A displacement vector, at an angle of `30^(@)` with y-axis has an x-component of `10` units. Then the magnitude of the vector is-

Options

A. 5
B. 10
C. 11.5
D. 20

Explanation

To find the magnitude of the displacement vector given its x-component and the angle it makes with the y-axis, we can follow these steps:

### Step 1: Understand the components of the vector
The displacement vector makes an angle of \(30^\circ\) with the y-axis. The x-component of the vector is given as \(10\) units.

### Step 2: Determine the angle with the x-axis
Since the vector makes an angle of \(30^\circ\) with the y-axis, the angle it makes with the x-axis can be calculated as:
\[ \text{Angle with x-axis} = 90^\circ - 30^\circ = 60^\circ \]

### Step 3: Use the cosine function to relate the components
The x-component of the vector can be expressed in terms of the magnitude of the vector \(r\) and the angle with the x-axis:
\[ \cos(60^\circ) = \frac{\text{x-component}}{\text{magnitude of the vector}} \]
Substituting the known values:
\[ \cos(60^\circ) = \frac{10}{r} \]

### Step 4: Solve for the magnitude of the vector
We know that \(\cos(60^\circ) = \frac{1}{2}\). Therefore, we can rewrite the equation:
\[ \frac{1}{2} = \frac{10}{r} \]
Cross-multiplying gives:
\[ r = 10 \times 2 = 20 \]

### Step 5: Conclusion
The magnitude of the displacement vector is \(20\) units.

### Final Answer
The magnitude of the vector is \(20\) units.

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