Test: Pressure & Its Applications - NEET 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.

Since the system is accelerating horizontally such that no component of acceleration in the vertical direction. Hence, the pressure in the vertical direction will remain unaffected.

For air trapped in tube, p₁V₁ = p₂V₂
p₁ = p_atm = pg76
V₁ = A.8 [ A = area of cross section]
p₂ = p_atm - ρg(54-x) = ρg(22+x)
V = A.x
ρg76 x 8A = ρg (22+x) (Ax)
x²  + 22x - 78 x 8 by solving, x = 16.

► 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.