The diagram of the problem is:
http://www.webassign.net/sf/p9_80.gif

As (a) shows, there is a U tube open at both ends. There is water in the tube. Air in both ends are at equal pressure Pa (atmospheric pressure). The water levels are equal.
Figure (b) shows that oil is added in the right tube so that there is oil column of length L. Oil level is height h above water level.
Then (figure c) air is blown on the top of the left tube so that the water level rises and the levels in the two tubes again become equal. Any air movement in the right tube is prevented with the help of a shield. It can be assumed that only the air in the left tube moves. Any movement of oil, water, air in the right tube can be neglected.
What is the speed with which the air is blown?
Density of oil = 800kg/m^3
Density of water = 1000kmg/m^3
Density of air = 1.29kg/m^3
Depth of oil L=0.04m

Relevant equation:
Bernoulli's equation: P + pgh + 1/2 *pv^2 = constant

Work done so far:
I calculated height h shown in figure b.
Take a point on the water surface and the left tube and a point on the oil surface on the right tube. Pressure at both = Pa. Speed = 0 for both.

Pa + pwater * g * hwater + 0 = Pa + poil * g * hoil
Subtracting Pa from both sides and then dividing both sides by g,
pwater * hwater = poil * hoil
Measuring heights from the horizontal plane through the water/oil interface, we get
hwater = L-h and hoil = L
Therefore
pwater * (L-h) = poil * L
By substituting the given values I found h = 0.008 m

Then I considered figure c.
Let P1 = air pressure in the left tube.
P1 + pwater*gL + 1/2 *pair*v^2 = Pa + poil*gL + 0

The problem is that now I get two unknowns P1 and v and only one equation, which contains these unknowns. How do I solve for v?
I calculated h (of figure c). But is that really needed to calculate v ?
Can I get any equation connecting h and P1 or h and v? Or any other way to solve for v? Thanks.

Two things:

1. First, you stated that you were analyzing the situation at the surface of the water and the surface of the oil, but that's not right. What you actually did - correctly - was to analyze the pressure at the interface between oil and water in the right tube, and at that same height in the water column of the left tube. See the attached figure - your analysis of figure B was conducted at the points where the blue line crosses the tubes - this is where you kow the pressures are equal, and from that your calculation of the value of h is correct.

2. In alayzing figure C, you should consider the point(s) where the blue line crosses the tube, and apply Bernoulli's there, as you did for Figure B. Also, the first term in Bernoulli's is P1, meaning the hydrostatic air pressure, which in this problem is simply P(air). There is no need to introduce a new variable.

Good luck!

If I do like that for figure (c), then
Pa + 1/2 *pa *v^2 + pwater *gL
= Pa + poil *gL
Subtracting Pa from both sides,
1/2 *pa *v^2 + pwater *gL = poil *gL
1/2 *pa *v^2 = (poil - pwater)gL
But poil < pwater
So the equation will give
1/2 *pa *v^2 < 0
Or v^2 < 0
But that is not possible. So how do I take care of that?

One more thing. Is it correct to say that in figure (c), P1 will be equal to Pa? When the air is blown, then will the air pressure in left tube not be reduced?

You are correct - I'm afraid I jumped a step. Let's try it this way:

Let P1 equal the air pressure at the top of the pipe on the left side. See the attached modified figure. If you compare this point against what's happening anywhere else outside in the atmosphere you have:

Pa = P1 + 1/2 p(air) v(air)^2

So now you can find P1, and with that then compare the pressures at the blue line between the left and right tubes. The net effect of all this is that the 1/2 p v^2 gets subtracted, not added. Sorry for the confusion.

Thank you, thank you!
So the equation becomes
Pa - 1/2 *pair * vair^2 + pwater * gL = Pa + poil *gL
- 1/2 *pair * vair^2 + pwater * gL = poil *gL
1/2 * pair * vair^2 = (pwater - poil)gL
Now vair can be easily calculated. Thanks.