The configuration of a planar four bar mechanism with frictionless joints is shown in the figure below. The length of the links are L₁ = 40 mm, L₂ = 15 mm, L₃ = 35 mm, and L₄ = 30 mm. The suffixes 1, 2, 3 and 4 represents the fixed, input, coupler and output link respectively. If $$T_i$$ and T₀ are the torque to the input and output link, what is the mechanical advantage $$(\frac{T_0}{T_i})$$​ of the mechanism? 

Correct option is A

​$$\begin{aligned}&\text{Using the \textbf{Freudenstein equation} for four-bar mechanisms, the angular velocity ratio is derived from the} \\&\text{\textbf{instantaneous centers of rotation}. For the given configuration:} \\[1em]&\textbf{1. Locate instant centers:} \\&\quad \circ \ I_{12}: \text{ Joint between fixed link (1) and input link (2).} \\&\quad \circ \ I_{14}: \text{ Joint between fixed link (1) and output link (4).} \\&\quad \circ \ I_{24}: \text{ Intersection point of links 2 and 4 (found graphically or analytically).} \\[1em]&\textbf{2. Velocity relationship:} \\&\text{The linear velocity at } I_{24} \text{ is the same for both links 2 and 4:} \\&\omega_i \cdot L_2 \cdot \sin(\theta_2) = \omega_o \cdot L_4 \cdot \sin(\theta_4) \\[0.5em]&\text{Thus:} \\&\frac{\omega_o}{\omega_i} = \frac{L_2 \cdot \sin(\theta_2)}{L_4 \cdot \sin(\theta_4)} \\[1em]&\text{However, without the current angles } (\theta_2 \text{ and } \theta_4), \text{ we use the \textbf{Grashof condition} to infer motion.}\end{aligned}$$

$$\begin{aligned}&\text{For a \textbf{Grashof crank-rocker mechanism} (input link is crank, output link rocks), the MA varies with position.} \\&\text{The \textbf{maximum MA} occurs when the input and output links are \textbf{perpendicular to the coupler}:} \\&MA_{\text{max}} = \frac{L_4}{L_2} \cdot \frac{\sin(\theta_4)}{\sin(\theta_2)} \\[1em]&\text{At the \textbf{toggle positions} (when links 2 and 3 are aligned), MA approaches infinity (theoretical),} \\&\text{but practically, it’s limited by friction and inertia.} \\[1em]&\text{For \textbf{general analysis}, we use the torque ratio directly:} \\&MA = \left| \frac{T_o}{T_i} \right| = \left| \frac{\omega_i}{\omega_o} \right| = \left| \frac{L_4 \cdot \sin(\theta_4)}{L_2 \cdot \sin(\theta_2)} \right| \\[1em]&\text{Without specific angles, we assume an \textbf{average configuration} where } \sin(\theta_2) \approx \sin(\theta_4): \\&MA \approx \frac{L_4}{L_2} = \frac{30}{15} = \boxed{2}\end{aligned}$$​​​