Why is the weight of the mercury column written as $10\text{ cm of Hg}$ in the pressure equation?

In fluid statics, pressure is often expressed in terms of the height of a liquid column (like cm of Hg). The weight of a mercury column of height $h$ and cross-sectional area $A$ is $W = \rho g h A$. The pressure exerted by this weight is $P = W/A = \rho g h$. Since our gas pressures are already given in 'cm of Hg', expressing the weight of the $10\text{ cm}$ mercury column as a pressure of $10\text{ cm of Hg}$ allows us to directly add and subtract these values without converting to Pascals.

JEE Advanced 1978

LEVELJEE Advanced

Animated Solution for Physics - Properties of Solids and Liquids: A column of mercury of length 10 cm is contained in the middle of a horizontal tube of length 1 m which is closed at both the ends. The two equal lengths contain air at standard atmospheric pressure of 0.76 m of mercury. The tube is now turned to vertical position. By what distance will the column of mercury be displaced ? Assume temperature to be constant.

Q: Why did we discard the negative root of the quadratic equation?

The variable $x$ represents the physical downward displacement of the mercury column. A negative value would imply a displacement upwards, which is physically impossible since gravity pulls the column downwards. Thus, only the positive root is physically meaningful.

Enter Numerical Value:

Visualized Solution

Visualizing the Initial State

Total length of the tube: $L = 1 m = 100 cm$
Length of the mercury column: $L_{Hg} = 10 cm$
Remaining length for air: $100 cm - 10 cm = 90 cm$
Since the mercury is in the middle, each air column has length: $L_{0} = 45 cm$

Initial Parameters of Air Columns

Let the cross-sectional area of the tube be $A$ .
Initial volume of air in each chamber: $V_{0} = 45 A$
Initial pressure of air in both chambers: $P_{0} = 76 cm of Hg$
Temperature is constant ( $T = constant$ ), so we can apply Boyle's Law.

Transition to the Vertical State

When the tube is turned vertical, the mercury column shifts downwards by a distance $x$ due to gravity.
This compresses the lower air column and expands the upper air column.
We need to find this displacement $x$ at equilibrium.

Expressing New Volumes

New length of the upper air column: $L_{1} = 45 + x$
New volume of the upper air column: $V_{1} = (45 + x) A$
New length of the lower air column: $L_{2} = 45 - x$
New volume of the lower air column: $V_{2} = (45 - x) A$

Applying Boyle's Law

For the upper chamber (isothermal expansion):
$P_{0} V_{0} = P_{2} V_{1} ⟹ 76 \times 45 A = P_{2} (45 + x) A$
$P_{2} = \frac{76 \times 45}{45 + x}$
For the lower chamber (isothermal compression):
$P_{0} V_{0} = P_{1} V_{2} ⟹ 76 \times 45 A = P_{1} (45 - x) A$
$P_{1} = \frac{76 \times 45}{45 - x}$

Equilibrium of the Mercury Column

For vertical equilibrium of the mercury column:
Upward force = Downward forces
$P_{1} A = P_{2} A + W_{Hg}$
Since the weight of a $10 cm$ mercury column is equivalent to a pressure of $10 cm of Hg$ :
$P_{1} - P_{2} = 10 cm of Hg$

Formulating the Master Equation

Substitute the expressions for $P_{1}$ and $P_{2}$ into the equilibrium equation:
$\frac{76 \times 45}{45 - x} - \frac{76 \times 45}{45 + x} = 10$
Simplify by dividing both sides by $10$ :
$7.6 \times 45 (\frac{1}{45 - x} - \frac{1}{45 + x}) = 1$
$342 (\frac{( 45 + x ) - ( 45 - x )}{4 5 ^{2} - x ^{2}}) = 1$

Solving the Quadratic Equation

Simplify the algebraic expression:
$342 (\frac{2 x}{2025 - x ^{2}}) = 1$
$684 x = 2025 - x^{2}$
Rearrange into standard quadratic form:
$x^{2} + 684 x - 2025 = 0$

Calculating the Displacement

Using the quadratic formula $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ :
$x = \frac{- 684 + 68 4 ^{2} - 4 ( 1 ) ( - 2025 )}{2}$
$x = \frac{- 684 + 467856 + 8100}{2}$
$x = \frac{- 684 + 475956}{2} \approx \frac{- 684 + 689.9}{2} \approx 2.95 cm$
The mercury column is displaced by $2.95 cm$ .

The Way Forward

What if the temperature of the system is raised by $Δ T$ ?
What if the tube is placed in an elevator accelerating upwards with $a$ ?
These variations will alter the pressure balance and the effective weight of the mercury column, leading to a new equilibrium displacement.

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The Sigma Insight: Fluid Pressure and Pascal's Law

Solution Diagram

Analyzing the Setup

Imagine a horizontal glass tube, closed at both ends, with a total length of

1 m

(

100 cm

Right in the middle of this tube sits a

10 cm

column of mercury.

Since the mercury is perfectly centered, it divides the remaining

90 cm

of space into two equal air columns, each of length

L_{0} = 45 cm

Both chambers contain air at standard atmospheric pressure, which is

P_{0} = 76 cm of Hg

L_{0} = \frac{100 cm - 10 cm}{2} = 45 cm

Let the cross-sectional area of the tube be

A

The initial volume of air in each chamber is:

V_{0} = 45 A

Transition to the Vertical State

Now, let us rotate the tube to a vertical position.

Due to gravity, the heavy mercury column will naturally tend to slide downwards.

As it moves down by a distance

x

, it compresses the air in the lower chamber and expands the air in the upper chamber.

This movement continues until the pressure difference between the two chambers, along with the weight of the mercury, reaches a new state of mechanical equilibrium.

Let us express the new geometry of the air columns after the displacement

x

Upper air column length: $L_{1} = 45 + x$ Upper air column volume:

V_{1} = (45 + x) A

Lower air column length: $L_{2} = 45 - x$ Lower air column volume:

V_{2} = (45 - x) A

Applying Boyle's Law

Since the temperature of the system is maintained constant, we can apply Boyle's Law (

P V = constant

) to both chambers.

For the upper chamber (isothermal expansion):

P_{0} V_{0} = P_{2} V_{1} ⟹ 76 \times 45 A = P_{2} (45 + x) A

P_{2} = \frac{76 \times 45}{45 + x}

For the lower chamber (isothermal compression):

P_{0} V_{0} = P_{1} V_{2} ⟹ 76 \times 45 A = P_{1} (45 - x) A

P_{1} = \frac{76 \times 45}{45 - x}

Equilibrium of the Mercury Column

For the mercury column to remain in vertical equilibrium, the upward force must balance the downward forces.

P_{1} A = P_{2} A + W_{Hg}

Since the weight of a

10 cm

mercury column is equivalent to a pressure of

10 cm of Hg

, we can write the pressure balance equation directly as:

P_{1} - P_{2} = 10 cm of Hg

Formulating and Solving the Master Equation

Substituting the expressions for

P_{1}

and

P_{2}

into the equilibrium equation:

\frac{76 \times 45}{45 - x} - \frac{76 \times 45}{45 + x} = 10

Divide both sides by

10

7.6 \times 45 (\frac{1}{45 - x} - \frac{1}{45 + x}) = 1

342 (\frac{( 45 + x ) - ( 45 - x )}{4 5 ^{2} - x ^{2}}) = 1

342 (\frac{2 x}{2025 - x ^{2}}) = 1

684 x = 2025 - x^{2} ⟹ x^{2} + 684 x - 2025 = 0

Using the quadratic formula

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

x = \frac{- 684 + 68 4 ^{2} - 4 ( 1 ) ( - 2025 )}{2}

x = \frac{- 684 + 467856 + 8100}{2}

x = \frac{- 684 + 475956}{2}

x \approx \frac{- 684 + 689.9}{2} \approx 2.95 cm

Thus, the mercury column is displaced by $2.95 cm$ .

Common Traps & FAQs

Why does the mercury column shift downwards when the tube is turned vertical?

When the tube is horizontal, gravity acts perpendicular to the length of the tube, so it doesn't push the mercury along the tube. When turned vertical, gravity acts along the tube, pulling the mercury down. This compression of the lower air column and expansion of the upper air column creates a pressure difference that eventually balances the weight of the mercury.

Why can we use Boyle's Law here?

The problem states that the temperature is constant. Since the mass of air trapped in both chambers is constant and the temperature does not change, the process is isothermal, allowing us to use Boyle's Law (

P V = constant

Why is the weight of the mercury column written as

10 cm of Hg

in the pressure equation?

In fluid statics, pressure is often expressed in terms of the height of a liquid column (like cm of Hg). The weight of a mercury column of height

h

and cross-sectional area

A

W = ρ g h A

. The pressure exerted by this weight is

P = W / A = ρ g h

. Since our gas pressures are already given in 'cm of Hg', expressing the weight of the

10 cm

mercury column as a pressure of

10 cm of Hg

allows us to directly add and subtract these values without converting to Pascals.

What would happen if the tube was not closed at both ends?

If the tube was open at both ends, the air inside would not be trapped, and the pressure at both ends would remain equal to atmospheric pressure. The mercury column would simply fall out of the tube due to gravity!

Why did we discard the negative root of the quadratic equation?

The variable

x

represents the physical downward displacement of the mercury column. A negative value would imply a displacement upwards, which is physically impossible since gravity pulls the column downwards. Thus, only the positive root is physically meaningful.

Enter Numerical Value:

Visualized Solution

Visualizing the Initial State

Initial Parameters of Air Columns

Transition to the Vertical State

Expressing New Volumes

Applying Boyle's Law

Equilibrium of the Mercury Column

Formulating the Master Equation

Solving the Quadratic Equation

Calculating the Displacement

The Way Forward

The Sigma Insight: Fluid Pressure and Pascal's Law

Analyzing the Setup

Transition to the Vertical State

Applying Boyle's Law

Equilibrium of the Mercury Column

Formulating and Solving the Master Equation

Similar Questions

A U-shaped tube contains a liquid of density $ρ$ and it is rotated about the line as shown in the figure. Find the difference in the levels of liquid column.

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d_{1}$ and $d_{2}$ are filled in the tube. Each liquid subtends $9 0^{\circ}$ angle at centre. Radius joining their interface makes an angle $α$ with vertical. Ratio $\frac{d _{1}}{d _{2}}$ is

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d_{1}$ and $d_{2}$ are filled in the tube. Each liquid subtends $9 0^{\circ}$ angle at centre. Radius joining their interface makes an angle $α$ with vertical. Ratio $d_{1} / d_{2}$ is

Enter Numerical Value:

Visualized Solution

Visualizing the Initial State

Initial Parameters of Air Columns

Transition to the Vertical State

Expressing New Volumes

Applying Boyle's Law

Equilibrium of the Mercury Column

Formulating the Master Equation

Solving the Quadratic Equation

Calculating the Displacement

The Way Forward

The Sigma Insight: Fluid Pressure and Pascal's Law

Analyzing the Setup

Transition to the Vertical State

Applying Boyle's Law

Equilibrium of the Mercury Column

Formulating and Solving the Master Equation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A U-shaped tube contains a liquid of density ρ and it is rotated about the line as shown in the figure. Find the difference in the levels of liquid column.

A U-shaped tube contains a liquid of density $ρ$ and it is rotated about the line as shown in the figure. Find the difference in the levels of liquid column.