IIT JEE Advanced 2009 Paper 1

Case (P): Since v = Constant, we have Mgsinθ = μMgcosθ = f ⇒ μ = tan θ Now, N = Mgcosθ. Therefore, R = $\sqrt{N^2+f^2}$ = g $\sqrt{\cos^2\theta+\sin^2\theta}$ Case (T): We have Mg = $F_v+F_b$, where $F_v$ is the viscous force and $F_b$ is the buoyant force. Case (S): We have Mg – $F_b$ = Ma. Case (R): We have Mg = T. Therefore, T + $M_0 g$ = $F_y$ (by the clamp) T = $F_X$ (by the clamp) Therefore, the force exerted by the clamp X on pulley Y is F = $\sqrt{F_X^2+F_Y^2}$ = $g\sqrt{M^2+(M+m_0)^2}$ Case (Q): We have Mg = $F_m$ (for Z) where $F_m$ is the magnetic repulsion. Also Mg + $F_m$ (for Y) Therefore, N = 2Mg. Note: Option (A): For option (T), if Fb is ignored, then Mg = $F_v$; otherwise, no case matches for option (A). Hence, (A)→(T), (P). Option (B): In option (Q), it is mentioned that the lift is moving up continuously; therefore, the gravitational potential energy of X goes on increasing. In option (B), as Y comes down, X goes up (displaced). The same is applicable for option (T). Hence, (B)→(Q), (S), (T). Option (C): For option (P), since Y moves down with a constant v, the gravitational potential energy of the system X + Y goes on decreasing; similar is the case in Options (R) and (T). Hence, (C)→(P), (R), (T). Option (D): For option (S), the mass moves down with acceleration. Therefore, the kinetic energy goes on increasing. Since the line of action of Mg of Y phases through point P, as mentioned in option (Q), its torque about P is zero. Hence, (D)→(Q).