#Resistivity

PHYSICS

A wire has a resistance of 30 Ω at 25° C and 31.5 Ω at 75° C . What is the temperature coefficient of resistivity?

Given: A wire has a resistance of 30 Ω at 25° C and 31.5 Ω at 75° C . What is the temperature coefficient of resistivity? These values define the system as per NCERT data. Formula: Use: R_t = R_0 [1 + α (T - T_0)]. This is standard NCERT relation. Substitution & Calculation: Substitute: 31.5 = 30 [1 + α (75 - 25)] . Solve: 31.5 = 30 + 1500α Rightarrow 1500α = 1.5 Rightarrow α = 1.5/1500 = 1.0 × 10⁻³°C^{-1 . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Current Electricity, Topic: Temperature dependence of resistance, R_t = R₀[1+α(T-T₀)], temperature coefficient α calculation. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

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PHYSICS

A wire of length 3 m and resistance 6 Ω is stretched to 6 m . What is the new resistance?

Given: A wire of length 3 m and resistance 6 Ω is stretched to 6 m . What is the new resistance? Formula: Volume constant: l A = l' A' Rightarrow A' = A/2. Substitution & Calculation: New resistance: R' = rho l'/A' = fracrho (2l)A/2 = 4 rho l/A = 4R = 4 × 6 = 24 Ω . 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), Chapter: Current Electricity (Latest NCERT 2026-27), Topic: Resistance R = ρl/A, volume constant lA = l'A', resistance becomes 4 times on doubling length. The section explains governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, SI units and.

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PHYSICS

A wire of length 15 m and cross-sectional area 2 × 10⁻⁶ m² has a resistance of 30 Ω . What is the resistivity of

Given: A wire of length 15 m and cross-sectional area 2 × 10⁻⁶ m² has a resistance of 30 Ω . What is the resistivity of the material? These values define the system as per NCERT data. Formula: Resistance: R = rho l/A. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Rearrange: rho = R A/l . Substitute: rho = frac30 × 2 × 10⁻⁶¹⁵= 4 × 10⁻⁶Ω m . 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: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.

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PHYSICS

A conductor has a resistivity of 6 × 10⁻⁸Ω m and α = 4 × 10⁻³°C^{-1 at 20° C . What is its resistivity at 9

Given: A conductor has a resistivity of 6 × 10⁻⁸Ω m and α = 4 × 10⁻³°C^{-1 at 20° C . What is its resistivity at 90° 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 = 6 × 10⁻⁸[1 + 4 × 10⁻³(90 - 20)] . Calculate: rho_t = 6 × 10⁻⁸[1 + 0.28] = 6 × 10⁻⁸ × 1.28 = 7.68 × 10⁻⁸Ω m . 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: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.

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