Question

Two non-viscous, incompressible and immiscible liquids of densities $\rho$ and $1.5\rho$ are poured into the two limbs of a circular tube of radius R and small cross-section kept fixed in a vertical plane as shown in figure. Each liquid occupies one fourth the circumference of the tube.
(a) Find the angle that the radius to the interface makes with the vertical in equilibrium position.
(b) If the whole liquid is given a small displacement from its equilibrium position, show that the resulting oscillations are simple harmonic. Find the time period of these oscillations.

• A. (a) $\theta ={\tan }^{-1}\frac{1}{5}$(b) $4.5\sqrt{R}$.
• B. (a) $\theta ={\tan }^{-1}\frac{1}{5}$(b) $3.5\sqrt{R}$.
• C. (a) $\theta ={\tan }^{-1}\frac{1}{5}$(b) $2.5\sqrt{R}$.
• D. (a) $\theta ={\tan }^{-1}\frac{2}{5}$(b) $2.5\sqrt{R}$.

Answer 1

Answer: (C)

(a) $\theta ={\tan }^{-1}\frac{1}{5}$

(b) $2.5\sqrt{R}$.

$\text{Solution}:$

(a) The pressure due to the liquid on the right is

$P_2(Rsin\theta +Rcos\theta )\rho g+(R-Rcos\theta )1.5\rho g$

Since the liquids are in equilibrium

$P_1=P_2$ or

$(R-Rsin\theta )1.5\rho g=(R-Rcos\theta )\rho g+(R-Rcos\theta )1.5\rho g$

Solving, we get

$tan\theta =0.2$ or

$\theta =ta{n}^{-1}0.2=ta{n}^{-1}\frac{1}{5}$

(b) Let the whole liquid be given a small angular displacement $\alpha$ towards right. Then the pressure difference between the right and the left limbs is

$dP=P_2-P_1$

$=\left[Rsin\right(\theta +\alpha )+Rcos(\theta +\alpha )\rho g+[R-Rcos(\theta +\alpha )1.5\rho g-[R-Rsin(\theta +\alpha \left)\right]1.5\rho g$

$=R\rho g\left[2.5sin\right(\theta +\alpha )-0.5cos(\theta +\alpha \left)\right]$

$=R\rho g\left[2.5\right(sin\theta cos\alpha +cos\theta sin\alpha )-0.5(cos\theta cos\alpha -sin\theta sin\alpha \left)\right]$

For small $\alpha$

$sin\alpha =\alpha ,cos\alpha \cong 1$

$\therefore dP=R\rho g[2.5sin\theta +2.5\alpha cos\theta -0.5cos\theta +0.5\alpha sin\theta ]$

As $tan\theta =0.2$,

$sin\theta \frac{0.2}{/sqrt1.04}\approx 0.2$ ,

$cos\theta =\frac{1}{\sqrt{1}.04}\approx 1$

$\therefore dP=R\rho g[2.5\times 0.2+2.5\alpha -0.5+0.5\times 0.2\alpha ]$

$=R\rho g\left[2.6\alpha \right]=236\rho g$

where $y=R\alpha$, the linear displacement

$\therefore$ Restoring force $F=2.6\rho gAy$

This shows that $F\propto y$

Hence the motion is simple harmonic with force constant $k=2.6\rho gA$

Now, total mass of the liquid

$m=\frac{2\pi R}{4}A\rho +\frac{2\pi R}{4}A\left(1.5p\right)=\frac{5\pi RA\rho }{4}$

$\therefore$ Time period $T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{5\pi rA\rho }{4\times 236\rho gA}}=\pi \sqrt{\frac{1.93\pi R}{9.8}}=2.47\sqrt{R}seconds.$

$\text{Hence C is the correct option}$