Consider a ring of mass m and radius r. Maximum gravitational intensity on the axis of the ring has value.

Options

A. `(Gm)/(r^(2))`
B. `(2Gm)/(sqrt(3)r^(2))`
C. `(2Gm)/(3sqrt(3)r^(2))`
D. `(Gm)/(2 r^(2))`

Answer

To find the maximum gravitational intensity on the axis of a ring of mass \( m \) and radius \( r \), we can follow these steps:

### Step 1: Understand Gravitational Intensity
Gravitational intensity (or gravitational field strength) \( g \) at a point in space due to a mass is given by the formula:
\[ g = \frac{G M}{r^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass creating the field, and \( r \) is the distance from the mass to the point where the field is being calculated.

### Step 2: Set Up the Problem
Consider a ring of mass \( m \) and radius \( r \). We want to find the gravitational intensity at a point along the axis of the ring. Let the distance from the center of the ring to the point on the axis be \( x \).

### Step 3: Calculate Gravitational Intensity on the Axis
The gravitational intensity \( g \) at a distance \( x \) from the center of the ring along the axis can be derived by integrating contributions from each infinitesimal mass element \( dm \) of the ring. The gravitational intensity due to a mass element \( dm \) at a distance \( d \) from the point on the axis is given by:
\[ dg = \frac{G \, dm}{d^2} \]
where \( d = \sqrt{r^2 + x^2} \) (using Pythagoras theorem).

### Step 4: Integrate Over the Ring
Since the ring has a uniform mass distribution, we can integrate \( dg \) around the entire ring:
\[ g = \int \frac{G \, dm}{r^2 + x^2} \]
The total mass \( m \) can be expressed in terms of \( dm \) as \( dm = \frac{m}{2\pi r} d\theta \) where \( d\theta \) is the angle subtended by the mass element at the center.

### Step 5: Simplify the Expression
The gravitational intensity \( g \) can be simplified to:
\[ g = \frac{G m}{(r^2 + x^2)^{3/2}} x \]

### Step 6: Find Maximum Gravitational Intensity
To find the maximum gravitational intensity, we need to differentiate \( g \) with respect to \( x \) and set the derivative to zero:
\[ \frac{dg}{dx} = 0 \]
After differentiation and simplification, we find that the maximum occurs at \( x = \frac{r}{\sqrt{2}} \).

### Step 7: Calculate Maximum Gravitational Intensity
Substituting \( x = \frac{r}{\sqrt{2}} \) back into the expression for \( g \):
\[ g_{max} = \frac{G m}{(r^2 + \left(\frac{r}{\sqrt{2}}\right)^2)^{3/2}} \cdot \frac{r}{\sqrt{2}} \]
This simplifies to:
\[ g_{max} = \frac{G m}{(2r^2)^{3/2}} \cdot \frac{r}{\sqrt{2}} = \frac{G m}{2\sqrt{2} r^3} \]

### Final Answer
Thus, the maximum gravitational intensity on the axis of the ring is:
\[ g_{max} = \frac{G m}{2\sqrt{2} r^2} \]