Two bodies whose masses are in the ratio 2:1 are dropped simultaneously at two places A and B where the accelerations due to gravity are `g_A` and `g_B` respectively. If they reach the ground simultaneously, the ratio of the heights from which they are dropped is

Options

A. `g_A : g_B`
B. `2g_A : g_B`
C. `g_A : 2g_B`
D. `sqrtg_a : sqrtg_b`

Answer

To solve the problem, we need to find the ratio of the heights from which two bodies are dropped, given that they reach the ground simultaneously and their masses are in the ratio 2:1. The accelerations due to gravity at the two locations are \( g_A \) and \( g_B \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two bodies with masses in the ratio \( 2:1 \). Let's denote the masses as \( m_A = 2m \) and \( m_B = m \).
- The bodies are dropped from heights \( H_A \) and \( H_B \) at locations A and B, where the accelerations due to gravity are \( g_A \) and \( g_B \) respectively.

2. **Using the Equation of Motion**:
- For free-falling objects, the distance fallen can be described by the equation:
\[ H = \frac{1}{2} g t^2 \]
- For body A:
\[ H_A = \frac{1}{2} g_A t_A^2 \]
- For body B:
\[ H_B = \frac{1}{2} g_B t_B^2 \]

3. **Setting the Time Equal**:
- Since both bodies are dropped simultaneously and reach the ground at the same time, we have:
\[ t_A = t_B = t \]
- Therefore, we can rewrite the equations as:
\[ H_A = \frac{1}{2} g_A t^2 \]
\[ H_B = \frac{1}{2} g_B t^2 \]

4. **Finding the Ratio of Heights**:
- Now, we can find the ratio of the heights \( H_A \) and \( H_B \):
\[ \frac{H_A}{H_B} = \frac{\frac{1}{2} g_A t^2}{\frac{1}{2} g_B t^2} \]
- The \( \frac{1}{2} t^2 \) cancels out:
\[ \frac{H_A}{H_B} = \frac{g_A}{g_B} \]

5. **Conclusion**:
- Thus, the ratio of the heights from which the bodies are dropped is:
\[ \frac{H_A}{H_B} = \frac{g_A}{g_B} \]

### Final Answer:
The ratio of the heights from which the bodies are dropped is \( \frac{H_A}{H_B} = \frac{g_A}{g_B} \).

---