Q140 / 150 · MHT CET 2022 (07 Aug Shift 2) · Physics

A U-tube of a uniform bore is with its arm vertical. The length of the liquid in the two arms or the U-tube is $L$; the time period $T$ of the oscillation of the liquid column when it is displaced by ' $Y$ ' is ( $g=$ acceleration due to gravity)

  • A
    $2 \pi \sqrt{\frac{Y}{g}}$
  • B
    $2 \pi \sqrt{\frac{2 L}{g}}$
  • C
    $2 \pi \sqrt{\frac{L}{g}}$
  • Correct answer
    D
    $2 \pi \sqrt{\frac{L}{2 g}}$

Solution

Due to the tiny displacement $y$, the force balance in the tube reads: $\begin{aligned} & \frac{d^2 y}{d t^2}=-\frac{F}{m}=-\left(\frac{2 y \rho g}{L A \rho}\right) \\ & \Rightarrow \frac{d^2 y}{d t^2}=-\left(\frac{2 g}{L}\right) y\end{aligned}$ or $\omega^2=\frac{2 g}{L} \Rightarrow T=2 \pi \sqrt{\frac{L}{2 g}}$

Hence, the correct answer is option: 'D'.

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