A simple open U-tube contains mercury. When $\mathbf{11}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{cm}$of water is poured into the right arm of the tube, how high above its initial level does the mercury rise in the left arm?

• Height of the water in the right arm $\left(h_w\right)=11.2cm.$
  
• The density of mercury is, ${\rho }_w=1g/c{m}^3.$
  
• The density of water is,${\rho }_m=13.65g/c{m}^3.$
  

Due to water, the mercury will be pushed down by height$\mathbf{\left(}\mathit{h}\mathbf{\right)}$in the right arm and pushed up by the same height $\mathbf{\left(}\mathit{h}\mathbf{\right)}$in the left arm from its initial level. Then the difference between the two mercury levelswill be$\mathbf{2}\mathbf{h}\mathbf{.}$. Using the concept that the pressure at two points in a fluid at the samehorizontal level is the same, we can find (h).

Formula:

$\mathrm{p}={\mathrm{p}}_0+\mathrm{\rho gh}$

From the figure, the pressure at point $A\left(P_a\right)$is due to atmospheric pressure $P_0$and the pressure of the water with height $h_w.$

Pressure at point $B\left(P_b\right)$is due to atmospheric pressure $P_0$and pressure of the mercury with height $2h.$

$P_a=P_b$

$P_0+{\rho }_{w}x{h}_{w}xg=P_0+{\rho }_{m}x2hxg$

By canceling equal terms and rearranging the equation, we get

$h=\frac{{\rho }_{w}x{h}_w}{2{\rho }_m}$

$=\frac{1\mathrm{g}/{\mathrm{cm}}^3\mathrm{x}11.2\mathrm{cm}}{2\mathrm{x}13.65\mathrm{g}/{\mathrm{cm}}^3}$

$=0.413\mathrm{cm}$

The increased height of the mercury is$0.413\mathrm{cm}.$