PHYSICS

PHYSICS

A material with susceptibility chi = -3 × 10⁻⁵ has a relative permeability μ_r of:

Given: A material with susceptibility chi = -3 × 10⁻⁵ has a relative permeability μ_r of: These values define the system as per NCERT data. Formula: μ_r = 1 + chi. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: chi = -3 × 10⁻⁵. Substitute: μ_r = 1 - 3 × 10⁻⁵= 0.99997 . 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 magnetic dipole experiences a torque of 0.02 N m in a field of 0.4 T at 90° . What is its magnetic moment?

Given: A magnetic dipole experiences a torque of 0.02 N m in a field of 0.4 T at 90° . What is its magnetic moment? These values define the system as per NCERT data. Formula: tau = m B sinθ, so m = tau/B sinθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: tau = 0.02 N m, B = 0.4 T, θ = 90°, sin 90° = 1 . m = 0.02/0.4 × 1 = 0.05 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/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 spring of k = 180 N/m has a 0.9 kg mass. If E = 0.9 J, what is the amplitude?

Given: A spring of k = 180 N/m has a 0.9 kg mass. If E = 0.9 J, what is the amplitude? These values define the system as per NCERT data. Formula: Total energy: E = 1/2 k A². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 0.9 = 0.5 × 180 × A² Rightarrow 0.9 = 90 A² Rightarrow A² = 0.01 Rightarrow A = 0.1 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

A 7 kg block on a horizontal surface ( μ_k = 0.3 ) is pulled by a 4 kg mass over a pulley. A 14 N force opposes the 7 k

Given: A 7 kg block on a horizontal surface ( μ_k = 0.3 ) is pulled by a 4 kg mass over a pulley. A 14 N force opposes the 7 kg block. What is the acceleration? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: For 4 kg : 4g - T = 4a Rightarrow 40 - T = 4a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For 7 kg : T - f_k - 14 = 7a . Normal: N = mg = 7 × 10 = 70 N . Friction: f_k = 0.3 × 70 = 21 N . Net force: T - 21 - 14 = 7a Rightarrow T - 35 = 7a . Solve: 40 - T = 4a, T - 35 = 7a . Substitute: 40 - (7a + 35) = 4a Rightarrow 40 - 35 - 7a = 4a Rightarrow 5 = 11a . a = 5/11 approx 0.45 m/s² . 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 coil of 80 turns is placed in a magnetic field that increases from 0 to 0.03 Wb/m² in 0.2 s. If the coil area is 0.05

Given: A coil of 80 turns is placed in a magnetic field that increases from 0 to 0.03 Wb/m² in 0.2 s. If the coil area is 0.05 m², what is the induced emf? These values define the system as per NCERT data. Formula: Flux change: Δ Phi = B A = 0.03 × 0.05 = 0.0015 Wb. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon = N Δ Phi/Δ t = 80 × 0.0015/0.2 = 80 × 0.0075 = 0.6 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: 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 gas mixture contains 4 g of helium and 32 g of oxygen. What is the total pressure if the volume is 22.4 litres at 273

Given: A gas mixture contains 4 g of helium and 32 g of oxygen. What is the total pressure if the volume is 22.4 litres at 273 K? (R = 8.31 J mol^{-1 K^{-1) These values define the system as per NCERT data. Formula: μ_{He = 4/4 = 1 mol, μ_{O_2 = 32/32 = 1 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Total moles = 1 + 1 = 2 mol, V = 22.4 × 10⁻³ m³. P = μ R T/V = frac2 × 8.31 × 27322.4 × 10⁻³= 2.02 × 10⁵ Pa approx 2 atm . 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 wire of length 6 m and cross-sectional area 5 × 10⁻⁶ m² has a resistance of 12 Ω . What is the resistivity of t

Given: A wire of length 6 m and cross-sectional area 5 × 10⁻⁶ m² has a resistance of 12 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 = frac12 × 5 × 10⁻⁶⁶= 10 × 10⁻⁶= 1.0 × 10⁻⁵Ω 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 steel rod has a length of 1 m at 20° C . What will be its length at 120° C if the coefficient of linear expansion of

Given: A steel rod has a length of 1 m at 20° C . What will be its length at 120° C if the coefficient of linear expansion of steel is 1.2 × 10⁻⁵ K^{-1 ? These values define the system as per NCERT data. Formula: Given: L_0 = 1 m, T_1 = 20° C, T_2 = 120° C, α_l = 1.2 × 10⁻⁵ K^{-1. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Δ T = 120 - 20 = 100° C . Linear expansion: Δ L = L_0 α_l Δ T = 1 × 1.2 × 10⁻⁵ × 100 = 1.2 × 10⁻³ m . New length: L = L_0 + Δ L = 1 + 0.0012 = 1.0012 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 wire of length 0.6 m carrying 9 A is at 60° to a magnetic field of 0.4 T . What is the force on the wire?

Given: A wire of length 0.6 m carrying 9 A is at 60° to a magnetic field of 0.4 T . What is the force on the wire? These values define the system as per NCERT data. Formula: F = I l B sin θ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: F = 9 × 0.6 × 0.4 × sin 60° = 2.16 × 0.866 = 1.8706 approx 1.87 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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PHYSICS

A photon’s energy is 3.0 eV . What is its momentum? (Take c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J )

Given: A photon’s energy is 3.0 eV . What is its momentum? (Take c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J ) These values define the system as per NCERT data. Formula: E = 3.0 × 1.6 × 10⁻¹⁹= 4.8 × 10⁻¹⁹ J. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For a photon, p = E/c = frac4.8 × 10⁻¹⁹³ × 10⁸= 1.6 × 10⁻²⁷ kg 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: 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 dipole p = 6 × 10⁻⁹ C m is rotated from θ = 0° to 90° in a field E = 3 × 10⁵ N/C . What is the work done?

Given: A dipole p = 6 × 10⁻⁹ C m is rotated from θ = 0° to 90° in a field E = 3 × 10⁵ N/C . What is the work done? These values define the system as per NCERT data. Formula: Work done: W = p E (cos θ_0 - cos θ_1) = 6 × 10⁻⁹ × 3 × 10⁵ × (cos 0° - cos 90°). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: W = 6 × 10⁻⁹ × 3 × 10⁵ × (1 - 0) = 1.8 × 10⁻³ 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

An electron moves with a speed of 2 × 10⁶ m/s perpendicular to a magnetic field of 0.7 T . What is the radius of its

Given: An electron moves with a speed of 2 × 10⁶ m/s perpendicular to a magnetic field of 0.7 T . What is the radius of its path? (Mass = 9.1 × 10⁻³¹ kg, charge = 1.6 × 10⁻¹⁹ C ) These values define the system as per NCERT data. Formula: Radius r = mv/qB. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r = frac9.1 × 10⁻³¹ × 2 × 10⁶¹.6 × 10⁻¹⁹ × 0.7 = frac1.82 × 10⁻²⁴¹.12 × 10⁻¹⁹= 1.625 × 10⁻⁵ m = 1.625 × 10⁻³ cm . 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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