NEET Class neet Answered
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Asked by brijk456 | 01 Oct, 2019, 11:24: AM
Expert Answer
Let us consider a mercury filled U-tube as shown in figure. Let L be the height of mercury colum on each limb.
If we displace mercury for a distance x downwards in left limb, excees height of mercury on right side limb
is 2x as shown in figure.
Hence force F due to excess pressure acting on whole mercury is F = - [ (2x) A ρ g ] = - (2 A ρ g) x .........................(1)
Eqn.(1) , shows that force F is proportional to displacement x , hence the motion is SHM with foce constant k = (2 A ρ g).
(-ve sign is for downward force )
period T of oscillation is given by,
Answered by Thiyagarajan K | 02 Oct, 2019, 12:31: PM
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