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#material science

12 public questions tagged with this topic.

Resistance, Resistivity and Ohm's Law

A wire of length \( 4 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) has a resistance of \

**Resistance** R = ρ l/A, ρ resistivity (Ω·m), l length (m), A area (m²), ρ = m/(n e² τ) from Drude model, τ average collision time. Ohm's law V = I R holds when ρ constant, independent of V. Volume constant stretching l→2l implies A→A/2, so R' = ρ·2l/(A/2)=4R, resistance quadruples when length doubles at constant volume. Resistance: R = (rho l/A) . Rearrange: rho = (R A/l) . Substitute: rho = (8 × 2 × 10⁻⁶/4) = 4 × 10⁻⁶ Ω m . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V =

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

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Resistance, Resistivity and Ohm's Law

A conductor has a resistivity of \( 9 \times 10^{-8} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\te

**Resistance** R = ρ l/A, ρ resistivity (Ω·m), l length (m), A area (m²), ρ = m/(n e² τ) from Drude model, τ average collision time. Ohm's law V = I R holds when ρ constant, independent of V. Volume constant stretching l→2l implies A→A/2, so R' = ρ·2l/(A/2)=4R, resistance quadruples when length doubles at constant volume. Use: rho_t = rho₀ [1 + α (T - T₀)] . Substitute: rho_t = 9 × 10⁻⁸ [1 + 4 × 10⁻³ (60 - 20)] . Calculate: rho_t = 9 × 10⁻⁸ [1 + 0.16] = 9 × 10⁻⁸ × 1.16 = 1.044 × 10⁻⁷ Ω m .

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

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Earth's Magnetism and Magnetic Declination

A material’s susceptibility becomes negative and large when:

**Elements of Earth's field** include declination D, inclination I, horizontal component B_H, total field B = √(B_H² + B_V²). B_H provides compass direction, declination varies with location, important for navigation, inclination 0° at magnetic equator, 90° at poles. A superconductor has a susceptibility of chi = -1 (large and negative) when it transitions to the superconducting state below its critical temperature, expelling all magnetic fields via the Meissner effect, a unique property among materials. Substituting values gives It becomes superconducting, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Earth's Magnetism and Magnetic Declination

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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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Isobaric and Isothermal Processes Work Calculation

Why does the specific heat capacity of a solid generally agree with 3R at ordinary temperatures?

**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. The law of equipartition predicts that each atom in a solid has 3 degrees of freedom (vibrational), contributing (3)/(2) k_B T kinetic and potential energy per atom. For a mole, U = 3 R T , so C = (Δ U)/(Δ T) =

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

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PHYSICS

An aluminium block of dimensions 0.3 m × 0.2 m × 0.05 m is subjected to a shearing force of 1.5 × 10⁴N . If the shear mo

Given: An aluminium block of dimensions 0.3 m × 0.2 m × 0.05 m is subjected to a shearing force of 1.5 × 10⁴N . If the shear modulus of aluminium is 2.5 × 10¹⁰N/m², what is the shear strain? Formula: Shear modulus: G = fracShear stressShear strain. Substitution & Calculation: Shear stress: Shear stress = F/A, A = 0.3 × 0.2 = 0.06 m² . Shear stress: 1.5 × 10⁴/0.06 = 2.5 × 10⁵N/m² . Shear strain: Shear strain = fracShear stressG = frac2.5 × 10⁵².5 × 10¹⁰= 1 × 10⁻⁵. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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PHYSICS

A brass block of dimensions 0.5 m × 0.2 m × 0.05 m is subjected to a shearing force of 2 × 10⁴ N . If the shear mod

Given: A brass block of dimensions 0.5 m × 0.2 m × 0.05 m is subjected to a shearing force of 2 × 10⁴ N . If the shear modulus of brass is 3.6 × 10¹⁰ N/m², what is the shear strain? These values define the system as per NCERT data. Formula: Shear modulus: G = fracShear stressShear strain. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Shear stress: Shear stress = F/A, A = 0.5 × 0.2 = 0.1 m² . Shear stress: 2 × 10⁴/0.1 = 2 × 10⁵ N/m² . Shear strain: Shear strain = fracShear stressG = frac2 × 10⁵³.6 × 10¹⁰ approx 5.56 × 10⁻⁶. 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 Physics Textbook for Class XI and XII, Chapter: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

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PHYSICS

A material’s weak repulsion from a magnetic field is due to:

Diamagnetic materials exhibit weak repulsion from a magnetic field because an external field induces small currents in their atoms that generate an opposing magnetic moment, per Lenz’s law, resulting in a slight reduction of the field inside the material.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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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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Shear Modulus and Rigidity

In designing bridges, why is it critical to understand the elastic properties of materials?

Understanding elastic properties ensures that materials can withstand various loads (traffic, wind, weight) without undergoing permanent deformation, maintaining structural integrity. As per NCERT, applying relevant law/formula with correct units and sign convention leads to To ensure structural integrity. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

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

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Bulk Modulus and Compressibility

What is the key difference between elastic and plastic deformation in a material?

Elastic deformation is reversible, meaning the material returns to its original shape after the load is removed, while plastic deformation results in permanent changes. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Elastic deformation is reversible, plastic is not. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

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

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Young's Modulus and Elasticity

What property of a material primarily determines its resistance to uniform compression?

The bulk modulus determines a material’s resistance to uniform compression by measuring how much it resists volume change under pressure applied in all directions. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Bulk modulus. 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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