The radius of gyration of a uniform circular ring of radius R, about an axis which is a chord of circle of length `sqrt3R` is ,

Options

A. (a)`R/2`
B. (b)`(sqrt3R)/2`
C. (c)`(2R)/(sqrt3)`
D. (d)`R/sqrt3`

Explanation

To find the radius of gyration of a uniform circular ring of radius \( R \) about an axis which is a chord of length \( \sqrt{3}R \), we can follow these steps:

### Step 1: Understand the Geometry
We have a circular ring with radius \( R \). A chord of length \( \sqrt{3}R \) divides the circle into two equal halves. The midpoint of the chord will be directly below the center of the circle.

### Step 2: Calculate the Distance from the Center to the Chord
The length of the chord is given as \( \sqrt{3}R \). The distance from the center of the circle to the chord can be calculated using the Pythagorean theorem.

Let \( d \) be the distance from the center of the circle to the chord. The relationship can be expressed as:
\[ R^2 = \left(\frac{\sqrt{3}R}{2}\right)^2 + d^2 \]
where \( \frac{\sqrt{3}R}{2} \) is half the length of the chord.

### Step 3: Solve for \( d \)
Substituting the values into the equation:
\[ R^2 = \frac{3R^2}{4} + d^2 \]
Rearranging gives:
\[ d^2 = R^2 - \frac{3R^2}{4} = \frac{1R^2}{4} \]
Thus,
\[ d = \frac{R}{2} \]

### Step 4: Use the Parallel Axis Theorem
The moment of inertia \( I \) of the ring about its center of mass is:
\[ I_{cm} = mR^2 \]
Using the parallel axis theorem, the moment of inertia about the chord is given by:
\[ I = I_{cm} + md^2 \]
Substituting the values we have:
\[ I = mR^2 + m\left(\frac{R}{2}\right)^2 \]
\[ I = mR^2 + m\frac{R^2}{4} = mR^2\left(1 + \frac{1}{4}\right) = mR^2\left(\frac{5}{4}\right) \]

### Step 5: Find the Radius of Gyration
The radius of gyration \( k \) is related to the moment of inertia by:
\[ I = mk^2 \]
Substituting for \( I \):
\[ mk^2 = mR^2\left(\frac{5}{4}\right) \]
Dividing both sides by \( m \):
\[ k^2 = R^2\left(\frac{5}{4}\right) \]
Taking the square root gives:
\[ k = R\sqrt{\frac{5}{4}} = \frac{R\sqrt{5}}{2} \]

### Final Result
Thus, the radius of gyration of the uniform circular ring about the axis which is a chord of length \( \sqrt{3}R \) is:
\[ k = \frac{R\sqrt{5}}{2} \]