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)         [CET 1998]

A)                 0.9659

B)                 2

C)                 1

D)                 \[\frac{1}{2}\]

Correct Answer: C

Solution :

                    \[\frac{P}{\sin {{\theta }_{1}}}=\frac{Q}{\sin {{\theta }_{2}}}=\frac{R}{\sin \,\,150{}^\circ }\]                 \[\Rightarrow \,\frac{1.93}{\sin {{\theta }_{1}}}=\frac{R}{\sin \,150{}^\circ }\]                 \[\Rightarrow \,R=\frac{1.93\times \sin \,150{}^\circ }{\sin {{\theta }_{1}}}=\frac{1.93\times 0.5}{0.9659}=1\]