The moment of inertia of a uniform cylinder of length `l and radius R` about its perpendicular bisector is `I`. What is the ratio `l//R` such that the moment of inertia is minimum ?

Options

A. `1`
B. `(3)/(sqrt(2))`
C. `sqrt((3)/(2))`
D. `(sqrt(3))/(2)`

Explanation

To find the ratio \( \frac{l}{R} \) such that the moment of inertia \( I \) of a uniform cylinder about its perpendicular bisector is minimized, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Moment of Inertia**:
The moment of inertia \( I \) of a cylinder about its perpendicular bisector can be expressed in terms of its mass \( M \), length \( l \), and radius \( R \).

2. **Moment of Inertia Calculation**:
The moment of inertia \( I \) for a cylinder can be derived using the formula:
\[ I = \frac{1}{12} M (3R^2 + l^2) \]
This formula accounts for the distribution of mass around the axis of rotation.

3. **Expressing Mass Density**:
The mass density \( \rho \) of the cylinder can be expressed as:
\[ \rho = \frac{M}{\pi R^2 l} \]
where \( M \) is the mass, \( R \) is the radius, and \( l \) is the length of the cylinder.

4. **Substituting Density into the Moment of Inertia**:
To minimize \( I \), we can express it in terms of \( l \) and \( R \). Substituting for \( M \) using the density gives:
\[ I = \frac{1}{12} \left( \frac{M}{\pi R^2 l} \right) (3R^2 + l^2) \]

5. **Differentiating to Find Minimum**:
To find the minimum moment of inertia, differentiate \( I \) with respect to \( l \) and set the derivative to zero:
\[ \frac{dI}{dl} = 0 \]
This will lead to a relationship between \( l \) and \( R \).

6. **Finding the Ratio**:
After differentiating and simplifying, we can find the optimal ratio \( \frac{l}{R} \). The calculations will yield:
\[ \frac{l}{R} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{3}}{2} \]

### Final Result
Thus, the ratio \( \frac{l}{R} \) such that the moment of inertia is minimized is:
\[ \frac{l}{R} = \frac{\sqrt{3}}{\sqrt{2}} \]