Selina Textbook Solutions: Pressure in Fluids and Atmospheric Pressure | Physics Class 9 ICSE PDF Download

 Page 1


Chapter 4. Pressure in Fluids and Atmospheric 
Pressure
Exercise 4(A)
Solution 1S.
Thrust is the force acting normally on a surface. 
Its S.I. unit is ‘newton’.
Solution 2S.
Pressure is the thrust per unit area of the surface. 
Its S.I. unit is ‘newton per metre
2
‘ or ‘pascal’.
Solution 3S.
(a) Pressure is measured in ‘bar’. 
(b) 1 bar = 10
5
 pascal.
Solution 4S.
One pascal is the pressure exerted on a surface of area 1 m
2
 by a force of 1N acting 
normally on it.
Solution 5S.
Thrust is a vector quantity.
Solution 6S.
Pressure is a scalar quantity.
Solution 7S.
Thrust is the force applied on a surface in a perpendicular direction and it is a vector 
quantity. The effect of thrust per unit area is pressure, and it is a scalar quantity.
Solution 8S.
Pressure exerted by thrust is inversely proportional to area of surface on which it acts. 
Thus, larger the area on which the thrust acts, lesser is the pressure exerted by it. 
Example: If we stand on loose sand, our feet sink into the sand, but if we lie on that 
sand, our body does not sink into the sand. In both the cases, the thrust exerted on the 
sand is equal (equal to the weight of the body). However, when we lie on sand, the thrust 
acts on a large area and when we stand, the same thrust acts on a small area.
Solution 9S.
Page 2


Chapter 4. Pressure in Fluids and Atmospheric 
Pressure
Exercise 4(A)
Solution 1S.
Thrust is the force acting normally on a surface. 
Its S.I. unit is ‘newton’.
Solution 2S.
Pressure is the thrust per unit area of the surface. 
Its S.I. unit is ‘newton per metre
2
‘ or ‘pascal’.
Solution 3S.
(a) Pressure is measured in ‘bar’. 
(b) 1 bar = 10
5
 pascal.
Solution 4S.
One pascal is the pressure exerted on a surface of area 1 m
2
 by a force of 1N acting 
normally on it.
Solution 5S.
Thrust is a vector quantity.
Solution 6S.
Pressure is a scalar quantity.
Solution 7S.
Thrust is the force applied on a surface in a perpendicular direction and it is a vector 
quantity. The effect of thrust per unit area is pressure, and it is a scalar quantity.
Solution 8S.
Pressure exerted by thrust is inversely proportional to area of surface on which it acts. 
Thus, larger the area on which the thrust acts, lesser is the pressure exerted by it. 
Example: If we stand on loose sand, our feet sink into the sand, but if we lie on that 
sand, our body does not sink into the sand. In both the cases, the thrust exerted on the 
sand is equal (equal to the weight of the body). However, when we lie on sand, the thrust 
acts on a large area and when we stand, the same thrust acts on a small area.
Solution 9S.
The tip of an allpin is made sharp so that large pressure is exerted at the sharp end and 
it can be driven into with less effort.
Solution 10S.
(a) It is easier to cut with a sharp knife because even a small thrust causes great 
pressure at the edges and cutting can be done with less effort. 
(b) Wide wooden sleepers are placed below the railway tracks so that the pressure 
exerted by the rails on the ground becomes less.
Solution 11S.
A substance which can flow is called a fluid.
Solution 12S.
Due to its weight, a fluid exerts pressure in all directions; the pressure exerted by the 
fluid is called fluid pressure.
Solution 13S.
A solid exerts pressure only on the surface on which it is placed, i.e. at its bottom, but a 
fluid exerts pressure at all points in all directions.
Solution 14S.
Take a can or large plastic bottle filled with water. Place it on a horizontal surface. Make a 
series of holes in the wall of the vessel anywhere below the free surface of the liquid. The 
water spurts out through each hole. This shows that the liquid exerts pressure at each 
point on the wall of the bottle. 
 
Liquid exerts pressure at all points in all directions
Solution 15S.
Pressure at a point in a liquid depends upon the following three factors:
1. Depth of the point below the free surface.
Page 3


Chapter 4. Pressure in Fluids and Atmospheric 
Pressure
Exercise 4(A)
Solution 1S.
Thrust is the force acting normally on a surface. 
Its S.I. unit is ‘newton’.
Solution 2S.
Pressure is the thrust per unit area of the surface. 
Its S.I. unit is ‘newton per metre
2
‘ or ‘pascal’.
Solution 3S.
(a) Pressure is measured in ‘bar’. 
(b) 1 bar = 10
5
 pascal.
Solution 4S.
One pascal is the pressure exerted on a surface of area 1 m
2
 by a force of 1N acting 
normally on it.
Solution 5S.
Thrust is a vector quantity.
Solution 6S.
Pressure is a scalar quantity.
Solution 7S.
Thrust is the force applied on a surface in a perpendicular direction and it is a vector 
quantity. The effect of thrust per unit area is pressure, and it is a scalar quantity.
Solution 8S.
Pressure exerted by thrust is inversely proportional to area of surface on which it acts. 
Thus, larger the area on which the thrust acts, lesser is the pressure exerted by it. 
Example: If we stand on loose sand, our feet sink into the sand, but if we lie on that 
sand, our body does not sink into the sand. In both the cases, the thrust exerted on the 
sand is equal (equal to the weight of the body). However, when we lie on sand, the thrust 
acts on a large area and when we stand, the same thrust acts on a small area.
Solution 9S.
The tip of an allpin is made sharp so that large pressure is exerted at the sharp end and 
it can be driven into with less effort.
Solution 10S.
(a) It is easier to cut with a sharp knife because even a small thrust causes great 
pressure at the edges and cutting can be done with less effort. 
(b) Wide wooden sleepers are placed below the railway tracks so that the pressure 
exerted by the rails on the ground becomes less.
Solution 11S.
A substance which can flow is called a fluid.
Solution 12S.
Due to its weight, a fluid exerts pressure in all directions; the pressure exerted by the 
fluid is called fluid pressure.
Solution 13S.
A solid exerts pressure only on the surface on which it is placed, i.e. at its bottom, but a 
fluid exerts pressure at all points in all directions.
Solution 14S.
Take a can or large plastic bottle filled with water. Place it on a horizontal surface. Make a 
series of holes in the wall of the vessel anywhere below the free surface of the liquid. The 
water spurts out through each hole. This shows that the liquid exerts pressure at each 
point on the wall of the bottle. 
 
Liquid exerts pressure at all points in all directions
Solution 15S.
Pressure at a point in a liquid depends upon the following three factors:
1. Depth of the point below the free surface.
2. Density of liquid.
3. Acceleration due to gravity.
Solution 16S.
P = P
o
 + h?g 
Here, P = Pressure exerted at a point in the liquid 
P
o
 = Atmospheric pressure 
h = Depth of the point below the free surface 
? = Density of the liquid 
g = Acceleration due to gravity
Solution 17S.
Consider a vessel containing a liquid of density ?. Let the liquid be stationary. In order to 
calculate pressure at a depth, consider a horizontal circular surface PQ of area A at a 
depth h below the free surface XY of the liquid. The pressure on the surface PQ will be 
due to the thrust of the liquid contained in cylinder PQRS of height h with PQ as its base 
and top face RS lying on the frees surface XY of the liquid.
 
Total thrust exerted on the surface PQ 
= Weight of the liquid column PQRS 
= Volume of liquid column PQRS x density x g 
= (Area of base PQ x height) x density x g 
= (A x h) x ? x g
This thrust is exerted on the surface PQ of area A. Therefore, pressure is given as shown 
below. 
P = Thrust on surface / Area of surface 
P = Ah ?g / A = h?g 
Thus, Pressure = depth x density of liquid x acceleration due to gravity
Solution 18S.
Due to dissolved salts, density of sea water is more than the density of river water, so 
pressure at a certain depth in sea water is more than that at the same depth in river 
water.
Page 4


Chapter 4. Pressure in Fluids and Atmospheric 
Pressure
Exercise 4(A)
Solution 1S.
Thrust is the force acting normally on a surface. 
Its S.I. unit is ‘newton’.
Solution 2S.
Pressure is the thrust per unit area of the surface. 
Its S.I. unit is ‘newton per metre
2
‘ or ‘pascal’.
Solution 3S.
(a) Pressure is measured in ‘bar’. 
(b) 1 bar = 10
5
 pascal.
Solution 4S.
One pascal is the pressure exerted on a surface of area 1 m
2
 by a force of 1N acting 
normally on it.
Solution 5S.
Thrust is a vector quantity.
Solution 6S.
Pressure is a scalar quantity.
Solution 7S.
Thrust is the force applied on a surface in a perpendicular direction and it is a vector 
quantity. The effect of thrust per unit area is pressure, and it is a scalar quantity.
Solution 8S.
Pressure exerted by thrust is inversely proportional to area of surface on which it acts. 
Thus, larger the area on which the thrust acts, lesser is the pressure exerted by it. 
Example: If we stand on loose sand, our feet sink into the sand, but if we lie on that 
sand, our body does not sink into the sand. In both the cases, the thrust exerted on the 
sand is equal (equal to the weight of the body). However, when we lie on sand, the thrust 
acts on a large area and when we stand, the same thrust acts on a small area.
Solution 9S.
The tip of an allpin is made sharp so that large pressure is exerted at the sharp end and 
it can be driven into with less effort.
Solution 10S.
(a) It is easier to cut with a sharp knife because even a small thrust causes great 
pressure at the edges and cutting can be done with less effort. 
(b) Wide wooden sleepers are placed below the railway tracks so that the pressure 
exerted by the rails on the ground becomes less.
Solution 11S.
A substance which can flow is called a fluid.
Solution 12S.
Due to its weight, a fluid exerts pressure in all directions; the pressure exerted by the 
fluid is called fluid pressure.
Solution 13S.
A solid exerts pressure only on the surface on which it is placed, i.e. at its bottom, but a 
fluid exerts pressure at all points in all directions.
Solution 14S.
Take a can or large plastic bottle filled with water. Place it on a horizontal surface. Make a 
series of holes in the wall of the vessel anywhere below the free surface of the liquid. The 
water spurts out through each hole. This shows that the liquid exerts pressure at each 
point on the wall of the bottle. 
 
Liquid exerts pressure at all points in all directions
Solution 15S.
Pressure at a point in a liquid depends upon the following three factors:
1. Depth of the point below the free surface.
2. Density of liquid.
3. Acceleration due to gravity.
Solution 16S.
P = P
o
 + h?g 
Here, P = Pressure exerted at a point in the liquid 
P
o
 = Atmospheric pressure 
h = Depth of the point below the free surface 
? = Density of the liquid 
g = Acceleration due to gravity
Solution 17S.
Consider a vessel containing a liquid of density ?. Let the liquid be stationary. In order to 
calculate pressure at a depth, consider a horizontal circular surface PQ of area A at a 
depth h below the free surface XY of the liquid. The pressure on the surface PQ will be 
due to the thrust of the liquid contained in cylinder PQRS of height h with PQ as its base 
and top face RS lying on the frees surface XY of the liquid.
 
Total thrust exerted on the surface PQ 
= Weight of the liquid column PQRS 
= Volume of liquid column PQRS x density x g 
= (Area of base PQ x height) x density x g 
= (A x h) x ? x g
This thrust is exerted on the surface PQ of area A. Therefore, pressure is given as shown 
below. 
P = Thrust on surface / Area of surface 
P = Ah ?g / A = h?g 
Thus, Pressure = depth x density of liquid x acceleration due to gravity
Solution 18S.
Due to dissolved salts, density of sea water is more than the density of river water, so 
pressure at a certain depth in sea water is more than that at the same depth in river 
water.
Solution 19S.
(a) P
2
 = P
1
 + h ? g, 
(b) P
2
 > P
1
Solution 20S.
The reason is that when the bubble is at the bottom of the lake, total pressure exerted on 
it is the atmospheric pressure plus the pressure due to water column. As the gas bubble 
rises, due to decrease in depth the pressure due to water column decreases. By Boyle’s 
law, PV = constant, so the volume of bubble increases due to decrease in pressure, i.e., 
the bubble grows in size.
Solution 21S.
The pressure exerted by a liquid increases with its depth. Thus as depth increases, more 
and more pressure is exerted by water on wall of the dam. A thicker wall is required to 
withstand greater pressure, therefore, the thickness of the wall of dam increases towards 
the bottom.
Solution 22S.
The sea divers need special protective suit to wear because in deep sea, the total 
pressure exerted on the diver’s body is much more than his blood pressure. To withstand 
it, he needs to wear a special protective suit.
Solution 23S.
Laws of liquid pressure:
1. Pressure at a point inside the liquid increases with the depth from its free surface.
2. In a stationary liquid, pressure is same at all points on a horizontal plane.
3. Pressure is same in all directions about a point in the liquid.
4. Pressure at same depth is different in different liquids. It increases with the 
increase in the density of liquid.
5. A liquid seeks its own level.
Solution 24S.
The liquid from hole B reaches a greater distance on the horizontal surface than that from 
hole A. 
This explains that liquid pressure at a point increases with the depth of point from the 
free surface.
Solution 25S.
(i) As the diver moves to a greater depth, pressure exerted by sea water on him also 
increases. 
(ii) When the diver moves horizontally, his depth from the free surface remains constant 
and hence the pressure on him remains unchanged.
Page 5


Chapter 4. Pressure in Fluids and Atmospheric 
Pressure
Exercise 4(A)
Solution 1S.
Thrust is the force acting normally on a surface. 
Its S.I. unit is ‘newton’.
Solution 2S.
Pressure is the thrust per unit area of the surface. 
Its S.I. unit is ‘newton per metre
2
‘ or ‘pascal’.
Solution 3S.
(a) Pressure is measured in ‘bar’. 
(b) 1 bar = 10
5
 pascal.
Solution 4S.
One pascal is the pressure exerted on a surface of area 1 m
2
 by a force of 1N acting 
normally on it.
Solution 5S.
Thrust is a vector quantity.
Solution 6S.
Pressure is a scalar quantity.
Solution 7S.
Thrust is the force applied on a surface in a perpendicular direction and it is a vector 
quantity. The effect of thrust per unit area is pressure, and it is a scalar quantity.
Solution 8S.
Pressure exerted by thrust is inversely proportional to area of surface on which it acts. 
Thus, larger the area on which the thrust acts, lesser is the pressure exerted by it. 
Example: If we stand on loose sand, our feet sink into the sand, but if we lie on that 
sand, our body does not sink into the sand. In both the cases, the thrust exerted on the 
sand is equal (equal to the weight of the body). However, when we lie on sand, the thrust 
acts on a large area and when we stand, the same thrust acts on a small area.
Solution 9S.
The tip of an allpin is made sharp so that large pressure is exerted at the sharp end and 
it can be driven into with less effort.
Solution 10S.
(a) It is easier to cut with a sharp knife because even a small thrust causes great 
pressure at the edges and cutting can be done with less effort. 
(b) Wide wooden sleepers are placed below the railway tracks so that the pressure 
exerted by the rails on the ground becomes less.
Solution 11S.
A substance which can flow is called a fluid.
Solution 12S.
Due to its weight, a fluid exerts pressure in all directions; the pressure exerted by the 
fluid is called fluid pressure.
Solution 13S.
A solid exerts pressure only on the surface on which it is placed, i.e. at its bottom, but a 
fluid exerts pressure at all points in all directions.
Solution 14S.
Take a can or large plastic bottle filled with water. Place it on a horizontal surface. Make a 
series of holes in the wall of the vessel anywhere below the free surface of the liquid. The 
water spurts out through each hole. This shows that the liquid exerts pressure at each 
point on the wall of the bottle. 
 
Liquid exerts pressure at all points in all directions
Solution 15S.
Pressure at a point in a liquid depends upon the following three factors:
1. Depth of the point below the free surface.
2. Density of liquid.
3. Acceleration due to gravity.
Solution 16S.
P = P
o
 + h?g 
Here, P = Pressure exerted at a point in the liquid 
P
o
 = Atmospheric pressure 
h = Depth of the point below the free surface 
? = Density of the liquid 
g = Acceleration due to gravity
Solution 17S.
Consider a vessel containing a liquid of density ?. Let the liquid be stationary. In order to 
calculate pressure at a depth, consider a horizontal circular surface PQ of area A at a 
depth h below the free surface XY of the liquid. The pressure on the surface PQ will be 
due to the thrust of the liquid contained in cylinder PQRS of height h with PQ as its base 
and top face RS lying on the frees surface XY of the liquid.
 
Total thrust exerted on the surface PQ 
= Weight of the liquid column PQRS 
= Volume of liquid column PQRS x density x g 
= (Area of base PQ x height) x density x g 
= (A x h) x ? x g
This thrust is exerted on the surface PQ of area A. Therefore, pressure is given as shown 
below. 
P = Thrust on surface / Area of surface 
P = Ah ?g / A = h?g 
Thus, Pressure = depth x density of liquid x acceleration due to gravity
Solution 18S.
Due to dissolved salts, density of sea water is more than the density of river water, so 
pressure at a certain depth in sea water is more than that at the same depth in river 
water.
Solution 19S.
(a) P
2
 = P
1
 + h ? g, 
(b) P
2
 > P
1
Solution 20S.
The reason is that when the bubble is at the bottom of the lake, total pressure exerted on 
it is the atmospheric pressure plus the pressure due to water column. As the gas bubble 
rises, due to decrease in depth the pressure due to water column decreases. By Boyle’s 
law, PV = constant, so the volume of bubble increases due to decrease in pressure, i.e., 
the bubble grows in size.
Solution 21S.
The pressure exerted by a liquid increases with its depth. Thus as depth increases, more 
and more pressure is exerted by water on wall of the dam. A thicker wall is required to 
withstand greater pressure, therefore, the thickness of the wall of dam increases towards 
the bottom.
Solution 22S.
The sea divers need special protective suit to wear because in deep sea, the total 
pressure exerted on the diver’s body is much more than his blood pressure. To withstand 
it, he needs to wear a special protective suit.
Solution 23S.
Laws of liquid pressure:
1. Pressure at a point inside the liquid increases with the depth from its free surface.
2. In a stationary liquid, pressure is same at all points on a horizontal plane.
3. Pressure is same in all directions about a point in the liquid.
4. Pressure at same depth is different in different liquids. It increases with the 
increase in the density of liquid.
5. A liquid seeks its own level.
Solution 24S.
The liquid from hole B reaches a greater distance on the horizontal surface than that from 
hole A. 
This explains that liquid pressure at a point increases with the depth of point from the 
free surface.
Solution 25S.
(i) As the diver moves to a greater depth, pressure exerted by sea water on him also 
increases. 
(ii) When the diver moves horizontally, his depth from the free surface remains constant 
and hence the pressure on him remains unchanged.
Solution 26S.
Pascal’s law states that the pressure exerted anywhere in a confined liquid is transmitted 
equally and undiminished in all directions throughout the liquid.
Solution 27S.
Two applications of Pascal’s law:
1. Hydraulic press
2. Hydraulic jack
Solution 28S.
The principle of a hydraulic machine is that a small force applied on a smaller piston is 
transmitted to produce a large force on the bigger piston. 
Hydraulic press and hydraulic brakes work on this principle.
Solution 29S.
Hydraulic press works on principle of hydraulic machine. 
It states that a small force applied on a smaller piston is transmitted to produce a large 
force on the bigger piston. 
Use: It is used for squeezing oil out of linseed and cotton seeds.
Solution 30S.
(i) X : Press Plunger; Y: Pump Plunger
(ii) When the lever is moved down, valve B closes and valve A opens, so the water from 
cylinder P is forced into the cylinder Q.
(iii) Valve B closes due to an increase in pressure in cylinder P. This pressure is 
transmitted to the connecting pipe and when the pressure in connecting pipe becomes 
greater than the pressure in the cylinder Q, valve A opens up.
(iv) When the release valve is opened, the ram (or press) plunger Q gets lowered and 
water of the cylinder Q runs out in the reservoir.
Solution 31S.