A circular disc fixed at its centre to a metal wire and the other end of the wire is clamped. The disc is rotated about the wire as axis and is released. The disc about this axis makes simple harmonic oscillations with a frequency of `(1)/(2)`Hz. If moment of inertia of the disc about this axis is 0.1 kg `-m^(2)`, the torsional constant of the wire is

Options

A. 0.99 kg `m^(2) s^(-2)`
B. 0.45 kg `m^(2) s^(-2)`
C. 09.9 kg `m^(2) s^(-2)`
D. 0.49 kg `m^(2) s^(-2)`

Explanation

To find the torsional constant \( K \) of the wire in the given problem, we can use the relationship between the frequency of oscillation, moment of inertia, and the torsional constant. Here’s a step-by-step solution:

### Step 1: Understand the relationship between frequency, moment of inertia, and torsional constant
The frequency \( f \) of a torsional oscillator is given by the formula:
\[ f = \frac{1}{2\pi} \sqrt{\frac{K}{I}} \]
where:
- \( K \) is the torsional constant,
- \( I \) is the moment of inertia.

### Step 2: Rearrange the formula to solve for \( K \)
We can rearrange the formula to express \( K \) in terms of \( f \) and \( I \):
\[ K = (2\pi f)^2 I \]

### Step 3: Substitute the given values
From the problem, we know:
- The frequency \( f = \frac{1}{2} \) Hz,
- The moment of inertia \( I = 0.1 \) kg·m².

Now substitute these values into the equation:
\[ K = (2\pi \cdot \frac{1}{2})^2 \cdot 0.1 \]

### Step 4: Calculate \( K \)
First, calculate \( 2\pi \cdot \frac{1}{2} \):
\[ 2\pi \cdot \frac{1}{2} = \pi \]
Now square it:
\[ (\pi)^2 \approx 3.14^2 \approx 9.8596 \]
Now multiply by \( 0.1 \):
\[ K = 9.8596 \cdot 0.1 \approx 0.98596 \text{ kg·m}^2/\text{s}^2 \]

### Step 5: Round the answer
Rounding this value gives:
\[ K \approx 0.99 \text{ kg·m}^2/\text{s}^2 \]

### Final Answer
The torsional constant \( K \) of the wire is approximately \( 0.99 \text{ kg·m}^2/\text{s}^2 \).

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