This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here are the solutions to the multiple-choice questions: 9. At resonance the phase angle equals At resonance in an RLC circuit, the inductive reactance (X_L) cancels out the capacitive reactance (X_C). The circuit behaves purely resistively. Therefore, the voltage and current are in phase, and the phase angle () is 0^. Answer: a) 0 Degree 10. The band width of resonant circuit is The bandwidth ( f) of a resonant circuit is inversely proportional to its quality factor (Q). The formula is f = (f_r)/(Q), where f_r is the resonant frequency. Answer: b) Inversely proportional to Q 11. In silicon atom distribution of electrons in cells is Silicon (Si) has an atomic number of 14. Its electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^2. The electron distribution in shells (K, L, M) is: K-shell (1st shell): 2 electrons L-shell (2nd shell): 2+6 = 8 electrons M-shell (3rd shell): 2+2 = 4 electrons So the distribution is 2, 8, 4. Answer: a) 2,8,4 12. An inductive circuit with L=100 mH and R=10 K has time constant of The time constant () for an RL circuit is given by the formula = (L)/(R). Given L = 100 mH = 100 × 10^-3 H = 0.1 H. Given R = 10 K = 10 × 10^3 . = 0.1 H10 × 10^3 = (0.1)/(10000) s = 0.00001 s = 10 × 10^-6 s = 10 µs Answer: d) 10 µsec 13. A capacitive circuit with R=15 K and C=0.01 µF has a time constant of The time constant () for an RC circuit is given by the formula = RC. Given R = 15 K = 15 × 10^3 . Given C = 0.01 µF = 0.01 × 10^-6 F = 1 × 10^-8 F. = (15 × 10^3 ) × (1 × 10^-8 F) = 15 × 10^-5 s = 0.00015 s = 150 × 10^-6 s = 150 µs Answer: c) 150 µsec 14. The development of voltage difference due to light intensity on the surface of a PN junction is known When light falls on a PN junction, it generates electron-hole pairs, leading to a voltage difference across the junction. This phenomenon is called the photovoltaic effect and is the basis for solar cells. Answer: b) Photo voltaic effect 15. The function of choke in fluorescent tube is A choke (inductor) in a fluorescent tube circuit performs two main functions: It provides a high voltage spike (strike voltage) to ionize the gas and start the discharge when the circuit is switched on. Once the tube is lit, it limits the current flowing through the tube to a safe operating level. Answer: c) Both a and b are correct That's 2 down. 3 left today — send the next one.