Test: Thrust And Pressure (Old Syllabus) - Class 9 MCQ

Thrust is defined as the force acting perpendicularly on a surface. Here the weight of the iron cuboid is the force which is acting normal or perpendicular to the surface of the sand.
The weight or thrust acting per unit surface area is nothing but pressure.
Therefore, we can say that, the weight of the iron cuboid will be the thrust exerted by it on sand.

The pressure exerted by a fluid in a tank is given by the formula:

P = ρgh

where

• ρ (rho) is the density of the fluid (water, 10³ kg/m³)


• g is the acceleration due to gravity (approximately 9.8 m/s²)


• h is the height of the fluid column (2 meters)

Substitute the values into the formula:

• P = 10³ kg/m³ × 9.8 m/s² × 2 m


• P = 1.96 × 10⁴ Pa

Therefore, the correct answer is:

D: 1.96 × 10⁴ Pa

Pressure exerted by a sharp needle on a surface is more than the pressure exerted by a blunt needle.

Diameter of the neck is = 2cm = 0.02m
So, the area of the cross section of the neck is, a = Πd² /4 = 0.000314 m²
Diameter of the bottom is = 20cm = 0.2m
So, the area of the cross section of bottom is,  A = Πd² /4 = 0.0314 m²
Force applied to the cork, f = 1.2 kgf
Let F be the force at the bottom, According to pascal’s law  F/A = f/a
⇒ F = (f/a)*A = (1.2/0.000314)*(0.0314) = 120 Kgf

The pressure acting on the surface is given by P = F/A
The force applied os perpendicular to the surface of objects per unit area. Here, P is the pressure, F is the force and A is the area. Pressure is scalar quality and its unit is pa 
(pascal). Thus, the pressure acting on the surface is given by P = F/A. 

Mass of the Wooden block, m = 5Kg
Thrust due to wooden block, Mg = 5kg x 9.8 m/s² = 49N

(a)    Surface area of 20 Cm x 10 Cm surface ,
A = (20 x 10) cm² = (0.2 x 0.1) m² = 0.02 m²
P = mg / A = 49 N / 0.02 m² = 49/0.02 N/m²  
Or Pressure, P = 2450 N/m²

(b)   Surface area of  40 cm x 20 cm surface,
A = ( 40 x 20) cm² = ( 0.4 x 0.2 ) m² = 0.08 m²
P = mg / A = 49 N / 0.08 m² = 49/0.08 N/m²
Or Pressure, P = 612.5 N/m²