Fig, shows a U-tube of uniform cross-sectional area `A` accelerated with acceleration `a` as shown. If `d` is the seperation between the limbs. Then the difference in the levels of the liquid in the `U-tube` is

Options

A. `(ad)/(g)`
B. `(g)/(ad)`
C. `adg`
D. `ad+g`

Answer

A

Explanation

Mass of liquid in horizontal portion of U-tube =`A d rho`
Pseudo force on this case = `Adrhoa`
force due to pressrue difference in the two limbs
`=(h_(1)rho g - h_(2)rho g)A`
Equating, `(h_(1)-h_(2))rho g A = A d rho a`
Equanting , `(h_(1)-h_(2))rhogA = A d rho a`
or `h_(1) - h_(2) = (Adrhoa)/(rho g A ) = (ad)/(g)`.