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PHYSICS

Latest questions in this category.

45 questions

PHYSICS

A solenoid of 1000 turns/m and area 0.02 m² has μ_r = 2 . What is its self-inductance? ( μ_0 = 4π × 10⁻⁷ H/m )

Given: A solenoid of 1000 turns/m and area 0.02 m² has μ_r = 2 . What is its self-inductance? ( μ_0 = 4π × 10⁻⁷ H/m ) These values define the system as per NCERT data. Formula: L = μ_r μ_0 n² A l, assume l = 1 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: L = 2 × 4π × 10⁻⁷ × (1000)² × 0.02 × 1 = 0.05024 H approx 0.05 H . 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: 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 proton moves at 1.5 × 10⁷ m/s perpendicular to a field of 0.2 T . What is the radius of its path? (Mass = 1.67 × 1

Given: A proton moves at 1.5 × 10⁷ m/s perpendicular to a field of 0.2 T . What is the radius of its path? (Mass = 1.67 × 10⁻²⁷ kg, charge = 1.6 × 10⁻¹⁹ C ) These values define the system as per NCERT data. Formula: r = mv/qB. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r = frac1.67 × 10⁻²⁷ × 1.5 × 10⁷¹.6 × 10⁻¹⁹ × 0.2 = frac2.505 × 10⁻²⁰³.2 × 10⁻²⁰= 0.7828 approx 0.78 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: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A paramagnetic material with chi = 2 × 10⁻³ is placed in H = 500 A m^{-1 . What is M ?

Given: A paramagnetic material with chi = 2 × 10⁻³ is placed in H = 500 A m^{-1 . What is 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 = 2 × 10⁻³, H = 500 A m^{-1 . M = 2 × 10⁻³ × 500 = 1 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 transverse wave travels on a string with tension 50 N and linear mass density 0.02 kg/m. What is the wavelength if the

Given: A transverse wave travels on a string with tension 50 N and linear mass density 0.02 kg/m. What is the wavelength if the frequency is 25 Hz? These values define the system as per NCERT data. Formula: Speed: v = sqrtT/μ = sqrt50/0.02 = sqrt2500 = 50 m/s. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Wavelength: lambda = v/v = 50/25 = 2 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: Oscillations and Waves, Topic: SHM, wave motion and superposition.

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PHYSICS

A circuit has a 20 V battery with 2 Ω internal resistance and three resistors 5 Ω, 10 Ω, 20 Ω in parallel. What is t

Given: A circuit has a 20 V battery with 2 Ω internal resistance and three resistors 5 Ω, 10 Ω, 20 Ω in parallel. What is the current through the 10 Ω resistor? These values define the system as per NCERT data. Formula: Parallel resistance: 1/R_p = 1/5 + 1/10 + 1/20 = 4 + 2 + 1/20 = 7/20 Rightarrow R_p = 20/7 approx 2.86 Ω. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Total resistance: R_{total = 2 + 2.86 = 4.86 Ω . Total current: I = fracvarepsilonR_{total = 20/4.86 approx 4.12 A . Voltage across parallel: V = I R_p = 4.12 × 2.86 approx 11.78 V . Current through 10 Ω : I_{10 = V/10 = 11.78/10 approx 1.18 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

The electric field of an EM wave oscillates with amplitude 48 V/m and frequency 2 × 10¹⁰ Hz . What is the amplitude

Given: The electric field of an EM wave oscillates with amplitude 48 V/m and frequency 2 × 10¹⁰ Hz . What is the amplitude of the magnetic field? (Given c = 3 × 10⁸ m/s ) These values define the system as per NCERT data. Formula: Using B_0 = E_0/c, we have B_0 = 48/3 × 10⁸= 1.6 × 10⁻⁷ T .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. 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: 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 submarine window ( 0.03 m² ) is at 500 m depth in seawater ( rho = 1.03 × 10³ kg/m³ ). What force acts on it if in

Given: A submarine window ( 0.03 m² ) is at 500 m depth in seawater ( rho = 1.03 × 10³ kg/m³ ). What force acts on it if inside pressure is atmospheric? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Gauge pressure: P_g = rho g h = 1.03 × 10³ × 10 × 500 = 5.15 × 10⁶ Pa. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: F = P_g A = 5.15 × 10⁶ × 0.03 = 1.545 × 10⁵ N . 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: 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

Light of wavelength 350 nm is incident on a metal with work function 2.0 eV . What is the stopping potential? (Take h c

Given: Light of wavelength 350 nm is incident on a metal with work function 2.0 eV . What is the stopping potential? (Take h c = 1240 eV nm ) These values define the system as per NCERT data. Formula: E = h c/lambda = 1240/350 approx 3.54 eV. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: K_{max = E - phi_0 = 3.54 - 2.0 = 1.54 eV . V_0 = fracK_{maxe = 1.54 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: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

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PHYSICS

Two parallel wires 0.02 m apart carry 10 A and 3 A in the same direction. What is the force per unit length? ( μ_0 = 4

Given: Two parallel wires 0.02 m apart carry 10 A and 3 A in the same direction. What is the force per unit length? ( μ_0 = 4 π × 10⁻⁷ T m/A ) These values define the system as per NCERT data. Formula: f = μ_0 I_1 I_2/2 π d. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: f = frac4 π × 10⁻⁷ × 10 × 32 π × 0.02 = frac120 × 10⁻⁷⁰.04 = 3 × 10⁻⁵ 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, 10⁻⁵ are properly used as per NCERT.

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

What is the pressure exerted by 1 mole of an ideal gas in a 10-litre container at 300 K? (R = 8.31 J mol^{-1 K^{-1)

Given: What is the pressure exerted by 1 mole of an ideal gas in a 10-litre container at 300 K? (R = 8.31 J mol^{-1 K^{-1) These values define the system as per NCERT data. Formula: PV = μ R T, P = μ R T/V. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: P = frac1 × 8.31 × 30010 × 10⁻³= 2.493 × 10⁵ Pa approx 2.5 atm (1 atm approx 10⁵ Pa). 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 plane sheet has a surface charge density sigma = 3.54 × 10⁻¹¹ C/m² . What is the electric field magnitude near i

Given: A plane sheet has a surface charge density sigma = 3.54 × 10⁻¹¹ C/m² . What is the electric field magnitude near it? These values define the system as per NCERT data. Formula: E = sigma/2 varepsilon_0. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E = frac3.54 × 10⁻¹¹² × 8.854 × 10⁻¹²= 2 N/C . 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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