Animated Solution for Physics - Laws of Motion: Column II shows five systems in which two objects are labelled as $X$ and $Y$. Also in each case a point $P$ is shown. Column I gives some statements about $X$ and/or $Y$. Match these statements to the appropriate system(s) from Column II

Imagine you are sitting in the JEE Advanced exam hall, and you are presented with a matrix match question that tests not one, but four different chapters of physics simultaneously. This 2009 classic is exactly that kind of beautiful beast. It seamlessly weaves together Newton's Laws, Work and Energy, Fluid Mechanics, and Rotational Dynamics. Let's break it down step by step and uncover the elegant physics hidden within each system.

The Hunt for Zero Net Force

Statement A asks us to identify the systems where the force exerted by object $X$ on object $Y$ has a magnitude of exactly $Mg$. The key to unlocking this is the phrase constant velocity or terminal velocity. According to Newton's First Law, if an object is moving with a constant velocity, the net force acting on it must be absolutely zero.

Let's look at system (p). The block $Y$ is sliding down the incline $X$ with a constant velocity. The Earth is pulling it down with a gravitational force $Mg$. For the net force to be zero, the incline $X$ must exert a combined contact force (Normal force plus Kinetic friction) that is exactly equal and opposite to gravity. Therefore, the force from $X$ is $Mg$ upwards. System (p) is a perfect match!

Similarly, in system (t), the sphere $Y$ is falling through a viscous liquid $X$ with terminal velocity. Again, the net force is zero. The liquid exerts an upward buoyant force and an upward viscous drag. Together, these forces perfectly balance the downward weight $Mg$. Thus, system (t) is also a match.

Why not system (q)? In the magnetic stand, the base $X$ has to support both magnets. Magnet $Z$ repels magnet $Y$ downwards with a force $Mg$ to maintain its own equilibrium. So, the total downward force on $Y$ is $2Mg$, meaning $X$ must exert $2Mg$ upwards.

The Rising Center of Mass

Statement B shifts our focus to gravitational potential energy. We are looking for systems where the potential energy of $X$ is continuously increasing. Remember, potential energy increases when the center of mass of an object moves higher against gravity.

In system (q), the entire setup is inside a lift that is moving upwards. Therefore, the base $X$ is continuously gaining height, and its potential energy is increasing.

The more fascinating cases are systems (s) and (t). Here, a heavy sphere $Y$ is sinking into a liquid $X$. As the sphere moves down, it displaces a volume of liquid equal to its own volume. This displaced liquid is effectively pushed from the bottom to the top surface. Because the liquid's mass is being shifted upwards, the overall center of mass of the liquid $X$ rises! Consequently, the potential energy of the liquid increases.

The Dissipation of Mechanical Energy

Statement C requires us to find where the mechanical energy of the combined system $(X+Y)$ is continuously decreasing. Mechanical energy is lost when non-conservative forces do negative work.

In system (p), the block slides against kinetic friction, dissipating energy as heat. In system (t), the sphere falls through a viscous liquid, and the viscous drag acts as fluid friction, again dissipating energy.

System (r) is an interesting kinematic case. The entire pulley and clamp system is in a lift moving downwards. Both the clamp $X$ and the pulley $Y$ are losing height at a constant rate. Their kinetic energy remains constant, but their potential energy is continuously decreasing. Hence, the total mechanical energy decreases.

A note on system (s): The problem states the liquid is 'non-viscous', which ideally implies no energy dissipation. However, the official JEE answer key includes (s) as a match. This is a known historical ambiguity, likely assuming that in any real macroscopic fluid displacement, some energy is inevitably lost to turbulence, or focusing strictly on the sphere's loss of mechanical energy.

The Zero Torque Alignment

Finally, Statement D asks where the torque of the weight of $Y$ about point $P$ is zero. Torque is defined as the cross product $\vec{\tau }=\vec{r}\times \vec{F}$. For torque to be zero, the line of action of the force must pass directly through the pivot point $P$, making the perpendicular lever arm zero.

If we examine systems (p), (r), (s), and (t), point $P$ is located off to the side, while the weight of $Y$ acts vertically downwards from its center. There is a clear perpendicular distance, so the torque is non-zero.

However, in system (q), point $P$ is located at the very top of the central vertical axis. The weight of the ring magnet $Y$ acts straight down along this exact same axis. The line of action of the weight passes perfectly through point $P$! The lever arm is zero, and therefore, the torque is zero.

By systematically applying fundamental principles—Newton's Laws, center of mass kinematics, energy conservation, and rotational mechanics—we can confidently conquer even the most intimidating matrix match problems.