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Pressure of 1 atm = 1.013 x 10^{5} Pa
In a car lift, compressed air exerts a force F_{1} on a small piston having a radius of 5cm. This pressure is transmitted to the second piston of a radius of 15cm. If the mass of the car to be lifted is 1350 kg. What is F_{1}?
From Pascal’s law: P_{1 }= P_{2}
⇒ F_{1}/A_{1 }= F_{2}/A_{2}
⇒ F_{1}/πr_{1}^{2} = F_{2}/πr_{2}^{2 }
⇒ F_{1} = (F_{2 }* r_{1}^{2} ) / r_{2}^{2 }= 1350 * 9.8 * (5 * 10^{2})^{2} / (15 * 10^{2})^{2} = 1470 N = 1.47 * 10^{3 }N
Which of the following is not an application of Pascal’s Law?
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.
The formula used to find the pressure on a swimmer h meters below the surface of a lake is: (where P_{a} is the atmospheric pressure)
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
The pressure at the bottom of a tank containing a liquid does not depend on:
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.
An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be the length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)
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, p_{1}V_{1} = p_{2}V_{2}
p_{1} = p_{atm} = pg76
V_{1 }= A.8 [ A = area of cross section]
p_{2} = p_{atm}  ρg(54x) = ρg(22+x)
V = A.x
ρg76 x 8A = ρg (22+x) (Ax)
x^{2} + 22x  78 x 8 by solving, x = 16.
Find the density when a liquid 5 m high in a column exerts a pressure of 80 Pa.
► Pressure = Density x Gravity x Height = ρgh
⇒ ρ = P/(g*h) = 80 Pa / (9.8 m/s^{2} x 5 m)
⇒ Density = 1.632 kg/m^{3}
Two vessels with equal base and unequal height have water filled to the same height. The force at the base of the vessels is:
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.
Water is flowing continuously from a tap having an internal diameter 8 x 10^{3} m. The water velocity as it leaves the tap is 0.4 ms^{1}. The diameter of the water stream at a distance 2 x 10^{1} m below the tap is close to:
If the terminal speed of a sphere of gold (density = 19.5 kg/m3 ) is 0.2 m/s in a viscous liquid (density = 1.5 kg/m3 ) of the same size in the same liquid.
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