FBISE Class 9th (SSC-l)
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FBISE Class 9th (SSC-l)
Answer: A reversible process must be quasi-static (infinitely slow) and free from dissipative forces like friction.
Answer: f₀ = φ/h = 3.0/(4.14 × 10⁻¹⁵) = 7.25 × 10^14 Hz
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Answer: E = hc/λ = (12400 eV·Å)/(4000 Å) = 3.1 eV. Vstop = E - φ = 3.1 - 2.0 = 1.1 V
Answer: The number of photoelectrons is directly proportional to the intensity of incident light (number of photons).
Answer: KE_max = hf - φ, so maximum kinetic energy increases linearly with frequency of incident light.
Answer: R = 1.097 × 10^7 m⁻¹ is the Rydberg constant used in the hydrogen spectral series formula.
Answer: Transitions to n=2 produce the Balmer series (visible light region).
Answer: ΔE = 13.6(1 - 1/4) = 13.6 × 3/4 = 10.2 eV
Answer: When the plane is parallel to the field, the area vector is perpendicular, giving maximum torque τ = M × B.
Answer: KE_max = q²B²R²/(2m), depending on magnetic field B and radius R of the dees.
Answer: Self-inductance produces a back EMF that opposes any change in current (Lenz\'s law applied to self-induction).
Answer: Energy stored in inductor = ½LI², analogous to energy stored in capacitor = ½CV².
Answer: Z = √(R² + XL²) = √(9 + 16) = √25 = 5 Ω
Answer: tan φ = XC/R = 10/10 = 1 → φ = tan⁻¹(1) = 45°. Current leads voltage.
Answer: At resonance, Z = R (minimum), so power factor cos φ = R/Z = 1. Circuit behaves as purely resistive.
Answer: Number of neutrons = Mass number - Atomic number = 235 - 92 = 143
Answer: Q-value = (mass of reactants - mass of products)c² = total energy released in the nuclear reaction.
Answer: Nuclei must have enough kinetic energy to overcome the Coulomb repulsion between positively charged nuclei.
Answer: In EM waves, E and B fields are perpendicular to each other and both are perpendicular to the direction of wave propagation.
Answer: All electromagnetic waves travel at c = 3 × 10^8 m/s in vacuum.