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PHYSICS

A set of physics questions organised by topic, spanning mechanics, thermodynamics, waves and sound, optics, electricity and magnetism, and modern physics. Numerical problems sit alongside conceptual questions so you can practise setting up a solution as well as recalling the underlying principle.

45 questions

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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