Two balls a `A` and `B` of same size are dropped from the same point under gravity. The mass of `A` is greater than that of `B`.If the air resistance acting on each ball is same, then

Options

A. both the balls reach the ground simultaneously
B. the ball `A` reaches earlier
C. the ball `B` reaches earlier
D. nothing can be said

Explanation

To solve the problem of two balls A and B being dropped from the same height under the influence of gravity, we will analyze the forces acting on each ball and determine their accelerations.

### Step-by-Step Solution:

1. **Identify the Forces Acting on Each Ball**:
- Both balls are dropped from the same height, so they are subjected to gravitational force (weight) acting downward and air resistance acting upward.
- Let the mass of ball A be \( m_A \) and the mass of ball B be \( m_B \). Given that \( m_A > m_B \).

2. **Write the Force Balance Equation**:
- For ball A:
\[ F_{net, A} = m_A g - F_{air} \]
- For ball B:
\[ F_{net, B} = m_B g - F_{air} \]
- Here, \( g \) is the acceleration due to gravity, and \( F_{air} \) is the air resistance acting on both balls, which is the same.

3. **Apply Newton's Second Law**:
- According to Newton's second law, \( F_{net} = m \cdot a \).
- For ball A:
\[ m_A a_A = m_A g - F_{air} \]
Rearranging gives:
\[ a_A = g - \frac{F_{air}}{m_A} \]
- For ball B:
\[ m_B a_B = m_B g - F_{air} \]
Rearranging gives:
\[ a_B = g - \frac{F_{air}}{m_B} \]

4. **Compare the Accelerations**:
- Since \( F_{air} \) is the same for both balls, we can compare their accelerations:
\[ a_A = g - \frac{F_{air}}{m_A} \]
\[ a_B = g - \frac{F_{air}}{m_B} \]
- Because \( m_A > m_B \), it follows that \( \frac{F_{air}}{m_A} < \frac{F_{air}}{m_B} \). Therefore:
\[ a_A > a_B \]

5. **Conclusion**:
- Since the acceleration of ball A is greater than that of ball B, ball A will reach the ground before ball B.
- Thus, the correct answer is that ball A reaches the ground earlier than ball B.

### Final Answer:
**Ball A reaches the ground earlier.**