The `U`-tube shown in the figure has a uniform cross section. A liquid is filled up to heights `h_(1)` and `h_(2)` in the two arms and they are allowed to move. Neglecting viscosity and surface tension, find the state of the liquid when the levels equalize in the two arms. Find the correct statement.

Options

A. be at rest
B. be moving with an acceleration of `g((h_1 - h_2)/(h_1 + h_2 + h))`
C. be moving with a velocity of `(h_1 - h_2) sqrt((g)/(2(h_1 + h_2 + h)))`
D. exert a net force to the right on the tube.

Answer

C

Explanation

When the levels equalize, let height is `h`
`rho a h_1 + rho a h_2 = rho ah + rho ah` (Volume remains same)
`h = (h_1 + h_2)/(2)`
`U_1 = (rho a h_1 g h_1)/(2)+(rho a h_2 g h_2)/(2) = (rho ag)/(2) (h_1^2 + h_2^2)`
`U_2 = (rho ah g h)/(2) xx 2 rho ag((h_1 + h_2)/(2))^2`
Loss in potential energy `= U_1 - U_2`
`Delta U = rho ag[((h_1^2)/(2) + (h_2^2)/(2))-((h_1^2)/(4) + (h_2^2)/(4) + (h_1 h_2)/(2))]`
=`rho ag[(h_1^2)/(4) + (h_2^2)/(4) -(h_1 h_2)/(2)] = rho ag((h_1 - h_2)/(2))^2`
Mass of liquid in the tube `m = rho ah_1 + rho ah_2 + rho ah`
If mass moves velocity `v`
`(1)/(2) mv^2 = Delta U rArr v = sqrt((2 Delta U)/(m))= (h_1 - h_2) sqrt((g)/(2(h_1 + h_2 + h)))`.