1) A rod l m long is acted upon by couples as shown in the figure. The moment of the couple is $\tau$ Nm. If the force at each end of the rod, then the magnitude of each force is $(\sin 30^{0}=\cos 60^{0}=0.5)$ A) $\frac{\tau}{l}$ B) $\frac{l}{2 \tau}$ C) $\frac{2 \tau}{l}$ D) $\frac{2 l}{ \tau}$ Answer: Option CExplanation:Length of rod =l Moment of couple, $\tau=Fl\sin \theta$ $\therefore$ $F= \frac{\tau}{l\sin\theta}=\frac{\tau}{l\sin 30^{0}}=\frac{\tau}{l\times\frac{1}{2}}$ Force, F= $\frac{2 \tau}{l}$
