A dancer is standing on a rotating platform taking two sphere on her hands. If she drops down the sphere on ground, then dancer's:

Options

A. angular velocity will increase
B. angular momentum & angular velocity both will unchange
C. angular momentum unchanged & angular velocity will increase
D. both will decrease

Answer

To solve the problem step-by-step, we will analyze the situation of the dancer on the rotating platform holding two spheres and what happens when she drops them.

### Step-by-Step Solution:

1. **Understand the System**:
- The dancer is standing on a rotating platform and is holding two spheres.
- The initial angular velocity of the system (dancer + spheres) is denoted as \( \omega_i \).

2. **Identify the Moment of Inertia**:
- The moment of inertia of the dancer alone is \( I_0 \).
- The moment of inertia contributed by the two spheres can be expressed as \( m r_1^2 + m r_2^2 \), where \( m \) is the mass of each sphere and \( r_1 \) and \( r_2 \) are their respective distances from the axis of rotation.
- Therefore, the total initial moment of inertia \( I_i \) is:
\[ I_i = I_0 + m r_1^2 + m r_2^2 \]

3. **Conservation of Angular Momentum**:
- Since there are no external torques acting on the system, angular momentum is conserved.
- The initial angular momentum \( L_i \) can be expressed as:
\[ L_i = I_i \cdot \omega_i \]
- After the dancer drops the spheres, the final moment of inertia \( I_f \) will be just the moment of inertia of the dancer:
\[ I_f = I_0 \]
- The final angular momentum \( L_f \) is:
\[ L_f = I_f \cdot \omega_f \]

4. **Setting Up the Equation**:
- By conservation of angular momentum, we have:
\[ L_i = L_f \]
- Substituting the expressions for angular momentum gives:
\[ I_i \cdot \omega_i = I_0 \cdot \omega_f \]

5. **Solving for Final Angular Velocity**:
- Rearranging the equation to solve for \( \omega_f \):
\[ \omega_f = \frac{I_i}{I_0} \cdot \omega_i \]
- Substituting \( I_i \):
\[ \omega_f = \frac{I_0 + m r_1^2 + m r_2^2}{I_0} \cdot \omega_i \]
- This can be simplified to:
\[ \omega_f = \left(1 + \frac{m r_1^2 + m r_2^2}{I_0}\right) \cdot \omega_i \]

6. **Analyzing the Result**:
- Since \( m r_1^2 + m r_2^2 \) is positive, the term \( \frac{m r_1^2 + m r_2^2}{I_0} \) is also positive.
- Therefore, \( \omega_f > \omega_i \), indicating that the angular velocity of the dancer increases after dropping the spheres.

7. **Conclusion**:
- The angular momentum of the dancer remains unchanged (conserved), while the angular velocity increases.

### Final Answer:
- The dancer's angular velocity increases, while her angular momentum remains unchanged.