A straight rod of length `L` extends from `x=a` to `x=L+a`. Find the gravitational force exerts on a point mass `m` at `x=0` is (if the linear density of rod`mu=A+Bx^(2)`)

Options

A. `Gm[A/a+BL]`
B. `Gm[A(1/a-1/(a+L))+BL]`
C. `Gm[BL+A/(a+L)]`
D. `Gm[BL-A/a]`

Explanation

To find the gravitational force exerted on a point mass \( m \) located at \( x = 0 \) by a straight rod of length \( L \) extending from \( x = a \) to \( x = L + a \) with a linear density given by \( \mu = A + Bx^2 \), we can follow these steps:

### Step-by-Step Solution

1. **Define the setup**:
- The rod extends from \( x = a \) to \( x = L + a \).
- The point mass \( m \) is located at \( x = 0 \).
- The linear density of the rod is given by \( \mu = A + Bx^2 \).

2. **Consider a small element of the rod**:
- Let \( dx \) be a small element of the rod at position \( x \).
- The mass of this small element \( dm \) can be expressed as:
\[ dm = \mu \cdot dx = (A + Bx^2) \cdot dx \]

3. **Calculate the gravitational force \( dF \) due to the small element**:
- The gravitational force exerted by the small mass \( dm \) on the point mass \( m \) is given by Newton's law of gravitation:
\[ dF = \frac{G m \, dm}{r^2} \]
- Here, \( r \) is the distance from the mass \( m \) to the element \( dx \), which is \( r = x \).
- Thus, we have:
\[ dF = \frac{G m \, (A + Bx^2) \, dx}{x^2} \]

4. **Express \( dF \) in terms of \( dx \)**:
- We can separate the terms:
\[ dF = Gm \left( \frac{A}{x^2} + B \right) dx \]

5. **Integrate to find the total force \( F \)**:
- The total force \( F \) on the mass \( m \) is the integral of \( dF \) from \( x = a \) to \( x = L + a \):
\[ F = \int_{a}^{L + a} Gm \left( \frac{A}{x^2} + B \right) dx \]

6. **Calculate the integral**:
- The integral can be split into two parts:
\[ F = Gm \left( A \int_{a}^{L + a} \frac{1}{x^2} dx + B \int_{a}^{L + a} dx \right) \]
- The first integral:
\[ \int \frac{1}{x^2} dx = -\frac{1}{x} \]
- Evaluating from \( a \) to \( L + a \):
\[ \int_{a}^{L + a} \frac{1}{x^2} dx = -\frac{1}{L + a} + \frac{1}{a} = \frac{1}{a} - \frac{1}{L + a} \]
- The second integral:
\[ \int dx = x \]
- Evaluating from \( a \) to \( L + a \):
\[ \int_{a}^{L + a} dx = (L + a) - a = L \]

7. **Combine the results**:
- Putting it all together:
\[ F = Gm \left( A \left( \frac{1}{a} - \frac{1}{L + a} \right) + B \cdot L \right) \]

### Final Expression for the Gravitational Force
\[ F = Gm \left( A \left( \frac{1}{a} - \frac{1}{L + a} \right) + B \cdot L \right) \]