Physics — Electricity, Magnetism, Optics & Modern Physics — Practice

Question 1

Light of wavelength 400 nm falls on a metal with work function 2.0 eV. Find (a) the maximum kinetic energy of emitted electrons and (b) the stopping potential. (h = 6.63×10⁻³⁴ J·s, c = 3×10⁸ m/s, 1 eV = 1.6×10⁻¹⁹ J)

Accepted Answer

(a) Energy of photon: E = hc/λ = (6.63×10⁻³⁴ × 3×10⁸)/(400×10⁻⁹) = 4.97×10⁻¹⁹ J = 4.97×10⁻¹⁹/(1.6×10⁻¹⁹) = 3.1 eV. KE_max = E − φ = 3.1 − 2.0 = 1.1 eV. (b) Stopping potential V₀: eV₀ = KE_max = 1.1 eV → V₀ = 1.1 V.

Question 2

A convex lens of focal length 20 cm forms a real image of an object placed 30 cm in front of it. Find the image distance, magnification, and state whether the image is erect or inverted.

Accepted Answer

Using sign convention: u = −30 cm (object to the left), f = +20 cm (convex lens). Lens formula: 1/v − 1/u = 1/f → 1/v − 1/(−30) = 1/20 → 1/v + 1/30 = 1/20 → 1/v = 1/20 − 1/30 = 3/60 − 2/60 = 1/60. v = +60 cm (real image, on the other side). Magnification m = v/u = 60/(−30) = −2. The image is REAL (v is positive), INVERTED (m is negative), and MAGNIFIED (|m| = 2).

Question 3

A rectangular coil of N=100 turns, area A=0.02 m², resistance R=20 Ω, rotates in a magnetic field B=0.5 T at ω=100π rad/s. Find (a) peak EMF, (b) rms EMF, (c) rms current, (d) average power dissipated.

Accepted Answer

(a) Peak EMF: ε₀ = NBAω = 100 × 0.5 × 0.02 × 100π = 100π ≈ 314.16 V. (b) rms EMF: ε_rms = ε₀/√2 = 100π/√2 = 100×3.14/1.414 ≈ 222 V. (c) rms current: I_rms = ε_rms/R = 222/20 = 11.1 A. (d) Average power: P = I_rms² × R = (11.1)² × 20 = 123.21 × 20 ≈ 2464 W ≈ 2.46 kW.

Practice — Physics — Electricity, Magnetism, Optics & Modern Physics

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