A 25kg uniform solid with a 20cm radius respectively by a verticle wire such then the point suspended is velocity about the center of the sphere torque of 0.10N-m and then montain the sphere then at angle of 1.0 rad if sphere is then released its period of the oscillation will be

Options

A. `pi` second
B. `sqrt(2) pi` second
C. `2 pi` second
D. `4 pi` second

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant physics concepts.

### Step 1: Understand the Given Information
- Mass of the sphere (m) = 25 kg
- Radius of the sphere (r) = 20 cm = 0.2 m
- Torque (τ) = 0.10 N·m
- Angle (θ) = 1.0 rad

### Step 2: Relate Torque to Torsional Constant
The relationship between torque (τ) and the angular displacement (θ) is given by:
\[ \tau = -k \theta \]
Where:
- \( k \) is the torsional constant of the wire.

Taking the magnitude, we can write:
\[ 0.10 = k \cdot 1.0 \]
Thus, we find:
\[ k = 0.10 \, \text{N·m/rad} \]

### Step 3: Calculate the Moment of Inertia (I)
The moment of inertia (I) for a solid sphere is given by:
\[ I = \frac{2}{5} m r^2 \]
Substituting the values:
\[ I = \frac{2}{5} \cdot 25 \, \text{kg} \cdot (0.2 \, \text{m})^2 \]
Calculating:
\[ I = \frac{2}{5} \cdot 25 \cdot 0.04 = \frac{2}{5} \cdot 1 = \frac{2}{5} \, \text{kg·m}^2 \]

### Step 4: Calculate the Time Period of Oscillation (T)
The time period (T) of oscillation for a torsional pendulum is given by:
\[ T = 2\pi \sqrt{\frac{I}{k}} \]
Substituting the values of \( I \) and \( k \):
\[ T = 2\pi \sqrt{\frac{\frac{2}{5}}{0.10}} \]
Calculating:
\[ T = 2\pi \sqrt{\frac{2}{5 \cdot 0.10}} = 2\pi \sqrt{\frac{2}{0.5}} = 2\pi \sqrt{4} = 2\pi \cdot 2 = 4\pi \, \text{seconds} \]

### Final Answer
The time period of the oscillation when the sphere is released is:
\[ T = 4\pi \, \text{seconds} \]

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