Pascal's Law

Pascal's Law

Pascal's law states that if the gravity effect is neglected then the pressure at a point in a fluid at rest is the same in all directions. This law helps us to understand the isotropic nature of pressure.

Pascal's law can be also stated as

The increase in pressure at one point of the enclosed liquid in the equilibrium of rest is transmitted equally to all other points of the liquid and also to the walls of the container, provided the effect of gravity is neglected.

The applications of this law can be seen in Hydraulic lifts, hydraulic presses, and hydraulic brakes, etc

Working of Hydraulic lift

A hydraulic lift is used to lift the heavy loads.

For the above figure

If a small force f is applied on the piston of C then the pressure exerted on the liquid

$P=\frac{f}{a}$

Where

a = Area of a cross-section of the piston in C

This pressure is transmitted equally to the piston of cylinder D.

So,

$\frac{f}{a}=\frac{F}{A} \Rightarrow F=\frac{f}{a} A$

Where F=Upward force acting on the piston of cylinder D.

A=Area of a cross-section of the piston in D

Condition of Hydraulic Lift-

$A \gg a$ therefore
$
F \gg>f
$

So heavy load placed on the larger Piston is easily lifted upward.

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Solved Example Based On Pascal's Law

Example 1: Two communicated cylindrical tubes contain mercury. The diameter of one vessel is four times larger than the diameter of the other. A column of water of height 70 cm is poured into the narrow vessel. How much will the mercury level rise in the other vessel and how much will it sink in the narrow one?

1) 4.8 cm and 0.3 cm respectively

2) 0.3 cm and 4.8 cm respectively

3) 4.8 cm in both

4) 0.3 cm for both

Solution:

Let x= rise of Hg level in the broad vessel, y= drop of Hg level in the narrow vessel.

Then, $x A_1=y A_2 \Rightarrow 16 x=y$

Equating pressures at A and B on the same horizontal plane,

we get,

$70 \times 1 \times g=(x+y) \times 13.6 \times g \Rightarrow x+y=\frac{70}{13.6}$

So, x= 0.3 cm and y = 4.8 cm

Hence, the answer is option (2).

Example 2: A hydraulic automobile lift is designed to lift vehicles of mass $5000 \mathrm{~kg}$. The area of cross-section of the cylinder carrying the load is $250 \mathrm{~cm}^2$. The maximum pressure the smaller piston would have to bear $\left[\right.$ Assume $\left.g=10 \mathrm{~m} / \mathrm{s}^2\right]$.

1) $2 \times 10^{+5} \mathrm{~Pa}$
2) $20 \times 10^{+6} \mathrm{~Pa}$
3) $200 \times 10^{+6} \mathrm{~Pa}$
4) $2 \times 10^{+6} \mathrm{~Pa}$

Solution:

From Pascal's law same $\Delta \mathrm{F}$ transmitted throughout liquid\begin{aligned}
& \Delta \mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}=\frac{5000 \times 10}{250 \times 10^{-4}} \\
& =2 \times 10^6 \mathrm{~Pa}
\end{aligned}

Hence, the answer is option (2).

Example 3: Isotropic pressure means

1) It is the same at all points

2) It is the same in one direction

3) At a point it is the same in all directions.

4) None of the above

Solution:

Isotropic pressure -

The pressure exerted by a liquid at a point is the same in all direction

wherein

Hence, the answer is option (3).

Example 4: A hydraulic press can lift 100kg when a mass 'm' is placed on the smaller piston. It can lift _______- kg when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times keeping the same mass 'm' on the smaller piston.

1) 25600

2) 51200

3) 12800

4) 6400

Solution:

$
\frac{100 \times g}{\mathrm{~A}_2}=\frac{\mathrm{mg}}{\mathrm{A}_1} \ldots
$

Let $\mathrm{m}$ mass can lift $\mathrm{M}_0$ in second case then
$
\frac{\mathrm{M}_0 \mathrm{~g}}{16 \mathrm{~A}_2}=\frac{\mathrm{mg}}{\mathrm{A}_1 / 16} \ldots \text { (2) }\left[\text { Since } \mathrm{A}=\frac{\pi \mathrm{d}^2}{4}\right]
$

From equation (1) and (2) we get
$
\frac{\mathrm{M}_0}{16 \times 100}=16 \Rightarrow \mathrm{M}_0=25600 \mathrm{~kg}
$

Hence, the answer is option (1).

Example 5: Two identical cylindrical vessels are kept on the ground and each contains the same liquid of density d. The area of the base of both vessels is S but the height of liquid in one vessel is x₁ and in the other, x₂. When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is:

1) $g d S\left(x_2^2+x_1^2\right)$
2) $g d S\left(x_2+x_1\right)^2$
3) $\frac{3}{4} g d S\left(x_2-x_1\right)^2$
4) $\frac{1}{4} g d S\left(x_2-x_1\right)^2$

Solution:

$
\begin{aligned}
& \mathrm{U}_{\mathrm{i}}=\left(\rho \mathrm{Sx}_1\right) \mathrm{g} \cdot \frac{\mathrm{x}_1}{2}+\left(\rho \mathrm{Sx}_2\right) \mathrm{g} \cdot \frac{\mathrm{x}_2}{2} \\
& \mathrm{U}_{\mathrm{f}}=\left(\rho \mathrm{Sx}_{\mathrm{f}}\right) \mathrm{g} \cdot \frac{\mathrm{x}_{\mathrm{f}}}{2} \times 2
\end{aligned}
$

By volume conservation
$
\begin{aligned}
& \mathrm{Sx}_1+\mathrm{Sx}_2=\mathrm{S}\left(2 \mathrm{x}_{\mathrm{f}}\right) \\
& \mathrm{x}_{\mathrm{f}}=\frac{\mathrm{x}_1+\mathrm{x}_2}{2}
\end{aligned}
$

$\begin{aligned}
\Delta U & =\rho \operatorname{Sg}\left[\left(\frac{x_1^1}{2}+\frac{x_2^2}{2}\right)-x_f^2\right] \\
& =\rho \operatorname{Sg}\left[\frac{x_1^2}{2}+\frac{x_2^2}{2}-\left(\frac{x_1+x_2}{2}\right)^2\right] \\
& =\frac{\rho S g}{2}\left[\frac{x_1^2}{2}+\frac{x_2^2}{2}-x_1 x_2\right] \\
& =\frac{\rho S g}{4}\left(x_1-x_2\right)^2 \\
& =\frac{1}{4} g d S\left(x_2-x_1\right)^2
\end{aligned}$

Hence, the answer is option (4).

Summary

In short, in the above article on Pascal's law, we learned about what Pascal's law is and how hydraulic lifts are commonly used in our daily lives. This law is critical if you are preparing for an engineering or medical entrance exam, or merely taking a state board exam.