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 given linear mass density on a particle of mass 'm' located at x=0. We will follow these steps:

### Step 1: Define the Linear Mass Density
The linear mass density of the rod is given by:
\[ \mu = \frac{\mu_0}{l^3} x^3 \]
where \( \mu_0 \) is a constant.

### Step 2: Set Up the Mass Element
Consider a small segment of the rod of length \( dx \) located at position \( x \). The mass of this small segment \( dm \) can be expressed as:
\[ dm = \mu \, dx = \frac{\mu_0}{l^3} x^3 \, dx \]

### Step 3: Calculate the Gravitational Force from the Mass Element
The gravitational force \( dF \) exerted by this small mass \( dm \) on the mass \( m \) located at \( x=0 \) is given by Newton's law of gravitation:
\[ dF = \frac{G m \, dm}{x^2} \]
Substituting \( dm \) into this equation gives:
\[ dF = \frac{G m \left(\frac{\mu_0}{l^3} x^3 \, dx\right)}{x^2} = \frac{G m \mu_0}{l^3} x \, dx \]

### Step 4: Integrate to Find the Total Force
To find the total gravitational force \( F_n \) exerted by the entire rod on the mass \( m \), we need to integrate \( dF \) from \( x=l \) to \( x=2l \):
\[ F_n = \int_{l}^{2l} dF = \int_{l}^{2l} \frac{G m \mu_0}{l^3} x \, dx \]

### Step 5: Perform the Integration
Calculating the integral:
\[ F_n = \frac{G m \mu_0}{l^3} \int_{l}^{2l} x \, dx \]
The integral of \( x \) is:
\[ \int x \, dx = \frac{x^2}{2} \]
Thus,
\[ F_n = \frac{G m \mu_0}{l^3} \left[ \frac{x^2}{2} \right]_{l}^{2l} = \frac{G m \mu_0}{l^3} \left( \frac{(2l)^2}{2} - \frac{l^2}{2} \right) \]
Calculating the limits:
\[ = \frac{G m \mu_0}{l^3} \left( \frac{4l^2}{2} - \frac{l^2}{2} \right) = \frac{G m \mu_0}{l^3} \left( \frac{4l^2 - l^2}{2} \right) = \frac{G m \mu_0}{l^3} \left( \frac{3l^2}{2} \right) \]

### Step 6: Simplify the Result
Finally, we simplify the expression:
\[ F_n = \frac{3 G m \mu_0}{2 l} \]

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