Question

A U-tube is filled with a single homogeneous liquid. The liquid is temporarily depressed on one side by a piston. The piston is removed and the level of the liquid in each side oscillates. The period of oscillation is (Where L is total length of liquid in the tube)

• A. $\pi \sqrt{\left(2L/g\right)}$
• B. $\pi \sqrt{\left(L/g\right)}$
• C. $\pi \sqrt{\left(2g/L\right)}$
• D. $\sqrt{\left(2g/L\right)}$

Answer 1

The correct option is A $\pi \sqrt{\left(2L/g\right)}$

Where L is total length of the liquid in the tube.
If the liquid is depressed by a distance x,
The pressure difference$=2\rho gx$
So,
As the restoring force i.e.,$2\rho gAx$ is proportional to $x$,
So,
$2\rho gAx=\left(\rho LA\right)w^{2}x$
$\Rightarrow \frac{2g}{L}x=w^{2}x$
So,
$w=\sqrt{\frac{2g}{L}}$
So,
the period of oscillation$=\frac{2\pi }{w}=2\pi \sqrt{\frac{L}{2g}}$
$\Rightarrow T=\pi \sqrt{\frac{2L}{g}}$