Matrix Match of Forces Energy and Torque in Mechanical Systems

To match each statement in Column I to the appropriate system(s) in Column II, we analyze each physical scenario in detail:

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Analysis of Statement (A): The force exerted by $\mathrm{X}$ on $\mathrm{Y}$ has a magnitude $Mg$.

1. System (p):

   • Block $\mathrm{Y}$ of mass $M$ slides down the fixed inclined plane $\mathrm{X}$ at a constant velocity, meaning its acceleration is zero ($\vec{a}_{\mathrm{Y}} = \vec{0}$).
   • From Newton's second law:
     $\sum \vec{F} = \vec{F}_{\mathrm{X}\text{ on }\mathrm{Y}} + M\vec{g} = \vec{0} \implies \vec{F}_{\mathrm{X}\text{ on }\mathrm{Y}} = -M\vec{g}$
   • The magnitude of the contact force exerted by $\mathrm{X}$ on $\mathrm{Y}$ (the resultant of normal reaction and friction) is:
     $|\vec{F}_{\mathrm{X}\text{ on }\mathrm{Y}}| = Mg$
   • Hence, (p) matches (A).

2. System (q):

   • Magnet $\mathrm{Z}$ of mass $M$ hangs in equilibrium above ring magnet $\mathrm{Y}$, so the magnetic repulsion between them is $F_{\text{mag}} = Mg$.
   • Ring magnet $\mathrm{Y}$ (mass $M$) rests on the base $\mathrm{X}$ and moves with constant velocity. The forces acting on $\mathrm{Y}$ in the vertical direction are its weight $Mg$ (downward), magnetic repulsion from $\mathrm{Z}$ of magnitude $Mg$ (downward), and the normal force $N$ from $\mathrm{X}$ (upward):
     $N - Mg - F_{\text{mag}} = 0 \implies N = 2Mg$
   • Thus, the force exerted by $\mathrm{X}$ on $\mathrm{Y}$ is $2Mg \neq Mg$.

3. System (r):

   • The pulley $\mathrm{Y}$ of mass $m_0$ experiences the downward tension $T = Mg$ from the hanging mass, a horizontal tension $T = Mg$ from the anchored string, and its own weight $m_0 g$. The net force exerted by clamp $\mathrm{X}$ on $\mathrm{Y}$ balances these forces and does not have a magnitude of $Mg$.

4. System (s):

   • The sphere $\mathrm{Y}$ accelerates downwards in the non-viscous liquid $\mathrm{X}$, meaning the buoyant force (the force exerted by the liquid $\mathrm{X}$ on $\mathrm{Y}$) is strictly less than its weight: $F_B < Mg$.

5. System (t):

   • The sphere $\mathrm{Y}$ falls with terminal velocity (constant velocity, $\vec{a} = \vec{0}$) in the viscous liquid $\mathrm{X}$.
   • By equilibrium of the sphere:
     $\vec{F}_{\text{net}} = \vec{F}_{\mathrm{X}\text{ on }\mathrm{Y}} + M\vec{g} = \vec{0} \implies |\vec{F}_{\mathrm{X}\text{ on }\mathrm{Y}}| = Mg$
   • Hence, (t) matches (A).

Conclusion for (A): $\mathbf{(A) \to (p), (t)}$

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Analysis of Statement (B): The gravitational potential energy of $\mathrm{X}$ is continuously increasing.

1. System (p):

   • Inclined plane $\mathrm{X}$ is fixed; its position does not change, so its potential energy is constant.

2. System (q):

   • The base $\mathrm{X}$ is inside a lift moving upwards with constant velocity. As its vertical coordinate $z_{\mathrm{X}}$ increases with time, its gravitational potential energy $U_{\mathrm{X}} = m_{\mathrm{X}} g z_{\mathrm{X}}$ continuously increases.
   • Hence, (q) matches (B).

3. System (r):

   • The lift is moving downwards, so the height and gravitational potential energy of clamp $\mathrm{X}$ are continuously decreasing.

4. System (s) and System (t):

   • In both cases, as the sphere $\mathrm{Y}$ moves downwards through the liquid $\mathrm{X}$, it displaces an equal volume of liquid upwards to occupy the space vacated above it.
   • Because liquid is transferred from lower levels to higher levels, the center of mass of the liquid $\mathrm{X}$ continuously shifts upwards.
   • Consequently, the gravitational potential energy of the liquid $\mathrm{X}$ is continuously increasing.
   • Hence, both (s) and (t) match (B).

Conclusion for (B): $\mathbf{(B) \to (q), (s), (t)}$

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Analysis of Statement (C): Mechanical energy of the system $\mathrm{X} + \mathrm{Y}$ is continuously decreasing.

1. System (p):

   • Block $\mathrm{Y}$ moves down at constant velocity (constant kinetic energy), so its gravitational potential energy decreases. The inclined plane $\mathrm{X}$ is stationary. Mechanical energy is dissipated into thermal energy via friction between $\mathrm{X}$ and $\mathrm{Y}$.
   • Hence, the mechanical energy of $\mathrm{X} + \mathrm{Y}$ is continuously decreasing.
   • Thus, (p) matches (C).

2. System (q):

   • The lift moves upwards, increasing the potential energy of both $\mathrm{X}$ and $\mathrm{Y}$ while keeping their kinetic energy constant. Thus, the mechanical energy is increasing.

3. System (r):

   • The lift moves downwards at a constant velocity. The kinetic energies of both $\mathrm{X}$ and $\mathrm{Y}$ remain constant, while their gravitational potential energies continuously decrease. Therefore, the total mechanical energy of $\mathrm{X} + \mathrm{Y}$ is continuously decreasing.
   • Thus, (r) matches (C).

4. System (s):

   • The liquid is non-viscous and no non-conservative forces act on the system. By conservation of mechanical energy, the mechanical energy of the system remains constant.

5. System (t):

   • Due to the viscous drag (a non-conservative dissipative force), mechanical energy is continuously dissipated as heat as the sphere falls at terminal velocity.
   • Thus, (t) matches (C).

Conclusion for (C): $\mathbf{(C) \to (p), (r), (t)}$

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Analysis of Statement (D): The torque of the weight of $\mathrm{Y}$ about point $\mathrm{P}$ is zero.

• The torque of the weight about point $\mathrm{P}$ is given by:
  $\vec{\tau}_{\mathrm{P}} = \vec{r}_{\mathrm{P} \to \mathrm{cm}} \times (M\vec{g})$
• This torque is zero if and only if the position vector of the center of mass of $\mathrm{Y}$ relative to $\mathrm{P}$ is parallel to $\vec{g}$, meaning that point $\mathrm{P}$ lies on the vertical line passing through the center of mass of $\mathrm{Y}$.

1. System (p): Point $\mathrm{P}$ is at the bottom corner of the incline and does not lie on the vertical line passing through the center of mass of block $\mathrm{Y}$. Thus, $\vec{\tau}_{\mathrm{P}} \neq \vec{0}$.
2. System (q): $\mathrm{P}$ lies on the central vertical axis of the stand, which is also the axis of symmetry of the annular ring magnet $\mathrm{Y}$. The center of mass of $\mathrm{Y}$ lies on this vertical axis directly below $\mathrm{P}$. Hence, the line of action of the weight of $\mathrm{Y}$ passes through $\mathrm{P}$, yielding $\vec{\tau}_{\mathrm{P}} = \vec{0}$.
   • Thus, (q) matches (D).
3. Systems (r), (s), (t): In all these systems, point $\mathrm{P}$ is located away from the vertical line passing through the center of mass of $\mathrm{Y}$. Hence, the torque is non-zero.

Conclusion for (D): $\mathbf{(D) \to (q)}$

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Final Matching:

• (A) $\to$ (p), (t)
• (B) $\to$ (q), (s), (t)
• (C) $\to$ (p), (r), (t)
• (D) $\to$ (q)