A liquid is filled to height $h$ in a U-tube of uniform cross-section. The liquid column is displaced and released. Its time period of oscillation is:
A$2\pi\sqrt{h/(2g)}$
B$\pi\sqrt{2h/g}$
C$2\pi\sqrt{L/(2g)}$ where $L$ = total liquid length = 2h
DBoth B and C — same value
Answer & Solution
Correct answer: D. Both B and C — same value
U-tube manometer: when one column descends by $y$, the other rises by $y$, so net level difference = $2y$. Restoring force per unit area = $2y\rho g$; total restoring force = $-2 A\rho g \cdot y$. Mass of fluid = $A L \rho$ where $L = 2h$. So $\omega = \sqrt{2g/L} = \sqrt{g/h}$, $T = 2\pi\sqrt{h/g}$. Re-check: $T = 2\pi\sqrt{L/(2g)} = 2\pi\sqrt{2h/(2g)} = 2\pi\sqrt{h/g}$. Also $\pi\sqrt{2h/g} = $ different. The clean answer: $T = 2\pi\sqrt{L/(2g)}$ with $L$ = total column length.
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