ICSE Class 9 Physics · Selina Solutions Concise Physics Class 9 ICSE · 65 solved questions
The thrust exerted by a body placed on a surface is :
equal to the weight of the body
Reason — The thrust exerted by a body placed on a surface is equal to its weight. Thrust is a vector quantity.
The pressure exerted on a surface depends on :
both the area of contact and the magnitude of the thrust applied
Reason — The pressure exerted on a surface depends on two factors :
Pressure and thrust are :
scalar and vector quantities respectively
Reason — Pressure is a scalar quantity whereas thrust is a vector quantity, because pressure is defined in terms of magnitude only, and thrust is defined in terms of both magnitude and direction.
The C.G.S. unit of pressure is :
dyne cm-2
Reason — C.G.S. unit of pressure is dyne cm-2 where 1 dyne cm-2 = 0.1 N m-2 or 1 N m-2 = 10 dyne cm-2
One bar is equal to :
105 N m-2
Reason — 1 bar = 105 N m-2.
1 atm of pressure is equal to :
760 torr
Reason — torr is a unit of atmospheric pressure after the name of the scientist Torricelli where, 1 torr = 1 mm of Hg and 1 atm = 760 mm of Hg = 760 torr.
The pressure on a surface is reduced by :
increasing the area of surface
Reason — Larger the area on which a given thrust acts, lesser is the pressure exerted by it, as pressure is inversely proportional to the area of surface.
$$\begin{aligned}\text{Pressure} = \frac{\text{Thrust}}{\text{Area}} \\[1em] \text{or} \\[1em] \text P = \frac{\text{F}}{\text{A}}\end{aligned}$$
Cutting tools have either sharp or pointed edges so that a ............... thrust may cause a ............... pressure at that edges and cutting can be done with less effort.
small, high
Reason — For a given thrust, the pressure on a surface is increased by reducing the area of surface on which it is acting, hence, cutting tools have either sharp or pointed edges so that a small thrust may cause a high pressure at that edges and cutting can be done with less effort.
The pressure inside a liquid of density ρ at a depth h is:
$\text {hρg}$
Reason — The three factors on which the pressure at a point in a liquid depends are —
depth of the point below the free surface (h)
density of liquid (ρ) and
acceleration due to gravity (g)
Pressure = hρg
Out of the following, which one is correct about the law of liquid pressure?
It increases with an increase in the density of the liquid.
Reason — As Pressure = hρg, so pressure is directly proportional to the density of the liquid. Hence, it increases with an increase in the density of the liquid.
The pressure P1 at a certain depth in river water and P2 at the same depth in sea water are related as:
P1 < P2
Reason — As the density of sea water is more than the density of river water, hence, the pressure at a certain depth in sea water is more than that at the same depth in river water because pressure increases with the increase in density of liquid.
The pressure P1 at the top of a dam and P2 at a depth h from the top inside water (density ρ) are related as:
P2 - P1 = hρg
Reason — The pressure at a point inside a liquid at a depth h is P2
P2 = pressure at the top of a dam (P1) + pressure due to liquid column
= P1 + hρg
So, P2 = P1 + hρg
Hence, P2 - P1 = hρg
The wall of a dam is made thicker at the bottom because :
the pressure exerted by a liquid increases with its depth.
Reason — A dam has broader walls at the bottom than at the top as the pressure exerted by the liquid increases with it's depth. Thus, as depth increases, more and more pressure is exerted by water on the walls of the dam.
Hydraulic machines work on the principle of :
Pascal's law
Reason — 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. This principle is based on Pascal's law.
Hydraulic machines act like a :
force multiplier
Reason — 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. Hence, it acts as a force multiplier.
A rectangular cement block has length, breadth and height of 60 cm, 30 cm and 15 cm respectively. Identify the correct position of the block and pressure exerted by it on the ground.
Maximum pressure is exerted when breadth and height form the base
Reason — Pressure is given by the formula : $\text {Pressure} = \frac {\text {Thrust}}{\text {Area}}$
For the same weight (thrust) of the block, the pressure is inversely proportional to the contact area. So, larger base area → smaller pressure, and smaller base area → greater pressure.
As, minimum area → side with minimum dimensions → 30 cm breadth and 15 cm height.
Hence, maximum pressure is exerted when breadth and height form the base because it has smallest contact area.
The incorrect statement from the following is :
The pressure exerted by the liquid filled in a vessel is same at all points.
Reason — Liquid pressure increases with depth so, pressure is not the same at all points in the vessel as it depends on how deep the point is from the free surface.
Define the term thrust. State its S.I. unit.
Thrust is the force acting normally on a surface.
The S.I. unit of thrust is newton (N).
Thrust is a vector quantity.
(a) What physical quantity is measured in bar ?
(b) How is the unit bar related to the S.I. unit pascal ?
(a) The physical quantity that is measured in bar is Pressure.
(b) The relation between bar and pascal is : 1 bar = 105 pascal.
Define one pascal (Pa), the S.I. unit of pressure.
One pascal is the pressure exerted on a surface of area 1 m2 by a force of 1 N acting normally on it.
State whether thrust is a scalar or vector?
Thrust is a vector quantity. Its direction of application is normal to the surface.
State whether pressure is a scalar or vector?
Pressure is a scalar quantity.
What is a fluid?
A substance which can flow is called a fluid. All liquids and gases are thus fluids.
Pressure at free surface of a water lake is P1, while at a point at depth h below its free surface is P2. (a) How are P1 and P2 related ? (b) Which is more P1 or P2 ?
(a) As we know,
Total pressure in a liquid at a depth h is P2
= atmospheric pressure acting on the surface (P1) + pressure due to liquid column (h ρ g)
= P1 + h ρ g
Hence,
P2 = P1 + h ρ g
(b) From the above expression we observe that,
P2 > P1
How does the liquid pressure on a diver change if —
(i) the diver moves to the greater depth, and
(ii) the diver moves horizontally ?
(i) When the diver moves to a greater depth the liquid pressure increases as liquid pressure at a point increases with the increase of depth from it's free surface.
(ii) When the diver moves horizontally, the liquid pressure remains unchanged as inside a liquid, pressure is same at all points on a horizontal plane.
State Pascal's law of transmission of pressure.
Pascal's law states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.
Name two applications of Pascal's law.
Applications of Pascal's law are —
Complete the following sentences —
(a) Pressure at a depth h in a liquid of density ρ is ............... .
(b) Pressure is ............... in all directions about a point in a liquid.
(c) Pressure at all points at the same depth is ............... .
(d) Pressure at a point inside a liquid is ............... to it's depth.
(e) Pressure of a liquid at a given depth is ............... to the density of liquid.
(a) Pressure at a depth h in a liquid of density ρ is h ρ g.
(b) Pressure is same in all directions about a point in a liquid.
(c) Pressure at all points at the same depth is same.
(d) Pressure at a point inside a liquid is directly proportional to its depth.
(e) Pressure of a liquid at a given depth is directly proportional to the density of liquid.
What is meant by pressure ? State its S.I. unit.
Pressure is the thrust per unit area of surface.
$$\begin{aligned}\text {Pressure} = \frac{\text {Thrust}}{\text {Area}} \\[1em] \text{or} \\[1em] \text P = \frac{\text F}{\text A}\end{aligned}$$
Pressure is a scalar quantity.
The S.I. unit of pressure is newton per metre2 . This unit is named as pascal (symbol Pa).
Differentiate between thrust and pressure.
| Thrust | Pressure |
|---|---|
| Thrust is the force acting normally on a surface. | Pressure is the thrust per unit area of surface. |
| It is a vector quantity | It is a scalar quantity |
| The S.I. unit is newton(N) | The S.I. is newton per metre2 |
How does the pressure exerted by a thrust depend on the area of surface on which it acts ? Explain with a suitable example.
The pressure exerted by a thrust is inversely proportional to the area of surface on which it acts. Larger the area on which a given thrust acts, lesser is the pressure exerted by it.
Example — A brick of weight 4 kgf having dimensions 20 cm x 10 cm x 5 cm, exerts maximum pressure on ground when it is placed with it's longest side (20 cm) vertical, as shown in fig below, while it exerts minimum pressure when it is placed with it's shortest side (5 cm) vertical, even though the thrust is same in each case.
Why is the tip of an allpin made sharp?
The tip of an allpin is made sharp so that large pressure is exerted through the pointed end and that they can be driven into, with less effort.
Explain the following —
(a) It is easier to cut with a sharp knife than with a blunt one.
(b) Sleepers are laid below the rails
(a) It is easier to cut with a sharp knife than with a blunt knife because in a sharp knife even a small thrust causes a great pressure at the edges and hence, cutting can be done with less effort.
(b) Sleepers are laid below the rails so that the pressure exerted by the iron nails on the ground becomes less.
What do you mean by the term fluid pressure?
A fluid contained in a vessel exerts pressure at all points and in all directions due to it's weight and this pressure is called fluid pressure.
How does the pressure exerted by a solid and a fluid differ ?
Both liquids and solids exert pressure due to it's weight, however, pressure exerted by a solid acts only on the surface on which it is placed i.e. at it's bottom, but pressure exerted by a fluid acts on the bottom as well as the walls of the container due to it's tendency to flow.
State three factors on which the pressure at a point in a liquid depends.
The three factors on which the pressure at a point in a liquid depends are —
depth of the point below the free surface (h),
density of liquid (ρ), and
acceleration due to gravity (g).
Write an expression for the pressure at a point inside a liquid. Explain the meaning of the symbols used.
The pressure at a point inside a liquid at a depth h
= Atmospheric pressure + pressure due to liquid column
= P0 + hρg
where,
P0 = atmospheric pressure acting on the free surface of liquid
h = depth of the point below the free surface
ρ = the density of the fluid
g = acceleration due to gravity.
How does the pressure at a certain depth in sea water differ from that at the same depth in river water ? Explain your answer.
The pressure at a certain depth in sea water is more than the same depth in river water because the density of sea water is more than the density of river water.
According to the formula,
Pressure = depth x density of liquid x acceleration due to gravity = hρg
Therefore, when density of a liquid is more then the pressure exerted is also more as density and pressure are directly proportional.
Explain why a gas bubble released at the bottom of a lake grows in size as it rises to the surface of lake.
It is noticed that as the gas bubble formed at the bottom of the lake rises, it grows in size. The reason is that when the bubble is at the bottom of the lake, total pressure exerted on it is the sum of the atmospheric pressure and the pressure due to the water column above it.
As the gas bubble rises, due to decrease in depth, the pressure due to water column decreases, so the total pressure on the bubble decreases.
According to Boyle's law, the volume of a gas is inversely proportional to the pressure on it. Therefore, the volume of bubble increases due to the decrease in pressure, i.e., the bubble grows in size.
When the bubble reaches the surface of liquid, total pressure exerted on it becomes minimum, just equal to the atmospheric pressure and so the size of bubble when touching the surface becomes maximum.
A dam has broader walls at the bottom than at the top. Explain.
A dam has broader walls at the bottom than at the top as the pressure exerted by the liquid increases with it's depth. Thus, as depth increases, more and more pressure is exerted by water on the walls of the dam.
A thicker wall is required to withstand a greater pressure, therefore, the wall of a dam is made with thickness increasing towards the base.
In the figure given below, the increasing length of arrows in water represents the increasing pressure on the wall of the dam towards the bottom.
Why do sea divers need special protective suit ?
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, they need to wear a special protective suit, made from glass reinforced plastic or cast aluminium. The pressure inside the suit is maintained at one atmosphere.
State the laws of liquid pressure.
The laws of liquid pressure are —
Inside the liquid, pressure increases with the increase in depth from it's free surface.
In a stationary liquid, pressure is same at all points on a horizontal plane.
Pressure is same in all directions about a point inside the liquid.
Pressure at same depth is different in different liquids. It increases with the increase in density of liquid.
A liquid seeks it's own level.
A tall vertical cylinder filled with water is kept on a horizontal table top. Two small holes A and B are made on the wall of the cylinder, A near the middle and B just below the free surface of water. State and explain your observation.
We observe that the throw of the liquid from hole A is more than the B i.e., the liquid reaches to a greater distance on the horizontal surface from hole A than hole B. This shows that liquid pressure at a point increases with the increase of depth from it's free surface.
Name and state the principle on which a hydraulic press works. Write one use of the hydraulic press.
A hydraulic press works on the principle of Pascal's law.
Principle — When a force F1 is applied on the piston A, it exerts a pressure on liquid contained in the cylinder P. According to Pascal's law, this pressure is transmitted through liquid in tube R to the piston B of the other cylinder Q due to which the piston B tends to move upwards.
Since, the area of cross section of cylinder P is less than that of the cylinder Q, therefore by applying a small force on the piston A, we can lift a large weight kept on the piston B.
When no weight is placed on the piston B, it rises up against a fixed roof with a force F2 (F2 > F1). If a bale of cotton is kept on the press plunger B, it gets compressed.
Use of hydraulic press — It is used for pressing cotton bales and goods such as books, quilts etc.
Describe a simple experiment to demonstrate that a liquid enclosed in a vessel exerts pressure in all directions.
Take a vessel filled with a liquid (say, water). Place it on a horizontal surface. Make several small holes in the wall of the vessel anywhere below the free surface of liquid. It is observed that liquid spurts out through each hole. This shows that the liquid exerts pressure at each point on the wall of the vessel.
Deduce an expression for the pressure at a depth inside a liquid.
Consider a vessel containing a liquid with density ρ. Let the liquid be stationary. In order to calculate pressure at a depth h, consider a horizontal circular surface PQ with area A at depth h below the free surface XY of the liquid as shown below.
The pressure on surface PQ will be due to the weight of the liquid column above the surface PQ, (i.e., the liquid contained in cylinder PQRS of height h with PQ as it's base and top face RS lying on the free surface XY of the liquid).
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) ρ x g = A h ρ g
This thrust is exerted on the surface PQ of area A. Therefore, pressure
P = $\frac{\text{Thrust on surface}}{\text{Area of surface}}$ = $\frac{\text{Ahρg}}{\text{A}} = \text {hρg}$
Hence, Pressure = depth x density of liquid x acceleration due to gravity = hρg
Explain the principle of a hydraulic machine. Name two devices which work on this principle.
The principle of hydraulic machine is that a small force applied on a smaller piston is transmitted to produce a large force on the bigger piston.
Two devices that work on this principle are —
The diagram in figure below shows a device which makes use of the principle of transmission of pressure.
(a) Name the parts labelled by the letters X and Y.
(b) Describe what happens to the valves A and B and to the quantity of water in the two cylinders when the lever arm is moved down.
(c) State one use of the above device.
(a) The parts are —
X → Press plunger
Y → Pump plunger
(b) When the lever arm is moved down, the valve B closes and the valve A opens, so water from cylinder P is forced into the cylinder Q.
(c) A hydraulic press is used for pressing cotton bales and goods like quilts, books, etc.
Draw a simple diagram of a hydraulic jack and explain its principle.
When handle H of lever is pressed down by applying an effort, the valve V opens because of increase in pressure in the cylinder P. The liquid runs out from the cylinder P to the cylinder Q.
As a result, the piston B rises up and it raises the car placed on the platform. When the car reaches the desired height, the handle H of lever is no longer pressed.
The valve V gets closed (since the pressure on either side of the valve becomes same) so that the liquid may not run back from the cylinder Q to the cylinder P.
Explain the principle of a hydraulic brake with a simple labelled diagram.
To apply brakes, the foot pedal is pressed due to which pressure is exerted on the liquid in the master cylinder P, so liquid runs out from the master cylinder P to the wheel cylinder Q.
As a result, the pressure is equally transmitted and undiminished through the liquid to the pistons B1 and B2 of the wheel cylinder Q. Therefore, the pistons B1 and B2 gets pushed outwards and brake shoes gets pressed against the rim of the wheel due to which the motion of the wheel retards.
Since, the area of cross section of piston A in the master cylinder P is less than that in the wheel cylinder Q, a small force applied at the foot pedal produces a large force on the pistons B1 and B2 of the wheel cylinder Q.
This is the force responsible for retarding the vehicle. It should be noted that due to transmission of pressure through liquid, equal pressure is exerted on all wheels of the vehicle connected to the pipe line R.
On releasing the pressure on the pedal, the spring pulls the brake shoes to it's original position and forces the piston's B1 and B2 to return back into the wheel cylinder Q. As a result the liquid runs back from the wheel cylinder Q to the master cylinder P and thus the brakes gets released.
A hammer exerts a force of 1.5 N on each of the two nails A and B. The area of cross section of tip of nail A is 2 mm2 while that of nail B is 6 mm2. Calculate pressure on each nail in pascal.
As we know,
Pressure (P) = $\frac{\text {Thrust (F)}}{\text {Area (A)}}$
Given,
Converting 2 mm2 in metre2
As 1 mm = $\frac{1}{1000}$ m
So, 1 mm2 = (1) mm x (1) mm
= $\left(\frac{1}{1000}\right) \text m \times \left(\frac{1}{1000}\right) \text m$
= 1 x 10 -6 m2
Hence,
2 mm2 = 2 x 10-6 m2
AA = 2 x 10 -6 m2
AB = 6 x 10 -6 m2
Substituting the values in the formula above, we get,
Pressure on A
$$\begin{aligned}\text P = \frac{1.5}{2 \times 10^{-6}} \\[1em] = 0.75 \times 10^{6} \\[1em] = 7.5 \times 10^{5} \text { Pa}\end{aligned}$$
Hence, pressure on A = 7.5 x 105 Pa
Pressure on B
$$\begin{aligned}\text P = \frac{1.5}{6 \times 10^{-6}} \\[1em] = 0.25 \times 10^{6} \\[1em] = 2.5 \times 10^{5} \text { Pa}\end{aligned}$$
Hence, pressure on B = 2.5 x 105 Pa
A block of iron of mass 7.5 kg and of dimensions 12 cm x 8 cm x 10 cm is kept on a table top on it's base of side 12 cm x 8 cm. Calculate : (a) thrust and (b) pressure exerted on the table top. Take 1 kgf = 10 N.
As we know,
(a) Force (F) = mass (m) x acceleration due to gravity (g)
Given,
Substituting the values in the formula above we get,
$$\begin{aligned}\text F = 7.5 \times 10 \\[1em] = 75\ \text N\end{aligned}$$
Hence, F = 75 N
(b) As we know,
Pressure (P) = $\frac{\text {Thrust (F)}}{\text {Area (A)}}$
Given,
Converting cm2 into m2
100 cm = 1 m
So, 100 cm x 100 cm = 1 m2
Hence, 96 cm2 = $\frac{1 \times 96}{10000}$
Therefore, A = 0.0096 m 2
Substituting the values in the formula above, we get,
$$\begin{aligned}\text P = \frac{75}{0.0096} \\[1em] = 7812.5\ \text Pa\end{aligned}$$
Hence, pressure = 7812.5 Pa
A vessel contains water up to a height of 1.5 m. Taking the density of water 103 kg m-3, acceleration due to gravity 9.8 m s-2 and area of base vessel 100 cm2, calculate : (a) the pressure and (b) the thrust, at the base of vessel.
(a) As we know,
Pressure due to water column of height h = h ρ g
Given,
Substituting the values in the formula above, we get,
$$\begin{aligned}\text P = 1.5 \times 1000 \times 9.8 \\[1em] = 14.7 \times 10^{3}\ \text {N m}^{-2}\end{aligned}$$
Hence, pressure = 1.47 x 104 N m-2
(b) As we know,
Pressure (P) = $\frac{\textbf {Thrust (F)}}{\textbf {Area (A)}}$
Given,
Converting cm2 into m2
100 cm = 1 m
And, 100 cm x 100 cm = 1 m2
Therefore, 100 cm2 = $\frac{1 \times 100}{10000}$ m2
Hence, A = 10-2 m2
Substituting the values in the formula above, we get,
$$\begin{aligned}1.47 \times 10^{4} = \frac{\text {Thrust (F)}}{10^{-2}} \\[1em] \Rightarrow \text {Thrust (F)} = 1.47 \times 10^{2} \\[1em] \Rightarrow \text {Thrust (F)} = 147\ \text N\end{aligned}$$
Hence, thrust = 147 N
The area of base of a cylindrical vessel is 300 cm2. Water (density = 1000 kg m-3) is poured into it up to a depth of 6 cm. Calculate : (a) the pressure and (b) the thrust of water on the base. (g = 10 m s-2).
(a) As we know,
Pressure due to water column of height h = hρg
Given,
Converting cm to m
100 cm = 1 m
So, 6 cm = $\frac{1}{100}$ x 6 = 0.06 m
Hence, h = 0.06 m
Substituting the values in the formula above we get,
P = 0.06 x 1000 x 10
⇒ P = 600 Pa
Hence, pressure = 600 Pa
(b) As we know,
Pressure (P) = $\frac{\textbf {Thrust (F)}}{\textbf {Area (A)}}$
Given,
Converting cm2 into m2
100 cm = 1 m
And, 100 cm x 100 cm = 1 m2
Therefore, 300 cm2 = $\frac{1 \times 300}{10000}$ m2
Hence, A = 3 x 10-2 m2
Substituting the values in the formula we get,
$$\begin{aligned}600 = \frac{\text {Thrust (F)}}{3 \times 10^{-2}} \\[1em] \Rightarrow \text {Thrust (F)} = 600 \times 3 \times 10^{-2}\\[1em] \Rightarrow \text {Thrust (F)} = 18\ \text N\end{aligned}$$
Hence, thrust = 18 N
(a) Calculate the height of a water column which will exert on it's base the same pressure as the 70 cm column of mercury. Density of mercury is 13.6 g cm-3.
(b) Will the height of the water column in part (a) change if the cross section of the water column is made wider ?
(a) As we know,
Pressure due to water column of height h = hρg
and
Pascal's law, states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.
Thus,
Pressure due to water column = Pressure due to mercury column
Hence,
hw ρw g = hm ρm g
Given,
From the above formula, we get,
hw = $\frac{\text h_\text m \text ρ_\text m}{\text ρ_\text w}$
Substituting the values, we get,
hw = $\left(\frac{13.6}{1}\right)$ x 70
⇒ hw = 952 cm
⇒ hw = 9.52 m
Hence, height of a water column = 9.52 m
(b) No, if the cross section of the water column is made wider, the height of the water column will be unaffected.
The pressure of water on the ground floor is 40,000 Pa and on the first floor is 10,000 Pa. Find the height of the first floor.
(Take : density of water = 1000 kg m-3, g = 10 m s-2)
As we know,
Pressure due to water column of height h = hρg
Given,
In order to know the height of the first floor, let us calculate the difference in pressure
P = Pg - Pf
= 40,000 – 10,000
= 30,000 Pa
Substituting the values in the formula,
P = h ρ g
⇒ 30,000 = 1000 x 10 x h
⇒ h = 3 m
Hence, the height of the first floor is 3 m.
A simple U tube contains mercury to the same level in both of it's arms. If water is poured to a height of 13.6 cm in one arm, how much will be the rise in mercury level in the other arm ?
Given: density of mercury = 13.6 x 103 kg m-3 and density of water = 103 kg m-3
Given,
By pouring 13.6 cm of water, the mercury level in the left arm goes down to point A by x cm, while in the right arm, it rises to point C by x cm. Therefore, BC = hm = 2x cm
By Pascal's law,
Pressure in the water column = pressure in the mercury column
Therefore, PA = PB
⇒ hw ρw g = hm ρm g
⇒ 13.6 x 103 x g = 2𝑥 x 13.6 x 103 x g
⇒ 1 = 2𝑥
⇒ 𝑥 = $\frac{1}{2}$ = 0.5 cm
Hence, the rise in mercury level = 0.5 cm
In a hydraulic machine, a force of 2 N is applied on the piston of area of cross section 10 cm2. What force is obtained on it's piston of area of cross section 100 cm2?
As we know,
Pressure (P) = $\frac{\textbf {Thrust (F)}}{\textbf {Area (A)}}$
Given,
Converting cm2 into m2
100 cm = 1 m
And, 100 cm x 100 cm = 1 m2
Therefore, 10 cm2 = $\frac{1 \times 10}{10000}$
Hence, A = 10-3 m2
F = 2 N at area of cross section 10 cm2
Substituting the values in the formula we get,
P = $\frac{2}{10^{-3}}$ [Equation 1]
Now when, A = 100 cm2
Converting cm2 into m2
A = 100 cm 2 = $\frac{1 \times 100}{10000}$ = 10-2 m 2
Substituting the value in the formula above we get,
Therefore, P = $\frac{\text F}{10^{-2}}$ [Equation 2]
Equating 1 and 2 we get,
$$\begin{aligned}\frac{2}{10^{-3}} = \frac{\text F}{10^{-2}} \\[0.5em] \Rightarrow \text F = \frac{2 \times 10^{-2} }{10^{-3}} \\[0.5em] \Rightarrow \text F = 20\ \text N\end{aligned}$$
Hence, force at 100 cm2 is 20 N.
What should be the ratio of area of cross section of the master cylinder and wheel cylinder of a hydraulic brake so that a force of 15 N can be obtained at each of it's brake shoe by exerting a force of 0.5 N on the pedal ?
As we know,
Pressure (P) = $\frac{\text {Thrust (F)}}{\text {Area (A)}}$
Let,
By the principle of hydraulic brakes which works on Pascal's law
Pressure on narrow piston = Pressure on broader piston
Thus,
$$\begin{aligned}\frac{\text F_{1}}{\text A_{1}} = \frac{\text F_{2}}{\text A_{2}} \\[1em] \frac{0.5}{\text A_{1}} = \frac{15}{\text A_{2}} \\[1em] \frac{\text A_{1}}{\text A_{2}} = \frac{0.5}{15} \\[1em] \Rightarrow \frac{\text A_{1}}{\text A_{2}} = \frac{5}{150} \\[1em] \Rightarrow \frac{\text A_{1}}{\text A_{2}} = \frac{1}{30}\end{aligned}$$
Hence, the ratio of area of cross section = 1 : 30
The areas of pistons in a hydraulic machine are 5 cm2 and 625 cm2. What force on the smaller piston will support a load of 1250 N on the larger piston? State any assumption which you make in your calculation.
As we know,
Pressure (P) = $\frac{\textbf {Thrust (F)}}{\textbf {Area (A)}}$
and by the principle of hydraulic machine
Pressure on narrow piston = Pressure on broader piston
So,
$$\begin{aligned}\frac{\text F_{1}}{\text A_{1}} = \frac{\text F_{2}}{\text A_{2}} \\[1em]\end{aligned}$$
Given,
Substituting the values in the formula above we get,
$$\begin{aligned}\frac{\text F_{1}}{5} = \frac{1250}{625} \\[1em] \text F_{1} = \frac{1250 \times 5 }{625} \\[1em] \Rightarrow \text F_{1} = 10\ \text N\end{aligned}$$
Hence, force acting on the smaller piston = 10 N
Assumption — There is no friction and no leakage of liquid.
(a) The diameter of neck and bottom of a bottle are 2 cm and 10 cm respectively. The bottle is completely filled with oil. If the cork in the neck is pressed in with a force of 1.2 kgf, what force is exerted on the bottom of the bottle ?
(b) Name the law/principle you have used to find the force in part (a)
(a) As we know,
Pressure (P) = $\frac{\textbf {Thrust (F)}}{\textbf {Area (A)}}$
and by the principle of hydraulic machine
Pressure on neck = pressure on bottom of bottle
So,
$$\begin{aligned}\frac{\text F_{1}}{\text A_{1}} = \frac{\text F_{2}}{\text A_{2}} \\[1em]\end{aligned}$$
Given,
Then,
A1 = 𝜋$(\frac{2}{2})^{2}$ = 𝜋
And
A2 = 𝜋$(\frac{10}{2})^{2}$ = 25 𝜋
Substituting the values in the formula above we get,
$$\begin{aligned}\frac{1.2}{\text π} = \frac{\text F_{2}}{25\text π} \\[1em] \text F_{2} = 1.2 \times 25 \\[1em] \text F_{2} = 30 \text{ kgf}\end{aligned}$$
Hence, force exerted at the bottom of the neck = 30 kgf.
(b) The Pascal's law is applied to solve the part (a) which states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.
Hence,
$$\begin{aligned}\frac{\text F_{1}}{\text A_{1}} = \frac{\text F_{2}}{\text A_{2}} \\[1em]\end{aligned}$$
A force of 50 kgf is applied to the smaller piston of a hydraulic machine. Neglecting friction, find the force exerted on the large piston, if the diameters of the pistons are 5 cm and 25 cm respectively.
As we know, by the principle of hydraulic machine
Pressure on the smaller piston = Pressure on the larger piston
So,
$$\begin{aligned}\frac{\text F_{1}}{\text A_{1}} = \frac{\text F_{2}}{\text A_{2}} \\[0.5em]\end{aligned}$$
Given,
Then,
A1 = 𝜋$(\frac{5}{2})^{2}$ = 6.25𝜋
And
A2 = 𝜋$(\frac{25}{2})^{2}$ = 156.25𝜋
Substituting the values in the formula above we get,
$$\begin{aligned}\frac{50}{6.25\text π} = \frac{\text F_{2}}{156.25\text π} \\[1em] \text F_{2} = \frac{50 \times 156.25}{6.25} \\[1em] \text F_{2} = 1250 \text { kgf}\end{aligned}$$
Hence, force exerted on the larger piston is 1250 kgf.
Two cylindrical vessels fitted with pistons A and B of area of cross section 8 cm2 and 320 cm2 respectively, are joined at their bottom by a tube and they are completely filled with water. When a mass of 4 kg is placed on piston A, find : (i) the pressure on piston A, (ii) the pressure on piston B, and (iii) the thrust on piston B.
(i) As we know,
Pressure (P) = $\frac{\textbf {Thrust (F)}}{\textbf {Area (A)}}$
Given,
Substituting the values in the formula above, we get,
$$\begin{aligned}\text P_\text A = \frac{4}{8} \\[1em] \Rightarrow \text P_\text A = 0.5 \text { kg cm}^{-2}\end{aligned}$$
Hence, PA = 0.5 kg cm-2
(ii) As we know, by the principle of hydraulic machine
Pressure on piston A = Pressure on piston B
Hence, PB = 0.5 kg cm-2
(iii) Thrust on piston B is acting in the upward direction, which is given by
Pressure (P) = $\frac{\textbf {Thrust (F)}}{\textbf {Area (A)}}$
Substituting the values, we get,
$$\begin{aligned}0.5 = \frac{\text{thrust}}{320} \\[1em] \Rightarrow \text{thrust} = 0.5 \times 320 \\[1em] \Rightarrow \text{thrust} = 160 \text{ kgf}\end{aligned}$$
Hence, thrust on piston B = 160 kgf
What force is applied on a piston of area of cross section 2 cm2 to obtain a force 150 N on the piston of area of cross section 12 cm2 in a hydraulic machine ?
As we know, by the principle of hydraulic machine
Pressure on piston A = Pressure on piston B
Hence,
$\frac{\text F_{1}}{\text A_{1}} = \frac{\text F_{2}}{\text A_{2}}$
Given,
Converting cm2 into m2
100 cm = 1 m
And, 100 cm x 100 cm = 1 m2
Therefore, 1 cm2 = $\frac{1}{10000}$ m2
Hence,
Substituting the values in the formula above we get,
$$\begin{aligned}\frac{\text F_1}{2 \times 10^{-4}} = \frac{150}{12 \times 10^{-4}} \\[1 em] \Rightarrow \text F_1 = \frac{150 \times 2 \times 10^{-4}}{12 \times 10^{-4}} \\[0.5em] \Rightarrow \text F_1 = 25\ \text N \\[0.5em]\end{aligned}$$
Hence, force applied = 25 N
A hydraulic jack is used in a car workshop to lift a car. The jack consists of two connected pistons A(area = 5 cm2) and B(area = 250 cm2), filled completely with an incompressible hydraulic fluid (such as hydraulic oil). The fluid transmits pressure uniformly throughout the system. A force of 120 N is applied vertically downward on piston A.
Assuming the system to be ideal, answer the following questions:
(a) Calculate the pressure produced in the liquid by piston A.
(b) Determine the force exerted on piston B. Name the law used.
(c) Does this hydraulic jack provide a gain in force or a gain in distance? Justify your answer.
(d) If piston B rises by a height of 2 cm, calculate the distance moved by piston A.
(e) What would happen if a small air bubble enters the liquid of the hydraulic jack?
Given,
(a) Converting cm2 into m2
100 cm = 1 m
And, 100 cm x 100 cm = 1 m2
Therefore, 1 cm2 = $\frac{1}{10000}$ m2
Hence,
As pressure on a surface is given by,
$\text {Pressure} = \frac{\text {Force}}{\text {Area}}$
Then,
$$\begin{aligned}\text {Pressure on piston A} = \frac{\text {Force applied on piston A}}{\text {Area of piston A}}\\[1em] = \frac{120}{5 \times 10^{-4}} \\[1em] = \frac{120 \times 10^4}{5} \\[1em] = 24 \times 10^4 \text { Pa} \\[1em] = 2.4 \times 10^5 \text { Pa}\end{aligned}$$
Hence, the pressure produced in the fluid by piston A is 2.4 x 105 Pa.
(b) As Pascal's law states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.
Thus,
Pressure on piston A = Pressure on piston B
Now,
$$\begin{aligned}\text {Pressure on piston B} = \frac{\text {Force applied on piston B}}{\text {Area of piston B}} \\[1em] \Rightarrow \text {Force applied on piston B} = \text {Pressure on piston B} \times \text {Area of piston B} \\[1em] = \text {Pressure on piston A} \times \text {Area of piston B} \\[1em] = 2.4 \times 10^5 \times 250 \times 10^{-4} \\[1em] = 24 \times 10^4 \times 250 \times 10^{-4} \\[1em] = 24 \times 250 \\[1em] = 6000\ \text N \\[1em]\end{aligned}$$
Hence, the force exerted on piston B is 6000 N and the law is Pascal's law.
(c) The hydraulic jack provides gain in force. This is because a small force of 120 N applied on the smaller piston produces a much larger force of 6000 N on the larger piston. Since the output force is greater than the input force, the machine acts as a force multiplier. It does not give a gain in distance.
(d) Given,
From conservation of volume.
Volume of fluid displaced in piston A = Volume of fluid displaced in piston B
⇒ Area of piston A × Distance moved by piston A = Area of piston B × Distance moved by piston B
⇒ Distance moved by piston A = $\frac{\text {Piston B Area} \times \text {Dist. moved Piston B}}{\text {Piston A Area}}$
$$\begin{aligned}= \frac{250 \times 2}{5} \\[1em] = \frac{250 \times 2}{5} \\[1em] = 50 \times 2 \\[1em] = 100 \text { cm}\end{aligned}$$
Hence, the distance moved by piston A is 100 cm.
(e) If a small air bubble enters the liquid, the hydraulic jack will not work efficiently.
This is because air is compressible, whereas hydraulic fluid is incompressible. When force is applied, some part of it will be used to compress the air bubble instead of being fully transmitted through the liquid.
As a result, the pressure transmitted to piston B will decrease, and the jack may lift the car less effectively or work sluggishly.
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