Question

Determine the period of oscillation of $200\text{g}$ of $Hg$ in a bent U-tube whose right arm forms an angle ${30}^{\circ }$ with the vertical as shown. The cross-sectional area of the tube is $0.5{\text{cm}}^2$. Neglect viscosity and density of mercury is $13.6{\text{g/cm}}^3$

• A. $0.68\text{s}$
• B. $0.74\text{s}$
• C. $0.80\text{s}$
• D. $0.88\text{s}$

Answer 1

The correct option is C $0.80\text{s}$

Suppose you depress the liquid in the left arm by a distance $y$. Then , the liquid in the right arm will rise by vertical distance $y\cos \theta$ where $\theta ={30}^{\circ }$

Then, restoring force $F=-\rho gAy(\cos \theta +1)=-m{\omega }^{2}y$
where $A$ is the area of cross section of the tube.

$\Rightarrow \omega =\sqrt{\frac{\rho gA(1+\cos \theta )}{m}}$

$\therefore$ Time period $T=\frac{2\pi }{\omega }$
$=2\pi \sqrt{\frac{m}{A\rho g(1+\cos \theta )}}$
$=2\pi \sqrt{\frac{0.2}{0.5\times {10}^{-4}\times 13.6\times {10}^3\times 10(1+\frac{\sqrt{3}}{2})}}$
$\approx 0.8\text{s}$