PHYSICS

PHYSICS

The magnetic potential energy of a dipole with m = 0.9 A m² in a field B = 0.2 T at 90° is:

Given: The magnetic potential energy of a dipole with m = 0.9 A m² in a field B = 0.2 T at 90° is: These values define the system as per NCERT data. Formula: U_m = -m B cosθ. This is standard NCERT relation. Substitution & Calculation: Given: m = 0.9 A m², B = 0.2 T, θ = 90°, cos 90° = 0 . Substitute: U_m = -0.9 × 0.2 × 0 = 0 J . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Magnetism and Matter, Topic: Magnetic potential energy U = -m·B, dipole in magnetic field at 180°. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

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PHYSICS

A 6 μF capacitor charged to 150 V is connected to an uncharged 9 μF capacitor. What is the energy lost?

Given: A 6 μF capacitor charged to 150 V is connected to an uncharged 9 μF capacitor. What is the energy lost? These values define the system as per NCERT data. Formula: Initial energy: U_i = 1/2 × 6 × 10⁻⁶ × (150)² = 0.0675 J. This is standard NCERT relation. Substitution & Calculation: Charge: Q = 6 × 10⁻⁶ × 150 = 9 × 10⁻⁴C . Total C = 6 + 9 = 15 μF, V = frac9 × 10⁻⁴¹⁵ × 10⁻⁶= 60 V . Final energy: U_f = 1/2 × 15 × 10⁻⁶ × (60)² = 0.027 J . Loss: U_i - U_f = 0.0675 - 0.027 = 0.0405 J . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A rod of length 0.8 m moves at 1.5 m/s in a 0.4 T field perpendicular to its length. What is the induced emf?

Given: A rod of length 0.8 m moves at 1.5 m/s in a 0.4 T field perpendicular to its length. What is the induced emf? These values define the system as per NCERT data. Formula: varepsilon = B l v = 0.4 × 0.8 × 1.5 = 0.48 V .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A 3.5 kg mass falls from 7 m onto a spring ( k = 1200 N/m ). What is the maximum compression? (Take g = 10 m/s² )

Given: A 3.5 kg mass falls from 7 m onto a spring ( k = 1200 N/m ). What is the maximum compression? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Potential energy mgh = 3.5 × 10 × 7 = 245 J. This is standard NCERT relation. Substitution & Calculation: Spring energy 1/2 k x_m² = 245 Rightarrow 600 x_m² = 245 Rightarrow x_m = sqrt0.4083 approx 0.639 m . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

What is the magnetic field at a point on the equatorial line of a bar magnet with magnetic moment 2 A m² at a distance o

Given: What is the magnetic field at a point on the equatorial line of a bar magnet with magnetic moment 2 A m² at a distance of 10 cm from its nter? (Take μ_0 = 4π × 10⁻⁷T m A^{-1 ). These values define the system as per NCERT data. Formula: The magnetic field on the equatorial line is B = μ_0/4π m/r³. This is standard NCERT relation. Substitution & Calculation: Given: m = 2 A m², r = 0.1 m, μ_0/4π = 10⁻⁷T m A^{-1 . Substitute: B = 10⁻⁷ × 2/(0.1)³ = 10⁻⁷ × 2/0.001 = 2 × 10⁻⁴T . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Magnetism and Matter, Topic: Bar magnet cut transversely, magnetic moment halves, m' = m/2. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

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PHYSICS

What is the nuclear density of a nucleus with mass 2.33 × 10⁻²⁷kg and radius 1.7 × 10⁻¹⁵m ? (Use π = 3.14 )

Given: What is the nuclear density of a nucleus with mass 2.33 × 10⁻²⁷kg and radius 1.7 × 10⁻¹⁵m ? (Use π = 3.14 ) These values define the system as per NCERT data. Formula: Density = fracmassvolume, Volume = 4/3 π R³. This is standard NCERT relation. Substitution & Calculation: R³ = (1.7 × 10⁻¹⁵)³ = 4.913 × 10⁻⁴⁵m³ . Volume = 4/3 × 3.14 × 4.913 × 10⁻⁴⁵approx 2.06 × 10⁻⁴⁴m³ . Density = frac2.33 × 10⁻²⁷².06 × 10⁻⁴⁴approx 1.13 × 10¹⁷kg/m³ . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A solenoid has 600 turns per meter and carries a current of 3 A . What is the magnetic field inside it? ( μ_0 = 4 π × 10

Given: A solenoid has 600 turns per meter and carries a current of 3 A . What is the magnetic field inside it? ( μ_0 = 4 π × 10⁻⁷T m/A ) These values define the system as per NCERT data. Formula: Magnetic field B = μ_0 n I. This is standard NCERT relation. Substitution & Calculation: B = 4 π × 10⁻⁷ × 600 × 3 = 7.2 π × 10⁻⁴approx 2.26 × 10⁻³T . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A wire of length 0.8 m carrying 3 A makes an angle of 30° with a magnetic field of 0.6 T . What is the force on the wire

Given: A wire of length 0.8 m carrying 3 A makes an angle of 30° with a magnetic field of 0.6 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 standard NCERT relation. Substitution & Calculation: F = 3 × 0.8 × 0.6 × sin 30° = 2.4 × 0.6 × 0.5 = 0.72 N . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A copper rod of length 2.0 m and cross-sectional area 2.5 × 10⁻⁶m² is subjected to a tensile force of 500 N . If the You

Given: A copper rod of length 2.0 m and cross-sectional area 2.5 × 10⁻⁶m² is subjected to a tensile force of 500 N . If the Young's modulus of copper is 1.1 × 10¹¹N/m², what is the strain produced? These values define the system as per NCERT data. Formula: Stress: Stress = F/A = frac5002.5 × 10⁻⁶= 2 × 10⁸N/m². This is standard NCERT relation. Substitution & Calculation: Young's modulus: Y = fracStressStrain . Strain: Strain = fracStressY = frac2 × 10⁸¹.1 × 10¹¹approx 1.82 × 10⁻³. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A 40 V battery with negligible internal resistance is connected to a cubical network of 12 resistors, each 4 Ω . What is

Given: A 40 V battery with negligible internal resistance is connected to a cubical network of 12 resistors, each 4 Ω . What is the total current? These values define the system as per NCERT data. Formula: Equivalent resistance: R_{eq = 5/6 R = 5/6 × 4 = 20/6 approx 3.33 Ω. This is standard NCERT relation. Substitution & Calculation: Total current: I = fracVR_{eq = frac4020/6 = 40 × 6/20 = 12 A . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A force F = 7 i + 3 j N acts at r = -2 i + 1 j m . What is the magnitude of the torque about the origin?

Given: A force F = 7 i + 3 j N acts at r = -2 i + 1 j m . What is the magnitude of the torque about the origin? These values define the system as per NCERT data. Formula: tau = r × F = beginvmatrix i & j & k -2 & 1 & 0 7 & 3 & 0 endvmatrix = k ((-2) × 3 - 1 × 7) = k (-6 - 7) = -13 k Nm. This is standard NCERT relation. Substitution & Calculation: Magnitude = 13 Nm . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A 1350 kg car turns on a banked road ( θ = 25° ) with radius 50 m at the optimum speed to avoid friction. What is the sp

Given: A 1350 kg car turns on a banked road ( θ = 25° ) with radius 50 m at the optimum speed to avoid friction. What is the speed? (Take g = 10 m/s², tan 25° approx 0.466 ) These values define the system as per NCERT data. Formula: Optimum speed: v_0 = sqrtrg tanθ (no friction contribution). This is standard NCERT relation. Substitution & Calculation: Substitute: r = 50 m, g = 10 m/s², tan 25° = 0.466 . v_0² = 50 × 10 × 0.466 = 500 × 0.466 = 233 . v_0 = sqrt233 approx 15.26 m/s . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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