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. `(1)/(sqrt(2)`
B. `(1)/sqrt(12)`
C. `(1)/sqrt(3)`
D. `(1)/sqrt (6)`

Explanation

To solve the problem step by step, we need to analyze the situation of a uniform stick rotating about a horizontal axis. We will derive the expression for the time period of small oscillations and find the condition under which this time period is minimized.

### Step 1: Understand the System
We have a uniform stick of length \( l \) that is mounted to rotate about a horizontal axis. The axis is located at a distance \( d \) from the center of mass of the stick.

### Step 2: Moment of Inertia Calculation
The moment of inertia \( I \) of a uniform stick about an axis through its end is given by:
\[ I = \frac{1}{3} m l^2 \]
However, since the axis is at a distance \( d \) from the center of mass, we need to use the parallel axis theorem:
\[ I = I_{cm} + m d^2 = \frac{1}{12} m l^2 + m d^2 \]
where \( I_{cm} = \frac{1}{12} m l^2 \) is the moment of inertia about the center of mass.

### Step 3: Time Period of Oscillation
The time period \( T \) for small oscillations is given by:
\[ T = 2\pi \sqrt{\frac{I}{m g d}} \]
Substituting the expression for \( I \):
\[ T = 2\pi \sqrt{\frac{\frac{1}{12} m l^2 + m d^2}{m g d}} = 2\pi \sqrt{\frac{\frac{1}{12} l^2 + d^2}{g d}} \]

### Step 4: Minimize the Time Period
To find the minimum value of the time period \( T \), we need to differentiate the expression under the square root with respect to \( d \) and set the derivative to zero:
\[ \frac{d}{dd}\left(\frac{\frac{1}{12} l^2 + d^2}{g d}\right) = 0 \]

### Step 5: Differentiate and Solve
Using the quotient rule:
\[ \frac{d}{dd}\left(\frac{A + d^2}{d}\right) = \frac{(2d)(d) - (A + d^2)(1)}{d^2} \]
Setting the numerator to zero:
\[ 2d^2 - (A + d^2) = 0 \implies d^2 = A \implies d = \sqrt{\frac{l^2}{12}} \text{ (where A = } \frac{1}{12} l^2\text{)} \]

### Step 6: Find \( \frac{d}{l} \)
Now, we can express \( \frac{d}{l} \):
\[ \frac{d}{l} = \frac{1}{\sqrt{12}} = \frac{1}{2\sqrt{3}} \]

### Conclusion
Thus, the minimum value of \( \frac{d}{l} \) occurs when:
\[ \frac{d}{l} = \frac{1}{\sqrt{12}} \]