PHYSICS

PHYSICS

A 5 kg block on a 37° incline ( μ_k = 0.2 ) is pulled upward by a 7 kg mass over a pulley. What is the tension? (Take

Given: A 5 kg block on a 37° incline ( μ_k = 0.2 ) is pulled upward by a 7 kg mass over a pulley. What is the tension? (Take g = 10 m/s², sin 37° = 0.6, cos 37° = 0.8 ) These values define the system as per NCERT data. Formula: For 7 kg : 7g - T = 7a Rightarrow 70 - T = 7a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For 5 kg : T - mg sin 37° - f_k = 5a . N = mg cos 37° = 5 × 10 × 0.8 = 40 N . Friction: f_k = 0.2 × 40 = 8 N . mg sin 37° = 50 × 0.6 = 30 N . Net force: T - 30 - 8 = 5a Rightarrow T - 38 = 5a . Solve: 70 - T = 7a, T - 38 = 5a . Substitute: 70 - (5a + 38) = 7a Rightarrow 70 - 38 - 5a = 7a Rightarrow 32 = 12a . a = 32/12 approx 2.67 m/s² . T - 38 = 5 × 2.67 Rightarrow T - 38 approx 13.35 Rightarrow T approx 51.35 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

Which process balances diffusion current in a p-n junction at equilibrium?

At equilibrium, the diffusion current due to carrier concentration gradients is balanced by the drift current caused by the electric field in the depletion region, resulting in no net current.

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 0.3 kg stone is whirled at 50 rev/min in a circle of radius 2 m with a 5 N tangential force. What is the net force at

Given: A 0.3 kg stone is whirled at 50 rev/min in a circle of radius 2 m with a 5 N tangential force. What is the net force at that instant? These values define the system as per NCERT data. Formula: Centripetal force: F_c = m omega² r, omega = 50 × 2π/60 = 5π/3 rad/s. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: omega² = (5π/3)² = 25π²/9 . F_c = 0.3 × 25π²/9 × 2 approx 0.3 × 54.74 approx 16.42 N . Tangential force = 5 N . Net force F = sqrt(F_c)² + (F_t)² = sqrt(16.42)² + (5)² approx sqrt269.62 + 25 approx 17.16 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

What is the nuclear density if a nucleus has a mass of 9.27 × 10⁻²⁶ kg and volume of 4.05 × 10⁻⁴⁴ m³ ?

Given: What is the nuclear density if a nucleus has a mass of 9.27 × 10⁻²⁶ kg and volume of 4.05 × 10⁻⁴⁴ m³ ? These values define the system as per NCERT data. Formula: Density = fracmassvolume. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: frac9.27 × 10⁻²⁶⁴.05 × 10⁻⁴⁴ approx 2.29 × 10¹⁷ kg/m³ . 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

An electron in a hydrogen atom is excited to n = 3 and then falls to n = 1 . What is the maximum energy of the emitted p

Given: An electron in a hydrogen atom is excited to n = 3 and then falls to n = 1 . What is the maximum energy of the emitted photon? (Use E_n = -13.6/n² eV ) These values define the system as per NCERT data. Formula: Maximum energy from n = 3 to n = 1. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E_3 = -1.51 eV, E_1 = -13.6 eV . Δ E = 12.09 eV . 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 projectile is launched at 10 m/s at 37° . What is its vertical velocity after 1 s ? (Take g = 10 m/s², sin 37° = 0.

Given: A projectile is launched at 10 m/s at 37° . What is its vertical velocity after 1 s ? (Take g = 10 m/s², sin 37° = 0.6 ) These values define the system as per NCERT data. Formula: Vertical velocity v_y = v_0 sin θ_0 - g t. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: v_0 = 10 m/s, sin 37° = 0.6, t = 1 s, g = 10 m/s² . v_y = 10 × 0.6 - 10 × 1 = 6 - 10 = -4 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

Light of wavelength 350 nm is incident on a metal with work function 2.0 eV . What is the stopping potential? (Take h c

Given: Light of wavelength 350 nm is incident on a metal with work function 2.0 eV . What is the stopping potential? (Take h c = 1240 eV nm ) These values define the system as per NCERT data. Formula: E = h c/lambda = 1240/350 approx 3.54 eV. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: K_{max = E - phi_0 = 3.54 - 2.0 = 1.54 eV . V_0 = fracK_{maxe = 1.54 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: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

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PHYSICS

Light of frequency 5.0 × 10¹⁴ Hz is incident on a metal surface with work function 1.8 eV . What is the maximum spee

Given: Light of frequency 5.0 × 10¹⁴ Hz is incident on a metal surface with work function 1.8 eV . What is the maximum speed of emitted electrons? (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: E = h v = 6.63 × 10⁻³⁴ × 5.0 × 10¹⁴= 3.315 × 10⁻¹⁹ J. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E = frac3.315 × 10⁻¹⁹¹.6 × 10⁻¹⁹ approx 2.07 eV . K_{max = E - phi_0 = 2.07 - 1.8 = 0.27 eV = 0.27 × 1.6 × 10⁻¹⁹= 4.32 × 10⁻²⁰ J . K_{max = 1/2 m v_{max² Rightarrow v_{max = sqrtfrac2 K_{maxm = sqrtfrac2 × 4.32 × 10⁻²⁰⁹.11 × 10⁻³¹ approx 3.08 × 10⁵ 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: Oscillations and Waves, Topic: SHM, wave motion and superposition.

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PHYSICS

In a double-slit experiment, what is the distance of the second bright fringe from the ntral maximum if lambda = 500 nm,

Given: In a double-slit experiment, what is the distance of the second bright fringe from the ntral maximum if lambda = 500 nm, d = 0.2 mm, and D = 1.0 m ? These values define the system as per NCERT data. Formula: Distance x_n = n lambda D/d. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For the second bright fringe, n = 2 . lambda = 5.0 × 10⁻⁷ m, d = 2.0 × 10⁻⁴ m, D = 1.0 m . x_2 = frac2 × 5.0 × 10⁻⁷ × 1.02.0 × 10⁻⁴= 5.0 × 10⁻³ m = 5.0 mm . 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 train moving at 36 km/h accelerates at 0.5 m/s² for 20 s, then decelerates at 1 m/s² to rest. What is the total dist

Given: A train moving at 36 km/h accelerates at 0.5 m/s² for 20 s, then decelerates at 1 m/s² to rest. What is the total distance covered? These values define the system as per NCERT data. Formula: Initial speed: 36 km/h = 10 m/s. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Phase 1: v = 10 + 0.5 · 20 = 20 m/s, x_1 = 10 · 20 + 1/2 · 0.5 · (20)² = 200 + 100 = 300 m . Phase 2: t = 20/1 = 20 s, x_2 = 20 · 20 - 1/2 · 1 · (20)² = 400 - 200 = 200 m . Total = 300 + 200 = 500 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

A transverse wave travels on a string with tension 50 N and linear mass density 0.02 kg/m. What is the wavelength if the

Given: A transverse wave travels on a string with tension 50 N and linear mass density 0.02 kg/m. What is the wavelength if the frequency is 25 Hz? These values define the system as per NCERT data. Formula: Speed: v = sqrtT/μ = sqrt50/0.02 = sqrt2500 = 50 m/s. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Wavelength: lambda = v/v = 50/25 = 2 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 tank of water (refractive index = 1.33) has a depth of 20 cm . What is the apparent depth when viewed normally?

Given: A tank of water (refractive index = 1.33) has a depth of 20 cm . What is the apparent depth when viewed normally? These values define the system as per NCERT data. Formula: Apparent depth = fracreal depthn. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Real depth = 20 cm, n = 1.33 . Apparent depth = 20/1.33 approx 15.04 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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