PHYSICS

PHYSICS

A 5 kg block on a horizontal surface ( μ_k = 0.2 ) is pulled by a 8 kg mass over a pulley. A 10 N force opposes the 5 k

Given: A 5 kg block on a horizontal surface ( μ_k = 0.2 ) is pulled by a 8 kg mass over a pulley. A 10 N force opposes the 5 kg block. What is the tension? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: For 8 kg : 8g - T = 8a Rightarrow 80 - T = 8a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For 5 kg : T - f_k - 10 = 5a . Normal: N = mg = 5 × 10 = 50 N . Friction: f_k = 0.2 × 50 = 10 N . Net force: T - 10 - 10 = 5a Rightarrow T - 20 = 5a . Solve: 80 - T = 8a, T - 20 = 5a . Substitute: 80 - (5a + 20) = 8a Rightarrow 80 - 20 - 5a = 8a Rightarrow 60 = 13a . a approx 4.62 m/s², T - 20 = 5 × 4.62 Rightarrow T - 20 approx 23.1 Rightarrow T approx 43.1 N . 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

The threshold wavelength of a metal is 550 nm . What is its work function in joules? (Take h = 6.63 × 10⁻³⁴ J s, c

Given: The threshold wavelength of a metal is 550 nm . What is its work function in joules? (Take h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s ) These values define the system as per NCERT data. Formula: v_0 = c/lambda_0 = frac3 × 10⁸⁵⁵⁰ × 10⁻⁹ approx 5.455 × 10¹⁴ Hz. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: phi_0 = h v_0 = 6.63 × 10⁻³⁴ × 5.455 × 10¹⁴ approx 3.617 × 10⁻¹⁹ 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: Oscillations and Waves, Topic: SHM, wave motion and superposition.

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PHYSICS

A 2 kg mass is moved from 2 R_E to 4 R_E from Earth’s nter. What is the change in potential energy? ( M_E = 6 × 10²â

Given: A 2 kg mass is moved from 2 R_E to 4 R_E from Earth’s nter. What is the change in potential energy? ( M_E = 6 × 10²⁴ kg, R_E = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m²/kg² ) These values define the system as per NCERT data. Formula: Δ V = -G M_E m (1/r_2 - 1/r_1). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r_1 = 1.28 × 10⁷ m, r_2 = 2.56 × 10⁷ m . Δ V = -6.67 × 10⁻¹¹ × 6 × 10²⁴ × 2 (1/2.56 × 10⁷- 1/1.28 × 10⁷) . Δ V = -8.004 × 10¹⁴(-3.906 × 10⁻⁸) approx 3.13 × 10⁷ 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: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

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PHYSICS

A string vibrates with a stationary wave y = 0.1 sin (2Ï€ x/3) cos (150Ï€ t) . What is the distance between a node and t

Given: A string vibrates with a stationary wave y = 0.1 sin (2π x/3) cos (150π t) . What is the distance between a node and the next antinode? These values define the system as per NCERT data. Formula: k = 2π/3, lambda = 2π/k = 3 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Distance between node and antinode: lambda/4 = 3/4 = 0.75 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: Oscillations and Waves, Topic: SHM, wave motion and superposition.

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PHYSICS

A plane sheet has sigma = 8.854 × 10⁻¹¹ C/m² . What is the electric field near it?

Given: A plane sheet has sigma = 8.854 × 10⁻¹¹ C/m² . What is the electric field near it? These values define the system as per NCERT data. Formula: E = sigma/2 varepsilon_0. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E = frac8.854 × 10⁻¹¹² × 8.854 × 10⁻¹²= 5 N/C . 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: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

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PHYSICS

A spring system has m = 0.1 kg, k = 40 N/m, A = 10 cm . What is the kinetic energy at x = 5 cm ?

Given: A spring system has m = 0.1 kg, k = 40 N/m, A = 10 cm . What is the kinetic energy at x = 5 cm ? These values define the system as per NCERT data. Formula: Total energy: E = 1/2 k A² = 0.5 × 40 × (0.1)² = 0.2 J. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Potential energy: U = 1/2 k x² = 0.5 × 40 × (0.05)² = 0.05 J . Kinetic energy: K = E - U = 0.2 - 0.05 = 0.15 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: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

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PHYSICS

A 2 kg object moves with a velocity of 4 m/s along the x-axis. What is the velocity of its nter of mass?

Given: A 2 kg object moves with a velocity of 4 m/s along the x-axis. What is the velocity of its nter of mass? These values define the system as per NCERT data. Formula: Given: v = 4 m/s along x-axis. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For a single object, the velocity of the nter of mass equals the velocity of the object. . Velocity of CM = 4 i 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 spring-mass system has m = 1 kg, k = 400 N/m . What is its frequency?

Given: A spring-mass system has m = 1 kg, k = 400 N/m . What is its frequency? These values define the system as per NCERT data. Formula: Angular frequency: omega = sqrtk/m = sqrt400/1 = 20 rad/s. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Frequency: v = omega/2π = 20/2 × 3.14 approx 3.18 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

In a single-slit diffraction pattern, what is the angular width of the ntral maximum if the slit width is 2.0 μm and th

Given: In a single-slit diffraction pattern, what is the angular width of the ntral maximum if the slit width is 2.0 μm and the wavelength is 400 nm ? These values define the system as per NCERT data. Formula: Angular width of the ntral maximum 2θ = 2lambda/a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: lambda = 4.0 × 10⁻⁷ m, a = 2.0 × 10⁻⁶ m . sin θ = lambda/a = frac4.0 × 10⁻⁷².0 × 10⁻⁶= 0.2, θ = sin^{-1(0.2) approx 11.5°, 2θ approx 23° . 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: Ray Optics and Optical Instruments and Wave Optics, Topic: Refraction, lenses and interference/diffraction.

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PHYSICS

What is the critical angle for a glass-air interface if the refractive index of glass is 1.52 ?

Given: What is the critical angle for a glass-air interface if the refractive index of glass is 1.52 ? These values define the system as per NCERT data. Formula: Critical angle: sin i_c = n_2/n_1. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Glass ( n_1 = 1.52 ), air ( n_2 = 1 ). sin i_c = 1/1.52 approx 0.658 . i_c = sin^{-1(0.658) approx 41.1° . 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 1100 kg car turns on a banked road ( θ = 20°, μ_s = 0.15 ) with radius 55 m . What is the maximum speed without sli

Given: A 1100 kg car turns on a banked road ( θ = 20°, μ_s = 0.15 ) with radius 55 m . What is the maximum speed without slipping? (Take g = 10 m/s², tan 20° approx 0.364 ) 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 = 0.15 + 0.364 = 0.514 . Denominator: 1 - 0.15 × 0.364 = 1 - 0.0546 = 0.9454 . v_{max² = 55 × 10 × 0.514/0.9454 approx 550 × 0.5436 approx 298.98 . v_{max = sqrt298.98 approx 17.29 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

The mean free path of a gas is 3 × 10⁻⁷ m at a rtain temperature. If the temperature is halved at constant pressure

Given: The mean free path of a gas is 3 × 10⁻⁷ m at a rtain temperature. If the temperature is halved at constant pressure, what is the new mean free path? These values define the system as per NCERT data. Formula: New l = frac3 × 10⁻⁷²= 1.5 × 10⁻⁷ m .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: l propto 1/n, n propto P/T. At constant P, n propto 1/T. If T halved, n doubles, l halves. 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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