The hour hand of a clock is `6 cm` long. The magnitude of the displacement of the tip of hour hand between `1:00` pm to `9:00` pm is `:`

Options

A. `6cm`
B. `6sqrt(3)cm`
C. `4picm`
D. `3sqrt(3)cm`

Explanation

To find the magnitude of the displacement of the tip of the hour hand of a clock between 1:00 PM and 9:00 PM, we can follow these steps:

### Step 1: Understand the positions of the hour hand
At 1:00 PM, the hour hand is pointing at the 1 on the clock, and at 9:00 PM, it is pointing at the 9. The clock is circular, and the hour hand moves in a circular path.

### Step 2: Determine the angle moved by the hour hand
The clock is divided into 12 hours, and each hour represents an angle of \(30^\circ\) (since \(360^\circ / 12 = 30^\circ\)). From 1 to 9, the hour hand moves through 8 hours:
\[ \text{Angle} = 8 \times 30^\circ = 240^\circ \]

### Step 3: Represent the movement as vectors
The displacement can be represented as a vector from the position at 1:00 PM to the position at 9:00 PM. The lengths of these vectors are equal to the length of the hour hand, which is \(6 \, \text{cm}\).

### Step 4: Use the cosine rule to find the displacement
The displacement \(d\) can be calculated using the cosine rule for the triangle formed by the two positions of the hour hand:
\[ d^2 = R_1^2 + R_2^2 - 2R_1R_2 \cos(\theta) \]
where \(R_1 = R_2 = 6 \, \text{cm}\) and \(\theta = 240^\circ\).

Substituting the values:
\[ d^2 = 6^2 + 6^2 - 2 \cdot 6 \cdot 6 \cdot \cos(240^\circ) \]
Since \(\cos(240^\circ) = -\frac{1}{2}\):
\[ d^2 = 36 + 36 + 2 \cdot 36 \cdot \frac{1}{2} \]
\[ d^2 = 36 + 36 + 36 = 108 \]
\[ d = \sqrt{108} = 6\sqrt{3} \, \text{cm} \]

### Step 5: Conclusion
The magnitude of the displacement of the tip of the hour hand between 1:00 PM and 9:00 PM is \(6\sqrt{3} \, \text{cm}\).

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