PHYSICS

PHYSICS

A neutron ( 1 u ) moving at 2 × 10⁶ m/s collides elastically with a carbon nucleus ( 12 u ). What fraction of its kin

Given: A neutron ( 1 u ) moving at 2 × 10⁶ m/s collides elastically with a carbon nucleus ( 12 u ). What fraction of its kinetic energy is retained? These values define the system as per NCERT data. Formula: Fraction retained f_1 = ( m_1 - m_2/m_1 + m_2 )² = ( 1 - 12/1 + 12 )² = ( -11/13 )² = 121/169 approx 0.716 .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10⁻⁵, 236 J kg⁻¹ K⁻¹, CH₃CH₂NH₂ etc. into the formula and simplifying step by step. 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

In a full-wave rectifier, the number of diodes typically used with a ntre-tap transformer is:

A full-wave rectifier with a ntre-tap transformer uses two diodes, each conducting during alternate half-cycles, to rectify both positive and negative halves of the AC input.

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 stone is thrown upwards with a speed of 18 m/s . What is its velocity after 2 s ? (Take g = 10 m/s² )

Given: A stone is thrown upwards with a speed of 18 m/s . What is its velocity after 2 s ? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Use v = v_0 + a t. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Here, v_0 = 18 m/s, a = -10 m/s², t = 2 s . Substitute: v = 18 - 10 · 2 = 18 - 20 = -2 m/s . The velocity is -2 m/s (downward). 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 solenoid of 900 turns/m and area 0.015 m² has μ_r = 2 . What is its self-inductance? ( μ_0 = 4π × 10⁻⁷ H/m )

Given: A solenoid of 900 turns/m and area 0.015 m² has μ_r = 2 . What is its self-inductance? ( μ_0 = 4π × 10⁻⁷ H/m ) These values define the system as per NCERT data. Formula: L = μ_r μ_0 n² A l, assume l = 1 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: L = 2 × 4π × 10⁻⁷ × (900)² × 0.015 × 1 = 0.0305 H approx 0.03 H . 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

An electron moves with a speed of 3.0 × 10⁶ m/s . What is its de Broglie wavelength? (Take h = 6.63 × 10⁻³⁴ J s

Given: An electron moves with a speed of 3.0 × 10⁶ m/s . What is its de Broglie wavelength? (Take h = 6.63 × 10⁻³⁴ J s, m_e = 9.11 × 10⁻³¹ kg ) These values define the system as per NCERT data. Formula: Momentum p = m v = 9.11 × 10⁻³¹ × 3.0 × 10⁶= 2.733 × 10⁻²⁴ kg m/s. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: lambda = h/p = frac6.63 × 10⁻³⁴².733 × 10⁻²⁴ approx 2.425 × 10⁻¹⁰ m = 0.2425 nm . 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 bar magnet of magnetic moment 0.5 A m² is placed at a distance of 20 cm from its nter along its axis. Calculate the m

Given: A bar magnet of magnetic moment 0.5 A m² is placed at a distance of 20 cm from its nter along its axis. Calculate the magnetic field B at that point. (Take μ_0 = 4π × 10⁻⁷ T m A^{-1 ). These values define the system as per NCERT data. Formula: The magnetic field along the axis of a bar magnet is given by B = μ_0/4π 2m/r³. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: m = 0.5 A m², r = 20 cm = 0.2 m, μ_0/4π = 10⁻⁷ T m A^{-1 . Substitute: B = 10⁻⁷ × 2 × 0.5/(0.2)³ = 10⁻⁷ × 1/0.008 = 10⁻⁷ × 125 = 1.25 × 10⁻⁵ 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 circular loop of radius 0.08 m with 60 turns carries a current of 0.75 A . What is the magnetic field at the nter? ( Î

Given: A circular loop of radius 0.08 m with 60 turns carries a current of 0.75 A . What is the magnetic field at the nter? ( μ_0 = 4 π × 10⁻⁷ T m/A ) These values define the system as per NCERT data. Formula: Magnetic field B = μ_0 N I/2 R. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: B = frac4 π × 10⁻⁷ × 60 × 0.752 × 0.08 = frac18 π × 10⁻⁶⁰.16 = 1.125 π × 10⁻⁴ approx 3.53 × 10⁻⁴ 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 5 kg particle moves with velocity v = 4 i m/s at r = -3 j m . What is the magnitude of its angular momentum about the

Given: A 5 kg particle moves with velocity v = 4 i m/s at r = -3 j m . What is the magnitude of its angular momentum about the origin? These values define the system as per NCERT data. Formula: L = r × p = beginvmatrix i & j & k 0 & -3 & 0 4 & 0 & 0 endvmatrix = k (0 × 0 - (-3) × 4) = 12 k kg m²/s. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Magnitude = 12 kg m²/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

How much heat is required to raise the temperature of 0.5 moles of neon by 25 K at constant volume? (R = 8.31 J mol^{-1

Given: How much heat is required to raise the temperature of 0.5 moles of neon by 25 K at constant volume? (R = 8.31 J mol^{-1 K^{-1) These values define the system as per NCERT data. Formula: For monatomic gas, C_v = 3/2 R. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Q = μ C_v Δ T = 0.5 × 3/2 × 8.31 × 25 = 155.8125 J approx 155.8 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: Dual Nature, Atoms, Nuclei and Electronic Devices, Topic: Atomic structure, nuclear binding and semiconductor physics.

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PHYSICS

A 1250 kg car turns on a banked road ( θ = 15°, μ_s = 0.3 ) with radius 45 m . What is the maximum speed without slip

Given: A 1250 kg car turns on a banked road ( θ = 15°, μ_s = 0.3 ) with radius 45 m . What is the maximum speed without slipping? (Take g = 10 m/s², tan 15° approx 0.268 ) These values define the system as per NCERT data. Formula: Maximum speed: v_{max = sqrtrg μ_s + tanθ/1 - μ_s tanθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Numerator: μ_s + tanθ = 0.3 + 0.268 = 0.568 . Denominator: 1 - 0.3 × 0.268 = 1 - 0.0804 = 0.9196 . v_{max² = 45 × 10 × 0.568/0.9196 approx 450 × 0.6175 approx 277.875 . v_{max = sqrt277.875 approx 16.67 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 torque of 12 Nm is applied to a disk with moment of inertia 4 kg m² . What is its angular acceleration?

Given: A torque of 12 Nm is applied to a disk with moment of inertia 4 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 = 12 Nm, I = 4 kg m² . α = tau/I = 12/4 = 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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PHYSICS

A capillary tube of radius 0.65 mm is dipped in water ( S = 0.0727 N/m, rho = 1000 kg/m³, cos θ = 1 ). What is the cap

Given: A capillary tube of radius 0.65 mm is dipped in water ( S = 0.0727 N/m, rho = 1000 kg/m³, cos θ = 1 ). What is the capillary rise? (Take g = 9.8 m/s² ) These values define the system as per NCERT data. Formula: h = 2 S cos θ/rho g a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: S = 0.0727 N/m, rho = 1000 kg/m³, g = 9.8 m/s², a = 0.65 × 10⁻³ m . h = frac2 × 0.0727 × 11000 × 9.8 × 0.65 × 10⁻³= 0.02285 m = 2.285 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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