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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 A. `(ad)/(g)`B. `(g)/(ad)`C. `adg`D. `ad+g` |
Answer» Correct Answer - A 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)`. |
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