A uniform thin rod AB of length L has linear mass density `mu(x) = a + (bx)/(L)`, where x is measured from A. If the CM of the rod lies at a distance of `((7)/(12)L)` from A, then a and b are related as :_

Options

A. `2a = b`
B. `3a = 2b`
C. `a = 2b`
D. `a = b`

Explanation

To solve the problem, we need to find the relationship between the constants \( a \) and \( b \) given the linear mass density of the rod and the position of the center of mass (CM).

### Step-by-Step Solution:

1. **Understanding the Linear Mass Density**:
The linear mass density of the rod is given by:
\[ \mu(x) = a + \frac{b x}{L} \]
where \( x \) is the distance from point A along the rod.

2. **Setting Up the Center of Mass Formula**:
The center of mass \( x_{cm} \) for a continuous mass distribution is given by:
\[ x_{cm} = \frac{\int_0^L x \, dm}{\int_0^L dm} \]
Here, \( dm = \mu(x) \, dx \).

3. **Calculating \( dm \)**:
Substituting the expression for \( \mu(x) \):
\[ dm = \left(a + \frac{b x}{L}\right) dx \]

4. **Calculating the Numerator**:
We need to compute the integral for the numerator:
\[ \int_0^L x \, dm = \int_0^L x \left(a + \frac{b x}{L}\right) dx \]
This expands to:
\[ \int_0^L (a x + \frac{b x^2}{L}) dx \]
We can split this into two integrals:
\[ = a \int_0^L x \, dx + \frac{b}{L} \int_0^L x^2 \, dx \]

5. **Evaluating the Integrals**:
- The integral \( \int_0^L x \, dx = \frac{L^2}{2} \)
- The integral \( \int_0^L x^2 \, dx = \frac{L^3}{3} \)

Substituting these results:
\[ = a \cdot \frac{L^2}{2} + \frac{b}{L} \cdot \frac{L^3}{3} = \frac{a L^2}{2} + \frac{b L^2}{3} \]

6. **Calculating the Denominator**:
Now, we compute the denominator:
\[ \int_0^L dm = \int_0^L \left(a + \frac{b x}{L}\right) dx = aL + \frac{b}{L} \cdot \frac{L^2}{2} = aL + \frac{bL}{2} \]

7. **Putting It All Together**:
Now substituting the numerator and denominator into the center of mass formula:
\[ x_{cm} = \frac{\frac{a L^2}{2} + \frac{b L^2}{3}}{aL + \frac{bL}{2}} \]
Simplifying this gives:
\[ x_{cm} = \frac{L^2 \left(\frac{a}{2} + \frac{b}{3}\right)}{L \left(a + \frac{b}{2}\right)} = \frac{L \left(\frac{a}{2} + \frac{b}{3}\right)}{a + \frac{b}{2}} \]

8. **Setting the Center of Mass**:
We know from the problem statement that:
\[ x_{cm} = \frac{7L}{12} \]
Thus, we set the two expressions equal:
\[ \frac{L \left(\frac{a}{2} + \frac{b}{3}\right)}{a + \frac{b}{2}} = \frac{7L}{12} \]

9. **Cross Multiplying**:
Removing \( L \) from both sides (assuming \( L \neq 0 \)):
\[ 12 \left(\frac{a}{2} + \frac{b}{3}\right) = 7 \left(a + \frac{b}{2}\right) \]

10. **Expanding and Simplifying**:
Expanding both sides:
\[ 6a + 4b = 7a + \frac{7b}{2} \]
Rearranging gives:
\[ 6a + 4b - 7a - \frac{7b}{2} = 0 \]
Combining like terms leads to:
\[ -a + \left(4 - \frac{7}{2}\right)b = 0 \implies -a - \frac{3b}{2} = 0 \implies 2a = 3b \]

11. **Final Relationship**:
Thus, we find the relationship between \( a \) and \( b \):
\[ 2a = 3b \quad \text{or} \quad b = \frac{2}{3}a \]