# U-tube, water and oil.

1. Dec 11, 2005

### JJMezzapesa

I really need help on this problem. I've been stuck forever

A U-TUBE AT BOTH ENDS IS PARTIALLY FILLED WITH WATER. OIL (DENSITY = 750KG/M2) IS THEN POURED INTO THE RIGHT ARM AND FORMS A COLUMN L=5.00 CM HIGH. DETERMINE THE DIFFERENCE h IN THE HEIGHTS OF THE TWO LIQIUD SURFACES. THE RIGHT ARM IS THEN SHIELDED FROM ANY AIR MOTION WHILE AIR IS BLOWN ACROSS THE TOP OF THE LEFT ARM UNTIL THE SURFACES OF THE TWO LIQUIDS ARE AT THE SAME HEIGHT. DETERMINE THE SPEED OF OF THE AIR BEING BLOWN ACROSS THE LEFT ARM. ASSUME THE DENSITY OF AIR IS 1.29 KG/M2

Last edited: Dec 12, 2005
2. Dec 11, 2005

### Pengwuino

What work have you done?

3. Dec 11, 2005

### JJMezzapesa

i dont know where to start

4. Dec 13, 2005

### siddharth

5. Dec 13, 2005

### andrevdh

The oil column will be higher than the water column since the density of the oil is less than that of water (1000 kg/m^3). The weight of the oil column can therefore be balanced by a smaller column of water. If one stays in the same liquid (water) then the pressure at the same level (or height) are equal. The pressure at pb is therefore the same at pc in the water. What you need to find is therefore the difference between 5 cm and hw. pa is the atmospheric pressure on top of the columns.

Last edited: Nov 29, 2006
6. Jun 4, 2006

### parttime

i was able to get the h without a problem but i am having trouble getting v
as im not sure what i am comparing to solve for v

7. Jun 6, 2006

### andrevdh

The drawing will basically be the same with the top of the water level with the top of the oil in the other arm, that is the top of the liquids are on the same height in the two arms. The pressure on top of the water will be reduced to say $$p_r$$ by the air rushing over it.

Last edited: Jun 6, 2006