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#conceptual question

9 public questions tagged with this topic.

Atomic Models - Rutherford, Thomson and Bohr

Which of the following statements is incorrect about the hydrogen atom’s energy levels in Bohr’s model?

**Rutherford's nuclear model** atom has small massive positively charged nucleus with electrons orbiting, size ratio atomic to nuclear ~10⁵, nucleus ~10⁻¹⁵ m atom ~10⁻¹⁰ m, most alpha particles with large impact parameter pass undeflected, small fraction >90° scatter from close approach, centripetal force provided by Coulomb attraction k Z e²/r², fails to explain stability because accelerating charge should radiate and collapse. Energy levels are discrete and negative, becoming less negative with increasing n , not positive or continuous as in classical models. Using E_n = -13.6/n² eV, r_n = n

Ref: NCERT > Physics Book > Atoms and Nuclei > Atomic Models - Rutherford, Thomson and Bohr

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Zeroth Law Thermal Equilibrium and Quasi-static

Which of the following statements is correct about internal energy?

**Quasi-static process** infinitely slow, system always near equilibrium, reversible, can be represented as continuous path on P-V diagram, non-quasi-static rapid process non-equilibrium, work W = ∫ P_ext dV, for quasi-static P_ext = P_system, work = ∫ P dV, zeroth law ensures temperature defined throughout quasi-static. Internal energy ( U ) is a state function, dependent on the system’s state (e.g., temperature for an ideal gas), not the path taken. Option A is correct. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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Zeroth Law Thermal Equilibrium and Quasi-static

Which of the following statements is incorrect about work in thermodynamics?

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. Work ( W = P Δ V ) is mechanical energy transfer, not requiring temperature differences (unlike heat). Option B is incorrect. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields It requires a temperature difference

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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Adiabatic Processes and Gamma Determination

Which of the following statements is incorrect regarding the First Law of Thermodynamics?

**Adiabatic process** no heat exchange Q=0, first law ΔU = -W, for ideal gas P V^γ = constant, T V^{γ-1}= constant, P^{1-γ} T^{γ}= constant, γ = C_p/C_v = (f+2)/f, monatomic γ=5/3, diatomic γ=7/5. Work done W = (P₁V₁ - P₂V₂)/(γ-1), temperature changes due to work. The First Law ( Δ Q = Δ U + Δ W ) is a conservation of energy principle, not requiring equilibrium or constant temperature. Option C is incorrect as it imposes an unnecessary condition. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁),

Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination

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Isobaric and Isothermal Processes Work Calculation

Which of the following statements is incorrect about an adiabatic expansion of an ideal gas?

**Isobaric and isothermal** are fundamental thermodynamic processes, isobaric P constant horizontal line on P-V diagram, isothermal hyperbolic P = n R T/V, work equals area under curve, isothermal work larger than adiabatic for same volume change because pressure higher. In adiabatic expansion ( Δ Q = 0 ), the gas does work ( W > 0 ), decreasing internal energy ( Δ U < 0 ) and temperature. Option B is incorrect; temperature decreases, not increases. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ =

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

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Work, Energy and Orbital Mechanics

Why does the gravitational force provide the necessary centripetal force for a circular orbit?

In a circular orbit, the gravitational force (GMEmr2) equals the centripetal force (mv2r), providing the inward force needed to maintain circular motion. As per NCERT, applying relevant law/formula with correct units and sign convention leads to It balances the required inward force. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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Potential Energy of Mass Systems

Why does an object need a minimum speed to escape Earth’s gravity?

Escape speed (ve = 2GMR) is the minimum speed where kinetic energy equals the gravitational potential energy’s magnitude, ensuring total energy is zero, allowing escape to infinity. As per NCERT, applying relevant law/formula with correct units and sign convention leads to To balance kinetic and potential energy. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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