Q.44

Expert-verifiedFound in: Page 385

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

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

The rise in the level of the mercury in the right arm, from its initial position is

We know that water and mercury are liquids that are both incompressible and immiscible.

The diagram which can support the fact is given below :

As we know that the water floats on top of the mercury in the left arm, pushing it down from its initial position.

This will happen because the mercury levels are identical at the point and in the figure, the pressure at the point and in the figure must be equal.

Hence, we can say that if the pressures at the sites and are not equal, the fluid will flow in a way that violates the static equilibrium requirement.

In the figure, the pressure at the point is due to water having depth of .

Here, is the atmospheric pressure is the pressure at point , is the density of the water, is the acceleration,

due to gravity, and is the depth of water.

Because mercury is an incompressible fluid, it will travel down a distance in the left arm while travelling up a distance in the right arm.

This implies that the pressure at point is due to the mercury having a depth of .

Now, we can see that the pressure at point is

In the above expression here, is the atmospheric pressure is the pressure at point , is the density of the mercury. is the acceleration due to gravity, and is the depth of mercury.

Substitute for

Equate the equation and

Rearrange the above equation in terms of

Substitute for , for , for

The temperature in the right arm has increased from its original

position.

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