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35 public questions tagged with this topic.

Radioactive Decay, Nuclear Forces and Stability

Which statement best describes the nuclear force at distances greater than 0.8 fm?

**Nuclear force** strong, short-range ~1 fm, attractive, charge independent, saturated in large nuclei because each nucleon interacts only with neighbors, not all others, so BE/A saturates ~8 MeV, primary factor limiting stable nuclei size is Coulomb repulsion between protons growing as Z² vs strong force saturating, beyond Z≈83 no stable nuclei, competition between Coulomb and strong. At distances greater than 0.8 fm, the nuclear force is attractive and dominates over the Coulomb force, binding nucleons together until it drops off rapidly beyond a few femtometres. Using E_n = -13.6/n² eV, r_n

Ref: NCERT > Physics Book > Atoms and Nuclei > Radioactive Decay, Nuclear Forces and Stability

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Capacitance of Parallel Plate and Spherical Capacitor

A spherical conductor of radius 10 cm has a charge of \( 5 \times 10^{-8} \, \text{C} \). What is the electric field at

**Parallel plate capacitor** capacitance C = ε₀ A/d, ε₀=8.85×10⁻¹² F/m, A plate area (m²), d separation (m), for air, with dielectric C = K ε₀ A/d. For A=0.08 m², d=0.4 mm=4×10⁻⁴ m, C=8.85×10⁻¹²×0.08/4×10⁻⁴=1.77×10⁻⁹ F=1.77 nF, illustrating small capacitance for cm separation. For r = 0.25 m > R = 0.1 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (5 × 10⁻⁸/(0.25)²) = 9 × 10⁹ × (5 × 10⁻⁸/0.0625) = 7.2 × 10³ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitance of Parallel Plate and Spherical Capacitor

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Torque on Current Loop, Magnetic Moment and Galvanometer

A long wire carries \( 14 \, \text{A} \). At what distance is the magnetic field \( 7 \times 10^{-6} \, \text{T} \)? (\(

**SI unit of magnetic field** is tesla (T), defined as force 1 N on 1 A·m wire perpendicular to field. Moving coil galvanometer uses torque τ = N I A B balanced by spring torque k φ, so deflection φ ∝ I, enabling current measurement, with radial field ensuring τ = N I A B always maximum. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 14/2 π × 7 × 10⁻⁶) = (56 × 10⁻⁷/14 × 10⁻⁶) = 0.4 m . Using F = q v B sinθ, F = I

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A long straight wire carries a current of \( 25 \, \text{A} \). What is the magnetic field at a distance of \( 0.25 \, \

**Biot-Savart law** dB = μ₀/4π·I dl × r̂/r² underlies both straight wire and loop formulas. For square loop side a, area A = a², moment m = N I A, field pattern similar to dipole at large distances, with superposition for N turns. Magnetic field B = (μ₀ I/2 π r) . B = (4 π × 10⁻⁷ × 25/2 π × 0.25) = (100 × 10⁻⁷/0.5) = 2 × 10⁻⁵ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I A B sinθ, result

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Field Due to Current - Straight Wire and Circular Loop

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A long wire carries \( 20 \, \text{A} \). At what distance from the wire is the magnetic field \( 4 \times 10^{-6} \, \t

**Effect of doubling velocity** on magnetic force F = q v B sinθ is linear increase, F doubles for same θ and B. Electron with charge 1.6×10⁻¹⁹ C, v = 4.5×10⁶ m/s, B = 0.35 T, θ = 90°, F = 1.6×10⁻¹⁹×4.5×10⁶×0.35 = 2.52×10⁻¹³ N, illustrating magnitude for typical lab values. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 20/2 π × 4 × 10⁻⁶) = (8 × 10⁻⁶/8 × 10⁻⁶) = 1 m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Diamagnetism, Paramagnetism and Ferromagnetism

A bar magnet’s field strength decreases with distance because:

**Ferromagnetism** shows large positive χ ≈ 10³ to 10⁵, strong attraction, domain structure with spontaneous magnetization, hysteresis, retentivity. Distinction based on sign and magnitude of χ and behaviour in non-uniform field, explaining attraction versus repulsion. The magnetic field strength of a bar magnet decreases with distance as the field lines spread out, reducing their density and thus the field intensity, following an inverse cube relationship ( B ∝ 1/r³ ) for a dipole at large distances. Substituting values gives The field lines spread out, which matches expected magnitude for th

Ref: NCERT > Physics Book > Magnetism and Matter > Diamagnetism, Paramagnetism and Ferromagnetism

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PHYSICS

A cyclist travels at a constant speed of 18 km/h for 10 minutes . What is the distance covered?

Given: A cyclist travels at a constant speed of 18 km/h for 10 minutes . What is the distance covered? These values define the system as per NCERT data. Formula: Convert speed: 18 km/h = 18 · 1000/3600 = 5 m/s. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Time = 10 min = 10 · 60 = 600 s . Distance x = v t = 5 · 600 = 3000 m = 3 km . The distance covered is 3 km . 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 stone is dropped from rest. What is the distance covered in the 5th second of its fall? (Take g = 10 m/s² )

Given: A stone is dropped from rest. What is the distance covered in the 5th second of its fall? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Distance in n -th second = g · 2n - 1/2. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For 5th second, n = 5 . Substitute: d = 10 · 2 · 5 - 1/2 = 10 · 9/2 = 45 m . The distance covered is 45 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

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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Vector Product and Rotational Kinematics

Two particles of masses 3kg and 7kg are at (1,4) and (9,2) respectively. What is the distance of their center of mass fr

X = (3×1)+(7×9)3+7 = 3+6310 = 6.6. Y = (3×4)+(7×2)3+7 = 12+1410 = 2.6. Distance = (6.6)2+(2.6)2 = 43.56+6.76 = 50.32≈7.09m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7.09 m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Rotational Motion Applications

Two particles of masses 8kg and 2kg are at (0,5) and (10,0) respectively. What is the distance of their center of mass f

X = (8×0)+(2×10)8+2 = 2010 = 2. Y = (8×5)+(2×0)8+2 = 4010 = 4. Distance = (2)2+(4)2 = 4+16 = 20≈4.47m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.47 m. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Motion in a Straight Line - Advanced Problems

A body accelerates uniformly from rest to a velocity of 20m/s over a distance of 50m. What is the acceleration?

Use v2=v02+2ax. Here, v0=0, v=20m/s, x=50m. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 4 m/s² as the result, so option C is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Motion in a Straight Line - Advanced Problems

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