A man in a lift feels an apparent weight W when the lift is moving up with a uniform acceleration of 1/3rd of the acceleration due to gravity. If the same man were in the same lift now moving down with a uniform acceleration that is 1/2 of the acceleration due to gravity, then his apparent weight is

Options

A. `(3W)/8`
B. `(3W)/4`
C. `W/8`
D. `(5W)/8`

Answer

To solve the problem step by step, we will analyze the situation when the man is in the lift moving upwards and downwards with different accelerations.

### Step 1: Understand the Forces Acting on the Man When the Lift Moves Upwards
When the lift is moving upwards with an acceleration of \( a = \frac{g}{3} \):
- The weight of the man (downward force) is \( W = mg \).
- The normal force (apparent weight) is \( R \) (upward force).

Using Newton's second law, we can write the equation of motion:
\[ R - mg = ma \]
Substituting \( a = \frac{g}{3} \):
\[ R - mg = m \left(\frac{g}{3}\right) \]

### Step 2: Rearranging the Equation for Apparent Weight When Moving Upwards
Rearranging the above equation gives:
\[ R = mg + m \left(\frac{g}{3}\right) \]
Factoring out \( mg \):
\[ R = mg \left(1 + \frac{1}{3}\right) = mg \left(\frac{4}{3}\right) \]
Thus, the apparent weight \( R \) when the lift is moving upwards is:
\[ R = \frac{4}{3} mg \]
Since \( W = mg \), we can express this as:
\[ R = \frac{4}{3} W \]

### Step 3: Understand the Forces Acting on the Man When the Lift Moves Downwards
Now, when the lift is moving downwards with an acceleration of \( a = \frac{g}{2} \):
- The weight of the man (downward force) is still \( mg \).
- The normal force (apparent weight) is \( R \) (upward force).

Using Newton's second law again, we write:
\[ mg - R = ma \]
Substituting \( a = \frac{g}{2} \):
\[ mg - R = m \left(\frac{g}{2}\right) \]

### Step 4: Rearranging the Equation for Apparent Weight When Moving Downwards
Rearranging this equation gives:
\[ R = mg - m \left(\frac{g}{2}\right) \]
Factoring out \( mg \):
\[ R = mg \left(1 - \frac{1}{2}\right) = mg \left(\frac{1}{2}\right) \]
Thus, the apparent weight \( R \) when the lift is moving downwards is:
\[ R = \frac{1}{2} mg \]
Expressing this in terms of \( W \):
\[ R = \frac{1}{2} W \]

### Step 5: Finding the Ratio of Apparent Weights
Now we have two expressions for apparent weight:
1. When moving upwards: \( R_{up} = \frac{4}{3} W \)
2. When moving downwards: \( R_{down} = \frac{1}{2} W \)

To find the relationship between the two apparent weights:
\[ \frac{R_{down}}{R_{up}} = \frac{\frac{1}{2} W}{\frac{4}{3} W} = \frac{1/2}{4/3} = \frac{3}{8} \]

### Final Step: Conclusion
Thus, the apparent weight of the man when the lift is moving down with an acceleration of \( \frac{g}{2} \) is:
\[ R_{down} = \frac{3}{8} W \]