PHYSICS

PHYSICS

A solenoid with 1000 turns per meter and current 2 A has a core with relative permeability μ_r = 200 . What is the magn

Given: A solenoid with 1000 turns per meter and current 2 A has a core with relative permeability μ_r = 200 . What is the magnetic field B inside? These values define the system as per NCERT data. Formula: Magnetic field B = μ_0 μ_r H, where H = n I. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: n = 1000 m^{-1, I = 2 A, μ_r = 200, μ_0 = 4π × 10⁻⁷ T m A^{-1 . First, H = 1000 × 2 = 2000 A m^{-1 . Then, B = 4π × 10⁻⁷ × 200 × 2000 = 0.5024 T approx 0.5 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 bar magnet with original m = 1.8 A m² is cut transversely into two equal parts. What is m of each part?

Given: A bar magnet with original m = 1.8 A m² is cut transversely into two equal parts. What is m of each part? These values define the system as per NCERT data. Formula: Given: m = 1.8 A m². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: When cut transversely, each part has half the original magnetic moment. . Each part: m' = 1.8/2 = 0.9 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 paramagnetic material with chi = 5 × 10⁻⁴ in H = 4000 A m^{-1 has magnetization M :

Given: A paramagnetic material with chi = 5 × 10⁻⁴ in H = 4000 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 = 5 × 10⁻⁴, H = 4000 A m^{-1 . M = 5 × 10⁻⁴ × 4000 = 2.0 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 square loop of side 30 cm rotates at 12 rad/s in a 0.1 T field. What is the maximum emf induced?

Given: A square loop of side 30 cm rotates at 12 rad/s in a 0.1 T field. What is the maximum emf induced? These values define the system as per NCERT data. Formula: A = (0.3)² = 0.09 m². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon_0 = N B A omega = 1 × 0.1 × 0.09 × 12 = 0.108 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 bar magnet with m = 1.4 A m² produces a field at 0.4 m on its equatorial line. What is B ? (Take μ_0 = 4Ï€ × 10⁻â

Given: A bar magnet with m = 1.4 A m² produces a field at 0.4 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 = 1.4 A m², r = 0.4 m, μ_0/4π = 10⁻⁷. B = 10⁻⁷ × 1.4/(0.4)³ = 10⁻⁷ × 1.4/0.064 approx 2.1875 × 10⁻⁶ T approx 2.19 × 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 bar magnet produces a field of 2 × 10⁻⁶ T at 0.5 m on its equatorial line. What is its magnetic moment? (Take μ_

Given: A bar magnet produces a field of 2 × 10⁻⁶ T at 0.5 m on its equatorial line. What is its magnetic moment? (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, so m = fracB r³μ_0/4π. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: B = 2 × 10⁻⁶ T, r = 0.5 m, μ_0/4π = 10⁻⁷. m = frac2 × 10⁻⁶ × (0.5)³¹⁰⁻⁷= frac2 × 10⁻⁶ × 0.12510⁻⁷= 2.5 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/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 1 kg mass is moved from Earth’s surface to a height of 3.2 × 10⁶ m . What is the change in gravitational potentia

Given: A 1 kg mass is moved from Earth’s surface to a height of 3.2 × 10⁶ m . What is the change in gravitational potential energy? ( M_E = 6 × 10²⁴ kg, R_E = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m²/kg² ) These values define the system as per NCERT data. Formula: Δ V = -G M_E m (1/r_2 - 1/r_1). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r_1 = R_E = 6.4 × 10⁶ m, r_2 = R_E + h = 9.6 × 10⁶ m . Δ V = -6.67 × 10⁻¹¹ × 6 × 10²⁴ × 1 (1/9.6 × 10⁶- 1/6.4 × 10⁶) . Δ V = -4.002 × 10¹⁴(1.0417 × 10⁻⁷- 1.5625 × 10⁻⁷) . Δ V = -4.002 × 10¹⁴ × (-5.208 × 10⁻⁸) approx 2.08 × 10⁷ 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

An aluminium block of dimensions 0.4 m × 0.2 m × 0.05 m is subjected to a shearing force of 2 × 10⁴ N . If the shea

Given: An aluminium block of dimensions 0.4 m × 0.2 m × 0.05 m is subjected to a shearing force of 2 × 10⁴ N . If the shear modulus of aluminium is 2.5 × 10¹⁰ N/m², what is the displacement of the top face? These values define the system as per NCERT data. Formula: Shear modulus: G = F / A/Δ x / L. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Rearrange: Δ x = F L/A G . Area: A = 0.4 × 0.2 = 0.08 m², L = 0.05 m . Substitute: Δ x = frac2 × 10⁴ × 0.050.08 × 2.5 × 10¹⁰= 1000/2 × 10⁹= 5 × 10⁻⁷ 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

Which electromagnetic waves are produced by bombarding a metal target with high-energy electrons?

The document mentions that X-rays are generated by bombarding a metal target with high-energy electrons. This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

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 10 μF capacitor is connected to a 220 V, 50 Hz source. What is the capacitive reactance?

Given: A 10 μF capacitor is connected to a 220 V, 50 Hz source. What is the capacitive reactance? These values define the system as per NCERT data. Formula: X_C = 1/omega C, omega = 2π × 50 = 314 rad/s. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: C = 10 × 10⁻⁶ F . X_C = frac1314 × 10 × 10⁻⁶ approx 318.5 Ω . 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 charges 4 × 10⁻⁸ C and -2 × 10⁻⁸ C are 20 cm apart. At what distance from the positive charge on the line

Given: Two charges 4 × 10⁻⁸ C and -2 × 10⁻⁸ C are 20 cm apart. At what distance from the positive charge on the line joining them is the potential zero? (Take 1/4 π varepsilon_0 = 9 × 10⁹ Nm² C^{-2 ). These values define the system as per NCERT data. Formula: Total potential: V = 1/4 π varepsilon_0 ( frac4 × 10⁻⁸ x + frac-2 × 10⁻⁸⁰.2 - x ) = 0. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Let distance from 4 × 10⁻⁸ C be x m, then distance from -2 × 10⁻⁸ C is 0.2 - x . . Simplify: 9 × 10⁹( frac4 × 10⁻⁸ x - frac2 × 10⁻⁸⁰.2 - x ) = 0 . 4/x = 2/0.2 - x Rightarrow 4 (0.2 - x) = 2x Rightarrow 0.8 - 4x = 2x Rightarrow 0.8 = 6x Rightarrow x = 0.8/6 = 0.133 m = 13.3 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: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

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PHYSICS

A torque of 15 Nm is applied to a disk with moment of inertia 5 kg m² . What is its angular acceleration?

Given: A torque of 15 Nm is applied to a disk with moment of inertia 5 kg m² . What is its angular acceleration? These values define the system as per NCERT data. Formula: tau = I α. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: tau = 15 Nm, I = 5 kg m² . α = tau/I = 15/5 = 3 rad/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: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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