Test: Pressure & Its Applications - JEE MCQ

Pressure of 1 atm = 1.013 x 10⁵ Pa

From Pascal’s law: P₁ ​= P₂​

⇒ F₁/A₁ ​​= ​F₂/A₂​​
⇒ F₁/πr_1​²​ ​= F₂/πr₂​² ​​
⇒ F₁ = (F₂ * r_1​² ) / r₂​²  = 1350 * 9.8 * (5 * 10^-2)² / (15 * 10^-2)² = 1470 N ​= 1.47 * 10³ N

Pascal's law states that the magnitude of pressure within the fluid is equal in all parts.

Option A: Brahma's press is a Hydraulic Press that works on the principle of Pascal's law.

Option B: Submarines don't work on the principle of Pascal's law.

Option C: Hydraulic lifts works on the principle of pascals law i.e. a force applied on a smaller cylinder is transmitted to lift heavy loads using larger cylinders.

Hence, option B is correct.

We know that the pressure at some point inside the water can be represented by: P_a + ρhg
where,
ρ = Density of the liquid
P_a = Atmospheric pressure
H = Depth at which the body is present
g = Gravitational acceleration

Pressure at the bottom of a tank containing liquid is given as P = hρg which is independent of the surface area bottom of the tank.

Step 1: Understand the Initial Conditions
We have an open glass tube immersed in mercury, with a length of 8 cm extending above the mercury level. The tube is then closed and sealed.

Step 2: Raise the Tube
The tube is raised vertically by an additional 46 cm. Therefore, the total length of the tube that is now above the mercury level is:
Total height above mercury = 8 cm + 46 cm = 54 cm

Step 3: Define Variables
Let x be the height of the mercury column that rises in the tube when it is raised. The new height of the air column above the mercury will then be:
Height of air column = 54 cm − x

Step 4: Apply Pressure Conditions
The pressure at the top of the air column inside the tube is equal to the atmospheric pressure minus the height of the mercury column:
P = P₀ − x
Where P₀ = 76 cm of Hg.

Step 5: Use the Ideal Gas Law
Using the ideal gas law, we can write:

Where:
- P₁ = 76 cm (initial pressure)
- V₁ = 8 cm (initial volume)
- P₂ = 76 − x cm (final pressure)
- V₂ = 54 − x cm (final volume)

Step 6: Set Up the Equation
Substituting the known values into the equation gives:
76 × 8= (76−x) (54−x)

Step 7: Expand and Rearrange
Expanding the right side:
608 = (76 × 54) − (76 x+ 54x − x²)
Calculating 76×54:
76 × 54 = 4104
So, we have:
608 = 4104 − (130 x−x²)
Rearranging gives:
x² − 130x +(4104−608)=0
This simplifies to:
x²−130 x+ 3496=0

Step 8: Solve the Quadratic Equation
Using the quadratic formula 
Here, a=1,b=−130,c=3496:

Calculating the discriminant:
(−130)2−4⋅1⋅3496=16900−13984=2916
Thus,

Calculating √2916=54:

This gives two possible solutions:
1. x = 184/2 = 92 cm
2. x = 76/2 = 38 cm

Step 9: Determine the Length of the Air Column
Using x=38cm:
Length of air column=54−x=54−38=16cm

► Pressure = Density x Gravity x Height = ρgh 
⇒  ρ = P/(g*h) = 80 Pa / (9.8 m/s² x 5 m)
⇒ Density = 1.632 kg/m³

Force at the bases of two vessels will be equal as force depends on height and area. Since it is the same here, force is equal.