The resultant of two forces acting at an angle of $120^{\circ}$ is $10 \mathrm{~kg}$-wt and is perpendicular to one of the forces. That force is

• A

  $10 \sqrt{3} \mathrm{~kg}$-wt
• B

  $20 \sqrt{3} \mathrm{~kg}$-wt
• C

  $10 \mathrm{~kg}-\mathrm{wt}$
• Correct answer
  D

  $\frac{10}{\sqrt{3}} \mathrm{~kg}-\mathrm{wt}$

Solution

Here, $$ \begin{gathered} \tan 30^{\circ}=\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}}=\frac{x}{10} \\ x=\frac{10}{\sqrt{3}} \mathrm{~kg}-\mathrm{wt} \end{gathered} $$

Hence, the correct answer is option: 'D'.