PHYSICS

PHYSICS

In an LCR circuit with R = 3 Ω, X_L = 8 Ω, X_C = 4 Ω, what is the power factor?

Given: In an LCR circuit with R = 3 Ω, X_L = 8 Ω, X_C = 4 Ω, what is the power factor? These values define the system as per NCERT data. Formula: Impedance: Z = sqrtR² + (X_L - X_C)² = sqrt3² + (8 - 4)² = sqrt9 + 16 = 5 Ω. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Power factor: cos phi = R/Z = 3/5 = 0.6 . 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 rectangular loop of 0.24 m × 0.38 m moves out of a 0.3 T field at 0.6 m/s along its shorter side. What is the emf?

Given: A rectangular loop of 0.24 m × 0.38 m moves out of a 0.3 T field at 0.6 m/s along its shorter side. What is the emf? These values define the system as per NCERT data. Formula: varepsilon = B l v, l = 0.38 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon = 0.3 × 0.38 × 0.6 = 0.0684 V approx 0.068 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 dipole with m = 0.35 A m² in B = 0.9 T at 30° has torque:

Given: A dipole with m = 0.35 A m² in B = 0.9 T at 30° has torque: These values define the system as per NCERT data. Formula: tau = m B sinθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: m = 0.35 A m², B = 0.9 T, θ = 30°, sin 30° = 0.5 . tau = 0.35 × 0.9 × 0.5 = 0.1575 N 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: 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 rod rotates at 15 rad/s in a 0.3 T field. If the length from the axis to the tip is 0.4 m, what is the emf induced?

Given: A rod rotates at 15 rad/s in a 0.3 T field. If the length from the axis to the tip is 0.4 m, what is the emf induced? These values define the system as per NCERT data. Formula: varepsilon = 1/2 B omega R². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon = 1/2 × 0.3 × 15 × (0.4)² = 0.36 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 spherical shell has a net flux of 1.13 × 10⁵ Nm²/C through it. What is the charge enclosed?

Given: A spherical shell has a net flux of 1.13 × 10⁵ Nm²/C through it. What is the charge enclosed? These values define the system as per NCERT data. Formula: phi = q/varepsilon_0. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: q = phi varepsilon_0 = 1.13 × 10⁵ × 8.854 × 10⁻¹²= 1.0 × 10⁻⁶ C = 1 μC . 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 rectangular loop of area 0.05 m² with 16 turns carries 2.2 A in a field of 0.7 T at 45° to the plane. What is the to

Given: A rectangular loop of area 0.05 m² with 16 turns carries 2.2 A in a field of 0.7 T at 45° 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 = 45° to plane means sin 45° with normal. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: tau = 16 × 2.2 × 0.05 × 0.7 × sin 45° = 1.232 × 0.707 = 0.871 approx 0.87 N 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

Two identical springs ( k = 75 N/m ) are attached to a 1.5 kg mass as in Fig. 13.14. What is the frequency?

Given: Two identical springs ( k = 75 N/m ) are attached to a 1.5 kg mass as in Fig. 13.14. What is the frequency? These values define the system as per NCERT data. Formula: Effective k_{eff = 2k = 2 × 75 = 150 N/m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: omega = sqrtfrack_{effm = sqrt150/1.5 = sqrt100 = 10 rad/s . v = omega/2π = 10/2 × 3.14 approx 1.59 Hz . 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: Oscillations and Waves, Topic: SHM, wave motion and superposition.

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PHYSICS

A 35 g bullet at 1200 m/s emerges from a block with 80% of its initial kinetic energy. What is its emergent speed?

Given: A 35 g bullet at 1200 m/s emerges from a block with 80% of its initial kinetic energy. What is its emergent speed? These values define the system as per NCERT data. Formula: Initial K = 1/2 × 0.035 × 1200² = 25200 J. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Final K = 0.8 × 25200 = 20160 J . 1/2 × 0.035 × v_f² = 20160 Rightarrow v_f² = 1152000 Rightarrow v_f = sqrt1152000 approx 1073.3 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 stone is dropped from a cliff 45 m high with an initial horizontal speed of 15 m/s. What is its speed on hitting the g

Given: A stone is dropped from a cliff 45 m high with an initial horizontal speed of 15 m/s. What is its speed on hitting the ground? (Take g = 10 m/s²) These values define the system as per NCERT data. Formula: Horizontal velocity: v_x = 15 m/s (constant). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Vertical velocity: v_y = √(2gh) = √(2 × 10 × 45) = √900 = 30 m/s. Speed v = √(v_x² + v_y²) = √(15² + 30²) = √(225 + 900) = √1125 ≈ 33.54 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: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A Wheatstone bridge has R_1 = 14 Ω, R_2 = 28 Ω, R_3 = 10 Ω . What is R_4 for balance?

Given: A Wheatstone bridge has R_1 = 14 Ω, R_2 = 28 Ω, R_3 = 10 Ω . What is R_4 for balance? These values define the system as per NCERT data. Formula: Balance condition: R_1/R_2 = R_3/R_4. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substitute: 14/28 = 10/R_4 . Solve: 0.5 = 10/R_4 Rightarrow R_4 = 10/0.5 = 20 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: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A glass slab ( n = 1.5 ) of thickness 7.5 cm is placed over a point. What is the apparent shift?

Given: A glass slab ( n = 1.5 ) of thickness 7.5 cm is placed over a point. What is the apparent shift? These values define the system as per NCERT data. Formula: Shift = t ( 1 - 1/n ). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: t = 7.5 cm, n = 1.5 . Shift = 7.5 ( 1 - 1/1.5 ) = 7.5 ( 1 - 2/3 ) = 7.5 × 1/3 = 2.5 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: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A circuit has a 6 V battery with 1 Ω internal resistance and two resistors 2 Ω and 3 Ω in parallel. What is the total

Given: A circuit has a 6 V battery with 1 Ω internal resistance and two resistors 2 Ω and 3 Ω in parallel. What is the total current? These values define the system as per NCERT data. Formula: Parallel resistance: 1/R_p = 1/2 + 1/3 = 3 + 2/6 = 5/6 Rightarrow R_p = 6/5 = 1.2 Ω. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Total resistance: R_{total = 1 + 1.2 = 2.2 Ω . Current: I = fracvarepsilonR_{total = 6/2.2 approx 2.73 A . 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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