A rod of uniform cross-section of mass `M` and length `L` is hinged about an end to swing freely in a vertical plane. Howerver, its density is non uniform and varies linearly from hinged end to the free end doubling its value. The moment of inertia of the rod, about the rotation axis passing through the hinge point

Options

A. `2(ML^(2))/(9)`
B. `(3ML^(2))/(16)`
C. `(7ML^(2))/(18)`
D. none of these

Explanation

To find the moment of inertia of a rod with non-uniform density, we can follow these steps:

### Step 1: Define the Problem
We have a rod of length \( L \) and mass \( M \) that is hinged at one end. The density of the rod varies linearly from the hinged end to the free end, doubling its value. We need to find the moment of inertia about the hinge point.

### Step 2: Express the Density Function
Let the density at the hinged end be \( \rho_0 \). Since the density doubles at the free end, the density at the free end (at \( x = L \)) will be \( 2\rho_0 \). The linear variation of density can be expressed as:
\[
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