other
\[ 100 - 10a = 90 - 9a \]

### Step 6: Solve for \( a \)
Rearranging gives:
\[ 100 - 90 = 10a - 9a \]
\[ 10 = a \]
\[ a = 10 \, \text{m/s}^2 \]

### Step 7: Find the acceleration of the 9 kg block
Since the 9 kg block is accelerating downward, its acceleration is:
\[ a_B = a = 10 \, \text{m/s}^2 \]

### Final Result
The acceleration of the 9 kg block is \( 10 \, \text{m/s}^2 \).

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