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

Options

A. `(1)/(2pi)sqrt((rhogAsin((theta_(1)+theta_(2))/(2)))/(m))`
B. `(1)/(2pi)sqrt((rhogA(sintheta_(1)-sintheta_(1)))/(m))`
C. `(1)/(2pi)sqrt((rhogA(sintheta_(1)+sintheta_(2)))/(m))`
D. `(1)/(2pi)sqrt((rhogAsin((theta_(1)-theta_(2))/(2)))/(m))`

Answer

C

Explanation

Initially the level of liquid in the two limbs will be at the same height. If the liquid is depressed by y in one limb, it will rise by `y` along the length of the tube in the other limb. The force that is responsible for restoring the liquid levels in the two arms of the tube is
`F=-DeltaPA=-(h_(1)+h_(2))rhogA`,
where `DeltaP` is the pressure difference and A is the area of cross section of the tube, `h_(1) and h_(2)` being the rise and fall of liquid levels in the two arms in vertical direction respectively.
`F=-(ysintheta_(1)+ysintheta_(2))rhogA`
`=-rhogA(sintheta_(1)+sintheta_(2))y` . . . (i)
`ma=-rhogA(sintheta_(1)+sintheta_(2))y`
or `a=-(rhogA(sintheta_(1)+sintheta_(2))y)/(m)`
For SHM
`a=-omega^(2)y`
`a=-omega^(2)y` . . . (ii)
Comparing (i) and (ii) , we get
`omega^(2)=(rhogA(sintheta_(1)+sintheta_(2)))/(m)`
`omega=sqrt((rhogA(sintheta_(1)+sintheta_(2)))/(m))`
Frequency, `upsilon=(omega)/(2pi)=(1)/(2pi)sqrt((rhogA(sintheta_(1)+sintheta_(2)))/(m))`