A disc of radius R=10 cm oscillates as a physical pendulum about an axis perpendicular to the plane of the disc at a distance r from its centre. If `r=(R)/(4)`, the approximate period of oscillation is (Take `g=10ms^(-2)`)

Options

A. 0.84 s
B. 0.94 s
C. 1.26 s
D. 1.42 s

Explanation

To find the approximate period of oscillation for a disc acting as a physical pendulum, we can follow these steps:

### Step 1: Identify the given values
- Radius of the disc, \( R = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Distance from the center to the pivot, \( r = \frac{R}{4} = \frac{0.1}{4} = 0.025 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the moment of inertia \( I \)
Using the parallel axis theorem, the moment of inertia \( I \) about the pivot point is given by:
\[ I = I_{cm} + m h^2 \]
Where:
- \( I_{cm} = \frac{1}{2} m R^2 \) (moment of inertia about the center of mass)
- \( h = r = \frac{R}{4} \)

Substituting the values:
\[ I_{cm} = \frac{1}{2} m (0.1)^2 = \frac{1}{2} m (0.01) = 0.005m \]
\[ h = \frac{0.1}{4} = 0.025 \, \text{m} \]
Now, substituting \( h \) into the equation for \( I \):
\[ I = 0.005m + m(0.025)^2 = 0.005m + m(0.000625) = 0.005m + 0.000625m = 0.005625m \]

### Step 3: Calculate the period of oscillation \( T \)
The formula for the period of oscillation of a physical pendulum is:
\[ T = 2\pi \sqrt{\frac{I}{mgh}} \]
Substituting \( I \) and \( h \):
\[ T = 2\pi \sqrt{\frac{0.005625m}{mg(0.025)}} \]
The mass \( m \) cancels out:
\[ T = 2\pi \sqrt{\frac{0.005625}{g(0.025)}} \]
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ T = 2\pi \sqrt{\frac{0.005625}{10 \times 0.025}} = 2\pi \sqrt{\frac{0.005625}{0.25}} = 2\pi \sqrt{0.0225} \]
Calculating the square root:
\[ \sqrt{0.0225} = 0.15 \]
Thus,
\[ T = 2\pi \times 0.15 = 0.3\pi \]
Calculating \( 0.3\pi \):
\[ T \approx 0.3 \times 3.14 \approx 0.942 \, \text{s} \]

### Final Answer
The approximate period of oscillation is \( T \approx 0.94 \, \text{s} \).

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