Physics MCQs for NEET — Practice Questions with Answers

From the NCERT text (Chapter: LAWS_OF_MOTION, after Example 4.6), 'Equation (4.11) implies that $F_{1x} + F_{2x} + F_{3x} = 0$, $F_{1y} + F_{2y} + F_{3y} = 0$, $F_{1z} + F_{2z} + F_{3z} = 0$'. This means the sum of force components along each of the x, y, and z axes must be zero for equilibrium.

The NCERT text (Chapter: LAWS_OF_MOTION, 4.8 EQUILIBRIUM OF A PARTICLE) explicitly states, 'According to the first law, this means that, the particle is either at rest or in uniform motion.'

The NCERT text (Chapter: SYSTEMS_OF_PARTICLES_AND_ROTATIONAL_MOTION, Equilibrium of a Rigid Body section) states, 'Since consideration of rotational motion does not apply to a particle, only the conditions for translational equilibrium (Eq. 6.30 a) apply to a particle.' A particle is considered a point mass, and torque (which causes rotational motion) requires an extended body with a moment arm.

As per the NCERT text (Chapter: LAWS_OF_MOTION, 4.8 EQUILIBRIUM OF A PARTICLE), 'If two forces $F_1$ and $F_2$, act on a particle, equilibrium requires $F_1 = -F_2$ i.e. the two forces on the particle must be equal and opposite.'

The NCERT text (Chapter: SYSTEMS_OF_PARTICLES_AND_ROTATIONAL_MOTION, Equilibrium of a Rigid Body section) states, 'A body may be in partial equilibrium, i.e., it may be in translational equilibrium and not in rotational equilibrium, or it may be in rotational equilibrium and not in translational equilibrium.'

The NCERT text (Chapter: SYSTEMS_OF_PARTICLES_AND_ROTATIONAL_MOTION, POINTS TO PONDER, point 11) clearly states: 'A rigid body is in mechanical equilibrium if (1) it is in translational equilibrium, i.e., the total external force on it is zero : $F_i = \sum F_i = 0$, and (2) it is in rotational equilibrium, i.e. the total external torque on it is zero : $\tau = \sum \tau_i = \sum r_i \times F_i = 0$.'

Based on Fig. 6.20(a) and its description in NCERT (Chapter: SYSTEMS_OF_PARTICLES_AND_ROTATIONAL_MOTION, Equilibrium of a Rigid Body section), when two parallel forces of equal magnitude act in the same direction at the ends of a rod, the net moment (torque) on the rod will be zero (rotational equilibrium) but the net force will not be zero ($\sum F \neq 0$), meaning it will not be in translational equilibrium.

The NCERT text (Chapter: SYSTEMS_OF_PARTICLES_AND_ROTATIONAL_MOTION, Equilibrium of a Rigid Body section) states, 'Since all these forces act on the single particle, they must be concurrent. Equilibrium under concurrent forces was discussed in the earlier chapters.'

The NCERT text (Chapter: SYSTEMS_OF_PARTICLES_AND_ROTATIONAL_MOTION, Equilibrium of a Rigid Body section) clarifies: 'Since consideration of rotational motion does not apply to a particle, only the conditions for translational equilibrium (Eq. 6.30 a) apply to a particle. Thus, for equilibrium of a particle the vector sum of all the forces on it must be zero.' For a rigid body, both zero net force and zero net torque are required (as per 'POINTS TO PONDER' point 11).

Rotational equilibrium means the net torque is zero. Not translational equilibrium means the net force is non-zero. This aligns with the definition of partial equilibrium mentioned in the NCERT text (Chapter: SYSTEMS_OF_PARTICLES_AND_ROTATIONAL_MOTION).