A uniform stick of length `l` is mounted so as to rotate about a horizontal axis perpendicular to the stick and at a distance `d` from the centre of mass. The time period of small oscillation has a minimum value when `d//l` is

Options

A. `(a)(1)/(sqrt(2)`
B. `(b)(1)/sqrt(12)`
C. `(c)(1)/sqrt(3)`
D. `(d)(1)/sqrt (6)`

Explanation

To solve the problem of finding the ratio \( \frac{d}{l} \) for the minimum time period of small oscillations of a uniform stick, we will follow these steps:

### Step 1: Understand the Problem
We have a uniform stick of length \( l \) that is rotating about a horizontal axis perpendicular to the stick, at a distance \( d \) from its center of mass. We need to find the ratio \( \frac{d}{l} \) when the time period of small oscillations is minimized.

### Step 2: Write the Formula for Time Period
The time period \( T \) of a physical pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{I}{mgd}} \]
where:
- \( I \) is the moment of inertia about the suspension point,
- \( m \) is the mass of the stick,
- \( g \) is the acceleration due to gravity,
- \( d \) is the distance from the suspension point to the center of mass.

### Step 3: Calculate the Moment of Inertia
The moment of inertia \( I \) of a uniform stick about its center of mass is:
\[ I_{cm} = \frac{ml^2}{12} \]
When the stick is rotated about a point \( d \) distance away from the center of mass, we use the parallel axis theorem:
\[ I = I_{cm} + md^2 = \frac{ml^2}{12} + md^2 \]

### Step 4: Substitute into the Time Period Formula
Substituting \( I \) into the time period formula, we get:
\[ T = 2\pi \sqrt{\frac{\frac{ml^2}{12} + md^2}{mgd}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{l^2}{12g d} + \frac{d}{g}} \]

### Step 5: Simplify the Expression
We can factor out \( d \) from the square root:
\[ T = 2\pi \sqrt{\frac{1}{g} \left( \frac{l^2}{12d} + 1 \right)} \]

### Step 6: Find the Minimum Time Period
To find the minimum time period, we need to minimize the expression inside the square root:
\[ \frac{l^2}{12d} + 1 \]
To do this, we can differentiate with respect to \( d \) and set the derivative to zero:
\[ \frac{d}{dd}\left(\frac{l^2}{12d} + 1\right) = -\frac{l^2}{12d^2} = 0 \]
This gives us:
\[ l^2 = 12d \]
Thus, we can express \( d \) in terms of \( l \):
\[ d = \frac{l^2}{12} \]

### Step 7: Calculate the Ratio \( \frac{d}{l} \)
Now we can find the ratio:
\[ \frac{d}{l} = \frac{\frac{l^2}{12}}{l} = \frac{l}{12} \]

### Step 8: Final Result
To express this in terms of \( l \):
\[ \frac{d}{l} = \frac{1}{12} \]

### Conclusion
The minimum value of \( \frac{d}{l} \) when the time period of small oscillation is minimized is:
\[ \frac{d}{l} = \frac{1}{\sqrt{12}} \]