A copper sphere attached to the bottom of a vertical spring is oscillating with time period 10 s. If the copper sphere is immersed in a fluid (assume the viscosity of the fluid is negligilbe) of specfic gravity `(1)/(4)` of that of the copper, then time period of the oscillation is

Options

A. 5 s
B. 10 s
C. 2.5 s
D. 20 s

Explanation

To solve the problem, we need to analyze the effect of immersing the copper sphere in a fluid on the time period of oscillation.

### Step-by-Step Solution:

1. **Understand the Time Period Formula**:
The time period \( T \) of a mass-spring system is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass of the bob (copper sphere) and \( k \) is the spring constant.

2. **Identify Given Values**:
- Initial time period \( T_1 = 10 \, \text{s} \)
- Specific gravity of the fluid = \( \frac{1}{4} \) of that of copper.

3. **Effect of Immersion in Fluid**:
When the copper sphere is immersed in a fluid, the buoyant force acts on it. However, the mass of the sphere and the spring constant remain unchanged. The buoyant force does not affect the time period of oscillation because it does not change the effective mass of the system in the context of simple harmonic motion.

4. **Conclusion**:
Since neither the mass \( m \) nor the spring constant \( k \) changes, the time period of the oscillation remains the same. Therefore, the time period after immersion in the fluid is still:
\[ T_2 = T_1 = 10 \, \text{s} \]

5. **Final Answer**:
The time period of the oscillation after immersing the copper sphere in the fluid is **10 seconds**.