The Q factor of a resonant LCR circuit is measure of MCQ with Answer and Explanation

Explanation:
54 km/h = 15 m/s. u=15, v=0, t=5. a = (0-15)/5 = -3 m/s². s = ut+½at² = 15×5 + ½×(-3)×25 = 75 - 37.5 = 37.5 m. Or s = (u+v)/2 × t = 7.5×5 = 37.5 m.

Explanation:
For parallel combination: 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ = 1/2 + 1/3 + 1/6 = (3+2+1)/6 = 6/6 = 1. Thus R_eq = 1Ω. Note: Equivalent resistance in parallel is always less than the smallest individual resistance (here 2Ω). Memory tip: 'Parallel: reciprocal sum; result < min(R)'. This standard calculation tests parallel resistance formula application, frequently appearing in competitive exams. Always simplify fractions carefully; common denominator method avoids errors. Verify: 1Ω < 2Ω, consistent with parallel behavior.

Explanation:
Work done to stretch spring from 0 to x: W₁ = ½kx². Work done from 0 to 2x: W₂ = ½k(2x)² = 2kx². Thus work for additional stretch from x to 2x: ΔW = W₂ - W₁ = 2kx² - ½kx² = ³/₂kx². Spring force is variable (F=kx), so work is integral of F·dx, yielding parabolic energy storage. Memory aid: Elastic potential energy U = ½kx²; always calculate difference for incremental work. This problem tests understanding of work done by variable forces, a common theme in energy conservation questions in competitive exams.