A thin rod of length L is lying along the x-axis with its ends at x = 0 and x = L. Its linear density (mass/length) varies with x as `k((x)/(L))^n` where n can be zero or any positive number. If the position `X_(CM)` of the centre of mass of the rod is plotted against n, which of the following graphs best approximates the dependence of `X_(CM)` on n?

Options

A.
B.
C.
D.

Explanation

To find the position of the center of mass \( X_{CM} \) of the thin rod with a varying linear density, we will follow these steps:

### Step 1: Define the linear density
The linear density of the rod is given by:
\[ \lambda(x) = k \left( \frac{x}{L} \right)^n \]
where \( k \) is a constant, \( L \) is the length of the rod, and \( n \) can be 0 or any positive number.

### Step 2: Express the mass element
The mass element \( dm \) of a small segment of the rod of length \( dx \) at position \( x \) is given by:
\[ dm = \lambda(x) \, dx = k \left( \frac{x}{L} \right)^n \, dx \]

### Step 3: Calculate the total mass of the rod
The total mass \( M \) of the rod can be calculated by integrating \( dm \) from 0 to \( L \):
\[ M = \int_0^L dm = \int_0^L k \left( \frac{x}{L} \right)^n \, dx \]
This simplifies to:
\[ M = k \int_0^L \left( \frac{x^n}{L^n} \right) \, dx = \frac{k}{L^n} \int_0^L x^n \, dx \]
The integral \( \int_0^L x^n \, dx \) evaluates to:
\[ \int_0^L x^n \, dx = \frac{L^{n+1}}{n+1} \]
Thus, the total mass becomes:
\[ M = \frac{k}{L^n} \cdot \frac{L^{n+1}}{n+1} = \frac{kL}{n+1} \]

### Step 4: Calculate the center of mass \( X_{CM} \)
The center of mass \( X_{CM} \) is given by:
\[ X_{CM} = \frac{1}{M} \int_0^L x \, dm = \frac{1}{M} \int_0^L x \cdot k \left( \frac{x}{L} \right)^n \, dx \]
This can be simplified to:
\[ X_{CM} = \frac{1}{M} \cdot k \int_0^L \frac{x^{n+1}}{L^n} \, dx = \frac{1}{M} \cdot \frac{k}{L^n} \cdot \frac{L^{n+2}}{n+2} \]
Substituting \( M \) from the previous step:
\[ X_{CM} = \frac{1}{\frac{kL}{n+1}} \cdot \frac{kL^{n+2}}{(n+2)L^n} = \frac{L^{n+2}}{L(n+1)(n+2)} = \frac{L(n+1)}{(n+2)} \]

### Step 5: Analyze the behavior of \( X_{CM} \) with respect to \( n \)
The expression for the center of mass is:
\[ X_{CM} = \frac{L(n+1)}{(n+2)} \]
As \( n \) increases, the ratio \( \frac{n+1}{n+2} \) approaches 1, meaning \( X_{CM} \) approaches \( L \).

### Step 6: Graphical representation
- For \( n = 0 \): \( X_{CM} = \frac{L(1)}{(2)} = \frac{L}{2} \)
- As \( n \) increases, \( X_{CM} \) approaches \( L \).

Thus, the graph of \( X_{CM} \) against \( n \) will start at \( \frac{L}{2} \) when \( n = 0 \) and will asymptotically approach \( L \) as \( n \) increases.

### Final Answer
The graph that best represents the dependence of \( X_{CM} \) on \( n \) will start at \( \frac{L}{2} \) and approach \( L \) as \( n \) increases.