A rod of length L is placed along the X-axis between `x=0 and x=L`. The linear density (mass/length) `rho` of the rod varies with the distance x from the origin as `rhoj=a+bx. ` a. Find the SI units of a and b. b. Find the mass of the rod in terms of a,b, and L.

Explanation

To solve the problem, we will break it down into two parts as specified in the question.

### Part a: Finding the SI units of a and b

1. **Understanding Linear Density (ρ)**:
The linear density (ρ) is defined as mass per unit length. Therefore, its SI unit is:
\[ \text{SI unit of } \rho = \frac{\text{kg}}{\text{m}} \quad (\text{kilograms per meter}) \]

2. **Finding the Unit of a**:
The linear density is given by the expression \( \rho = a + bx \). Since \( \rho \) has the unit of kg/m, the unit of \( a \) must also be kg/m to maintain dimensional consistency. Thus:
\[ \text{SI unit of } a = \frac{\text{kg}}{\text{m}} \]

3. **Finding the Unit of b**:
The term \( bx \) must also have the same unit as \( \rho \). Here, \( x \) is a distance measured in meters (m). Therefore, to find the unit of \( b \), we can set up the equation:
\[ \text{Unit of } bx = \text{Unit of } b \times \text{Unit of } x = \frac{\text{kg}}{\text{m}} \]
Rearranging gives us:
\[ \text{Unit of } b = \frac{\text{kg/m}}{\text{m}} = \frac{\text{kg}}{\text{m}^2} \]
Thus:
\[ \text{SI unit of } b = \frac{\text{kg}}{\text{m}^2} \]

### Part b: Finding the mass of the rod in terms of a, b, and L

1. **Expression for Mass**:
The mass of the rod can be calculated by integrating the linear density over the length of the rod. The mass \( M \) can be expressed as:
\[ M = \int_{0}^{L} \rho \, dx \]
Substituting the expression for \( \rho \):
\[ M = \int_{0}^{L} (a + bx) \, dx \]

2. **Performing the Integration**:
We can split the integral:
\[ M = \int_{0}^{L} a \, dx + \int_{0}^{L} bx \, dx \]
The first integral evaluates to:
\[ \int_{0}^{L} a \, dx = a \cdot x \bigg|_{0}^{L} = aL \]
The second integral evaluates to:
\[ \int_{0}^{L} bx \, dx = b \cdot \frac{x^2}{2} \bigg|_{0}^{L} = b \cdot \frac{L^2}{2} \]

3. **Combining the Results**:
Therefore, the total mass \( M \) is:
\[ M = aL + \frac{bL^2}{2} \]

### Final Answer:
- The SI units are:
- \( a: \frac{\text{kg}}{\text{m}} \)
- \( b: \frac{\text{kg}}{\text{m}^2} \)
- The mass of the rod is:
\[ M = aL + \frac{bL^2}{2} \]