#material property

Potentiometer, Conductivity and Special Cases

A wire of length \( 9 \, \text{m} \) and cross-sectional area \( 1.5 \times 10^{-6} \, \text{m}^2 \) has a resistance of

**Potentiometer** measures potential difference without drawing current, using null deflection, principle V ∝ l, l balance length, accurate because no I r drop. Potential drop across resistor V = I R arises because electric field does work on charges, energy converted to heat, maintaining E = -dV/dx along wire. Resistance: R = (rho l/A) . Rearrange: rho = (R A/l) . Substitute: rho = (18 × 1.5 × 10⁻⁶/9) = 3 × 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 = ε - I r and P = I²R,

Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases

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Potentiometer, Conductivity and Special Cases

A wire has a resistance of \( 18 \, \Omega \) at \( 20^\circ \text{C} \) and \( 19.8 \, \Omega \) at \( 80^\circ \text{C

**Potential difference drop** across resistor when current flows because charges lose potential energy qV = I² R t as heat, field E = V/l drives drift, maintaining current. At very high E, velocity saturation or heating changes τ, causing non-ohmic behaviour. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: 19.8 = 18 [1 + α (80 - 20)] . Solve: 19.8 = 18 + 1080α ⇒ 1080α = 1.8 ⇒ α = (1.8/1080) ≈ 1.67 × 10⁻³ °C⁻¹ . 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 = ε

Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases

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Wheatstone Bridge and Meter Bridge

A wire has a resistance of \( 10 \, \Omega \) at \( 20^\circ \text{C} \) and \( 12 \, \Omega \) at \( 100^\circ \text{C}

**Wheatstone bridge balance** condition R₁/R₂ = R₃/R₄, R₄ = R₂ R₃/R₁, when galvanometer current zero, potentials at midpoints equal. At balance, no current through galvanometer, enabling precise resistance measurement independent of source voltage. Use: R_t = R₀ [1 + α (T - T₀)] . Given: R₀ = 10 Ω , R_t = 12 Ω , T = 100° C , T₀ = 20° C . Substitute: 12 = 10 [1 + α (100 - 20)] . Solve: 12 = 10 + 80α ⇒ 80α = 2 ⇒ α = (2/80) = 0.025 × 10⁻² = 2.5 × 10⁻⁴ °C⁻¹ . Applying I = n e

Ref: NCERT > Physics Book > Current Electricity > Wheatstone Bridge and Meter Bridge

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Temperature Dependence of Resistance and Resistivity

A wire of length \( 10 \, \text{m} \) and cross-sectional area \( 1 \times 10^{-6} \, \text{m}^2 \) has a resistance of

**Temperature dependence** of resistance R_t = R₀[1+α(T-T₀)], α temperature coefficient (per °C), R₀ resistance at T₀ (Ω). For metals α positive ≈10⁻³ /°C, resistance increases with temperature because τ decreases due to increased phonon scattering, n nearly constant. Resistance: R = (rho l/A) . Rearrange: rho = (R A/l) . Substitute: rho = (20 × 1 × 10⁻⁶/10) = 2 × 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 = ε - I r and P = I²R, evaluation yields 2.0 × 10⁻⁶ Ω m,

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

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Temperature Dependence of Resistance and Resistivity

A conductor has a resistivity of \( 4 \times 10^{-8} \, \Omega \text{m} \) at \( 20^\circ \text{C} \) and a temperature

**Temperature dependence** of resistance R_t = R₀[1+α(T-T₀)], α temperature coefficient (per °C), R₀ resistance at T₀ (Ω). For metals α positive ≈10⁻³ /°C, resistance increases with temperature because τ decreases due to increased phonon scattering, n nearly constant. Use: rho_t = rho₀ [1 + α (T - T₀)] . Given: rho₀ = 4 × 10⁻⁸ Ω m , α = 4 × 10⁻³ °C⁻¹ , T = 60° C , T₀ = 20° C . Substitute: rho_t = 4 × 10⁻⁸ [1 + 4 × 10⁻³ (60 - 20)] . Calculate: rho_t = 4 × 10⁻⁸ [1 + 0.16] = 4 × 10⁻⁸ × 1.16

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

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Torque on Magnetic Dipole and Potential Energy

The magnetic field contribution \( B_m \) due to a material with \( M = 3 \times 10^5 \, \text{A m}^{-1} \) is: (Take \(

**Potential energy of magnetic dipole** U = -m B cosθ explains stability. Given m = 0.9 A·m², B = 0.5 T, θ = 90°, sin90° = 1, τ = 0.45 N·m. For 60°, sin60° = √3/2 ≈0.866, reducing torque proportionally. B_m = μ₀ M . Given: M = 3 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B_m = 4π × 10⁻⁷ × 3 × 10⁵ = 0.3768 T ≈ 0.38 T . Substituting values gives 0.38 T, 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 > Torque on Magnetic Dipole and Potential Energy

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Transverse and Longitudinal Waves

What is the key requirement for a medium to support both transverse and longitudinal waves?

**Wave classification** depends on particle vibration relative to propagation. Longitudinal waves have particle oscillation parallel to propagation, creating compressions and rarefactions as in sound in air; transverse have perpendicular oscillation. Tuning fork generates longitudinal sound because air cannot sustain shear. A medium must be rigid (have shear modulus) to support transverse waves and compressible (have bulk modulus) for longitudinal waves, properties typical of solids. Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields Rigidity, il

Ref: NCERT > Physics Book > Waves > Transverse and Longitudinal Waves

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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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