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

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(D) If the level of liquid is depressed by $y$ on one side, then the level of liquid in the other column is $2y$ higher than the first side, as shown

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

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(D) If the level of liquid is depressed by $y$ on one side, then the level of liquid in the other column is $2y$ higher than the first side, as shown in the figure.The weight of the extra liquid column of height $2y$ is the restoring force.Restoring force $F = -(	ext{Volume} 	imes 	ext{density} 	imes g) = -(A 	imes 2y 	imes d 	imes g) = -2Adgy$.Since $F = Ma$, we have $Ma = -2Adgy$, which gives acceleration $a = -(\frac{2Adg}{M})y$.This is the equation of simple harmonic motion $(a = -\omega^2 y)$, where $\omega^2 = \frac{2Adg}{M}$.The time period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{1}{\omega^2}}$.Substituting the value of $\omega^2$, we get $T = 2\pi \sqrt{\frac{M}{2Adg}}$.

$A$ $U$-tube of uniform bore of cross-sectional area $A$ has been set up vertically with open ends facing up. Now $M$ grams of a liquid of density $d$ is poured into it. The column of liquid in this tube will oscillate with a period $T$ such that

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