Question

A vertical U-tube has two liquid 1 and 2. The height of liquids columns in both the limbs are h and 2h, as shown in the figure. If the density of the liquid 1 is $2\rho$.
a. Find the density of liquid 2.
b. If we accelerate the tube towards right till the heights of liquid columns will be the same, find the acceleration of the tube

Answer 1

Consider figure (a) :

Let the density of liquid 1 be $d$.

As the tube is at rest, thus $P_A=P_B$

OR $P_o+\left(2\rho \right)gh=P_o+dg\left(2h\right)$ $⟹d=\rho$

Consider figure (b) :

Let the tube moves with an acceleration $a$ and the height of liquid in each vertical column be $y$.

Thus the lenght of liquid 1 in horizontal pipe is $(2h-y)$ and that of liquid 2 is $(3h-y)$

$\therefore$ $2h=(2h-y)+(3h-y)$ $⟹y=\frac{3h}{2}$

Thus the lengths of liquid 1 and 2 in horizontal pipe are $\frac{h}{2}$ and $\frac{3h}{2}$ respectively.

Using $P_A-P_B=\left(2\rho \right)a(2h-y)+\rho a(3h-y)$

$\therefore$ $\left(2\rho \right)gy-\rho gy=2\rho a(2h-y)+\rho a(3h-y)$

$\therefore$ $y=\frac{3h}{2}$ $⟹$ $\frac{3}{2}\rho gh=\frac{5}{2}\rho ah$

$⟹$ $a=\frac{3g}{5}$