Two bodies of different masses are dropped simultaneously from same height. If air friction acting on them is directly proportional to the square of their mass, then,

Options

A. lighter body reaches the ground earlier
B. heavier body reaches the earlier
C. both reach the ground at the same time
D. Any of the above

Answer

To solve the problem, we need to analyze the forces acting on both bodies when they are dropped from the same height.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Bodies:**
When both bodies are dropped, they experience two main forces:
- The gravitational force (weight) acting downward, which is \( F_g = m \cdot g \) where \( m \) is the mass and \( g \) is the acceleration due to gravity.
- The air friction force acting upward, which is given to be directly proportional to the square of their mass.

2. **Express the Air Friction Forces:**
Let’s denote the masses of the two bodies as \( M_1 \) and \( M_2 \).
- The air friction force on body 1 (with mass \( M_1 \)) can be expressed as:
\[ F_1 = k \cdot M_1^2 \]
- The air friction force on body 2 (with mass \( M_2 \)) can be expressed as:
\[ F_2 = k \cdot M_2^2 \]
Here, \( k \) is a constant of proportionality.

3. **Set Up the Equations of Motion:**
For each body, we can write the net force acting on them:
- For body 1:
\[ F_{net1} = F_g - F_1 = M_1 \cdot g - k \cdot M_1^2 \]
- For body 2:
\[ F_{net2} = F_g - F_2 = M_2 \cdot g - k \cdot M_2^2 \]

4. **Determine the Acceleration of Each Body:**
Using Newton's second law, \( F_{net} = m \cdot a \):
- For body 1:
\[ M_1 \cdot a_1 = M_1 \cdot g - k \cdot M_1^2 \]
Simplifying gives:
\[ a_1 = g - \frac{k \cdot M_1}{M_1} = g - \frac{k \cdot M_1}{M_1} = g - k \]
- For body 2:
\[ M_2 \cdot a_2 = M_2 \cdot g - k \cdot M_2^2 \]
Simplifying gives:
\[ a_2 = g - \frac{k \cdot M_2}{M_2} = g - k \]

5. **Compare the Accelerations:**
Since \( k \) is a constant, the acceleration of each body will depend on their respective masses. If \( M_1 > M_2 \), then:
\[ a_1 < a_2 \]
This means that the heavier mass (body 1) experiences a greater air resistance relative to its weight, resulting in a lower acceleration compared to the lighter mass (body 2).

6. **Conclusion:**
Therefore, the lighter body (mass \( M_2 \)) will reach the ground earlier than the heavier body (mass \( M_1 \)).

### Final Answer:
The lighter body reaches the ground earlier.