A straight rod of length 'l' extends from x=l to x=2l. If linear mass density of rod is `mu=((mu_0)/(l^3)X^3)`, then the gravitational force exerted by rod on a particle of mass m at x=0 , is

Options

A. `1/2(Gmu_0m)/l`
B. `(Gmu_0m)/l`
C. `3/2(Gmu_0m)/l`
D. `(2Gmu_0m)/l`

Answer

To solve the problem, we need to find the gravitational force exerted by a straight rod of length 'l' with a varying linear mass density on a particle of mass 'm' located at x = 0.

### Step-by-Step Solution:

1. **Identify the rod and particle position**:
- The rod extends from \( x = l \) to \( x = 2l \).
- The particle of mass \( m \) is located at \( x = 0 \).

2. **Define the linear mass density**:
- The linear mass density of the rod is given as:
\[ \mu(x) = \frac{\mu_0}{l^3} x^3 \]

3. **Consider a small element of the rod**:
- Take a small element of the rod of length \( dx \) at position \( x \).
- The mass \( dm \) of this small element is:
\[ dm = \mu(x) \cdot dx = \frac{\mu_0}{l^3} x^3 \cdot dx \]

4. **Calculate the gravitational force due to the small element**:
- The distance from the small element at position \( x \) to the particle at \( x = 0 \) is \( r = x \).
- The gravitational force \( dF \) exerted by this small mass \( dm \) on the particle is given by Newton's law of gravitation:
\[ dF = \frac{G \cdot m \cdot dm}{r^2} = \frac{G \cdot m \cdot dm}{x^2} \]

5. **Substitute \( dm \) into the force equation**:
- Substitute \( dm \) into the expression for \( dF \):
\[ dF = \frac{G \cdot m \cdot \left(\frac{\mu_0}{l^3} x^3 \cdot dx\right)}{x^2} = \frac{G \cdot m \cdot \mu_0}{l^3} x \cdot dx \]

6. **Integrate to find the total force**:
- To find the total gravitational force \( F \) exerted by the entire rod on the particle, integrate \( dF \) from \( x = l \) to \( x = 2l \):
\[ F = \int_{l}^{2l} dF = \int_{l}^{2l} \frac{G \cdot m \cdot \mu_0}{l^3} x \, dx \]

7. **Calculate the integral**:
- The integral of \( x \) is:
\[ \int x \, dx = \frac{x^2}{2} \]
- Therefore, the limits from \( l \) to \( 2l \) give:
\[ F = \frac{G \cdot m \cdot \mu_0}{l^3} \left[ \frac{x^2}{2} \right]_{l}^{2l} = \frac{G \cdot m \cdot \mu_0}{l^3} \left( \frac{(2l)^2}{2} - \frac{l^2}{2} \right) \]
- Simplifying this:
\[ F = \frac{G \cdot m \cdot \mu_0}{l^3} \left( \frac{4l^2}{2} - \frac{l^2}{2} \right) = \frac{G \cdot m \cdot \mu_0}{l^3} \left( \frac{3l^2}{2} \right) \]
- Thus, the total gravitational force is:
\[ F = \frac{3G \cdot m \cdot \mu_0}{2l} \]

### Final Answer:
The gravitational force exerted by the rod on the particle of mass \( m \) at \( x = 0 \) is:
\[ F = \frac{3G \cdot m \cdot \mu_0}{2l} \]