Two blocks of mass 5 kg and 3 kg are attached to the ends of a string passing over a smooth pulley fixed to the ceiling of an elevator. A man inside the elevator accelerated upwards, finds the acceleration of the blocks to be `(9)/(32)g`. The acceleration of the elevator is

Options

A. `(g)/(3)`
B. `(g)/(4)`
C. `(g)/(8)`
D. `(g)/(6)`

Explanation

To solve the problem, we need to analyze the forces acting on the two blocks and relate them to the acceleration of the elevator. Let's break it down step by step.

### Step 1: Understand the System
We have two blocks:
- Block 1 (mass \( m_1 = 5 \, \text{kg} \))
- Block 2 (mass \( m_2 = 3 \, \text{kg} \))

These blocks are connected by a string over a pulley. The elevator is accelerating upwards, and a man inside the elevator measures the acceleration of the blocks to be \( \frac{9}{32}g \).

### Step 2: Identify Forces Acting on Each Block
For Block 1 (5 kg):
- Weight \( W_1 = m_1 g = 5g \)
- Tension in the string \( T \)

For Block 2 (3 kg):
- Weight \( W_2 = m_2 g = 3g \)
- Tension in the string \( T \)

### Step 3: Write the Equations of Motion
Since the elevator is accelerating upwards, we need to consider the effective acceleration due to gravity. The effective gravitational force acting on each block is modified by the acceleration of the elevator \( a_e \).

For Block 1 (5 kg):
\[ T - 5g - 5a_e = -5a \]
(Here, \( a \) is the acceleration of the block relative to the elevator.)

For Block 2 (3 kg):
\[ 3g + 3a_e - T = 3a \]

### Step 4: Substitute Known Values
We know that the acceleration of the blocks as observed by the man in the elevator is \( a = \frac{9}{32}g \).

### Step 5: Solve the Equations
Now we can add the two equations to eliminate \( T \):

1. From Block 1:
\[ T = 5g + 5a_e - 5a \]

2. From Block 2:
\[ T = 3g + 3a_e + 3a \]

Setting them equal to each other:
\[ 5g + 5a_e - 5a = 3g + 3a_e + 3a \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ 5g - 3g + 5a_e - 3a_e = 5a + 3a \]
\[ 2g + 2a_e = 8a \]
\[ a_e = 4a - g \]

### Step 7: Substitute the Value of \( a \)
Substituting \( a = \frac{9}{32}g \):
\[ a_e = 4\left(\frac{9}{32}g\right) - g \]
\[ a_e = \frac{36}{32}g - \frac{32}{32}g \]
\[ a_e = \frac{4}{32}g = \frac{1}{8}g \]

### Final Answer
The acceleration of the elevator is:
\[ \boxed{\frac{1}{8}g} \]