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in Physics
Answer: Average power = V_rms I_rms cos φ, where φ is the phase angle between voltage and current.
Answer: The number of photoelectrons is directly proportional to the intensity of incident light (number of photons).
Answer: Second law of thermodynamics: entropy of an isolated system always increases or remains constant in reversible processes.
244
The stopping potential for photoelectrons emitted by light of wavelength 4000 Å is: (φ = 2.0 eV)
Answer: E = hc/λ = (12400 eV·Å)/(4000 Å) = 3.1 eV. Vstop = E - φ = 3.1 - 2.0 = 1.1 V
Answer: Carnot cycle: two isothermal (at TH and TL) and two adiabatic processes forming a reversible cycle.
Answer: KE_max = hf - φ, so maximum kinetic energy increases linearly with frequency of incident light.
Answer: Fringe width β = λD/d. If d is halved, β doubles.
Answer: f₀ = φ/h = 3.0/(4.14 × 10⁻¹⁵) = 7.25 × 10^14 Hz
Answer: At resonance, XL = XC, so impedance Z = √(R² + (XL-XC)²) = R (minimum value).
Answer: Q = ωL/R = 1/(ωCR) at resonance. Higher Q means sharper resonance.
251
In a transformer, if the primary voltage is 220 V and turns ratio is 10:1, the secondary voltage is:
Answer: Vs/Vp = Ns/Np → Vs = Vp × (Ns/Np) = 220 × (1/10) = 22 V (step-down transformer).
Answer: Silicon has a band gap of about 1.1 eV at room temperature, making it a semiconductor.
Answer: An ideal transformer has no losses (no copper loss, no iron loss), so efficiency = 100%.
Answer: TIR occurs when angle of incidence > critical angle when light travels from denser to rarer medium.
Answer: A reversible process must be quasi-static (infinitely slow) and free from dissipative forces like friction.
Answer: Carnot efficiency η = 1 - TL/TH is always less than 100% unless TL = 0 K (impossible).
Answer: sin ic = 1/μ = 1/1.5 = 0.667 → ic = sin⁻¹(0.667) ≈ 41.8° ≈ 42°
Answer: Resolving power = D/(1.22λ), where D is the aperture diameter and λ is the wavelength of light used.
Answer: For a simple microscope (magnifying glass), magnifying power M = 1 + D/f, where D is the least distance of distinct vision.
Answer: When object is between F and optical center of convex lens, image is virtual, erect, and magnified on the same side.