A u-shaped tube, open to the air on both ends,contains mercury. Wather is poured into the left arm until the water column is 10.0 cm deep. How far upward from its initial position does the mercury in the right arm rise?

(1/2)*10.0 cm/13.6 = 0.37 cm

13.6 is the specific gravity of mercury.

The added weight of water on one side will then be balanced by the higher column of Hg on the other side. The 1/2 factor is there because it rises on one side and falls on the other

To determine how far upward the mercury in the right arm rises, we can use the principles of hydrostatic pressure and Pascal's law.

Pascal's law states that the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the container.

In this scenario, the u-shaped tube is open to the air on both ends. Since the pressure from the air is the same on both sides, we can assume that the pressure due to the air cancels out.

When water is poured into the left arm, it exerts pressure on the column of water and the mercury in the tube. The pressure exerted by a fluid at a certain depth is given by the formula:

P = ρgh

Where P represents the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth of the fluid.

Since the density of water (ρw) is greater than the density of mercury (ρm), the water column will exert a greater pressure on the mercury in the right arm than the mercury column on the left arm.

To find the difference in height between the two arms (Δh), we can equate the pressures:

Pwater = Pmercury

ρwghwater = ρmghmercury

Since ρw > ρm, we can cancel the densities:

ghwater = ghmercury

Since the acceleration due to gravity (g) and the height of the water column (hwater) are known, we can find the height of the mercury column (hmercury).

Given that hw = 10.0 cm, and assuming the density of mercury is 13.6 g/cm³, we can proceed with the calculations:

hmercury = hw

hmercury = 10.0 cm

Therefore, the mercury in the right arm will rise 10.0 cm upward from its initial position.

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