Mass of 20 kg is distributed uniformly over a ring of radius 2m. Find the the grvitational field at a point lies an th axis of the ring at a distance of `2 sqrt(3)` m from the centre.

Options

A. `2.1xx10^(-12)"Nkg"^(-1)`
B. `2.1xx10^(12)"Nkg"^(-1)`
C. `4.2xx10^(-12)"Nkg"^(-1)`
D. Zero

Explanation

To find the gravitational field at a point on the axis of a ring with a uniform mass distribution, we can use the formula for the gravitational field due to a ring of mass at a point along its axis.

### Given:
- Mass of the ring, \( M = 20 \, \text{kg} \)
- Radius of the ring, \( R = 2 \, \text{m} \)
- Distance from the center of the ring to the point on the axis, \( z = 2\sqrt{3} \, \text{m} \)

### Step 1: Use the formula for the gravitational field along the axis of the ring
The gravitational field \( E \) at a point on the axis of a ring is given by the formula:

\[ E = \frac{G M z}{(R^2 + z^2)^{3/2}} \]

where:
- \( G \) is the gravitational constant, \( G \approx 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)

### Step 2: Calculate \( R^2 + z^2 \)
First, we need to calculate \( R^2 + z^2 \):

\[ R^2 = (2 \, \text{m})^2 = 4 \, \text{m}^2 \]
\[ z^2 = (2\sqrt{3} \, \text{m})^2 = 4 \times 3 = 12 \, \text{m}^2 \]
\[ R^2 + z^2 = 4 + 12 = 16 \, \text{m}^2 \]

### Step 3: Calculate \( (R^2 + z^2)^{3/2} \)
Now, we calculate \( (R^2 + z^2)^{3/2} \):

\[ (R^2 + z^2)^{3/2} = (16)^{3/2} = 16^{1.5} = 64 \, \text{m}^3 \]

### Step 4: Substitute values into the gravitational field formula
Now we can substitute the values into the gravitational field formula:

\[ E = \frac{G M z}{(R^2 + z^2)^{3/2}} = \frac{(6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2)(20 \, \text{kg})(2\sqrt{3} \, \text{m})}{64 \, \text{m}^3} \]

### Step 5: Calculate the numerator
Calculating the numerator:

\[ G M z = (6.67 \times 10^{-11})(20)(2\sqrt{3}) = 6.67 \times 10^{-11} \times 40\sqrt{3} \]

Calculating \( 40\sqrt{3} \):

\[ 40\sqrt{3} \approx 40 \times 1.732 = 69.28 \]

Thus,

\[ G M z \approx 6.67 \times 10^{-11} \times 69.28 \approx 4.62 \times 10^{-9} \, \text{N m}^2/\text{kg} \]

### Step 6: Calculate the gravitational field \( E \)
Now substituting back into the equation for \( E \):

\[ E \approx \frac{4.62 \times 10^{-9}}{64} \approx 7.22 \times 10^{-11} \, \text{N/kg} \]

### Final Answer
The gravitational field at the point on the axis of the ring is approximately:

\[ E \approx 7.22 \times 10^{-11} \, \text{N/kg} \]