A semicircular wire hs a length L and mass M. A paricle of mass m is placed at the centre of the circle. Find the gravitational attraction on the particle due to the wire.

Options

A. `(GM)/(l^(2))` along x- axis
B. `(GM)/(pi l^(2))` along y-axis
C. `(2 pi GM)/(l^(2))` along x-axis
D. `2 pi GMm/(l^(2))` along y-axis

Explanation

To find the gravitational attraction on a particle of mass \( m \) placed at the center of a semicircular wire of length \( L \) and mass \( M \), we can follow these steps:

### Step 1: Define the Problem
We have a semicircular wire with length \( L \) and mass \( M \). A particle of mass \( m \) is located at the center of the semicircle. We need to calculate the gravitational force acting on the particle due to the wire.

### Step 2: Consider a Small Element of the Wire
Let's take a small element of the wire of length \( dx \). The mass of this small element can be expressed as:
\[ dm = \frac{M}{L} \cdot dx \]
where \( \frac{M}{L} \) is the mass per unit length of the wire.

### Step 3: Relate the Length Element to the Angle
The length element \( dx \) can be related to the angle \( d\theta \) subtended at the center of the semicircle. The relationship is given by:
\[ dx = R \, d\theta \]
where \( R \) is the radius of the semicircle.

### Step 4: Calculate the Gravitational Force from the Small Element
The gravitational force \( dF \) exerted by the small mass \( dm \) on the particle of mass \( m \) is given by Newton's law of gravitation:
\[ dF = \frac{G \, dm \, m}{r^2} \]
Substituting \( dm \) and \( r \) (which is equal to \( R \)):
\[ dF = \frac{G \left(\frac{M}{L} \cdot dx\right) m}{R^2} \]
Now substituting \( dx = R \, d\theta \):
\[ dF = \frac{G \left(\frac{M}{L} \cdot R \, d\theta\right) m}{R^2} = \frac{G M m}{L R} d\theta \]

### Step 5: Resolve the Force into Components
The gravitational force \( dF \) acts along the line connecting the mass \( dm \) and the mass \( m \). We need to resolve this force into vertical and horizontal components. The vertical component \( dF_y \) is given by:
\[ dF_y = dF \sin \theta \]
Since the horizontal components will cancel out due to symmetry, we only need to consider the vertical components.

### Step 6: Integrate the Vertical Component
To find the total vertical force \( F_y \), we integrate \( dF_y \) from \( \theta = 0 \) to \( \theta = \pi \):
\[ F_y = \int_0^{\pi} dF \sin \theta = \int_0^{\pi} \frac{G M m}{L R} \sin \theta \, d\theta \]
This integral evaluates to:
\[ F_y = \frac{G M m}{L R} \left[-\cos \theta \right]_0^{\pi} = \frac{G M m}{L R} [(-\cos(\pi)) - (-\cos(0))] = \frac{G M m}{L R} [1 + 1] = \frac{2 G M m}{L R} \]

### Step 7: Substitute the Radius in Terms of Length
The length of the semicircular wire \( L \) is related to the radius \( R \) by:
\[ L = \pi R \implies R = \frac{L}{\pi} \]
Substituting this into the expression for \( F_y \):
\[ F_y = \frac{2 G M m}{L \left(\frac{L}{\pi}\right)} = \frac{2 G M m \pi}{L^2} \]

### Final Result
Thus, the gravitational attraction on the particle due to the wire is:
\[ F = \frac{2 G M m \pi}{L^2} \]

---