Question

The acceleration of an U-tube containing a liquid, is $ {{a}_{0}} $ . If the diameter of circular part of U-tube is $ I $ then the difference in the heights of liquid in two arms will be

• A. $ \sqrt{\frac{A{{a}_{0}}gI}{2}} $
• B. $ \sqrt{\frac{{{a}_{0}}I}{g}} $
• C. $ \frac{A{{a}_{0}}g}{{{I}^{2}}} $
• D. $ \frac{{{a}_{0}}I}{g} $

Answer 1

Answer: (D)

The diameter of circular part of U-tube is $ l $ . This means the distance between vertical arms of U-tube is $ l $ . Suppose, the atmospheric pressure is $ {{p}_{a}}. $ Since, U-tube is accelerated horizontally. So, inertia force will be experienced in horizontal direction. As, net force = ma, gives $ {{p}_{a}}A+Al\times \rho \times {{a}_{0}} $ $ ={{p}_{a}}A+\rho gh\times A $ where, A = area of x-section of tube. and $ \rho $ = density. $ \Rightarrow $ $ hg={{a}_{0}}l $ $ \Rightarrow $ $ h=\frac{{{a}_{0}}l}{g} $