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
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:
A. 2.5%
B. 5%
C. 20%
D. 10%
Answer: Option D
Solution (By JKSSB Mock Tests)
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.
Explanation:
Electric potential V at a point is defined as work done per unit charge to bring a test charge from infinity to that point: V = W/q. Thus potential difference between two points is work per unit charge to move between them. Electric field intensity is force per unit charge; potential energy is work to assemble charges; capacitance is charge storage ability. Memory tip: 'Potential = work per unit charge; field = force per unit charge'. This definition-based question tests electrostatics fundamentals, frequently appearing in competitive exams. Always distinguish potential (scalar, work/charge) from field (vector, force/charge).
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