PHYSICS

PHYSICS

A mixture contains 1 mole of helium and 1 mole of oxygen at 300 K. What is the ratio of their partial pressures?

Given: A mixture contains 1 mole of helium and 1 mole of oxygen at 300 K. What is the ratio of their partial pressures? These values define the system as per NCERT data. Formula: For ideal gases, P = μ RT/V, partial pressure propto μ (since V and T are same). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Ratio fracP_{HeP_{O_2 = fracμ_{Heμ_{O_2 = 1/1 = 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

The dimensional formula of work is [M L² T^{-2] . What is the dimensional formula of power?

Given: The dimensional formula of work is [M L² T^{-2] . What is the dimensional formula of power? These values define the system as per NCERT data. Formula: Power = Work/Time. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: [P] = [M L² T^{-2] / [T] = [M L² T^{-3] . 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

A rod of length 0.2 m moves at 5 m/s in a 0.3 T field perpendicular to its length. What is the induced emf?

Given: A rod of length 0.2 m moves at 5 m/s in a 0.3 T field perpendicular to its length. What is the induced emf? These values define the system as per NCERT data. Formula: Motional emf: varepsilon = B l v. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon = 0.3 × 0.2 × 5 = 0.3 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

A dipole with m = 0.5 A m² in B = 0.7 T at 60° has torque:

Given: A dipole with m = 0.5 A m² in B = 0.7 T at 60° has torque: These values define the system as per NCERT data. Formula: tau = m B sinθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: m = 0.5 A m², B = 0.7 T, θ = 60°, sin 60° = fracsqrt32 approx 0.866 . tau = 0.5 × 0.7 × 0.866 approx 0.3031 N m approx 0.30 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: 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

The magnetic potential energy of a dipole with m = 0.4 A m² in a field B = 0.7 T at 180° is:

Given: The magnetic potential energy of a dipole with m = 0.4 A m² in a field B = 0.7 T at 180° is: These values define the system as per NCERT data. Formula: U_m = -m B cosθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: m = 0.4 A m², B = 0.7 T, θ = 180°, cos 180° = -1 . Substitute: U_m = -0.4 × 0.7 × (-1) = 0.28 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: 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 steel rod of radius 0.005 m and length 1 m is compressed by a force producing a stress of 2 × 10 ⁷ N/m ² . If the

Given: A steel rod of radius 0.005 m and length 1 m is compressed by a force producing a stress of 2 × 10 ⁷ N/m ² . If the Young's modulus of steel is 2 × 10 ¹¹ N/m ², what is the strain? These values define the system as per NCERT data. Formula: Young's modulus: Y = Stress / Strain. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Strain: Strain = Stress / Y = (2 × 10 ⁷ ) / (2 × 10 ¹¹ ) = 1 × 10 ⁻⁴ . 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

A solenoid produces B = 0.75 T with a core of μ_r = 500 and n = 1000 m^{-1 . What is the current I ? (Take μ_0 = 4Ï€ Ã

Given: A solenoid produces B = 0.75 T with a core of μ_r = 500 and n = 1000 m^{-1 . What is the current I ? (Take μ_0 = 4π × 10⁻⁷ T m A^{-1 ). These values define the system as per NCERT data. Formula: B = μ_0 μ_r n I, so I = B/μ_0 μ_r n. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: B = 0.75 T, μ_r = 500, n = 1000 m^{-1, μ_0 = 4π × 10⁻⁷. I = frac0.754π × 10⁻⁷ × 500 × 1000 = frac0.756.283 × 10⁻¹ approx 1.194 A approx 1.2 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: 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 particle in SHM has a = -16 x (in SI units). What is its period?

Given: A particle in SHM has a = -16 x (in SI units). What is its period? These values define the system as per NCERT data. Formula: For SHM, a = -omega² x. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given a = -16 x, so omega² = 16 Rightarrow omega = 4 rad/s . T = 2π/omega = 2π/4 = π/2 approx 1.57 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 bar magnet with m = 2.0 A m² produces a field at 0.5 m on its equatorial line. What is B ? (Take μ_0 = 4Ï€ × 10⁻â

Given: A bar magnet with m = 2.0 A m² produces a field at 0.5 m on its equatorial line. What is B ? (Take μ_0 = 4π × 10⁻⁷ T m A^{-1 ). These values define the system as per NCERT data. Formula: B = μ_0/4π m/r³. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: m = 2.0 A m², r = 0.5 m, μ_0/4π = 10⁻⁷. B = 10⁻⁷ × 2.0/(0.5)³ = 10⁻⁷ × 2.0/0.125 = 1.6 × 10⁻⁶ T . 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 wave on a string has an amplitude of 2 cm and a wavelength of 0.5 m. What is the maximum transverse speed of a particl

Given: A wave on a string has an amplitude of 2 cm and a wavelength of 0.5 m. What is the maximum transverse speed of a particle on the string if its frequency is 20 Hz? These values define the system as per NCERT data. Formula: Amplitude a = 0.02 m, frequency v = 20 Hz. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Angular frequency: omega = 2π v = 2π × 20 = 40π rad/s . Maximum speed: v_{max = a omega = 0.02 × 40π approx 2.51 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

Four 1 kg masses are at the corners of a square of side 2 m . What is the potential energy of the system? ( G = 6.67 ×

Given: Four 1 kg masses are at the corners of a square of side 2 m . What is the potential energy of the system? ( G = 6.67 × 10⁻¹¹ N m²/kg² ) These values define the system as per NCERT data. Formula: Pairs: 4 sides ( r = 2 m ), 2 diagonals ( r = 2sqrt2 m ). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: V = -4 G m²/2 - 2 fracG m²²sqrt2 . V = -4 frac6.67 × 10⁻¹¹ × 12 - 2 frac6.67 × 10⁻¹¹ × 12sqrt2 . V = -1.334 × 10⁻¹⁰- 0.471 × 10⁻¹⁰ approx -1.805 × 10⁻¹⁰ 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 wire of length 1.8 m carrying 2 A is at 30° to a magnetic field of 0.45 T . What is the force on the wire?

Given: A wire of length 1.8 m carrying 2 A is at 30° to a magnetic field of 0.45 T . What is the force on the wire? These values define the system as per NCERT data. Formula: Force F = I l B sin θ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: F = 2 × 1.8 × 0.45 × sin 30° = 3.6 × 0.45 × 0.5 = 0.81 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: 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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