PHYSICS

PHYSICS

A force F = -5i + 2j N acts on a particle moving along d = 3i + 4j m . What is the work done?

Given: A force F = -5i + 2j N acts on a particle moving along d = 3i + 4j m . What is the work done? These values define the system as per NCERT data. Formula: Work W = F · d. This is standard NCERT relation. Substitution & Calculation: F = -5i + 2j, d = 3i + 4j . Scalar product: W = (-5 × 3) + (2 × 4) = -15 + 8 = -7 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: Relevant Physics topic covering fundamental principles,

View Details

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,

View Details

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,

View Details

PHYSICS

What does the Ampere-Maxwell law relate in Maxwell's equations?

The Ampere-Maxwell law relates the magnetic field to both the conduction current and the displacement current, given by oint B · d l = μ_0 i_c + μ_0 varepsilon_0 d Phi_E/dt .

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

View Details

PHYSICS

In a circuit with resistors in parallel, what happens to the voltage across each resistor?

In a parallel configuration, all resistors share the same two nodes, so the potential difference (voltage) across each is equal to the source voltage, per Kirchhoff’s loop rule applied to each branch.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

View Details

PHYSICS

Why does steam at 100° C cause more severe burns than water at 100° C ?

Steam releases additional latent heat of vaporization when it condenses on skin, transferring more energy than liquid water at the same temperature. This follows from NCERT principle where relation explains outcome clearly for students.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

View Details

PHYSICS

Which of the following is an elemental semiconductor?

Elemental semiconductors are pure elements like Si and Ge, while compound semiconductors (e.g., GaAs, CdS) consist of multiple elements. Among the options, Ge is elemental.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

View Details

PHYSICS

What is the primary source of charge carriers in an intrinsic semiconductor at room temperature?

In an intrinsic semiconductor, charge carriers (electrons and holes) are generated by thermal excitation at temperatures above 0 K, breaking covalent bonds and creating electron-hole pairs.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

View Details

PHYSICS

A system of two particles, 3 kg at (1, 2) and 1 kg at (5, 6), moves as a rigid body. What is the distance of the nter of

Given: A system of two particles, 3 kg at (1, 2) and 1 kg at (5, 6), moves as a rigid body. What is the distance of the nter of mass from the origin? These values define the system as per NCERT data. Formula: X = (3 × 1) + (1 × 5)/3 + 1 = 3 + 5/4 = 2. This is standard NCERT relation. Substitution & Calculation: Y = (3 × 2) + (1 × 6)/3 + 1 = 6 + 6/4 = 3 . Distance = sqrtX² + Y² = sqrt2² + 3² = sqrt4 + 9 = sqrt13 approx 3.6 m . 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,

View Details

PHYSICS

Two capacitors 10 μF and 20 μF are in parallel. What is the total capacitance?

Given: Two capacitors 10 μF and 20 μF are in parallel. What is the total capacitance? These values define the system as per NCERT data. Formula: C = C_1 + C_2 = 10 + 20 = 30 μF .. 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,

View Details

PHYSICS

A circular loop of radius 0.2 m with 10 turns carries 1.5 A . What is the magnetic field at its nter? ( μ_0 = 4 π × 10⁻⁷

Given: A circular loop of radius 0.2 m with 10 turns carries 1.5 A . What is the magnetic field at its nter? ( μ_0 = 4 π × 10⁻⁷T m/A ) These values define the system as per NCERT data. Formula: Magnetic field B = μ_0 N I/2 R. This is standard NCERT relation. Substitution & Calculation: B = frac4 π × 10⁻⁷ × 10 × 1.52 × 0.2 = frac6 π × 10⁻⁶⁰.4 = 1.5 π × 10⁻⁵approx 4.71 × 10⁻⁵T . 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,

View Details

PHYSICS

In an n-type semiconductor, the minority carriers are:

In an n-type semiconductor, electrons are majority carriers due to pentavalent doping, while holes, generated intrinsically or reduced by recombination, are minority carriers.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

View Details