A non -uniform rod of length l having mass density `lamda(x) = (A+Bx^2)` is placed - x - axis with its ends at , `(a,0) and (a+ l, 0)` . The force it would exert on a point mass m kept at the origin is

Options

A. `Gm[A(l/(a(a+2)))+Bl]`
B. `Gm[A(l/(a(a-l)))-Bl]`
C. `Gm[A(l/(a(a+l)))+Bl]`
D. `Gm[A(l/(a(a+l)))-Bl]`

Answer

To solve the problem of finding the gravitational force exerted by a non-uniform rod on a point mass \( m \) located at the origin, we can follow these steps:

### Step 1: Define the mass density and the rod's position
The mass density of the rod is given by:
\[ \lambda(x) = A + Bx^2 \]
The rod is placed along the x-axis from \( x = a \) to \( x = a + l \).

### Step 2: Express the mass element
To find the gravitational force, we need to consider a small mass element \( dm \) of the rod. The mass element can be expressed as:
\[ dm = \lambda(x) \, dx = (A + Bx^2) \, dx \]
where \( dx \) is an infinitesimal length of the rod.

### Step 3: Determine the force due to the mass element
The gravitational force \( dF \) exerted by the mass element \( dm \) at a distance \( x \) from the point mass \( m \) located at the origin is given by Newton's law of gravitation:
\[ dF = \frac{G m \, dm}{x^2} \]
Substituting \( dm \) from the previous step:
\[ dF = \frac{G m (A + Bx^2) \, dx}{x^2} \]

### Step 4: Simplify the expression for \( dF \)
We can simplify \( dF \):
\[ dF = G m \left( \frac{A}{x^2} + B \right) dx \]

### Step 5: Set up the integral for total force
To find the total force \( F \) exerted by the entire rod on the mass \( m \), we need to integrate \( dF \) from \( x = a \) to \( x = a + l \):
\[ F = \int_{a}^{a+l} dF = \int_{a}^{a+l} G m \left( \frac{A}{x^2} + B \right) dx \]

### Step 6: Evaluate the integral
This integral can be split into two parts:
\[ F = G m \left( \int_{a}^{a+l} \frac{A}{x^2} \, dx + \int_{a}^{a+l} B \, dx \right) \]

1. **First integral**:
\[ \int \frac{A}{x^2} \, dx = -\frac{A}{x} \]
Evaluating from \( a \) to \( a+l \):
\[ -\frac{A}{a+l} + \frac{A}{a} = A \left( \frac{1}{a} - \frac{1}{a+l} \right) = A \left( \frac{(a+l) - a}{a(a+l)} \right) = \frac{Al}{a(a+l)} \]

2. **Second integral**:
\[ \int B \, dx = Bx \]
Evaluating from \( a \) to \( a+l \):
\[ B(a+l) - Ba = Bl \]

### Step 7: Combine the results
Now, substituting back into the expression for \( F \):
\[ F = G m \left( \frac{Al}{a(a+l)} + Bl \right) \]

### Final Result
Thus, the total gravitational force exerted by the rod on the mass \( m \) at the origin is:
\[ F = G m \left( \frac{Al}{a(a+l)} + Bl \right) \]