Question regarding the differential height of mercury in a manometer

[Physics345]

Homework Statement

Water flows upward through an inclined pipe with a 20-cm diameter at a rate of 1.5 m^3/min. The diameter of the pipe is then reduced to 10-cm. The pressure difference between the two pipe sections is measured using a mercury manometer. The elevation difference between the two points on the pipe where the two arms of the mercury between the two points on the pipe where the two arms of the manometer are attached is 0.2-m. Neglecting frictional effects, determine the differential height of mercury between the two pipe sections.

Relevant Equations

Bernoulli Eq:$p_{1} + \frac{1}{2}Pv_{1}^{2} + Pgy_{1}=p_{2} + \frac{1}{2}Pv_{2}^{2} + Pgy_{2}$
Where P= Density and p= pressure
Continuity: $A_{1}v_{1}=A_{1}v_{2}= Q$

Given:
$y_{2} - y_{1}= 0.2m$
$Q= 1.5\frac{m^{3}}{min}$
$Q= 0.025\frac{m^{3}}{s}$ After conversion
$D_{1}= 0.2m$ After conversion
$D_{2}= 0.1m$ After conversion
$r_{1}= 0.1m$
$r_{2}= 0.05m$
$p_{1} - p_{2} = \frac{1}{2}P(v_{2}^{2}-v_{1}^{2}) + Pg(y_{2}-y_{1})$

Calculating Velocity using circular Area of the pipe
$v_{1}= \frac{Q}{A_{1}}$
$v_{2}= \frac{Q}{A_{2}}$
$A_{1}=0.03142m^{2}$
$A_{2}=0.007854m^{2}$
$v_{1}= 0.7957\frac{m}{s}$
$v_{2}= 3.183 \frac{m}{s}$

Inputting

$p_{1} - p_{2} = \frac{1}{2}(1000\frac{kg}{m^{3}})((3.183\frac{m}{s})^{2}-(0.7957\frac{m}{s})^{2}) + (1000\frac{kg}{m^{3}})(9.81\frac{m}{s^{2}})(0.2m)$
$p_{1} - p_{2} =6710.9\frac{kg}{ms}$

My question is where do I go from here to find the height?

 

[Physics345]

I found the height, and its 0.05m