The density of a non-uniform rod of length `1m` is given by `rho (x) = a (1 + bx^(2))`
where a and b are constants and `0 le x le 1`.
The centre of mass of the rod will be at

Options

A. `(3(2+b))/(4(3+b))`
B. `(4(2+b))/(3(3+b))`
C. `(3(3+b))/(4(2+b))`
D. `(4(3+b))/(3(2+b))`

Explanation

To find the center of mass of a non-uniform rod with a given density function, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Density Function**:
The density of the rod is given by:
\[ \rho(x) = a(1 + bx^2) \]
where \( a \) and \( b \) are constants, and \( 0 \leq x \leq 1 \).

2. **Set Up the Expression for Center of Mass**:
The center of mass \( x_{cm} \) for a rod can be calculated using the formula:
\[ x_{cm} = \frac{\int_0^L x \cdot \rho(x) \, dx}{\int_0^L \rho(x) \, dx} \]
Here, \( L = 1 \) meter.

3. **Calculate the Numerator**:
Substitute the density function into the numerator:
\[ \int_0^1 x \cdot \rho(x) \, dx = \int_0^1 x \cdot a(1 + bx^2) \, dx \]
Expanding this gives:
\[ = a \int_0^1 (x + bx^3) \, dx \]
Now, calculate the integrals:
\[ = a \left( \int_0^1 x \, dx + b \int_0^1 x^3 \, dx \right) \]
The integrals evaluate to:
\[ \int_0^1 x \, dx = \frac{1}{2} \quad \text{and} \quad \int_0^1 x^3 \, dx = \frac{1}{4} \]
Thus, the numerator becomes:
\[ = a \left( \frac{1}{2} + b \cdot \frac{1}{4} \right) = a \left( \frac{1}{2} + \frac{b}{4} \right) = \frac{2a + b}{4} \]

4. **Calculate the Denominator**:
Now, calculate the denominator:
\[ \int_0^1 \rho(x) \, dx = \int_0^1 a(1 + bx^2) \, dx \]
Expanding this gives:
\[ = a \left( \int_0^1 1 \, dx + b \int_0^1 x^2 \, dx \right) \]
The integrals evaluate to:
\[ \int_0^1 1 \, dx = 1 \quad \text{and} \quad \int_0^1 x^2 \, dx = \frac{1}{3} \]
Thus, the denominator becomes:
\[ = a \left( 1 + b \cdot \frac{1}{3} \right) = a \left( 1 + \frac{b}{3} \right) = a \cdot \frac{3 + b}{3} \]

5. **Combine the Results**:
Now substitute the results of the numerator and denominator into the center of mass formula:
\[ x_{cm} = \frac{\frac{2a + b}{4}}{\frac{a(3 + b)}{3}} = \frac{3(2a + b)}{4a(3 + b)} \]

6. **Final Result**:
Therefore, the center of mass of the rod is located at:
\[ x_{cm} = \frac{3(2a + b)}{4a(3 + b)} \]