Three identical rods, each of length `L`, are joined to form a rigid equilateral triangle. Its radius of gyration about an axis passing thorugh a corner and perpendicular to plane of triangle is

Options

A. `l/(sqrt(3))`
B. `l/(sqrt(2))`
C. `l/(sqrt(5))`
D. `l/(sqrt(7))`

Explanation

To find the radius of gyration of the equilateral triangle formed by three identical rods of length \( L \) about an axis passing through one corner and perpendicular to the plane of the triangle, we can follow these steps:

### Step 1: Understand the Configuration
We have an equilateral triangle \( ABC \) formed by three rods \( AB, BC, \) and \( CA \) each of length \( L \). We need to calculate the moment of inertia about an axis through point \( A \) and perpendicular to the plane of the triangle.

### Step 2: Moment of Inertia of Each Rod
The moment of inertia of a rod about an axis through its end (perpendicular to its length) is given by:
\[ I = \frac{ML^2}{3} \]
where \( M \) is the mass of the rod.

### Step 3: Moment of Inertia of Rods
1. **Rod \( AB \)**: The moment of inertia about point \( A \) is:
\[ I_{AB} = \frac{ML^2}{3} \]

2. **Rod \( AC \)**: The moment of inertia about point \( A \) is also:
\[ I_{AC} = \frac{ML^2}{3} \]

3. **Rod \( BC \)**: We need to calculate the moment of inertia of rod \( BC \) about point \( A \). We will use the parallel axis theorem:
\[ I_{BC} = I_{BC, \text{cm}} + Md^2 \]
where \( I_{BC, \text{cm}} \) is the moment of inertia about its center of mass and \( d \) is the distance from point \( A \) to the center of mass of rod \( BC \).

The center of mass of rod \( BC \) is at a distance of \( \frac{L}{2} \) from \( B \) and \( \frac{L}{2} \) from \( C \). The distance \( d \) from point \( A \) to the center of mass of rod \( BC \) can be calculated using geometry:
\[ d = \sqrt{L^2 - \left(\frac{L}{2}\right)^2} = \sqrt{L^2 - \frac{L^2}{4}} = \sqrt{\frac{3L^2}{4}} = \frac{L\sqrt{3}}{2} \]

Now, substituting \( I_{BC, \text{cm}} = \frac{ML^2}{12} \) (moment of inertia about its center of mass):
\[ I_{BC} = \frac{ML^2}{12} + M\left(\frac{L\sqrt{3}}{2}\right)^2 = \frac{ML^2}{12} + M\frac{3L^2}{4} \]
\[ I_{BC} = \frac{ML^2}{12} + \frac{3ML^2}{4} = \frac{ML^2}{12} + \frac{9ML^2}{12} = \frac{10ML^2}{12} = \frac{5ML^2}{6} \]

### Step 4: Total Moment of Inertia
Now, we can sum the moments of inertia of all three rods:
\[ I_{\text{total}} = I_{AB} + I_{AC} + I_{BC} \]
\[ I_{\text{total}} = \frac{ML^2}{3} + \frac{ML^2}{3} + \frac{5ML^2}{6} \]
Finding a common denominator (which is 6):
\[ I_{\text{total}} = \frac{2ML^2}{6} + \frac{2ML^2}{6} + \frac{5ML^2}{6} = \frac{9ML^2}{6} = \frac{3ML^2}{2} \]

### Step 5: Radius of Gyration
The radius of gyration \( K \) is given by:
\[ K = \sqrt{\frac{I_{\text{total}}}{M_{\text{total}}}} \]
Here, \( M_{\text{total}} = 3M \) (since there are three rods):
\[ K = \sqrt{\frac{\frac{3ML^2}{2}}{3M}} = \sqrt{\frac{L^2}{2}} = \frac{L}{\sqrt{2}} \]

### Final Answer
The radius of gyration about the axis passing through a corner and perpendicular to the plane of the triangle is:
\[ K = \frac{L}{\sqrt{2}} \]