Question

An open-ended U-tube of uniform cross-sectional area contains water (density $10$${}^3$ ${\text{kg.m}}^{-3}$). Initially, the water level stands at $0.29\text{m}$ from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of density $800{\text{kg.m}}^{-3}$ is added to the left arm until its length is $0.1\text{m}$ , as shown in the schematic figure below. The ratio$\left(\frac{h_1}{h_2}\right)$ of the heights of the liquid in the two arms is

• A. $\frac{15}{14}$
• B. $\frac{35}{33}$
• C. $\frac{7}{6}$
• D. $\frac{5}{4}$

Answer 1

The correct option is B $\frac{35}{33}$

As two liquids are immiscible, total height of limbs in U tube manometer can be considered as $h_1+h_2=0.29\left(2\right)+0.1$

$h_1+h_2=\mathrm{0.68.......}\left(1\right)$

Pressure at bottom of left limb is
$\Rightarrow p_o+{\rho }_k$$g$ $\left(0.1\right)$ $+$ ${\rho }_w$$g$$(h_1-0.1$)

[ ${\rho }_k$ = density of kerosene and ${\rho }_w$ $=$ density of water ]
pressure at bottom of right limb is $p_o+{\rho }_w{h}_{2}g$

Using pascals law,
$\Rightarrow {\rho }_{k}g\left(0.1\right)+{\rho }_{w}g{h}_1-{\rho }_{w}g\times \left(0.1\right)=p_{w}g{h}_2$

$\Rightarrow$ $800\times 10\times 0.1\times 1000\times 10\times h_1$$-1000\times 10\times 0.1=1000\times 10\times h_2$

$\Rightarrow h_1-h_2=\mathrm{0.02.......}\left(2\right)$

On solving $\left(1\right)\&\left(2\right)$
$\Rightarrow h_1=0.35$

$\Rightarrow h_2=0.33$

So,$\frac{h_1}{h_2}=\frac{35}{33}$