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PHYSICS

Physics deals with matter, energy and the laws that connect them. This category collects questions across areas such as mechanics, waves and sound, heat, electricity and magnetism, and modern physics. Expect a mix of conceptual questions and numerical problems that test how well you can apply a formula, not just remember it.

45 questions

PHYSICS

A uniform square plate of side 6 m and mass 9 kg has one corner at (2, 2) along the x- and y-axes. What is the position

Given: A uniform square plate of side 6 m and mass 9 kg has one corner at (2, 2) along the x- and y-axes. What is the position of its nter of mass? These values define the system as per NCERT data. Formula: CM: X = 2 + 6/2 = 5, Y = 2 + 6/2 = 5. This is standard NCERT relation. Substitution & Calculation: For a uniform square, the nter of mass is at the ntroid. Vertices: (2, 2), (8, 2), (2, 8), (8, 8) . . Position: (5, 5) 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

A 7 μF capacitor is charged to 400 V . What is the energy stored in it?

Given: A 7 μF capacitor is charged to 400 V . What is the energy stored in it? These values define the system as per NCERT data. Formula: U = 1/2 C V² = 1/2 × 7 × 10⁻⁶ × (400)². This is standard NCERT relation. Substitution & Calculation: U = 1/2 × 7 × 10⁻⁶ × 160000 = 0.56 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 8.5 kg block on a horizontal surface ( μ_k = 0.25 ) is pulled by a 4.5 kg mass over a pulley. A 15 N force opposes the

Given: A 8.5 kg block on a horizontal surface ( μ_k = 0.25 ) is pulled by a 4.5 kg mass over a pulley. A 15 N force opposes the 8.5 kg block. What is the acceleration? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: For 4.5 kg : 4.5g - T = 4.5a Rightarrow 45 - T = 4.5a. This is standard NCERT relation. Substitution & Calculation: For 8.5 kg : T - f_k - 15 = 8.5a . Normal: N = mg = 8.5 × 10 = 85 N . Friction: f_k = 0.25 × 85 = 21.25 N . Net force: T - 21.25 - 15 = 8.5a Rightarrow T - 36.25 = 8.5a . Solve: 45 - T = 4.5a, T - 36.25 = 8.5a . Substitute: 45 - (8.5a + 36.25) = 4.5a Rightarrow 45 - 36.25 - 8.5a = 4.5a Rightarrow 8.75 = 13a . a = 8.75/13 approx 0.67 m/s² . 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 wheel with 10 spokes of 0.7 m each rotates at 60 rpm in a 0.5 T field. What is the induced emf?

Given: A wheel with 10 spokes of 0.7 m each rotates at 60 rpm in a 0.5 T field. What is the induced emf? These values define the system as per NCERT data. Formula: omega = 2π × 60/60 = 2π rad/s. This is standard NCERT relation. Substitution & Calculation: varepsilon = 1/2 B omega R² = 1/2 × 0.5 × 2π × (0.7)² = 0.7697 V approx 0.77 V . 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 wheel with 6 spokes of 0.6 m each rotates at 45 rpm in a 0.5 T field. What is the induced emf?

Given: A wheel with 6 spokes of 0.6 m each rotates at 45 rpm in a 0.5 T field. What is the induced emf? These values define the system as per NCERT data. Formula: omega = 2π × 45/60 = 1.5π rad/s. This is standard NCERT relation. Substitution & Calculation: varepsilon = 1/2 B omega R² = 1/2 × 0.5 × 1.5π × (0.6)² = 0.8478 V approx 0.85 V . 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 with 1200 turns per meter and current 2.5 A has a core with μ_r = 200 . What is B inside? (Take μ_0 = 4π × 10

Given: A solenoid with 1200 turns per meter and current 2.5 A has a core with μ_r = 200 . What is B inside? (Take μ_0 = 4π × 10⁻⁷T m A^{-1 ). These values define the system as per NCERT data. Formula: B = μ_0 μ_r n I. This is standard NCERT relation. Substitution & Calculation: Given: n = 1200 m^{-1, I = 2.5 A, μ_r = 200, μ_0 = 4π × 10⁻⁷. B = 4π × 10⁻⁷ × 200 × 1200 × 2.5 = 0.7536 T approx 0.75 T . 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 rectangular loop of area 0.06 m² with 15 turns carries 2.5 A in a field of 0.8 T at 60° to the plane. What is the torq

Given: A rectangular loop of area 0.06 m² with 15 turns carries 2.5 A in a field of 0.8 T at 60° to the plane. What is the torque? These values define the system as per NCERT data. Formula: tau = N I A B sin θ, where θ = 60° to plane means sin 30° with normal. This is standard NCERT relation. Substitution & Calculation: tau = 15 × 2.5 × 0.06 × 0.8 × sin 60° = 1.8 × 0.866 = 1.5588 approx 1.56 N 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

A particle moves with an initial velocity of 4 i m/s and an acceleration of (2 i - 3 j) m/s² . What is its speed after 3

Given: A particle moves with an initial velocity of 4 i m/s and an acceleration of (2 i - 3 j) m/s² . What is its speed after 3 s ? These values define the system as per NCERT data. Formula: Velocity v = v_0 + a t. This is standard NCERT relation. Substitution & Calculation: Given: v_0 = 4 i, a = 2 i - 3 j, t = 3 s . v = 4 i + (2 i - 3 j) × 3 = 4 i + 6 i - 9 j = 10 i - 9 j m/s . Speed v = sqrt10²+ (-9)² = sqrt100 + 81 = sqrt181 approx 13.45 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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PHYSICS

A 55 mH inductor is connected to a 230 V, 50 Hz AC source. What is the rms current?

Given: A 55 mH inductor is connected to a 230 V, 50 Hz AC source. What is the rms current? These values define the system as per NCERT data. Formula: X_L = omega L, omega = 2π × 50 = 314 rad/s. This is standard NCERT relation. Substitution & Calculation: L = 55 × 10⁻³H . X_L = 314 × 0.055 = 17.27 Ω . RMS current: I = V/X_L = 230/17.27 approx 13.32 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 vehicle accelerates uniformly from rest at 1.2 m/s² for 10 s . What is the final velocity?

Given: A vehicle accelerates uniformly from rest at 1.2 m/s² for 10 s . What is the final velocity? These values define the system as per NCERT data. Formula: Use the equation v = v_0 + a t. This is standard NCERT relation. Substitution & Calculation: Here, initial velocity v_0 = 0, acceleration a = 1.2 m/s², time t = 10 s . Substitute: v = 0 + 1.2 · 10 = 12 m/s . The final velocity is 12 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: Motion in a Straight Line, Topic: Kinematic equation v = v₀ + at, uniform acceleration and final velocity calculation. 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 bus starts from rest and moves with a uniform acceleration of 1.5 m/s² for 8 s . What is the final velocity of the bus

Given: A bus starts from rest and moves with a uniform acceleration of 1.5 m/s² for 8 s . What is the final velocity of the bus? These values define the system as per NCERT data. Formula: Use the equation v = v_0 + a t. This is standard NCERT relation. Substitution & Calculation: Here, initial velocity v_0 = 0, acceleration a = 1.5 m/s², time t = 8 s . Substitute: v = 0 + 1.5 · 8 = 12 m/s . The final velocity is 12 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: Motion in a Straight Line, Topic: Kinematic equation v = v₀ + at, uniform acceleration and final velocity calculation. 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 nucleus has a mass defect of 0.136 u . What is its binding energy in MeV? (Given 1 u = 931.5 MeV/c² )

Given: A nucleus has a mass defect of 0.136 u . What is its binding energy in MeV? (Given 1 u = 931.5 MeV/c² ) These values define the system as per NCERT data. Formula: Binding energy = Δ M · c². This is standard NCERT relation. Substitution & Calculation: Δ M = 0.136 u . Energy = 0.136 × 931.5 approx 126.7 MeV . 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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