A wire frame in the shape of an equilateral triangle is hinged at one vertex so that it can swing freely in a vertical plane, with the plane of the `Delta` always remaining vertical. The side of the frame is `1//sqrt(3)m`. The time period in seconds of small oscillations of the frame will be-

Options

A. `pi//sqrt(2)`
B. `pi//sqrt(3)`
C. `pi//sqrt(6)`
D. `pi//sqrt(5)`

Explanation

To solve the problem of finding the time period of small oscillations of a wire frame in the shape of an equilateral triangle hinged at one vertex, we can follow these steps:

### Step 1: Understand the Geometry of the System
The wire frame is an equilateral triangle with each side measuring \( \frac{1}{\sqrt{3}} \) m. The triangle is hinged at one vertex, allowing it to swing freely in a vertical plane.

### Step 2: Identify the Moment of Inertia
To find the time period of oscillation, we need to calculate the moment of inertia \( I \) of the triangular frame about the hinge point. The moment of inertia for each side of the triangle can be calculated using the parallel axis theorem.

1. The moment of inertia of a rod about one end is given by:
\[ I_{\text{rod}} = \frac{1}{3} m l^2 \]
where \( l \) is the length of the rod.

2. For the equilateral triangle, we have three sides. The moment of inertia for each side about the hinge point needs to be calculated.

### Step 3: Calculate the Moment of Inertia for Each Side
1. For the two sides meeting at the hinge:
\[ I_1 = \frac{1}{3} m \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} m \cdot \frac{1}{3} = \frac{m}{9} \]
(Each of these contributes \( \frac{m}{9} \) to the total moment of inertia.)

2. For the opposite side (using the parallel axis theorem):
- The distance from the hinge to the center of mass of this side is \( \frac{1}{\sqrt{3}} \).
- The moment of inertia about the center of mass is:
\[ I_{\text{cm}} = \frac{1}{3} m \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{m}{9} \]
- Applying the parallel axis theorem:
\[ I_2 = I_{\text{cm}} + m d^2 = \frac{m}{9} + m \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{m}{9} + \frac{m}{3} = \frac{4m}{9} \]

### Step 4: Total Moment of Inertia
The total moment of inertia \( I \) about the hinge is:
\[ I = I_1 + I_1 + I_2 = \frac{m}{9} + \frac{m}{9} + \frac{4m}{9} = \frac{6m}{9} = \frac{2m}{3} \]

### Step 5: Calculate the Time Period
The time period \( T \) of a compound pendulum is given by:
\[ T = 2\pi \sqrt{\frac{I}{mgh}} \]
where \( h \) is the distance from the hinge to the center of mass of the triangle. The center of mass of an equilateral triangle is located at a distance of \( \frac{a}{\sqrt{3}} \) from the base, which is \( \frac{1}{3\sqrt{3}} \) m from the hinge.

Substituting \( I \) and \( h \):
\[ T = 2\pi \sqrt{\frac{\frac{2m}{3}}{mg \cdot \frac{1}{3}}} = 2\pi \sqrt{\frac{2}{g}} \]

### Step 6: Substitute \( g \)
Assuming \( g = 10 \, \text{m/s}^2 \):
\[ T = 2\pi \sqrt{\frac{2}{10}} = 2\pi \sqrt{\frac{1}{5}} = \frac{2\pi}{\sqrt{5}} \]

### Final Answer
Thus, the time period of small oscillations of the frame is:
\[ T = \frac{2\pi}{\sqrt{5}} \text{ seconds} \]