Moment of inertia of a cylinder of mass M, length L and radius R about an axis passing through its centre and perpendicular to the axis of the cylinder is `I = M ((R^2)/4 + (L^2)/12)`. If such a cylinder is to be made for a given mass of a material, the ratio `L//R` for it to have minimum possible I is :

Options

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

Explanation

To find the ratio \( \frac{L}{R} \) for a cylinder to have the minimum possible moment of inertia \( I \), we start with the given formula for the moment of inertia:

\[ I = M \left( \frac{R^2}{4} + \frac{L^2}{12} \right) \]

### Step 1: Express Mass in Terms of Density and Volume

The mass \( M \) of the cylinder can be expressed in terms of its density \( \rho \) and volume \( V \). The volume \( V \) of a cylinder is given by:

\[ V = \pi R^2 L \]

Thus, we can write:

\[ M = \rho V = \rho \pi R^2 L \]

### Step 2: Substitute Mass into the Moment of Inertia Formula

Substituting \( M \) into the moment of inertia formula gives:

\[ I = \rho \pi R^2 L \left( \frac{R^2}{4} + \frac{L^2}{12} \right) \]

### Step 3: Simplify the Moment of Inertia Expression

Expanding this, we have:

\[ I = \rho \pi L \left( \frac{R^4}{4} + \frac{L^2 R^2}{12} \right) \]

### Step 4: Express \( R^2 \) in Terms of \( L \)

From the mass equation, we can express \( R^2 \) in terms of \( L \):

\[ R^2 = \frac{M}{\rho \pi L} \]

### Step 5: Substitute \( R^2 \) Back into the Moment of Inertia Expression

Now substituting \( R^2 \) back into the moment of inertia expression:

\[ I = \rho \pi L \left( \frac{1}{4} \left(\frac{M}{\rho \pi L}\right)^2 + \frac{L^2}{12} \left(\frac{M}{\rho \pi L}\right) \right) \]

### Step 6: Differentiate to Find Minimum

To find the minimum moment of inertia, we differentiate \( I \) with respect to \( L \) and set the derivative to zero:

\[ \frac{dI}{dL} = 0 \]

### Step 7: Solve the Derivative Equation

After differentiating and simplifying, we will find a relationship between \( L \) and \( R \).

### Step 8: Find the Ratio \( \frac{L}{R} \)

After performing the necessary algebra, we will arrive at:

\[ \frac{L^2}{R^2} = \frac{6}{4} \implies \frac{L}{R} = \sqrt{\frac{3}{2}} \]

### Final Answer

Thus, the ratio \( \frac{L}{R} \) for the cylinder to have the minimum possible moment of inertia is:

\[ \frac{L}{R} = \sqrt{\frac{3}{2}} \]

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