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 the non-uniform rod with the given density function, we can follow these steps:

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

### Step 2: Express the mass element
The mass element \( dm \) can be expressed in terms of the density and a small length element \( dx \):
\[ dm = \rho(x) \, dx = a(1 + bx^2) \, dx \]

### Step 3: Set up the center of mass formula
The center of mass \( x_{cm} \) of the rod is given by the formula:
\[ x_{cm} = \frac{\int_0^L x \, dm}{\int_0^L dm} \]
where \( L \) is the length of the rod (1 meter in this case).

### Step 4: Calculate the numerator
First, we need to calculate the integral for the numerator:
\[ \int_0^1 x \, dm = \int_0^1 x \cdot a(1 + bx^2) \, dx \]
This expands to:
\[ \int_0^1 (ax + abx^3) \, dx \]
Now, we can integrate term by term:
\[ = a \int_0^1 x \, dx + ab \int_0^1 x^3 \, dx \]
Calculating these integrals:
\[ = a \left[ \frac{x^2}{2} \right]_0^1 + ab \left[ \frac{x^4}{4} \right]_0^1 \]
\[ = a \cdot \frac{1}{2} + ab \cdot \frac{1}{4} = \frac{a}{2} + \frac{ab}{4} \]

### Step 5: Calculate the denominator
Next, we calculate the integral for the denominator:
\[ \int_0^1 dm = \int_0^1 a(1 + bx^2) \, dx \]
This expands to:
\[ = a \int_0^1 1 \, dx + ab \int_0^1 x^2 \, dx \]
Calculating these integrals:
\[ = a \left[ x \right]_0^1 + ab \left[ \frac{x^3}{3} \right]_0^1 \]
\[ = a \cdot 1 + ab \cdot \frac{1}{3} = a + \frac{ab}{3} \]

### Step 6: Combine results to find the center of mass
Now we can substitute the results from the numerator and denominator into the center of mass formula:
\[ x_{cm} = \frac{\frac{a}{2} + \frac{ab}{4}}{a + \frac{ab}{3}} \]

### Step 7: Simplify the expression
To simplify, we can multiply the numerator and denominator by 12 to eliminate the fractions:
\[ x_{cm} = \frac{6a + 3ab}{12a + 4ab} \]
This can be simplified further:
\[ x_{cm} = \frac{3(2 + b)}{4(3 + b)} \]

### Final Result
Thus, the center of mass of the rod is given by:
\[ x_{cm} = \frac{3(2 + b)}{4(3 + b)} \]