#electricity

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 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 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

A copper wire carries 4.5 A with a drift speed of 1.8 × 10⁻⁴ m/s . If n = 8.5 × 10²⁸ m^{-3 and e = 1.6 × 10⁻

Given: A copper wire carries 4.5 A with a drift speed of 1.8 × 10⁻⁴ m/s . If n = 8.5 × 10²⁸ m^{-3 and e = 1.6 × 10⁻¹⁹ C, what is the cross-sectional area? These values define the system as per NCERT data. Formula: Drift speed: v_d = I/n e A. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Rearrange: A = I/n e v_d . Substitute: A = frac4.58.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.8 × 10⁻⁴. Calculate: A = 4.5/2.448 × 10⁵ approx 1.84 × 10⁻⁵ m² . 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: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A conductor has a resistivity of 3 × 10⁻⁸Ω m and α = 4 × 10⁻³°C^{-1 at 25° C . What is its resistivity at 7

Given: A conductor has a resistivity of 3 × 10⁻⁸Ω m and α = 4 × 10⁻³°C^{-1 at 25° C . What is its resistivity at 75° 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 = 3 × 10⁻⁸[1 + 4 × 10⁻³(75 - 25)] . Calculate: rho_t = 3 × 10⁻⁸[1 + 0.2] = 3 × 10⁻⁸ × 1.2 = 3.6 × 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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PHYSICS

A 4 Ω resistor dissipates 16 W of power. What is the voltage across it?

Given: A 4 Ω resistor dissipates 16 W of power. What is the voltage across it? These values define the system as per NCERT data. Formula: Power: P = V²/R. This is standard NCERT relation. Substitution & Calculation: Rearrange: V = sqrtP R . Substitute: V = sqrt16 × 4 = sqrt64 = 8 V . 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

Two lls of emf 4 V and 6 V with internal resistances 1 Ω and 2 Ω respectively are connected in series. What is the equiv

Given: Two lls of emf 4 V and 6 V with internal resistances 1 Ω and 2 Ω respectively are connected in series. What is the equivalent emf and internal resistance of the combination? These values define the system as per NCERT data. Formula: For series: varepsilon_{eq = varepsilon_1 + varepsilon_2 = 4 + 6 = 10 V. This is standard NCERT relation. Substitution & Calculation: Internal resistance: r_{eq = r_1 + r_2 = 1 + 2 = 3 Ω . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Electrostatic Potential and Capacitance, Topic: Capacitors in series, equivalent capacitance 1/C_eq = 1/C₁ + 1/C₂. 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 12 V battery with negligible internal resistance is connected to a 3 Ω and 6 Ω resistor in series. What is the power

Given: A 12 V battery with negligible internal resistance is connected to a 3 Ω and 6 Ω resistor in series. What is the power dissipated in the 6 Ω resistor? These values define the system as per NCERT data. Formula: Total resistance: R = 3 + 6 = 9 Ω. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Current: I = V/R = 12/9 = 4/3 A . Power: P = I² R = (4/3)² × 6 = 16/9 × 6 = 32/3 approx 10.67 W . 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

Why does a conductor’s current density increase when its length is halved while keeping the potential difference const

Resistance R = rho l / A . Halving length ( l' = l/2 ) halves R ( R' = R/2 ). Current I = V / R, so I' = V / (R/2) = 2I . Current density j = I / A, so j' = 2I / A = 2j .

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

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PHYSICS

Why does a conductor exhibit zero net current in the absence of an electric field?

Without an electric field, electrons move randomly due to thermal energy, with no preferred direction. The average velocity of electrons cancels out, resulting in zero net current.

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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