The resultant of two forces acting an anlge of `120^(@)` is 10 kg wt and is perpendicular to one of the forces. That force is

Options

A. `10sqrt(3) kg wt`
B. `20sqrt(3) kg wt`
C. `10 kg wt`
D. `(10)/(sqrt(3)) kg wt`

Answer

To solve the problem step by step, we will analyze the forces acting at an angle and use trigonometric relationships to find the required force.

### Step 1: Understand the Problem
We have two forces, \( F_1 \) and \( F_2 \), acting at an angle of \( 120^\circ \). The resultant force \( F_R \) is given as \( 10 \, \text{kg wt} \) and is perpendicular to one of the forces (let's assume it is \( F_1 \)).

### Step 2: Draw a Diagram
Draw a diagram with \( F_1 \) and \( F_2 \) forming an angle of \( 120^\circ \). The resultant \( F_R \) is perpendicular to \( F_1 \). This means that the angle between \( F_R \) and \( F_2 \) is \( 30^\circ \) (since \( 120^\circ - 90^\circ = 30^\circ \)).

### Step 3: Resolve \( F_2 \) into Components
The horizontal component of \( F_2 \) can be expressed as:
\[ F_{2x} = F_2 \cos(30^\circ) \]
And since \( F_R \) is equal to this horizontal component:
\[ F_R = F_2 \cos(30^\circ) \]
Given \( F_R = 10 \, \text{kg wt} \):
\[ 10 = F_2 \cos(30^\circ) \]

### Step 4: Calculate \( F_2 \)
We know that \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ 10 = F_2 \cdot \frac{\sqrt{3}}{2} \]
Rearranging gives:
\[ F_2 = \frac{10 \cdot 2}{\sqrt{3}} = \frac{20}{\sqrt{3}} \, \text{kg wt} \]

### Step 5: Resolve the Vertical Component of \( F_2 \)
Now, we need to find the vertical component of \( F_2 \):
\[ F_{2y} = F_2 \sin(30^\circ) \]
We know that \( \sin(30^\circ) = \frac{1}{2} \):
\[ F_{2y} = F_2 \cdot \frac{1}{2} = \frac{20}{\sqrt{3}} \cdot \frac{1}{2} = \frac{10}{\sqrt{3}} \, \text{kg wt} \]

### Step 6: Relate \( F_{2y} \) to \( F_1 \)
Since \( F_{2y} \) is equal to \( F_1 \):
\[ F_1 = F_{2y} = \frac{10}{\sqrt{3}} \, \text{kg wt} \]

### Conclusion
Thus, the force \( F_1 \) is:
\[ F_1 = \frac{10}{\sqrt{3}} \, \text{kg wt} \]