Two racing cars of masses `m` and `4m` are moving in circles of radii `r` and `2r` respectively. If their speeds are such that each makes a complete circle in the same time, then the ratio of the angular speeds of the first to the second car is

Options

A. `8 : 1`
B. `4 : 1`
C. `2 : 1`
D. `1 : 1`

Answer

To solve the problem, we need to find the ratio of the angular speeds of two racing cars moving in circular paths. The first car has mass \( m \) and moves in a circle of radius \( r \), while the second car has mass \( 4m \) and moves in a circle of radius \( 2r \). Both cars complete a full circle in the same time.

### Step-by-step Solution:

1. **Identify the given information:**
- Mass of the first car, \( m_1 = m \)
- Mass of the second car, \( m_2 = 4m \)
- Radius of the first car's circular path, \( r_1 = r \)
- Radius of the second car's circular path, \( r_2 = 2r \)
- Time taken to complete one circle by both cars is the same, \( T_1 = T_2 \).

2. **Write the formula for angular speed:**
The angular speed \( \omega \) is defined as:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the time period.

3. **Calculate the angular speeds for both cars:**
- For the first car:
\[ \omega_1 = \frac{2\pi}{T_1} \]
- For the second car:
\[ \omega_2 = \frac{2\pi}{T_2} \]

4. **Set up the ratio of angular speeds:**
To find the ratio of the angular speeds of the first car to the second car:
\[ \frac{\omega_1}{\omega_2} = \frac{\frac{2\pi}{T_1}}{\frac{2\pi}{T_2}} = \frac{T_2}{T_1} \]

5. **Substitute the time periods:**
Since we know that \( T_1 = T_2 \) (both cars take the same time to complete a circle):
\[ \frac{\omega_1}{\omega_2} = \frac{T_2}{T_1} = \frac{T_1}{T_1} = 1 \]

6. **Final ratio:**
Therefore, the ratio of the angular speeds of the first car to the second car is:
\[ \omega_1 : \omega_2 = 1 : 1 \]

### Conclusion:
The ratio of the angular speeds of the first car to the second car is \( 1 : 1 \).