Three identical thin rods each of length `l` and mass `M` are joined together to from a letter. `H`. What is the moment of inertia of the system about one of the sides of `H` ?

Options

A. `(Ml^2)/(4)`
B. `(M l^2)/(3)`
C. `(2 M l^2)/(3)`
D. `(4 M l^2)/(3)`

Explanation

To find the moment of inertia of the system formed by three identical thin rods arranged in the shape of the letter "H" about one of the sides (let's say side AB), we can follow these steps:

### Step 1: Identify the rods and their arrangement
We have three rods:
- Rod AB (vertical)
- Rod CD (horizontal)
- Rod EF (horizontal)

### Step 2: Calculate the moment of inertia of rod CD about AB
Rod CD is perpendicular to rod AB. The moment of inertia \( I \) of a rod about an axis through one end and perpendicular to its length is given by the formula:
\[ I = \frac{1}{3}ML^2 \]
For rod CD:
\[ I_{CD} = \frac{1}{3}M l^2 \]

### Step 3: Calculate the moment of inertia of rod EF about AB
Rod EF is parallel to rod AB and is at a distance \( l \) from it. The moment of inertia \( I \) of a rod about an axis parallel to it and at a distance \( d \) is given by the parallel axis theorem:
\[ I = I_{cm} + Md^2 \]
where \( I_{cm} \) is the moment of inertia about its center of mass. For rod EF:
\[ I_{cm} = \frac{1}{3}Ml^2 \]
The distance \( d \) from AB to EF is \( l \):
\[ I_{EF} = \frac{1}{3}Ml^2 + M(l)^2 = \frac{1}{3}Ml^2 + Ml^2 = \frac{1}{3}Ml^2 + \frac{3}{3}Ml^2 = \frac{4}{3}Ml^2 \]

### Step 4: Total moment of inertia about AB
Now, we can find the total moment of inertia \( I_{total} \) about the axis through AB:
\[ I_{total} = I_{CD} + I_{EF} = \frac{1}{3}Ml^2 + \frac{4}{3}Ml^2 = \frac{5}{3}Ml^2 \]

### Step 5: Final expression
Thus, the moment of inertia of the system about one of the sides of H (AB) is:
\[ I_{total} = \frac{5}{3}Ml^2 \]