PHYSICS

PHYSICS

A stationary wave on a string fixed at both ends has a frequency of 150 Hz and a wave speed of 45 m/s. What is the wavel

Given: A stationary wave on a string fixed at both ends has a frequency of 150 Hz and a wave speed of 45 m/s. What is the wavelength? These values define the system as per NCERT data. Formula: Wavelength: lambda = v/v = 45/150 = 0.3 m .. 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

The work function of a metal is 3.0 eV . Light of frequency 8.0 × 10¹⁴Hz is incident on it. What is the stopping potenti

Given: The work function of a metal is 3.0 eV . Light of frequency 8.0 × 10¹⁴Hz is incident on it. What is the stopping potential? (Take h = 6.63 × 10⁻³⁴J s, e = 1.6 × 10⁻¹⁹C ) These values define the system as per NCERT data. Formula: E = h v = 6.63 × 10⁻³⁴ × 8.0 × 10¹⁴= 5.304 × 10⁻¹⁹J. This is standard NCERT relation. Substitution & Calculation: E = frac5.304 × 10⁻¹⁹¹.6 × 10⁻¹⁹approx 3.315 eV . K_{max = E - phi_0 = 3.315 - 3.0 = 0.315 eV . V_0 = fracK_{maxe = 0.315 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: Dual Nature of Radiation and Matter, Topic: Photoelectric effect, stopping potential, Kmax = eV₀, work function and photon energy. 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 spring-mass system has m = 0.6 kg, k = 240 N/m . If displaced by 7 cm, what is the total energy?

Given: A spring-mass system has m = 0.6 kg, k = 240 N/m . If displaced by 7 cm, what is the total energy? These values define the system as per NCERT data. Formula: Total energy: E = 1/2 k A². This is standard NCERT relation. Substitution & Calculation: A = 0.07 m, k = 240 N/m . E = 0.5 × 240 × (0.07)² = 0.5 × 240 × 0.0049 = 0.588 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: Gravitation, Topic: Total energy of satellite E = -GMm/2r, orbital energy at 5R_E and negative sign. 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 mercury barometer shows a height of 75 cm at sea level. What is the atmospheric pressure? ( rho = 13.6 × 10³kg/m³, g =

Given: A mercury barometer shows a height of 75 cm at sea level. What is the atmospheric pressure? ( rho = 13.6 × 10³kg/m³, g = 10 m/s² ) These values define the system as per NCERT data. Formula: P_a = rho g h. This is standard NCERT relation. Substitution & Calculation: rho = 13.6 × 10³kg/m³, g = 10 m/s², h = 0.75 m . P_a = 13.6 × 10³ × 10 × 0.75 = 1.02 × 10⁵Pa . 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 neutron ( 1 u ) at 6 × 10⁶m/s collides elastically with a deuterium ( 2 u ). What fraction of its kinetic energy is tr

Given: A neutron ( 1 u ) at 6 × 10⁶m/s collides elastically with a deuterium ( 2 u ). What fraction of its kinetic energy is transferred? These values define the system as per NCERT data. Formula: Fraction transferred f_2 = 4 m_1 m_2/(m_1 + m_2)² = 4 × 1 × 2/(1 + 2)² = 8/9 approx 0.889 .. 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

In a purely inductive AC circuit, the current lags the voltage by what phase angle?

For a pure inductor, i = i_m sin (omega t - π/2), while v = v_m sin omega t . Phase difference: phi = -π/2, meaning current lags voltage by π/2 or 90° .

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Wave Optics, Topic: Interference, phase difference φ = (2π/λ)Δ, path difference 7λ/4 and double-slit experiment. 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 bar magnet with original m = 2.8 A m² is cut transversely into two equal parts. What is m of each part?

Given: A bar magnet with original m = 2.8 A m² is cut transversely into two equal parts. What is m of each part? These values define the system as per NCERT data. Formula: Given: m = 2.8 A m². This is standard NCERT relation. Substitution & Calculation: When cut transversely, each part has half the original magnetic moment. . Each part: m' = 2.8/2 = 1.4 A m² . 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

An aluminium block of dimensions 0.5 m × 0.3 m × 0.1 m is subjected to a shearing force of 7 × 10⁴N . If the shear modul

Given: An aluminium block of dimensions 0.5 m × 0.3 m × 0.1 m is subjected to a shearing force of 7 × 10⁴N . If the shear modulus of aluminium is 2.5 × 10¹⁰N/m², what is the displacement of the top face? These values define the system as per NCERT data. Formula: Shear modulus: G = F / A/Δ x / L. This is standard NCERT relation. Substitution & Calculation: Rearrange: Δ x = F L/A G . Area: A = 0.5 × 0.3 = 0.15 m², L = 0.1 m . Substitute: Δ x = frac7 × 10⁴ × 0.10.15 × 2.5 × 10¹⁰= 7000/3.75 × 10⁹approx 1.87 × 10⁻⁶m . 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 4 kg mass rotates in a circle of radius 0.25 m with a linear speed of 2 m/s . What is its angular momentum about the n

Given: A 4 kg mass rotates in a circle of radius 0.25 m with a linear speed of 2 m/s . What is its angular momentum about the nter? These values define the system as per NCERT data. Formula: L = m v r. This is standard NCERT relation. Substitution & Calculation: m = 4 kg, v = 2 m/s, r = 0.25 m . L = 4 × 2 × 0.25 = 2 kg m²/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: System of Particles and Rotational Motion, Topic: Angular momentum L = m v r, rotating mass and moment of momentum. 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 energy equivalent of a neutron with mass 1.6749 × 10⁻²⁷kg in Joules? (Given c = 3 × 10⁸m/s )

Given: What is the energy equivalent of a neutron with mass 1.6749 × 10⁻²⁷kg in Joules? (Given c = 3 × 10⁸m/s ) These values define the system as per NCERT data. Formula: E = m c². This is standard NCERT relation. Substitution & Calculation: m = 1.6749 × 10⁻²⁷kg, c² = (3 × 10⁸)² = 9 × 10¹⁶m²/s² . E = 1.6749 × 10⁻²⁷ × 9 × 10¹⁶approx 1.507 × 10⁻¹⁰J . 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 thin spherical shell of radius 24 cm has a charge of 16 μC . What is the electric field at a point 30 cm from the nter

Given: A thin spherical shell of radius 24 cm has a charge of 16 μC . What is the electric field at a point 30 cm from the nter? These values define the system as per NCERT data. Formula: Outside shell ( r > R ): E = k q/r². This is standard NCERT relation. Substitution & Calculation: k = 9 × 10⁹Nm²/C², q = 16 × 10⁻⁶C, r = 0.3 m . E = 9 × 10⁹ × frac16 × 10⁻⁶(0.3)² = 9 × 10⁹ × frac16 × 10⁻⁶⁰.09 = 1.6 × 10⁶N/C . 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 coil of self-inductance 0.8 H has its current decreased from 6 A to 3 A in 0.2 s. What is the magnitude of the induced

Given: A coil of self-inductance 0.8 H has its current decreased from 6 A to 3 A in 0.2 s. What is the magnitude of the induced emf? These values define the system as per NCERT data. Formula: varepsilon = L Δ I/Δ t. This is standard NCERT relation. Substitution & Calculation: Δ I = 3 - 6 = -3 A, Δ t = 0.2 s . varepsilon = 0.8 × -3/0.2 = 0.8 × (-15) = -12 V, magnitude = 12 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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