A coin is placed on a gramophone record rotating at a speed of 45 rpm, it flies away when the rotational speed is 50 rpm. If two such coins are placed one over the other on the same record both of them will fly away when rotational speed is

Options

A. 100 rpm
B. 25 rpm
C. 12.5 rpm
D. 50 rpm

Explanation

To solve the problem, we need to analyze the forces acting on the coins placed on the rotating gramophone record.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a gramophone record rotating at a speed of 45 rpm. A coin placed on it flies off when the speed increases to 50 rpm.
- We need to find out at what speed two coins stacked on top of each other will fly off.

2. **Identifying Forces**:
- When the gramophone rotates, the coin experiences a centripetal force that keeps it in circular motion. This force is provided by the friction between the coin and the record.
- The centripetal force \( F_c \) required to keep the coin in circular motion is given by:
\[ F_c = m \cdot a_c = m \cdot \omega^2 \cdot r \]
where \( m \) is the mass of the coin, \( \omega \) is the angular velocity in radians per second, and \( r \) is the radius of the circular path.

3. **Frictional Force**:
- The maximum frictional force \( F_f \) that can act on the coin is given by:
\[ F_f = \mu \cdot N = \mu \cdot m \cdot g \]
where \( \mu \) is the coefficient of friction, and \( g \) is the acceleration due to gravity.

4. **Condition for the Coin to Fly Off**:
- The coin will fly off when the centripetal force exceeds the maximum frictional force:
\[ m \cdot \omega^2 \cdot r > \mu \cdot m \cdot g \]
- Simplifying this, we can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \omega^2 \cdot r > \mu \cdot g \]
- This implies that the angular velocity \( \omega \) must be greater than a certain threshold determined by \( \mu \), \( g \), and \( r \).

5. **Calculating Angular Velocity**:
- The angular velocity \( \omega \) in terms of revolutions per minute (rpm) is given by:
\[ \omega = \frac{2\pi n}{60} \]
- For the first coin, it flies off at 50 rpm. We can convert this to radians per second:
\[ \omega_{50} = \frac{2\pi \cdot 50}{60} \text{ rad/s} \]

6. **Considering Two Coins**:
- When two coins are placed on top of each other, the total mass becomes \( 2m \). However, the frictional force still depends on the coefficient of friction and the normal force, which is now \( 2mg \).
- The condition for flying off remains the same, as the frictional force will also double:
\[ 2m \cdot \omega^2 \cdot r > \mu \cdot (2m) \cdot g \]
- Thus, the mass cancels out again, and the condition remains the same:
\[ \omega^2 \cdot r > \mu \cdot g \]

7. **Conclusion**:
- Since the condition for flying off does not depend on the mass of the coins, both coins will fly off at the same angular velocity of 50 rpm.

### Final Answer:
Both coins will fly away when the rotational speed is **50 rpm**.