Latest questions in this category.
Semiconductors have intermediate conductivity ( 10âµ to 10â»â¶ S m^{-1 ), between metals ( 10² to 10⸠S m^{-1 ) and insulators ( 10â»Â¹Â¹ to 10â»Â¹â¹ S m^{-1 ).
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.
Charges on the conductor’s surface redistribute to cancel any external field inside, creating an electrostatic shield. This shielding effect ensures the internal field is zero, as charges adjust to maintain equilibrium within the conductor.
Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.
Kirchhoff’s junction rule, based on charge conservation, ensures that charge does not accumulate at a junction in steady state, so the sum of currents in equals the sum out.
Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.
A concave lens diverges light rays, preventing them from converging to a point on the opposite side. The rays appear to diverge from a virtual focal point on the same side as the object, resulting in a virtual image that cannot be projected, regardless of object position.
Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Ray Optics and Optical Instruments and Wave Optics, Topic: Refraction, lenses and interference/diffraction.
Given: A spring of k = 180 N/m has a 0.9 kg mass. If E = 0.9 J, what is the amplitude? These values define the system as per NCERT data. Formula: Total energy: E = 1/2 k A². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 0.9 = 0.5 × 180 × A² Rightarrow 0.9 = 90 A² Rightarrow A² = 0.01 Rightarrow A = 0.1 m . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ 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.
Given: A 7 kg block on a horizontal surface ( μ_k = 0.3 ) is pulled by a 4 kg mass over a pulley. A 14 N force opposes the 7 kg block. What is the acceleration? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: For 4 kg : 4g - T = 4a Rightarrow 40 - T = 4a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For 7 kg : T - f_k - 14 = 7a . Normal: N = mg = 7 × 10 = 70 N . Friction: f_k = 0.3 × 70 = 21 N . Net force: T - 21 - 14 = 7a Rightarrow T - 35 = 7a . Solve: 40 - T = 4a, T - 35 = 7a . Substitute: 40 - (7a + 35) = 4a Rightarrow 40 - 35 - 7a = 4a Rightarrow 5 = 11a . a = 5/11 approx 0.45 m/s² . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ 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.
Given: A wire of length 0.6 m carrying 9 A is at 60° to a magnetic field of 0.4 T . What is the force on the wire? These values define the system as per NCERT data. Formula: F = I l B sin θ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: F = 9 × 0.6 × 0.4 × sin 60° = 2.16 × 0.866 = 1.8706 approx 1.87 N . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ are properly used as per NCERT.
Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.
Given: A dipole p = 6 × 10â»â¹ C m is rotated from θ = 0° to 90° in a field E = 3 × 10âµ N/C . What is the work done? These values define the system as per NCERT data. Formula: Work done: W = p E (cos θ_0 - cos θ_1) = 6 × 10â»â¹ × 3 × 10ⵠ× (cos 0° - cos 90°). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: W = 6 × 10â»â¹ × 3 × 10ⵠ× (1 - 0) = 1.8 × 10â»Â³ J . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ are properly used as per NCERT.
Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.
Power loss in lines is P_{loss = I² R . For a given power ( P = V I ), increasing V reduces I ( I = P / V ), significantly lowering I² R losses. 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.
Given: A transformer has N_p = 800, N_s = 400 . If V_s = 110 V (rms), what is the primary voltage? These values define the system as per NCERT data. Formula: V_s/V_p = N_s/N_p. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: V_p = V_s × N_p/N_s = 110 × 800/400 = 220 V . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ 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.
Given: A magnetic dipole experiences a torque of 0.02 N m in a field of 0.4 T at 90° . What is its magnetic moment? These values define the system as per NCERT data. Formula: tau = m B sinθ, so m = tau/B sinθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: tau = 0.02 N m, B = 0.4 T, θ = 90°, sin 90° = 1 . m = 0.02/0.4 × 1 = 0.05 A m² . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ are properly used as per NCERT.
Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.
Given: A 10 kg mass at 8 m/s collides inelastically with a stationary 5 kg mass. What is the final speed? These values define the system as per NCERT data. Formula: Momentum conservation: 10 × 8 = (10 + 5) v_f Rightarrow 80 = 15 v_f Rightarrow v_f = 80/15 approx 5.33 m/s .. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substituting values like 1.2 × 10â»âµ, 236 J kgâ»Â¹ Kâ»Â¹, CH₃CHâ‚‚NHâ‚‚ 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 kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ 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.
