A V-shaped glass tube of uniform cross section is kept in a vertical plane as shown. A liquid is poured in the tube. In equilibrium the level of liquid in both limbs of tube are equal. Find the angular frequency of small oscillation of liquid.

Options

A. `pi sqrt((2h)/(g sin alpha . sin beta))`
B. `2 pi sqrt((h)/(g sin alpha . sin beta))`
C. `2 pi sqrt((h)/(2g sin alpha . sin beta))`
D. None of these

Answer

B

Explanation

Let us displace the liquid of left limb by a distance x, the liquid of right limb will rise relative equilibrum line by same distance x

Excess length of the liquid on the right limb
`Delta l = x + x`
(i)
From the figure ,
`y = x sin alpha = x' sin beta`
(ii)
from eq. (i)
`x = (x sin alpha)/(sin beta)`
from eqs (i) and (iii) , `Delta l = ((sin alpha + sin beta)/(sin beta)) alpha`
This excess length `Delta l` in right limb will proved restoring force to the liquid in tube
Weight of the liquid column
`W = p A. Delta l s = ((sin alpha + sin beta)/(sin beta)) rho Axg` ltbr `rho` is the density of liquid and A is the area of cross section of tube. The component of `W` along the tube will be equal to restoring force
`F = W sin beta = rho (sin alpha + sin beta)Axg`
This force helps the entire liquid to restore its original position by moving it which an acceleration `a = F//m, m = mass` of total liquids
`a = (rho (sin alpha + sin beta) Axg)/(m)`
Here `m =` mass of liquid in the tube `= A rho l l = `Length of total liquid column
`= (h)/(sin alpha) + (h)/( sin beta) = h [(sin alpha + sin beta)/(sin alpha + sin beta)]`
Hence `a = (rho (sin alpha + sin beta) Axg)/(A rho [h (sin alpha + sin beta)/(sin alpha. sin beta)])`
`bara = - ((g sin alpha. sin beta)/(h)) barx [("with proper"),("vector sign")]`
Compare with `bara = - omega^(2) bar x`
which gives
`omega = sqrt((g sin alpha . sin beta)/(h)) implies T = 2 pi sqrt((h)/(g sin alpha . sin beta))`