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#copper

14 public questions tagged with this topic.

Wave Equation and Displacement Relation

A copper rod has a density of 8900 kg/m³ and a longitudinal wave speed of 3560 m/s. What is its Young’s modulus?

**Wave equation** y(x,t) = A sin(kx - ωt + φ) describes displacement of progressive harmonic wave, where k = 2π/λ wave number (rad/m), ω = 2πf angular frequency (rad/s), v = ω/k wave speed (m/s). Sign of ωt indicates direction, amplitude A is maximum displacement. v = √((Y/rho)) . 3560 = √((Y/8900)) ⇒ 3560² = (Y/8900) . Y = 3560² × 8900 ≈ 1.13 × 10¹¹ Pa . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 1.13 × 10¹¹ Pa, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Wave Equation and Displacement Relation

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Wave Equation and Displacement Relation

A copper rod of density 8960 kg/m³ has a speed of sound of 3600 m/s. What is its Young’s modulus?

**Sinusoidal wave form** represents harmonic wave where each particle executes SHM. Coefficients of x and t give spatial and temporal periodicities, allowing wavelength and period extraction, basis for wave analysis in NCERT. Speed: v = √((Y/rho)) . 3600 = √((Y/8960)) ⇒ 3600² = (Y/8960) . Y = 3600² × 8960 = 1.16 × 10¹¹ Pa . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 1.16 × 10¹¹ Pa, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Wave Equation and Displacement Relation

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Isochoric Processes and Pressure-Temperature

How much heat is required to raise the temperature of 1 kg of copper from 20^circ C to 50^circ C ? (Specific heat of cop

**Pressure-temperature relation** at constant volume Gay-Lussac law P ∝ T, for V constant, P₁/T₁ = P₂/T₂, if T doubles from 300 K to 600 K P doubles, e.g., P₁=1 atm at 300 K P₂=2 atm at 600 K, no work done, ΔU = n C_v ΔT = Q. Heat capacity: Δ Q = m s Δ T . m = 1 kg , s = 386.4 J kg⁻¹ K⁻¹ , Δ T = 50 - 20 = 30 K . Δ Q = 1 × 386.4 × 30 = 11592 J . Using first law ΔU = Q - W, W = ∫ P dV,

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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Second Law Heat Engines and Kelvin-Planck

How much heat is required to raise the temperature of 0.6 kg of copper from 20^circ C to 50^circ C ? (Specific heat of c

**Second law Kelvin-Planck statement** no process possible whose sole result is absorption of heat from reservoir and complete conversion to work, heat engine must have at least two reservoirs hot and cold, efficiency η = W/Q_h =1 - Q_c/Q_h

Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck

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CHEMISTRY

How many electrons are involved in the reduction of 1 mol of Cu²⁺ to Cu(s)?

Given: How many electrons are involved in the reduction of 1 mol of Cu²⁺ to Cu(s)? Formula: Reaction: Cu²+ + 2e- -> Cu(s). Substitution & Calculation: 2 electrons are required per mole of Cu²⁺. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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Latent Heat and Phase Transitions

A copper block of mass 0.5kg at 100∘C is dropped into 1kg of water at 20∘C. Find the final temperature. (Specific heat o

Heat lost by copper = Heat gained by water. mcsc(100−T) = mwsw(T−20). 0.5×386×(100−T) = 1×4186×(T−20). 19300−193T = 4186T−83720. 19300+83720 = 4186T+193T. 103020 = 4379T⇒T≈23.53∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 23.5°C. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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Elastic Properties and Applications

A copper wire of length 2.0m and cross-sectional area 2×10−6m2 is stretched by a force producing a strain of 1×10−4. If

Young's modulus: Y = StressStrain. Stress: Stress = Y×Strain = 1.1×1011×1×10−4 = 1.1×107N/m2. Force: F = Stress×A = 1.1×107×2×10−6 = 22N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 22N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

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

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Elastic Properties and Applications

A copper wire of length 1.2 m and cross-sectional area 1.5 × 10-6 m2 is stretched by a force producing a stress of 2 × 1

Young's modulus: Y = Stress / Strain. Strain: Strain = Stress / Y = (2 × 107) / (1.1 × 1011) ≈ 1.82 × 10-4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.82 × 10-4. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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