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A simple pendulum of length 1 m on the earth's surface (g = π² m/s²) has period
A1 s
Bπ s
C2 s
D0.5 s
Answer & Solution
Correct answer: C. 2 s
1. For a simple pendulum T = 2π √(L/g).
2. Substituting L = 1, g = π²: T = 2π √(1/π²) = 2π × (1/π).
3. T = 2 s.
_Source: Maharashtra Balbharati Std XII Physics, Ch 5 "Oscillations" §5.7_
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