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Let the pendulums be in phase after t sec of start. Within this time, if the bigger pendulum executes n oscillations, the smaller one will have executed (n+1) oscillations. The time of n oscillations is:

1. 2π√(5/g) × n
2. 2π√(20/g) × n
3. 2π√(5/g) × (n+1)
4. 2π√(20/g) × (n+1)

Correct answer: 2π√(5/g) × n

Solution

The time period of the bigger pendulum is T = 2π√(5/g). For n oscillations, the total time is nT = 2π√(5/g) × n.

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