Calculate time period of oscillation of an ideal liquid oscillating in a Ushaped tube
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Answered by
1
T=2π√l/√2g
Where the two angles of the arms are same, i.e. 90°
Answered by
0
Answer:
Time period is directly proportinal to length and acceleration due to gravity.
Thus combining both above concepts we derive
T=2pie × square root of l/g
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