Case (P): Since v = Constant, we have
Mgsinθ = μMgcosθ = f ⇒ μ = tan θ
Now, N = Mgcosθ. Therefore,
R =
√N2+f2 = g
√cos2θ+sin2θ Case (T): We have Mg =
Fv+Fb, where
Fv is the viscous force and
Fb is the buoyant force.
Case (S): We have Mg –
Fb = Ma.
Case (R): We have Mg = T. Therefore,
T +
M0g =
Fy (by the clamp)
T =
FX (by the clamp)
Therefore, the force exerted by the clamp X on pulley Y is
F =
√FX2+FY2 =
g√M2+(M+m0)2 Case (Q): We have
Mg =
Fm (for Z)
where
Fm is the magnetic repulsion. Also
Mg +
Fm (for Y)
Therefore, N = 2Mg.
Note: Option (A): For option (T), if Fb is ignored, then Mg =
Fv;
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).