Question

A non viscous liquid of density $\rho$ is filled in a tube with A as the area of cross section, as shown in the figure. If the liquid is slightly depressed in one of the arms, the liquid column oscillates with a frequency

• A. $\frac{1}{2\pi }\sqrt{\frac{\rho gAsin\left[\frac{{\theta }_1+{\theta }_2}{2}\right]}{m}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{\rho gA(sin{\theta }_1-sin{\theta }_2)}{m}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{\rho gA(sin{\theta }_1+sin{\theta }_2)}{m}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{\rho gAsin\left[\frac{{\theta }_1-{\theta }_2}{2}\right]}{m}}$

Answer 1

The correct option is C $\frac{1}{2\pi }\sqrt{\frac{\rho gA(sin{\theta }_1+sin{\theta }_2)}{m}}$

Liquid of density $=\rho$

Area$=A$

Let, the level of liquid displaced by $x$ in one column and in other falls by $x$ in another column, then the difference in levels of liquid.

Vertical displacement:

$h=x(\sin {\theta }_1+\sin {\theta }_2)$

$\therefore$ Pressure differences$P=h\rho g=\rho g(\sin {\theta }_1+\sin {\theta }_2)x$

Now using hook's law in SHM

$F=-kx$,

Frequency, $=\frac{1}{2\pi }\sqrt{\frac{k}{m}}\Rightarrow f=\frac{1}{2\pi }\sqrt{\frac{\rho gA(\sin {\theta }_1+\sin {\theta }_2)}{m}}$