A U-tube of uniform bore of cross-sectional a | vedclass

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(B) When the liquid column is depressed by a displacement '$y$' on one side, the level of the liquid rises by '$y$' on the other side. Thus, the total

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$2 \pi \sqrt{\frac{M}{2 Adg}}$

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(B) When the liquid column is depressed by a displacement '$y$' on one side, the level of the liquid rises by '$y$' on the other side. Thus, the total difference in the height of the liquid levels in the two arms is '$2y$'.The weight of this extra liquid column acts as the restoring force.Restoring force $F = -(	ext{Volume} 	imes 	ext{density} 	imes g) = -(A 	imes 2y 	imes d 	imes g) = -2Adgy$.Since $F = Ma$, where '$M$' is the total mass of the liquid, we have $Ma = -2Adgy$.Therefore, the acceleration $a = -(\frac{2Adg}{M})y$.Comparing this with the standard $SHM$ equation $a = -\omega^2 y$, we get $\omega^2 = \frac{2Adg}{M}$.The time period '$T$' is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{M}{2Adg}}$.

$A$ $U$-tube of uniform bore of cross-sectional area '$A$' is set up vertically. '$M$' grams of a liquid of density '$d$' is poured into it. The column of liquid in this tube will oscillate with a period '$T$', which is equal to [$g$ = acceleration due to gravity]

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