The time period of a simple pendulum is measured as 2.0 s with an error of 0.1 s. The percentage error in the calculated value of g is approximately: MCQ with Answer and Explanation

Pendulum period T = 2π√(l/g) ⇒ g = 4π²l/T². For error propagation, % error in g = % error in l + 2 × % error in T. Assuming l measured precisely, % error in g ≈ 2 × (% error in T). % error in T = (0.1/2.0)×100% = 5%. Thus % error in g ≈ 2×5% = 10%. Memory tip: 'g ∝ 1/T² ⇒ %Δg = 2 × %ΔT'. This error analysis problem tests experimental physics skills, frequently appearing in competitive exams. Always apply power rule for error propagation: if Q ∝ xⁿ, %ΔQ = |n| × %Δx.