Q: Question 9How much thrust will be required to exert a pressure of 20,000 Pa on an area of 1 cm2?

Given:Pressure = 20000 PaArea = 1 cm2 = 110000\dfrac{\text{1}}{\text{10000}}100001 = 0.0001 m2 or 10-4 m2Thrust = ?Pressure = ThrustArea\dfrac{\text{Thrust}}{\text{Area}}AreaThrustThrust = Pressure x Area= 20000 x 10-4= 2 NSo Thrust = 2 N.

Question 1

Question 2The moment of a force of 25 N about a point is 2.5 N m. Find the perpendicular distance of force from that point.

Accepted Answer

Given:Force f = 25 NMoment of force = 2.5 N mPerpendicular distance d = ?Moment of force=force (f)×distance (d)Distance (d)=Moment of forceForce (f)=2.525=0.1 m\text{Moment of force} = \text{force (f}) \times \text{distance (d)} \[1em] \text{Distance (d)}= \dfrac{\text{Moment of force}}{\text{Force (f)}} \[1em] = \dfrac{\text{2.5}}{\text{25}} \[1em] = 0.1 \text{ m}Moment of force=force (f)×distance (d)Distance (d)=Force (f)Moment of force​=252.5​=0.1 mSo, perpendicular distance of force from the point = 0.1 m or 10 cm.

Question 2

Question 4A wheel of diameter 2 m can be rotated about an axis passing through its centre by a moment of force equal to 2.0 N m. What minimum force must be applied on its rim?

Accepted Answer

Given:Moment of force = 2.0 N mDiameter = 2 mRadius = 1 mTherefore perpendicular distance (d) = radius = 1 mForce (f) = ?Moment of force=force (f)×distance (d)Force (f)=Moment of forcedistance (d)=2.01=2 N\text{Moment of force} = \text{force (f)} \times \text{distance (d)} \[1em] \text{Force (f)}= \dfrac{\text{Moment of force}}{\text{distance (d)}} \[1em] = \dfrac{\text{2.0}}{\text{1}} \[1em] = 2 \text{ N}Moment of force=force (f)×distance (d)Force (f)=distance (d)Moment of force​=12.0​=2 NSo, minimum force required is 2 N.

Question 3

Question 5A normal force of 200 N acts on an area 0.02 m2. Find the pressure in pascal.

Accepted Answer

Given:Thrust (f) = 200 NArea (A) = 0.02 m2Pressure = ?Pressure=ThrustArea=2000.02=10000 Pa\text{Pressure} = \dfrac{\text{Thrust}}{\text{Area}} \[1em] = \dfrac{\text{200}}{\text{0.02}} \[1em] = 10000 \text{ Pa}Pressure=AreaThrust​=0.02200​=10000 PaSo Pressure = 10000 Pa.

Question 4

Question 6Find the thrust required to exert a pressure of 50,000 Pa on an area of 0.05 m2.

Accepted Answer

Given:Pressure = 50000 PaArea = 0.05 m2Thrust = ?Pressure = ThrustArea\dfrac{	ext{Thrust}}{	ext{Area}}AreaThrust​Thrust = Pressure x Area = 50000 x 0.05 = 2500 NSo, Thrust = 2500 N.

Question 5

Question 7Find the area of a body which experiences a pressure of 50,000 Pa by a thrust of 100 N.

Accepted Answer

Given:Pressure = 50,000 PaThrust = 100 NArea = ?Pressure=ThrustAreaArea=ThrustPressure=10050000=0.002 m2\text{Pressure}= \dfrac{\text{Thrust}}{\text{Area}} \[1em] \text{Area}= \dfrac{\text{Thrust}}{\text{Pressure}} \[1em] = \dfrac{100}{50000} \[1em] = 0.002 \text{ m}^{2}Pressure=AreaThrust​Area=PressureThrust​=50000100​=0.002 m2So Area of body = 0.002 m2 or 2 x 10-3 m2.

Question 6

Question 8Calculate the pressure in pascal exerted by a force of 300 N acting normally on an area of 30 cm2.

Accepted Answer

Given:Thrust = 300 NArea = 30 cm2 = 3010000\dfrac{30}{10000}1000030​ = 0.003 m2Pressure = ?Pressure=ThrustArea=3000.003=1,00,000 Pa\text{Pressure}= \dfrac{\text{Thrust}}{\text{Area}} \[1em] = \dfrac{300}{0.003} \[1em] = 1,00,000 \text{ Pa}Pressure=AreaThrust​=0.003300​=1,00,000 PaSo Pressure = 1,00,000 Pa or 105 Pa.

Question 7

Question 9How much thrust will be required to exert a pressure of 20,000 Pa on an area of 1 cm2?

Accepted Answer

Given:Pressure = 20000 PaArea = 1 cm2 = 110000\dfrac{	ext{1}}{	ext{10000}}100001​ = 0.0001 m2 or 10-4 m2Thrust = ?Pressure = ThrustArea\dfrac{	ext{Thrust}}{	ext{Area}}AreaThrust​Thrust = Pressure x Area= 20000 x 10-4= 2 NSo Thrust = 2 N.

Question 8

Question 10The base of a container measures 15 cm x 20 cm. It is placed on a table top. If the weight of the container is 60 N, what is the pressure exerted by the container on the table top?

Accepted Answer

Given:Weight = Thrust = 60 NArea = 15 cm x 20 cm = 300 cm2 = 30010000\dfrac{\text{300}}{\text{10000}}10000300​ = 0.03 m2Pressure = ?Pressure=ThrustArea=600.03=2000 Pa\text{Pressure}= \dfrac{\text{Thrust}}{\text{Area}} \[1em] = \dfrac{\text{60}}{\text{0.03}} \[1em] = 2000 \text{ Pa}Pressure=AreaThrust​=0.0360​=2000 PaSo, Pressure = 2000 Pa.

Question 9

Question 11Calculate the pressure exerted on a surface of 0.5 m2 by a thrust of 100 kgf.

Accepted Answer

Given:Thrust = 100 kgfArea = 0.5 m2Pressure = ?Pressure=ThrustArea=1000.5=200 kgf m−2\text{Pressure}= \dfrac{\text{Thrust}}{\text{Area}} \[1em] = \dfrac{\text{100}}{\text{0.5}} \[1em] = 200 \text{ kgf m}^{-2}Pressure=AreaThrust​=0.5100​=200 kgf m−2So Pressure = 200 kgf m-2.

Question 10

Question 12A boy weighing 60 kgf stands on a platform of dimensions 2.5 cm x 0.5 cm. What pressure in pascal does he exert?

Accepted Answer

Given:Thrust = Weight = 60 kgf1 kgf = 10 N∴ 60 kgf = 60 x 10 = 600 NSo, Thrust = 600 NArea = 2.5 cm x 0.5 cm = 1.25 cm2 = 1.2510000\dfrac{\text{1.25}}{\text{10000}}100001.25​ = 0.000125 m2Pressure =?Pressure=ThrustArea=6000.000125=48,00,000 Pa\text{Pressure}= \dfrac{\text{Thrust}}{\text{Area}} \[1em] = \dfrac{\text{600}}{\text{0.000125}} \[1em] = 48,00,000 \text{ Pa}Pressure=AreaThrust​=0.000125600​=48,00,000 PaSo, Pressure = 48,00,000 or 4.8 x 106 Pa.

Question 11

Question 13Figure below shows a brick of weight 2 kgf and dimensions 20 cm x 10 cm x 5 cm placed in three different positions on the ground. Find the pressure exerted by the brick in each case.

Accepted Answer

First case :Thrust = Weight = 2 kgfArea = 20 cm x 10 cm = 200 cm2Pressure = ?Pressure=ThrustArea=2200=0.01 kgf cm−2\text{Pressure}= \dfrac{\text{Thrust}}{\text{Area}} \[1em] = \dfrac{\text{2}}{\text{200}} \[1em] = 0.01 \text{ kgf cm}^{-2}Pressure=AreaThrust​=2002​=0.01 kgf cm−2So Pressure exerted by brick in first figure is 0.01 kgf cm-2.Second case :Thrust = Weight = 2 kgfArea = 5 cm x 10 cm = 50 cm2Pressure = ?Pressure=ThrustArea=250=0.04 kgf cm−2\text{Pressure}= \dfrac{\text{Thrust}}{\text{Area}} \[1em] = \dfrac{\text{2}}{\text{50}} \[1em] = 0.04 \text{ kgf cm}^{-2}Pressure=AreaThrust​=502​=0.04 kgf cm−2So Pressure exerted by brick in second figure is 0.04 kgf cm-2.Third case :Thrust = Weight = 2 kgfArea = 20 cm x 5 cm = 100 cm2Pressure = ?Pressure=ThrustArea=2100=0.02 kgf cm−2\text{Pressure}= \dfrac{\text{Thrust}}{\text{Area}} \[1em] = \dfrac{\text{2}}{\text{100}} \[1em] = 0.02 \text{ kgf cm}^{-2}Pressure=AreaThrust​=1002​=0.02 kgf cm−2So, Pressure exerted by brick in third figure is 0.02 kgf cm-2.

Question 12

Question 14On what factors does the effect of thrust on a surface depend?

Accepted Answer

The effect of thrust depends on the area of the surface on which it acts. Smaller the area of the surface on which a thrust acts larger is its effect and effect of thrust is less on a larger surface area.

Question 13

Question 15Define the term pressure and state its unit.

Accepted Answer

Pressure is defined as the thrust acting per unit area.Pressure = ThrustArea\dfrac{	ext{Thrust}}{	ext{Area}}AreaThrust​Its unit is pascal (Pa) or newtonmetre2\dfrac{	ext{newton}}{	ext{metre}^2}metre2newton​ (N m-2).

Question 14

Question 16How is thrust related to pressure?

Accepted Answer

Pressure is thrust acting per unit area. Thrust is directly proportional to pressure. Greater the thrust, greater is the pressure and smaller the thrust, smaller is the pressure.Pressure = ThrustArea\dfrac{	ext{Thrust}}{	ext{Area}}AreaThrust​

Question 15

Question 17Name two factors on which the pressure on a surface depends.

Accepted Answer

The two factors on which the pressure on a surface depends are:Area of the surface on which thrust acts.Magnitude of thrust acting on the surface.

Question 16

Question 18Why is the bottom part of the foundation of a building made wider?

Accepted Answer

The bottom part of the foundation of a building is made wider so that the weight of the building will act on larger surface area exerting less pressure on the ground (as pressure is inversely proportional to surface area). This avoids sinking of buildings into the earth.

Question 17

Question 19It is easier to cut with a sharp knife than with a blunt one. Explain.

Accepted Answer

A sharp knife has sharp edges that provides less surface area of contact as compared to a blunt knife which has more surface area of contact. So pressure applied by the force is more in sharp knife which makes it easier to cut than a blunt one.

Question 18

Question 20A gum bottle rests on its base. If it is placed upside down,(i) How does the thrust change?(ii) How does the pressure change?

Accepted Answer

(i) A gum bottle has broader base and narrower neck and when it is placed upside down there will be no change in thrust as thrust is the weight of the body that is exerted on the surface.(ii) When gum bottle is placed upside down, surface area is less so pressure will be more as pressure is inversely proportional to surface area.

Question 19

Question 21State two factors on which the pressure at a point in a liquid depends.

Accepted Answer

The two factors on which the pressure at a point in a liquid depends are:The height of the liquid column — Liquid pressure increases with the height of the liquid column above the point.The density of the liquid — Liquid pressure increases with the increase in the density of the liquid.

Question 20

Question 22What do you mean by atmospheric pressure?

Accepted Answer

The thrust on unit area of the earth surface due to the column of air is called the atmospheric pressure.

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