Determine the period of oscillations of mercury of mass `m=200g` poured into a bent tube (figure) whose right arm forms an angle `theta=30^(@)` with vertical. The cross-sectional area of the tube is `S=0.50 cm^(2)`. The viscosity of mercury is to be neglected.

Explanation

Let A be the uniform cross-sectional area of the tube. Let the level of liquid of density `rho` be depressed in the vertical arm of tube by distance y . The verticall rise of liquid in the slant portion of the tube `=ycostheta.`
The vertical difference in liquid level in two arms of bent tuve `-y+ycostheta=(1+costheta)y` in two arms of bent tube will provide a restoring force to the liquid. So restoring force on the liquid is
`F=-[` weight of the liquid column of height `(1+costheta(y]`
`=-(1-costheta)yrhogA`
`=-(1+costheta)Arhogy` ...(i)
Since `Fpropy` and F is directed towards equilibrium position of liquid in two arms of bent tube, hence if the force is removed or liquid is left free, it will execute SHM in bent tube.
Comparing (i), with , `F=-ky`
We have, spring factore, `k=(1+costheta)Arhog.`
Here, inertia factore `=` mass of mercury `=m=200g`
Period of oscillation, `T=2pisqrt((i n ertia fact o r)/(spr i n g fact o r))`
`=2xx(22)/70sqrt((200)/((1+cos30^(@))xx0.5xx13.6xx980))`
`=0.8 s`