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#mechanical equilibrium

2 public questions tagged with this topic.

What is the primary condition for a system to be in mechanical equilibrium?

**Isobaric process** constant pressure, work W = P ΔV = P(V₂ - V₁) = n R ΔT, for expansion ΔV positive W positive, for compression negative, heat Q = n C_p ΔT, ΔU = n C_v ΔT. Isothermal process constant temperature ΔU=0, work W = n R T ln(V₂/V₁) = n R T ln(P₁/P₂), Q = W, heat absorbed equals work done. Mechanical equilibrium in thermodynamics requires no net force or pressure difference across the system, ensuring no macroscopic motion (e.g., piston movement). This is distinct from thermal equilibrium (equal temperature). Using first law ΔU = Q - W, W = ∫ P dV, isobaric

Ref: NCERT > Physics Book > Thermodynamics > Isobaric and Isothermal Processes Work Calculation

Which of the following is a requirement for a rigid body to be in mechanical equilibrium?

For mechanical equilibrium, both translational and rotational equilibrium are required: the net force (∑F = 0) and net torque (∑τ = 0) must be zero. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Both net force and net torque must be zero. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.