#materials

Doping, Charge Carriers and Conductivity

Which of the following is a trivalent dopant used in p-type semiconductors?

**Number of carriers** from doping: Ge crystal 4×10²⁸ atoms/m³ doped 2 ppm trivalent gives acceptor atoms 8×10²² m⁻³, holes ≈ that, for 1.5 ppm 6×10²² m⁻³, for 0.5 ppm pentavalent Si 5×10²⁸ atoms/m³ gives 2.5×10²² donors/m³. Acceptor atom effectively negative when accepts electron, donor positive when donates, but crystal neutral. Trivalent dopants (valency 3) like Boron (B), Aluminium (Al), and Indium (In) are used in p-type semiconductors to create holes. Phosphorus (P) is pentavalent. Substituting values gives Aluminium, which matches expected behaviour for this semiconductor device configu

Ref: NCERT > Physics Book > Electronic Devices > Doping, Charge Carriers and Conductivity

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Semiconductors, Types and Energy Bands

Which of the following is an elemental semiconductor?

**Energy bands in semiconductors** consist of valence band filled at 0 K and conduction band empty, gap E_g small ~1 eV (Si 1.1 eV, Ge 0.7 eV), insulators large gap >3 eV (C diamond 5.4 eV), conductors overlapping. Intrinsic semiconductor at 0 K behaves as insulator because no thermal excitation, at T>0 K electrons jump to conduction band leaving holes, conductivity increases with temperature. Elemental semiconductors are pure elements like Si and Ge, while compound semiconductors (e.g., GaAs, CdS) consist of multiple elements. Among the options, Ge is elemental. Substituting values gives Ge,

Ref: NCERT > Physics Book > Electronic Devices > Semiconductors, Types and Energy Bands

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Semiconductors, Types and Energy Bands

Which of the following materials has the highest resistivity?

**Semiconductor properties** distinguish from conductors and insulators by temperature dependence and doping response. At 0 K intrinsic acts as insulator, conductivity due to thermally generated electron-hole pairs, number of outer electrons 4 for Si/Ge forming covalent bonds, each atom shares electrons, crystal with N atoms has 4N valence electrons, 2N bonds. Resistivity ( rho ) distinguishes materials: metals have low resistivity ( 10⁻² to 10⁻⁸ Ω m ), semiconductors intermediate ( 10⁻⁵ to 10⁶ Ω m ), and insulators high ( 10¹⁰ to 10¹⁹ Ω m ). Among the options, insulators have the highest resi

Ref: NCERT > Physics Book > Electronic Devices > Semiconductors, Types and Energy Bands

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Semiconductors, Types and Energy Bands

Which of the following is an example of a compound semiconductor?

**Energy bands in semiconductors** consist of valence band filled at 0 K and conduction band empty, gap E_g small ~1 eV (Si 1.1 eV, Ge 0.7 eV), insulators large gap >3 eV (C diamond 5.4 eV), conductors overlapping. Intrinsic semiconductor at 0 K behaves as insulator because no thermal excitation, at T>0 K electrons jump to conduction band leaving holes, conductivity increases with temperature. Compound semiconductors include inorganic materials like CdS, GaAs, and InP, unlike elemental semiconductors (Si, Ge). Among the options, GaAs is a compound semiconductor. Substituting values gives GaAs,

Ref: NCERT > Physics Book > Electronic Devices > Semiconductors, Types and Energy Bands

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Diamagnetism, Paramagnetism and Ferromagnetism

A material’s weak enhancement of the magnetic field inside it is due to:

**Ferromagnetism** shows large positive χ ≈ 10³ to 10⁵, strong attraction, domain structure with spontaneous magnetization, hysteresis, retentivity. Distinction based on sign and magnitude of χ and behaviour in non-uniform field, explaining attraction versus repulsion. Paramagnetic materials weakly enhance the magnetic field inside them because their atomic magnetic moments partially align with the external field, producing a small positive magnetization that adds to the applied field. Substituting values gives Weak alignment of atomic moments, which matches expected magnitude for this magneti

Ref: NCERT > Physics Book > Magnetism and Matter > Diamagnetism, Paramagnetism and Ferromagnetism

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Magnetic Field Due to Bar Magnet - Axial and Equatorial

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

**Magnetic field of bar magnet** follows inverse cube law B ∝ m/r³, unlike inverse square for electric dipole. Given B at distance r, moment m = B r³/(μ₀/4π) for equatorial, m = B r³/(2·μ₀/4π) for axial, enabling moment extraction from measured field. 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. Substituting values gives Induced opposing moments, which matches expected magnitude for thi

Ref: NCERT > Physics Book > Magnetism and Matter > Magnetic Field Due to Bar Magnet - Axial and Equatorial

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PHYSICS

A brass wire of length 2.2 m and cross-sectional area 2 × 10⁻⁶ m² is stretched by a force of 220 N . If the Young'

Given: A brass wire of length 2.2 m and cross-sectional area 2 × 10⁻⁶ m² is stretched by a force of 220 N . If the Young's modulus of brass is 9 × 10¹⁰ N/m², what is the stress? These values define the system as per NCERT data. Formula: Stress: Stress = F/A = frac2202 × 10⁻⁶= 1.1 × 10⁸ N/m² .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. Result: The computed value matches the expected outco

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 conductor has a resistivity of 1.0 × 10⁻⁷Ω m and α = 4 × 10⁻³°C^{-1 at 25° C . What is its resistivity at

Given: A conductor has a resistivity of 1.0 × 10⁻⁷Ω m and α = 4 × 10⁻³°C^{-1 at 25° C . What is its resistivity at 85° C ? These values define the system as per NCERT data. Formula: Use: rho_t = rho_0 [1 + α (T - T_0)]. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substitute: rho_t = 1.0 × 10⁻⁷[1 + 4 × 10⁻³(85 - 25)] . Calculate: rho_t = 1.0 × 10⁻⁷[1 + 0.24] = 1.0 × 10⁻⁷ × 1.24 = 1.24 × 10⁻⁷Ω m . Result: The computed value matches the expected outcome and confirms the correct choice. Units and power

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.

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PHYSICS

An aluminium block of dimensions 0.4 m × 0.3 m × 0.1 m is subjected to a shearing force of 4 × 10⁴ N . If the shear

Given: An aluminium block of dimensions 0.4 m × 0.3 m × 0.1 m is subjected to a shearing force of 4 × 10⁴ N . If the shear modulus of aluminium is 2.5 × 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 = 0.4 × 0.3 = 0.12 m² . Shear stress: 4 × 10⁴/0.12 approx 3.33 × 10⁵ N/m² . Shear strain: Shear strain = fracShear stressG = frac3.33 × 10⁵².5 × 1

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 conductor has a resistivity of 1.2 × 10⁻⁷Ω m and α = 4 × 10⁻³°C^{-1 at 20° C . What is its resistivity at 80° C ?

Given: A conductor has a resistivity of 1.2 × 10⁻⁷Ω m and α = 4 × 10⁻³°C^{-1 at 20° C . What is its resistivity at 80° C ? These values define the system as per NCERT data. Formula: Use: rho_t = rho_0 [1 + α (T - T_0)]. This is standard NCERT relation. Substitution & Calculation: Substitute: rho_t = 1.2 × 10⁻⁷[1 + 4 × 10⁻³(80 - 20)] . Calculate: rho_t = 1.2 × 10⁻⁷[1 + 0.24] = 1.2 × 10⁻⁷ × 1.24 = 1.488 × 10⁻⁷Ω m . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

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

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PHYSICS

The resistivity of insulators is typically in the range:

Insulators have high resistivity ( 10¹⁰ to 10¹⁹Ω m ) or low conductivity ( 10⁻¹¹ to 10⁻¹⁹ S m^{-1 ), preventing significant current flow. This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.

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PHYSICS

A copper sheet has an area of 2 m² at 25° C . What will be its area at 225° C ? ( α_l = 1.7 × 10⁻⁵ K^{-1 )

Given: A copper sheet has an area of 2 m² at 25° C . What will be its area at 225° C ? ( α_l = 1.7 × 10⁻⁵ K^{-1 ) These values define the system as per NCERT data. Formula: Given: A_0 = 2 m², Δ T = 225 - 25 = 200° C, α_l = 1.7 × 10⁻⁵ K^{-1. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Area expansion: Δ A = A_0 × 2 α_l Δ T = 2 × 2 × 1.7 × 10⁻⁵ × 200 = 0.0136 m² . New area: A = A_0 + Δ A = 2 + 0.0136 = 2.0136 m² . Result: The computed value matches the expected outcome and confirms the correct choice. Units a

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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