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14 public questions tagged with this topic.

Capacitance of Parallel Plate and Spherical Capacitor

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

**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. Potential at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (5 × 10⁻⁸/0.05) = 9 × 10⁹ × 10⁻⁶ = 9000 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 9000 V follows, reflecting potential-capacitance relations.

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

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A proton moves with a speed of \( 1.8 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.5 \, \text{

**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. Radius r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 1.8 × 10⁶/1.6 × 10⁻¹⁹ × 0.5) = (3.006 × 10⁻²¹/8 × 10⁻²⁰) = 3.7575 × 10⁻² m = 3.76 cm . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀

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

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Magnetic Field Due to Current - Straight Wire and Circular Loop

A circular coil of 60 turns and radius \( 8 \, \text{cm} \) carries a current of \( 0.9 \, \text{A} \). What is the magn

**Field at centre of circular loop** with N turns is B = μ₀ N I/(2R), R radius (m), direction along axis via right-hand rule, magnitude proportional to N I/R. For R = 0.09 m, N = 45, I = 1.2 A, B = 4π×10⁻⁷×45×1.2/(2×0.09) = 3.77×10⁻⁴ T, showing N enhancement. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 60 × 0.9/2 × 0.08) = (21.6 π × 10⁻⁶/0.16) = 1.35 π × 10⁻⁴ ≈ 4.24 × 10⁻⁴ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r),

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

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Displacement, Velocity and Acceleration in SHM

The projection of a particle in circular motion on the x-axis is \( x = 3 \cos (2\pi t) \) (in m). What is the radius of

**Sinusoidal description** links amplitude A and angular frequency ω to instantaneous values. Given x(t) = A cos(ωt), max acceleration ω²A quantifies force requirement F_max = m ω²A, and velocity at arbitrary x is v = ±ω√(A² - x²) from energy conservation. For SHM as projection of circular motion, radius = amplitude. Here, x = A cos (ω t) , so A = 3 m . Radius = 3 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 3.0 m follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Displacement, Velocity and Acceleration in SHM

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Equation of Continuity and Flow Rate

A soap bubble of radius 7mm has a surface tension of 0.028N/m. What is the excess pressure inside?

Excess pressure in a bubble: ΔP = 4Sr. S = 0.028N/m, r = 7×10−3m. ΔP = 4×0.0287×10−3 = 16Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 16 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Moment of Inertia and Radius of Gyration

What is the effect of doubling the radius of a rotating disk on its moment of inertia, assuming mass remains constant?

For a disk, I = 12MR2. Doubling R increases R2 by a factor of 4, so I increases by a factor of 4 if M is constant. As per NCERT, applying relevant law/formula with correct units and sign convention leads to It quadruples. 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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Acceleration and Velocity Analysis

A particle moves in a circle of radius 5m with an angular speed of 2rad/s. What is its centripetal acceleration?

Centripetal acceleration ac=ω2R. 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 20 m/s² as the result, so option B is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Acceleration and Velocity Analysis - Part 8

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Straight Line Motion - Mixed Concepts

An object moves in a circular path of radius 4m with a frequency of 5Hz. What is the magnitude of its centripetal accele

Angular speed ω=2πf=2π×5=10πrad/s. 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 3947.8 m/s² as the result, so option A is correct.

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

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Straight Line Motion - Mixed Concepts

A particle moves in a circular path of radius 6m with a frequency of 2Hz. What is its centripetal acceleration?

Angular speed ω=2πf=2π×2=4πrad/s. 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 as the result, so option A is correct.

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

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