A $ U$ tube of uniform bore of cross-sectional area $A$ has been set up vertically with open ends facing up. Now $m$ gm 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
A$T = 2\pi \sqrt {\frac{M}{g}} $
B$T = 2\pi \sqrt {\frac{{MA}}{{gd}}} $
C$T = 2\pi \sqrt {\frac{M}{{gdA}}} $
D$T = 2\pi \sqrt {\frac{M}{{2Adg}}} $
Diffcult
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D$T = 2\pi \sqrt {\frac{M}{{2Adg}}} $
d (d) If the level of liquid is depressed by $y\, cm$ on one side, then the level of liquid in column $P$ is $2y\, cm $ higher than $B$ as shown.
The weight of extra liquid on the side $P = 2Aydg$.
This becomes the restoring force on mass $M$.
$\therefore $ Restoring acceleration $ = \frac{{ - \,2Aydg}}{M}$
This relation satisfies the condition of $SHM$ i.e. $a \propto - y$.
Hence time period $T = 2\pi \sqrt {\frac{{{\rm{Displacement}}}}{{{\rm{|Acceleration|}}}}} $
$ = 2\pi \sqrt {\frac{y}{{\frac{{2Aydg}}{M}}}} $
$ \Rightarrow T = 2\pi \sqrt {\frac{M}{{2Adg}}} $
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