question_answer

P, Q and R are three coplanar forces acting at a point and are in equilibrium. Given \[P=1.9318\,kg\,wt,\] \[sin{{\theta }_{1}}=0.9659,\] the value of R is (in kg wt) 

A) 0.9659

B) 2

C) 1

D) \[\frac{1}{2}\]

Correct Answer: C

Solution :

[c] \[\frac{P}{\sin {{\theta }_{1}}}=\frac{Q}{\sin {{\theta }_{2}}}=\frac{R}{\sin {{150}^{o}}}\]

\[\Rightarrow \]   \[\frac{1.93}{\sin {{\theta }_{1}}}=\frac{R}{\sin {{150}^{o}}}\]

\[\Rightarrow \]   \[R=\frac{1.93\times \sin {{150}^{o}}}{\sin {{\theta }_{1}}}=\frac{1.93\times 0.5}{0.9659}=1\]