Ten one-rupee coins are put on top each other on a table. Each coin has a mass m. The rection of the `6^(th)` coin (counted from the bottom) on the `7^(th)` coin is

Options

A. 4mg
B. 6mg
C. 7mg
D. 3mg

Explanation

To solve the problem of finding the reaction force of the 6th coin on the 7th coin in a stack of 10 one-rupee coins, each with mass \( m \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Coins and Their Arrangement**:
- We have a total of 10 coins stacked on top of each other.
- Each coin has a mass \( m \).
- We need to focus on the 6th coin (from the bottom) and its reaction force on the 7th coin.

2. **Determine the Mass Above the 6th Coin**:
- The 6th coin is at the bottom of the top 4 coins (7th, 8th, 9th, and 10th).
- Therefore, the total mass of the coins above the 6th coin is \( 4m \) (since there are 4 coins above it).

3. **Draw the Free Body Diagram (FBD)**:
- In the FBD of the 6th coin, we have:
- The weight of the 4 coins above it acting downwards, which is \( 4mg \).
- The reaction force \( R \) exerted by the 6th coin on the 7th coin acting upwards.

4. **Apply Newton's Second Law**:
- Since the system is in equilibrium (the coins are at rest), the sum of forces acting on the 6th coin must be zero.
- Therefore, we can set up the equation:
\[ R - 4mg = 0 \]
- Rearranging gives:
\[ R = 4mg \]

5. **Conclusion**:
- The reaction force of the 6th coin on the 7th coin is \( R = 4mg \).

### Final Answer:
The reaction of the 6th coin on the 7th coin is \( 4mg \).

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