Question

A U tube with both end open to the atmosphere, is partially filled with water, Oil, which is immiscible with water, is poured into one side until it stands at a distance of $10mm$ above the water level on the other side. Meanwhile the water rises by $65mm$ from its original level (see diagram). The density of the oil is:

• A. $928kg{m}^{-3}$
• B. $650kg{m}^{-3}$
• C. $425kg{m}^{-3}$
• D. $800kg{m}^{-3}$

Answer 1

The correct option is A $928kg{m}^{-3}$

From the above diagram,
We get $FE=10mm$

Also, $AB=65+65+10=140mm=0.14m$

$EC=65+65=130mm=0.13m$
Pressure at point B $P_B=P_o+{\rho }_{oil}g\left(AB\right)$

where $P_o$ is the atmospheric pressure.

$\therefore$ $P_B=P_o+{\rho }_{oil}g\left(0.14\right)$

Similarly, pressure at point C $P_C=P_o+{\rho }_{water}g\left(EC\right)$ where ${\rho }_{water}=1000kg/m^3$

$P_C=P_o+{\rho }_{water}g\left(0.13\right)$

Pressure inside tube from both side will be same i.e $P_B=P_c$

$\therefore$ $P_o+{\rho }_{oil}g\left(0.14\right)$ $=P_o+{\rho }_{water}g\left(0.13\right)$
Or ${\rho }_{oil}g\left(0.14\right)$ $={\rho }_{water}g\left(0.13\right)$
Density of oil ${\rho }_{oil}=\frac{1000\times 0.13}{0.14}$
$⟹{\rho }_{oil}=928kg{m}^{-3}$