Test: Thrust And Pressure (Old Syllabus)

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

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## 10 Questions MCQ Test Physics for Class 9 - Test: Thrust And Pressure (Old Syllabus)

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Test: Thrust And Pressure (Old Syllabus) - Question 1

### Thrust exerted by an iron cuboid when placed on sand is equal to

Detailed Solution for Test: Thrust And Pressure (Old Syllabus) - Question 1

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.

Test: Thrust And Pressure (Old Syllabus) - Question 2

### If weight of an object is equal or less than upthrust acting on it then object would

Test: Thrust And Pressure (Old Syllabus) - Question 3

### The pressure exerted by water (density = 103 kg/m3) on the bottom surface (2m × 4m) of tank having dimensions 2m × 4m × 2m is​

Test: Thrust And Pressure (Old Syllabus) - Question 4

The SI unit of pressure is​

Test: Thrust And Pressure (Old Syllabus) - Question 5

A sharp edge blade is more effective in cutting than a blunt blade due to

Test: Thrust And Pressure (Old Syllabus) - Question 6

For a given applied thrust, pressure exerted on a surface by a sharp pin is

Detailed Solution for Test: Thrust And Pressure (Old Syllabus) - Question 6

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

Test: Thrust And Pressure (Old Syllabus) - Question 7

The neck and bottom of a bottle are 2cm and 20 cm in diameter respectively. The cork is pressed with same pressure. If the cork is pressed with a force of 1.2 kgf on the neck of the bottle, the force exerted on the bottom of the bottle will be

Detailed Solution for Test: Thrust And Pressure (Old Syllabus) - Question 7

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

Test: Thrust And Pressure (Old Syllabus) - Question 8

An object exerts a force F on a surface of surface area A. The pressure P acting on the surface is given by

Detailed Solution for Test: Thrust And Pressure (Old Syllabus) - Question 8

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.

Test: Thrust And Pressure (Old Syllabus) - Question 9

To observe and compare the pressure exerted by three different faces of a cuboid on sand, the following Cuboid is available to you:
(A) Wooden cuboid of dimension 20 cm x 30 cm x 50 cm
(B) Aluminium cuboid of dimension 3 cm x 6 cm x 12 cm
(C) Iron cuboid of dimension 5 cm x 10 cm x 15 cm
(D) Iron cuboid of dimension 20 cm x 30 cm x 15 cm
The best choice from the practical point of view would be:

Detailed Solution for Test: Thrust And Pressure (Old Syllabus) - Question 9
It will give us a correct measure of high and low pressure on different faces like 20×30 cm will give lowest pressure and 20×15 will give highest pressure and iron has quite large weight so it will be easy for us to determine pressure and collect information.
Test: Thrust And Pressure (Old Syllabus) - Question 10

A rectangular block of wood of dimensions 40cm x 20cm x 10cm is kept on a table top. The pressure is greatest when it is rested on

Detailed Solution for Test: Thrust And Pressure (Old Syllabus) - Question 10

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

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## Physics for Class 9

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