Question

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?

Short answer

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

Step-by-step solution

Listing the given quantities.

• 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,${\rho }_m=13.65g/c{m}^3.$
  

Understanding the concept of pressure.

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:

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

(a) Calculation of the amount of the increased height of mercury from its initial level in the left arm of the simple U tube.

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$0.413\mathrm{cm}.$