Question

Determine height $h$ of oil in the U tube as shown in figure. Density of oil $=0.9\text{g/cc}$; density of liquid is $1.6\text{g/cc}$ and density of mercury $=13.6\text{g/cc}$.

• A. $9.45\text{cm}$
• B. $18.9\text{cm}$
• C. $27.2\text{cm}$
• D. $16.1\text{cm}$

Answer 1

The correct option is B $18.9\text{cm}$

Given:
density of oil $={\rho }_o=0.9\text{g/cc}$
density of mercury $={\rho }_m=13.6\text{g/cc}$
density of liquid $={\rho }_L=1.6\text{g/cc}$

In equilibrium state, the pressure due to the same level of the same liquid must be equal.


Taking pressure at level C in both arms of the U tube, we have
$\text{Pressure due to}h\text{cm}\text{of oil + pressure due to}(20-h)\text{cm}\text{of mercury = pressure due to}20\text{cm}\text{of liquid}$
i.e. ${\rho }_{0}gh+{\rho }_{m}g(20-h)={\rho }_{L}g\left(20\right)$
$\Rightarrow 0.9\times g\times h+13.6\times g\times (20-h)=1.6\times g\times 20$
$\Rightarrow 0.9h+272-13.6h=32$
$\Rightarrow 12.7h=240$
$\therefore h=\frac{240}{12.7}=18.9\text{cm}$