A thin uniform rod of mass 1 kg and length 12 cm is suspended by a wire that passes through its centre and is perpendicular to its length.The wire is twisted and the rod is set oscillating. Time period of oscillation is found to be 3 s. When a flat triangular plate is suspended in same way through its centre of mass, the time period is found to be 6 s. The moment of inertia of the tringular plate about this axis is

Options

A. (a)`0.12kg-m^2`
B. (b)`0.24kg-m^2`
C. (c)`0.48kg-m^2`
D. (d)information insufficient

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the time period formula for a torsional pendulum
The time period \( T \) of a torsional pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{k}} \]
where \( I \) is the moment of inertia and \( k \) is the torsional spring constant.

### Step 2: Calculate the moment of inertia of the rod
For a thin uniform rod of mass \( m \) and length \( l \), the moment of inertia about an axis through its center is given by:
\[ I_{\text{rod}} = \frac{1}{12} m l^2 \]
Given \( m = 1 \, \text{kg} \) and \( l = 12 \, \text{cm} = 0.12 \, \text{m} \):
\[ I_{\text{rod}} = \frac{1}{12} \times 1 \times (0.12)^2 = \frac{1}{12} \times 1 \times 0.0144 = 0.0012 \, \text{kg m}^2 \]

### Step 3: Set up the ratio of the time periods
From the time period formula, we can set up the ratio of the time periods for the rod and the triangular plate:
\[ \frac{T_{\text{rod}}}{T_{\text{plate}}} = \sqrt{\frac{I_{\text{rod}}}{I_{\text{plate}}}} \]
Given \( T_{\text{rod}} = 3 \, \text{s} \) and \( T_{\text{plate}} = 6 \, \text{s} \):
\[ \frac{3}{6} = \sqrt{\frac{I_{\text{rod}}}{I_{\text{plate}}}} \]
This simplifies to:
\[ \frac{1}{2} = \sqrt{\frac{I_{\text{rod}}}{I_{\text{plate}}}} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{1}{2}\right)^2 = \frac{I_{\text{rod}}}{I_{\text{plate}}} \]
\[ \frac{1}{4} = \frac{I_{\text{rod}}}{I_{\text{plate}}} \]

### Step 5: Rearrange to find the moment of inertia of the plate
Rearranging gives:
\[ I_{\text{plate}} = 4 \times I_{\text{rod}} \]
Substituting \( I_{\text{rod}} = 0.0012 \, \text{kg m}^2 \):
\[ I_{\text{plate}} = 4 \times 0.0012 = 0.0048 \, \text{kg m}^2 \]

### Step 6: Final answer
Thus, the moment of inertia of the triangular plate about the axis is:
\[ I_{\text{plate}} = 0.0048 \, \text{kg m}^2 \]