One end of V-tube containing mercury is connected to a suction grouup and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of `45^(@)` each. A small pressure difference is created between two columns when the ssuction pump is removed. Will the column of mercuty in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Fing the time period of oscillation.

Explanation

Refer. Let l be the length of liquid in each arm of V tube in equilibrium position. Let A be the area of crosssection of liquid and `rho` be the density of liquid.
Mass of liquid in V-tube, `m=2lArho`.
Let the liquid be pushed down through small distance x is one arm of V tube. Its level will be raised by distance x in other arm of V tube.
Pressure difference due to difference of levels in V tube is
`Delta P=(2x si n45^(@))rhog`
Thrust perpendicular to the liquid meniscus in the tube is
`F=-(DeltaP)A//cos45^(@)=-2xsin45^(@)rhogA//cos45^(@)=-2rhogAx` ...(i)
As `Fpropx` and `-ve` sign shows that F is directed towards equilibrium position, hence if liquid is left free it will execute linear SHM. Comparing (i) with `F=-ky`. We have , spring factor `k=2rhogA`.
Inertia factor `=` mass of liquid in V tube `=2lArho`
Time period,`T=2pisqrt((i n ertia fact o r)/(spri ng fact o r))=2pisqrt((2lArho)/(2rhogA))==2pisqrt((l)/(g))`