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. $\Rightarrow$ $hg=a_{0}l$ $\Rightarrow$ $h=\frac{a_{0}l}{g}$