PHYSICS

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 brass block of dimensions 0.5 m × 0.2 m × 0.05 m is subjected to a shearing force of 2 × 10⁴ N . If the shear mod

Given: A brass block of dimensions 0.5 m × 0.2 m × 0.05 m is subjected to a shearing force of 2 × 10⁴ N . If the shear modulus of brass is 3.6 × 10¹⁰ N/m², what is the shear strain? These values define the system as per NCERT data. Formula: Shear modulus: G = fracShear stressShear strain. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Shear stress: Shear stress = F/A, A = 0.5 × 0.2 = 0.1 m² . Shear stress: 2 × 10⁴/0.1 = 2 × 10⁵ N/m² . Shear strain: Shear strain = fracShear stressG = frac2 × 10⁵³.6 × 10¹⁰ approx 5.56 × 10⁻⁶. 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: 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

Check the dimensional consistency of W = F s cosθ, where W is work, F is force, s is distance, and θ is an angle.

LHS: [W] = [M L² T^{-2] . RHS: [F s cosθ] = [M L T^{-2] [L] × dimensionless = [M L² T^{-2] . Consistent. This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

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 1.8 kg pendulum bob completes a vertical circle of radius 2 m . What is the speed at the top? (Take g = 10 m/s² )

Given: A 1.8 kg pendulum bob completes a vertical circle of radius 2 m . What is the speed at the top? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: At top, minimum speed v_C = sqrtgL = sqrt10 × 2 = sqrt20 approx 4.47 m/s .. 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: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

What is the frequency of light with a wavelength of 480 nm in air, given the speed of light in air is 3.0 × 10⁸ m/s ?

Given: What is the frequency of light with a wavelength of 480 nm in air, given the speed of light in air is 3.0 × 10⁸ m/s ? These values define the system as per NCERT data. Formula: Frequency nu = c/lambda. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: lambda = 4.8 × 10⁻⁷ m, c = 3.0 × 10⁸ m/s . nu = frac3.0 × 10⁸⁴.8 × 10⁻⁷= 6.25 × 10¹⁴ 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 nucleus with mass number 28 has a binding energy of 224 MeV . What is its binding energy per nucleon?

Given: A nucleus with mass number 28 has a binding energy of 224 MeV . What is its binding energy per nucleon? These values define the system as per NCERT data. Formula: E_{bn = E_b/A. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E_b = 224 MeV, A = 28 . E_{bn = 224/28 = 8.0 MeV . 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 of the following statements is correct about a satellite in circular orbit?

The gravitational force provides the ntripetal force ( G M_E m/r² = m v²/r ), making option 2 correct. This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

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 Si crystal with 5 × 10²⁸ atoms m^{-3 is doped with 0.5 ppm of pentavalent impurity. The number of donor electrons

Given: A Si crystal with 5 × 10²⁸ atoms m^{-3 is doped with 0.5 ppm of pentavalent impurity. The number of donor electrons per cubic meter is: These values define the system as per NCERT data. Formula: 0.5 ppm = 0.5 × 10⁻⁶. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Number of donor atoms = 0.5 × 10⁻⁶ × 5 × 10²⁸= 2.5 × 10²² m^{-3, each contributing one electron. 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

The conductivity of an extrinsic semiconductor increases due to:

Doping introduces impurities (pentavalent or trivalent) that provide additional charge carriers (electrons or holes), significantly enhancing conductivity compared to intrinsic semiconductors.

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 7 kg object moves with a velocity of 3 i - 4 j m/s . What is the magnitude of the velocity of its nter of mass?

Given: A 7 kg object moves with a velocity of 3 i - 4 j m/s . What is the magnitude of the velocity of its nter of mass? These values define the system as per NCERT data. Formula: Magnitude = sqrt(3)² + (-4)² = sqrt9 + 16 = sqrt25 = 5 m/s .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For a single object, the nter of mass velocity equals the object’s velocity. 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

In a rigid body, why does the point of application of force affect torque?

Torque depends on the perpendicular distance from the axis to the line of action of the force ( tau = r F sin θ ), so the point of application changes r, altering the torque.

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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