Question

The vertical limbs of a $U$ shaped tube are filled with a liquid of density $p$ upto a height $h$ on each side. The horizontal portion of the $U$ tube having length $2h$ contains a liquid of density $2p$. The $U$ is moved horizontally with an accelerator $g/2$ parallel to the horizontal arm. The difference in heights in liquid levels in th two vertical limbs, at steady state will be

• A. $\frac{2h}{7}$
• B. $\frac{8h}{7}$
• C. $\frac{4h}{7}$
• D. None of these

Answer 1

Answer: (B)

$\frac{8h}{7}$

Given

$P_1=$ pressure at left end base of u-tube

$P_2=$ pressure at right end base of u-tube

Since the u-tube is accelerating

$(P_1-P_2)A=ma$

Here. $A$ is the area of cross-section of u-tube.

We know that the pressure at depth is given by:

$P=hdg$

Here, $d$ is the density, $h$ is the height of the liquid column and $g$ is the acceleration due to gravity.

Substitute the values of pressure:

$\Rightarrow \left(dhg+2d{h}_{1}g-d(h-h_1\right)g)A=\left(d{V}_1+2d{V}_2\right)\frac{g}{2}$

$V_1\&V_2$ are volume of liquids with density $d\&2d$ respectively in each part of tube.

$\Rightarrow \left(dhg+2d{h}_{1}g-d(h-h_1\right)g)A=\left(dA{h}_1+2dA(2h-h_1\right)\frac{g}{2}$ ------(i)

Now solving the equation (i)

$\Rightarrow dhgA+2d{h}_{1}gA-dhgA+d{h}_{1}gA=\frac{d{h}_{1}gA}{2}+2dhgA-d{h}_{1}gA$

$\Rightarrow 3d{h}_{1}gA=\frac{d{h}_{1}gA}{2}+2dhgA-d{h}_{1}gA$

$\Rightarrow \frac{7d{h}_{1}gA}{2}=2dhgA$

$\therefore h_1=\frac{4h}{7}$

Now, differnce in height of 2 vertical limbs,

$\Rightarrow 2\times h_1=2\times \frac{4h}{7}=\frac{8h}{7}$

So the correct option is $B.$