#cross-sectional area

Potentiometer, Conductivity and Special Cases

A copper wire carries \( 2 \, \text{A} \) with a drift speed of \( 8 \times 10^{-5} \, \text{m/s} \). If \( n = 8.5 \tim

**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. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (2/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 8 × 10⁻⁵) . Calculate: A = (2/1.088 × 10⁵) ≈ 1.84 × 10⁻⁵ m² . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT],

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

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Potentiometer, Conductivity and Special Cases

A copper wire carries \( 2.72 \, \text{A} \) with a drift speed of \( 8 \times 10^{-5} \, \text{m/s} \). If \( n = 8.5 \

**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. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (2.72/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 8 × 10⁻⁵) . Calculate: A = (2.72/1.088 × 10⁵) ≈ 2.5 × 10⁻⁵ m² . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT],

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

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Potentiometer, Conductivity and Special Cases

A copper wire carries \( 4.5 \, \text{A} \) with a drift speed of \( 1.8 \times 10^{-4} \, \text{m/s} \). If \( n = 8.5

**Conductivity** σ = 1/ρ = n e² τ/m (S/m), τ relaxation time, m electron mass. Resistivity deviation at high fields occurs when τ depends on E or n changes due to impact ionization, breaking Ohm's law, seen in varistors, gas discharge. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (4.5/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.8 × 10⁻⁴) . Calculate: A = (4.5/2.448 × 10⁵) ≈ 1.84 × 10⁻⁵ m² . 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 copper wire carries a current of \( 3 \, \text{A} \) with a drift speed of \( 1.2 \times 10^{-4} \, \text{m/s} \). If

**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. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (3/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.2 × 10⁻⁴) . Calculate: A = (3/1.632 × 10⁶) ≈ 1.84 × 10⁻⁶ m² . 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

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

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Electric Current, Drift Velocity and Mobility

A copper wire carries \( 3.4 \, \text{A} \) with a drift speed of \( 1.25 \times 10^{-4} \, \text{m/s} \). If \( n = 8.5

**Current and drift relation** I = n e A v_d shows current proportional to drift velocity and area. For A=6×10⁻⁷ m², I=1.8 A, n=8.5×10²⁸ m⁻³, v_d =1.8/(8.5×10²⁸×1.6×10⁻¹⁹×6×10⁻⁷)=2.2×10⁻⁴ m/s, illustrating small drift speed even for ampere currents. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (3.4/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.25 × 10⁻⁴) . Calculate: A = (3.4/1.7 × 10⁵) = 2.0 × 10⁻⁵ m² . 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

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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Specific Heat and Calorimetry

A brass rod of length 2.5m at 20∘C is heated to 220∘C. If its cross-sectional area increases by 0.018cm2, what was its o

Given: ΔT = 220−20 = 200∘C, ΔA = 0.018cm2, αl = 1.8×10−5K−1. Area expansion: ΔA = A0×2αlΔT. 0.018 = A0×2×1.8×10−5×200. 0.018 = A0×7.2×10−3⇒A0 = 0.0187.2×10−3 = 2.5cm2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.5 cm². 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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Temperature Scales and Triple Point

A silver rod of length 1m at 10∘C is heated to 110∘C. If its cross-sectional area increases by 0.0076cm2, what was its o

Given: ΔT = 110−10 = 100∘C, ΔA = 0.0076cm2, αl = 1.9×10−5K−1. ΔA = A0×2αlΔT. 0.0076 = A0×2×1.9×10−5×100. 0.0076 = A0×3.8×10−3⇒A0 = 0.00763.8×10−3 = 2cm2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2 cm². This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

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

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Stress, Strain and Hooke's Law

A copper wire of length 2.8m and cross-sectional area 2×10−6m2 is stretched by a force of 160N. If the Young's modulus o

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 160×2.82×10−6×1.1×1011 = 4482.2×105≈2.04×10−3m = 2.04mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.04mm. 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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Stress, Strain and Hooke's Law

A steel wire of length 2.0m and cross-sectional area 2.5×10−6m2 is stretched by a force of 250N. If the Young's modulus

Stress: Stress = FA = 2502.5×10−6 = 1×108N/m2. Young's modulus: Y = StressStrain. Strain: Strain = StressY = 1×1082×1011 = 5×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5×10−4. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: University Physics, Chapter 17: Thermal Expansion.

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Shear Modulus and Rigidity

A cylindrical rod of length 0.5 m and radius 0.01 m is compressed by a force of 5000 N. If the compressive stress is 5 ×

Stress: Stress = F / A. Rearrange: A = F / Stress. Substitute: A = 5000 / (5 × 106) = 10-3 m2. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10-3 m2. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

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

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Young's Modulus and Elasticity

An aluminium wire of length 2.0m and cross-sectional area 1.5×10−6m2 is stretched by a force of 150N. If the Young's mod

Young's modulus: Y = FLAΔL. Rearrange: ΔL = FLAY. Substitute: ΔL = 150×2.01.5×10−6×7×1010 = 3001.05×105≈2.86×10−3m = 2.86mm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.86mm. 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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