#equilibrium

Equations of SHM, Phase and Angular Frequency

In SHM, if the displacement is zero, what can be inferred about the acceleration?

**General equation of SHM** x = A sin(ωt + φ) or A cos(ωt + φ) includes amplitude A (m), angular frequency ω = √(k/m) (rad/s) for spring system, and initial phase φ (rad) setting t=0 position. Phase (ωt + φ) determines instantaneous state, phase difference Δφ governs interference of two SHM motions. In SHM, acceleration is given by a = -ω² x . When x = 0 (mean position), a = 0 , as the restoring force vanishes. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It is zero follows,

Ref: NCERT > Physics Book > Oscillations > Equations of SHM, Phase and Angular Frequency

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Energy in SHM - Kinetic, Potential and Total

Which periodic motion lacks a restoring force directed towards a fixed equilibrium point?

**Energy distribution** shows maximum kinetic at equilibrium and maximum potential at extremes, sum constant. This relation enables calculation of amplitude, velocity at any displacement via v = ±√(2(E-U)/m), and understanding of energy storage in oscillating system for NEET problems. The rotation of a carousel is periodic but not oscillatory, as it involves continuous circular motion without a restoring force towards a fixed point, unlike SHM systems. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result The rotation of a carousel follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Energy in SHM - Kinetic, Potential and Total

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Spring-Mass System and Combination of Springs

In SHM, which quantity is maximized when the particle’s displacement is zero?

**Oscillation of mass attached to spring** follows Hooke's law F = -k x, SHM with ω = √(k/m). Effective stiffness for two identical springs in parallel doubles, k_eff = 2k, increasing frequency by √2, while series halves stiffness to k/2, lowering frequency. Energy E = ½ k_eff A². At x = 0 (mean position), velocity is maximum ( vₘₐₓ = ω A ), and thus kinetic energy ( K = (1/2) m v² ) reaches its peak. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result Kinetic energy follows, reflecting SHM

Ref: NCERT > Physics Book > Oscillations > Spring-Mass System and Combination of Springs

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Energy in SHM - Kinetic, Potential and Total

A spring system has \( m = 1 \, \text{kg}, k = 100 \, \text{N/m}, A = 20 \, \text{cm} \). What is the potential energy a

**Conservation of mechanical energy** in undamped SHM implies total energy proportional to amplitude squared A² and spring constant k. E = ½ k A² allows amplitude determination from known E and k via A = √(2E/k), with k = m ω² linking dynamical and energetic descriptions for spring-mass system. Potential energy: U = (1/2) k x² . At x = 0 m , U = 0 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.0 J follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Energy in SHM - Kinetic, Potential and Total

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Specific Heat Capacity and Latent Heat

In a thermodynamic process, what indicates that a variable is not in equilibrium?

**Specific heat capacity** c = Q/(m ΔT) (J/kg·K), molar C = Q/(n ΔT), heat required to raise temperature, Q = m c ΔT, for water c=4186 J/kg·K, latent heat L = Q/m for phase change at constant temperature, fusion L_f and vaporization L_v, Q = m L, e.g., ice melting L_f=3.34×10⁵ J/kg, water vaporization 2.26×10⁶ J/kg. A system not in equilibrium has macroscopic variables (e.g., pressure, temperature) that change with time or vary across the system, such as during rapid expansion or explosive reactions, where uniformity is lost. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

Ref: NCERT > Physics Book > Thermodynamics > Specific Heat Capacity and Latent Heat

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Adiabatic Processes and Gamma Determination

What is the significance of a quasi-static process in thermodynamics?

**Gamma determination** γ = C_p/C_v, C_p - C_v = R, for monatomic f=3 C_v=3/2 R C_p=5/2 R γ=1.67, diatomic f=5 C_v=5/2 R C_p=7/2 R γ=1.4, adiabatic relation P V^γ = const allows γ determination from P-V measurements, slope of log P vs log V = -γ. A quasi-static process is infinitely slow, ensuring the system remains in thermal and mechanical equilibrium with its surroundings at every stage. This allows well-defined state variables (e.g., P , T ) and is an idealized condition for reversible processes. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W

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

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CHEMISTRY

For the reaction 2COâ‚‚(g) 2CO(g) + Oâ‚‚(g), K_c = 0.01 at 1000 K. If initial [COâ‚‚] = 0.5 M, what is [CO] at equilibri

Given: For the reaction 2CO₂(g) 2CO(g) + O₂(g), K_c = 0.01 at 1000 K. If initial [CO₂] = 0.5 M, what is [CO] at equilibrium? These values define the system as per NCERT data. Formula: Let [CO] = 2x, [O₂] = x, [CO₂] = 0.5 - 2x. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: K_c = frac[CO]² [O₂][CO₂]² = (2x)² · x/(0.5 - 2x)² = 0.01 . 4x³/(0.5 - 2x)² = 0.01, solve: x approx 0.027, 2x approx 0.054 M. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Thermodynamics, Equilibrium and Chemical Kinetics, Topic: Energetics, equilibrium constants and reaction rates.

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CHEMISTRY

The K_{sp of BaCrO₄ is 1.2 × 10⁻¹⁰. What is its solubility in 0.01 M Na₂CrO₄ ?

Given: The K_{sp of BaCrO₄ is 1.2 × 10⁻¹⁰. What is its solubility in 0.01 M Na₂CrO₄ ? These values define the system as per NCERT data. Formula: K_{sp = [Ba²+][CrO₄²-], [CrO₄²-] = 0.01, 1.2 × 10⁻¹⁰= S · 0.01. This is standard NCERT relation. Substitution & Calculation: S = 1.2 × 10⁻⁸M. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

The K_{sp of Fe(OH)2 is 4.9 × 10⁻¹⁷. What is its solubility in water?

Given: The K_{sp of Fe(OH)2 is 4.9 × 10⁻¹⁷. What is its solubility in water? These values define the system as per NCERT data. Formula: Fe(OH)2(s) Fe²+ + 2OH-, K_{sp = S · (2S)² = 4S³. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 4S³ = 4.9 × 10⁻¹⁷, S³ = 1.225 × 10⁻¹⁷, S approx 2.3 × 10⁻⁶ M. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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CHEMISTRY

For Ca(OH)2(s) Ca²+ + 2OH-, K_{sp = 5.5 × 10⁻⁶. What is [Ca²+] in 0.01 M NaOH ?

Given: For Ca(OH)2(s) Ca²+ + 2OH-, K_{sp = 5.5 × 10⁻⁶. What is [Ca²+] in 0.01 M NaOH ? These values define the system as per NCERT data. Formula: K_{sp = [Ca²+][OH-]², [OH-] = 0.01 + 2S approx 0.01. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 5.5 × 10⁻⁶= [Ca²+] · (0.01)², [Ca²+] = 5.5 × 10⁻² M. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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CHEMISTRY

For Nâ‚‚(g) + 2Oâ‚‚(g) 2NOâ‚‚(g), increasing temperature for this endothermic reaction will:

Increasing temperature favors the endothermic direction (products). This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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CHEMISTRY

For NH₃(g) + HCl(g) NH₄Cl(s), increasing temperature will:

This is an exothermic reaction (solid formation). Increasing temperature shifts equilibrium to reactants (dissociation). This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Thermodynamics, Equilibrium and Chemical Kinetics, Topic: Energetics, equilibrium constants and reaction rates.

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