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(B) Let $T_Q$ and $T_S$ be the tension forces in rods $PQ$ and $RS$ respectively.From the force equilibrium of the rigid rod: $T_Q + T_S = F$.Taking t

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$\frac{9FL}{8AY}$

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(B) Let $T_Q$ and $T_S$ be the tension forces in rods $PQ$ and $RS$ respectively.From the force equilibrium of the rigid rod: $T_Q + T_S = F$.Taking torque about point $S$: $T_Q 	imes (2L) = F 	imes (4L) \implies T_Q = 2F$.Substituting $T_Q$ in the force equation: $2F + T_S = F \implies T_S = -F$ (The negative sign indicates that rod $RS$ is under compression).The deflection of rod $PQ$ is $\delta_Q = \frac{T_Q L}{AY} = \frac{(2F)L}{AY} = \frac{2FL}{AY}$ (upwards).The deflection of rod $RS$ is $\delta_S = \frac{|T_S| (3L/2)}{(2A)(2Y)} = \frac{F(3L/2)}{4AY} = \frac{3FL}{8AY}$ (downwards).Since the rod is rigid,the deflections $\delta_Q$ and $\delta_S$ follow a linear relationship based on the geometry of the rod. Let the rod rotate about a point at distance $x$ from $Q$. Then $\frac{\delta_Q}{x} = \frac{\delta_S}{2L-x}$.However,the question asks for the deflection of end $S$ directly,which is the compression of rod $RS$,calculated as $\delta_S = \frac{3FL}{8AY}$.

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