PHYSICS

PHYSICS

A gas mixture contains equal numbers of neon and argon molecules at 300 K. What is the ratio of their average kinetic en

Given: A gas mixture contains equal numbers of neon and argon molecules at 300 K. What is the ratio of their average kinetic energies? These values define the system as per NCERT data. Formula: Average KE per molecule = 3/2 k_B T, independent of mass. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Since T is same, fracKE_{NeKE_{Ar = 1:1. 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

Two small charged spheres with charges 5 × 10⁻⁷ C and 7 × 10⁻⁷ C are placed 50 cm apart in air. What is the fo

Given: Two small charged spheres with charges 5 × 10⁻⁷ C and 7 × 10⁻⁷ C are placed 50 cm apart in air. What is the force between them? These values define the system as per NCERT data. Formula: Using Coulomb’s law: F = k |q_1 q_2|/r². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: k = 9 × 10⁹ Nm²/C², q_1 = 5 × 10⁻⁷ C, q_2 = 7 × 10⁻⁷ C, r = 0.5 m . |q_1 q_2| = 5 × 7 × 10⁻¹⁴= 35 × 10⁻¹⁴ C² . r² = (0.5)² = 0.25 m² . F = 9 × 10⁹ × frac35 × 10⁻¹⁴⁰.25 = 9 × 10⁹ × 1.4 × 10⁻¹²= 0.0126 N .

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 square loop of side 25 cm rotates at 8 rad/s in a 0.15 T field. What is the maximum emf induced?

Given: A square loop of side 25 cm rotates at 8 rad/s in a 0.15 T field. What is the maximum emf induced? These values define the system as per NCERT data. Formula: A = (0.25)² = 0.0625 m². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon_0 = N B A omega = 1 × 0.15 × 0.0625 × 8 = 0.075 V . 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: Electromagnetic Induction, Alternating Current and Electromagnetic Waves, Topic: Induced emf, inductance and EM wave properties.

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PHYSICS

In a new system, the unit of mass is 2 kg, length is 0.5 m, and time is 3 s . What is the value of 1 J ( kg m² s^{-2 )

Given: In a new system, the unit of mass is 2 kg, length is 0.5 m, and time is 3 s . What is the value of 1 J ( kg m² s^{-2 ) in this system? These values define the system as per NCERT data. Formula: 1 J = 1 kg m² s^{-2. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: New units: kg = 2 α, m = 0.5 β, s = 3 γ . 1 J = (2 α) (0.5 β)² (3 γ)^{-2 = 2 × 0.25 × 1/9 = 0.5/9 = 0.0556 α β² γ^{-2 . 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⁻â

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 magnetic dipole of moment 0.45 A m² is in a uniform field of 0.6 T at 45° . What is the torque on it?

Given: A magnetic dipole of moment 0.45 A m² is in a uniform field of 0.6 T at 45° . What is the torque on it? These values define the system as per NCERT data. Formula: Torque is tau = m B sinθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: m = 0.45 A m², B = 0.6 T, θ = 45°, sin 45° = frac1sqrt2 approx 0.707 . Substitute: tau = 0.45 × 0.6 × 0.707 approx 0.19089 N m approx 0.191 N 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⁻â

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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PHYSICS

A bar magnet is cut along its length into two equal parts. If the original m = 1.0 A m², what is m of each part?

Given: A bar magnet is cut along its length into two equal parts. If the original m = 1.0 A m², what is m of each part? These values define the system as per NCERT data. Formula: Given: m = 1.0 A m². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: When cut along its length, each part retains the full magnetic moment, but here it’s implied as halved in typical problems. . Each part: m' = 1.0/2 = 0.5 A m² . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/sÂ

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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PHYSICS

A spring system has m = 3 kg, k = 1200 N/m, A = 5 cm . What is the potential energy at x = 2.5 cm ?

Given: A spring system has m = 3 kg, k = 1200 N/m, A = 5 cm . What is the potential energy at x = 2.5 cm ? These values define the system as per NCERT data. Formula: Potential energy: U = 1/2 k x². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: k = 1200 N/m, x = 0.025 m . U = 0.5 × 1200 × (0.025)² = 0.5 × 1200 × 0.000625 = 0.375 J . 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: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

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PHYSICS

A paramagnetic material with chi = 6 × 10⁻⁴ in H = 1500 A m^{-1 has magnetization M :

Given: A paramagnetic material with chi = 6 × 10⁻⁴ in H = 1500 A m^{-1 has magnetization M : These values define the system as per NCERT data. Formula: M = chi H. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: chi = 6 × 10⁻⁴, H = 1500 A m^{-1 . M = 6 × 10⁻⁴ × 1500 = 0.9 A m^{-1 . 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 particle in SHM has an amplitude of 10 cm and a period of 0.5 s . What is its maximum velocity? (Take π = 3.14 )

Given: A particle in SHM has an amplitude of 10 cm and a period of 0.5 s . What is its maximum velocity? (Take π = 3.14 ) These values define the system as per NCERT data. Formula: Maximum velocity: v_{max = A omega. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: omega = 2π/T = 2 × 3.14/0.5 = 12.56 rad/s . A = 0.1 m . v_{max = 0.1 × 12.56 = 1.256 m/s . 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: Oscillations and Waves, Topic: SHM, wave motion and superposition.

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PHYSICS

A stone is dropped from a height of 80 m while another is thrown upwards at 20 m/s from the ground at the same instant.

Given: A stone is dropped from a height of 80 m while another is thrown upwards at 20 m/s from the ground at the same instant. When do they meet? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Stone 1: y_1 = 80 - 1/2 · 10 · t² = 80 - 5 t². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Stone 2: y_2 = 20 t - 1/2 · 10 · t² = 20 t - 5 t² . Total height = 80 m, so y_1 + y_2 = 80 . Substitute: (80 - 5 t²) + (20 t - 5 t²) = 80 Rightarrow 80 + 20 t - 10 t² = 80 Rightarrow 20 t - 10 t² = 0 Rightarrow t (2 - t

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 ll of emf 8 V and internal resistance 2 Ω is connected to a 6 Ω resistor. What is the current in the circuit?

Given: A ll of emf 8 V and internal resistance 2 Ω is connected to a 6 Ω resistor. What is the current in the circuit? These values define the system as per NCERT data. Formula: Total resistance: R_{total = R + r = 6 + 2 = 8 Ω. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Current: I = fracvarepsilonR_{total = 8/8 = 1 A . 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 ll of emf 9 V and internal resistance 3 Ω is connected to a 6 Ω resistor. What is the current in the circuit?

Given: A ll of emf 9 V and internal resistance 3 Ω is connected to a 6 Ω resistor. What is the current in the circuit? These values define the system as per NCERT data. Formula: Total resistance: R_{total = 6 + 3 = 9 Ω. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Current: I = fracvarepsilonR_{total = 9/9 = 1 A . 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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