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

Options

A.
B.
C.
D.

Answer

To solve the problem of finding the center of mass \( x_{CM} \) of a thin rod with a varying linear density, we will follow these steps:

### Step 1: Define the linear density
The linear density \( \lambda(x) \) 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
A small element of the rod \( dx \) at position \( x \) has mass \( dm \) given by:
\[ dm = \lambda(x) \, dx = k \left( \frac{x}{L} \right)^n dx \]

### Step 3: Set up the center of mass formula
The center of mass \( x_{CM} \) is defined as:
\[ x_{CM} = \frac{\int_0^L x \, dm}{\int_0^L dm} \]

### Step 4: Calculate the numerator
Substituting \( dm \) into the numerator:
\[ \int_0^L x \, dm = \int_0^L x \cdot k \left( \frac{x}{L} \right)^n dx = k \int_0^L x^{n+1} \frac{1}{L^n} dx \]
This simplifies to:
\[ \frac{k}{L^n} \int_0^L x^{n+1} dx \]
Calculating the integral:
\[ \int_0^L x^{n+1} dx = \left[ \frac{x^{n+2}}{n+2} \right]_0^L = \frac{L^{n+2}}{n+2} \]
Thus, the numerator becomes:
\[ \int_0^L x \, dm = \frac{k L^{n+2}}{L^n (n+2)} = \frac{k L^2}{n+2} \]

### Step 5: Calculate the denominator
Now, we calculate the denominator:
\[ \int_0^L dm = \int_0^L k \left( \frac{x}{L} \right)^n dx = k \int_0^L \frac{x^n}{L^n} dx = \frac{k}{L^n} \int_0^L x^n dx \]
Calculating the integral:
\[ \int_0^L x^n dx = \left[ \frac{x^{n+1}}{n+1} \right]_0^L = \frac{L^{n+1}}{n+1} \]
Thus, the denominator becomes:
\[ \int_0^L dm = \frac{k L^{n+1}}{L^n (n+1)} = \frac{k L}{n+1} \]

### Step 6: Combine to find \( x_{CM} \)
Now substituting back into the formula for \( x_{CM} \):
\[ x_{CM} = \frac{\frac{k L^2}{n+2}}{\frac{k L}{n+1}} = \frac{L^2}{n+2} \cdot \frac{n+1}{L} = \frac{L(n+1)}{n+2} \]

### Step 7: Analyze the relationship with \( n \)
From the expression \( x_{CM} = \frac{L(n+1)}{n+2} \), we can analyze how \( x_{CM} \) behaves as \( n \) changes:
- When \( n = 0 \):
\[ x_{CM} = \frac{L(0+1)}{0+2} = \frac{L}{2} \]
- As \( n \to \infty \):
\[ x_{CM} \to L \]
This indicates that as \( n \) increases, \( x_{CM} \) increases from \( \frac{L}{2} \) to \( L \).

### Conclusion
The graph of \( x_{CM} \) versus \( n \) will start at \( \frac{L}{2} \) when \( n = 0 \) and approach \( L \) as \( n \) increases. Thus, the graph will be an increasing function of \( n \).