Biology For Class X - Chapter No. 7 - Man And His Environment - Text Book Chapter

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CHAPTER 7: MAN AND HIS ENVIRONMENT

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Islamiat (اسلامیات) For Class IX -New - Baab So'um: Mozuaati Mutalah - (ج) Siraat e Tayyabah - Da'awat o Tableegh - Short Question-Answers

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باب سوم: موضوعاتی مطالعہ

(ج) سیرت طیبہ 2- دعوت و تبلیغ

مختصر سوال و جواب

Islamiat (اسلامیات) For Class IX -New - Baab So'um: Mozuaati Mutalah - (ج) Siraat e Tayyabah - Da'awat o Tableegh - Detailed Question-Answers

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باب سوم: موضوعاتی مطالعہ

(ج) سیرت طیبہ 2- دعوت و تبلیغ

تفصیلی سوال و جواب

Islamiat (اسلامیات) For Class IX -New - Baab So'um: Mozuaati Mutalah - (ج) Siraat e Tayyabah - Da'awat o Tableegh - Multiple Choice Questions (MCQs) کثیرالانتخابی سوالات

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باب سوم: موضوعاتی مطالعہ

(ج) سیرت طیبہ 2- دعوت و تبلیغ

کثیرالانتخابی سوالات

Islamiat (اسلامیات) For Class IX -New - Baab So'um: Mozuaati Mutalah - (ج) Siraat e Tayyabah - Baasat e Nabvi (صَلَّی اللهُ عَلَیهِ وَّعَلٰٓی آلِہٖ وَ آصحَابِہٖ وَسَلِّمُ) - Short Question-Answers

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باب سوم: موضوعاتی مطالعہ

(ج) سیرت طیبہ 1- بعثتِ نبوی (صَلَّی اللهُ عَلَیهِ وَّعَلٰٓی آلِہٖ وَ آصحَابِہٖ وَسَلِّمُ)

مختصر سوال و جواب

سوال نمبر 2: بعثت نبوی کا مفہوم بیان کریں۔
جواب:

سوال نمبر 3: بعثت نبوی کے چند مقاصد بیان کریں۔
جواب:

Islamiat (اسلامیات) For Class IX -New - Baab So'um: Mozuaati Mutalah - (ج) Siraat e Tayyabah - Baasat e Nabvi (صَلَّی اللهُ عَلَیهِ وَّعَلٰٓی آلِہٖ وَ آصحَابِہٖ وَسَلِّمُ) - Detailed Question-Answers

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باب سوم: موضوعاتی مطالعہ

(ج) سیرت طیبہ 1- بعثتِ نبوی (صَلَّی اللهُ عَلَیهِ وَّعَلٰٓی آلِہٖ وَ آصحَابِہٖ وَسَلِّمُ)

تفصیلی سوال و جواب

سوال نمبر 3: بعثت نبوی کے آثار بیان کریں۔
جواب: بعثت نبوی کے آثار:

Islamiat (اسلامیات) For Class IX -New - Baab So'um: Mozuaati Mutalah - (ج) Siraat e Tayyabah - Baasat e Nabvi (صَلَّی اللهُ عَلَیهِ وَّعَلٰٓی آلِہٖ وَ آصحَابِہٖ وَسَلِّمُ) - Multiple Choice Questions (MCQs) کثیرالانتخابی سوالات

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باب سوم: موضوعاتی مطالعہ

(ج) سیرت طیبہ 1- بعثتِ نبوی (صَلَّی اللهُ عَلَیهِ وَّعَلٰٓی آلِہٖ وَ آصحَابِہٖ وَسَلِّمُ)

کثیرالانتخابی سوالات

Forces And Matter - Physics For Class IX (Science Group) - Numericals

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Physics For Class IX (Science Group)
UNIT 5: FORCES AND MATTER
Numericals

Worked Example 1

1. A spring has spring constant k = 30 Nm–1. What load is required to produce an extension of 4 m?
Solution:
Step 1: Write down known quantities and the quantities to be found.

• k = 30 Nm^-1
• x = 4 m
• F = ?

Step 2: Write down formula and rearrange if necessary

F = kx

Step 3: Put the values in formula and calculate.

F = 30 Nm^-1 x 4 m

F = 120 N

Ans: Hence, 120 N load is required to stretch the spring by 4m.

Worked Example 2

2. Calculate the pressure at a depth of 3m in a swimming pool?

(density of water = 1000 kgm^-3).

Solution:
Step 1: Write down known quantities and quantities to be found.

• d = 3 m
• ρ = 1000 kgm^-3
• g = 10 ms^-2
• p = ?

Step 2: Write down formula and rearrange if necessary

p = dρg

Step 3: Put the values in formula and calculate

p = 3 m x 1000 kgm^-3 x 10 ms^-2

p = 30000 pa = 3.0 x 10⁴ pa

Ans: Hence 3.0 x 10⁴ pa pressure will be observed at a depth of 3 m in the swimming pool.

Worked Example 3

Worked Example 4

4. In a hydraulic lift system, what must be the surface area of a piston. If a pressure of 300 kpa is used to provide an upward force of 2000 N?
Solution:
Step 1: Write down known quantities and quantities to be found.

• P = 300 kPa = 300 × 1000 Pa = 300000 Pa = 300000 Nm^-2
• F = 2000 N
• A = ?

Step 2: Write down formula and rearrange if necessary

P = ^F/_A

A = ^F/_P

Step 3: Put the values in formula and calculate

A = 2000 N/300000 Nm^–2 = 0.00667 m²

A = 6.67 x 10^-3 m²

Ans: Hence the surface area of piston is 6.67 x 10^-3 m².

Text Book Exercise (Numericals)

Section (B) Structured Questions
Stretching of Spring
1. Some students experimented to find out how a spring stretched when loads were added to it.
Table For Load and Extension

Load (N)	Extension (cm)
0	0
2	15
4	30
6	45
8	60
10	75
12	90
14	100

a) Use these results to plot a graph. (Plot x = load, y = extension in spring).
b) Use your graph to find
i) The extension when the load is 3 N;
ii) The load which produces an extension of 40 mm.
Solution:
a) Graph between force load and extension

b) (i): The extension when the load is 3 N:

Ans: The extension in the spring is 2.3 mm when the load is 3 N.

b) (ii) The load which produces an extension of 40 mm:

Ans: The load which produces extension of 40 mm is 5.2 N.


2. The variation in extension x of the force F for a spring is shown in Fig 5.14. The point L on the graph is the elastic limit of the spring.
a) Describe the meaning of elastic limit.
b) Calculate the force in extending the spring to its elastic limit ‘L’.

Ans 2 (b): The value of force:
According to above graph,

• Extension (x) is given On Y-axis and 1 square = 1 cm. So, the elastic limit 'L' has value 7 cm.
• Force (N) is given On X-axis and 1 square = 2 N and 1 small square = 0.4 N. The point 'L' is meeting on X-axis at the value 7.6 N.

Therefore, the force in extending the spring to its elastic limit ‘L’ is 7.6 N

Hooke's Law
4. Calculate the spring constant for a spring which extends by a distance of 3.5cm when a load of 14 N is hung from its end.
Ans: Solution:
Data:

• Distance x = 3.5 cm = ^3.5/₁₀₀ =0.035 m
• Force = F = 14 N
• Spring constant = k = ?

Working Formula:

• F = kx

Calculation:

14 N = k (0.035 m)

0.035 k = 14

k = ¹⁴/_0.035
k = 0.0025 Nm^-1 = 2.5 x 10^-3 Nm^-1

Ans: Therefore, the spring constant for a spring is 2.5 x 10^-3 Nm^-1.

5. Table 5.4 shows the results of an experiment to stretch a spring.

Table 5.4 For load and extension in spring

Load (N)	Extension (cm)
0.0	0.0
2.0	80.0
4.0	83.0
6.0	86.0
8.0	89.0
10.0	92.0
12.0	93.0

a) Use the result to plot an extension against load graph.
b) On the graph mark the limit of proportionality and state the value of the load at this point.
c) Calculate the spring constant k.
d) Show maximum force at which hooke's law is applicable.
Ans 5 (a): Graph Between Extension (mm) and Load (N)

Ans 5 (b): Graph Showing The Limit Of Proportionality:

• The limit of proportionality is 92 mm.
• The value of the load at this point is 10 N

Ans 5 (c):The spring constant k:
Solution:
Data:

• Force = F = 10 N
• Distance = x = 92 mm
• Spring constant = k =?

Working Formula:

F = kx

Calculation:

10 = k (92)

92 k = 10

k = ¹⁰ /₉₂

k = 0.1086 Nm^-1 = 1.086 x 10^-1 Nm^-1

Ans: Therefore, the spring constant k is 1.086 x 10^-1 Nm^-1

Ans 5 (d):The maximum force at which Hooke's law is applicable is 10 N .

8. A boy is pressing a thumbtack into a piece of wood with a force of 20 N. The surface area of head of thumbtack is 1 cm² and the cross-section area of the tip of the thumbtack is 0.01 cm². Calculate:
a) The pressure exerted by boy's thumb on the head of thumbtack.
b) The pressure of the tip of the thumbtack on the wood.
c) What conclusion can be drawn from answers of part (a) and (b)?
Ans: Solution:

Ans 8 (c):
The pressure exerted on a surface by an object increases as the weight of the object increases or the surface area of contact decreases as the weight of the object decreases or the surfsce area of the contact increases.

9. The Fig 5.15 shows a basic hydraulic system that has small and large pistons of cross section area of 0.005 m² and 0.1 m² respectively. A force of 20 N is applied to small piston. Calculate:
a) The pressure transmitted into hydraulic fluid.
b) The force at large piston.
c) Discuss the distance travelled by small and large pistons.

Ans 9 (a): Solution:
Data:

• Area of small piston: A₁ = 0.005 m²
• Area of large piston: A₂ = 0.1 m²
• Force on small piston = F₁ = 20 N
• Pressure = P = ?

Ans: The pressure transmitted into hydraulic fluid is 4000 Nm^-2 or 4.0 x 10³ Nm^-2.

Ans 9 (b): Solution:
Data:

• Area of small piston: A₁ = 0.005 m²
• Area of large piston: A₂ = 0.1 m²
• Force on small piston = F₁ = 20 N
• Force on large piston = F₂ = ?

Ans: The force at large piston is 400 N.

Ans 9 (c): We know that according to Pascal's law:

P₁ = P₂

This allow the lifting of a heavy load with a small force but there can be no multiplication of work, so in an ideal case with no frictional loss.

W_input = W_output

W₁ = W₂

We know that, Work done = Force x distance = F x d

W₁ = W₂

Therefore, F₁d₁ = F₂d₂

We have to pay for the multiplied output force by exerting the smaller input force through a large distance. Distance covered by the large piston is smaller as compared to the small piston.

Self Assessment Questions (Numericals)

Q.1: An elastic spring is 70 cm long. When it is stretched by hanging some load its length increases to 100 cm. Calculate its extension?
Solution:
Data:

• Original length of the spring = 70 cm
• Length of the stretched spring = 100 cm
• Extension = ?

Formula:

Length of stretched spring = Original length + Extension

Calculation:

100 = 70 + Extension

Extension = 100 - 70 = 30 cm

Ans:Therefore, its extension is 30 cm.

Q.2: Following table shows the results of an activity to stretch an elastic spring. Complete the table and draw a graph to represent this data.

Table for load and extension

Load (N)	Length (cm)	Extension (cm)
0.0	00	0.0
1.0	30	1.0
2.0	32	2.0
3.0	34	3.0
8.0	36	4.0
4.0	38	5.0
5.0	40	6.0
6.0	41.5	8.0
7.0	42	10
8.0	43	10

Solution:
Given Data:

• Load (N) = 0.0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0
• Length (cm) = 30, 32, 34, 36, 38, 40, 41.5, 42, 43

Required Data:

• Extension (cm) = ?
• Graph = ?

Table for load and extension

Load (N)	Length (cm)	Extension (cm)
0.0	30	0.0
1.0>	32	2.0
2.0	34	4.0
3.0	36	6.0
4.0	38	8.0
5.0>	40	10
6.0	41.5	11.5
7.0	42	12
8.0	43	13

GRAPH:

Result: The graph shows that extension depends on the load, the extension increases in equal steps as the load increases.

Q.3: How much force is needed to pull a spring to a distance of 30 cm, the spring constant is 15 Nm^-1?
Solution:
Data:

• x = 30 cm = ³⁰/₁₀₀  = 0.3 m
• k = 15 Nm^-1
• F = ?

Formula:

F = kx

Calculation:

F = 15 x 0.3 = 4.5 N

Ans:Therefore, 4.5 N force is needed to pull a spring to a distance of 30 cm.

Q.8: A wooden block of dimensions 0.5 m × 0.6 m × 1.0 m kept on the ground has a mass of 200 kg. Calculate the maximum pressure acting on the ground.
Ans: Solution:
Data:

• Dimensions of the wooden block = 0.5 m × 0.6 m × 1.0 m
• Area = A = l x b x h = 0.5 m × 0.6 m × 1.0 m = 0.3 m
• mass = 200 kg
• Acceleration due to gravity = 9.8 m/s²
• Force = weight = mg = (200) x (9.8) = 1960 N

Required Data:

• Maximum Pressure = ?

Ans:Therefore, the maximum pressure acting on the ground is 6533.33 pa

Q.9: If the density of sea water is 1150 kgm^-3, calculate the pressure on a body of 50 m below the surface of sea?
Solution:
Data:

• Density of sea water = ρ = 1150 kgm^-3
• Depth = 50 m
• Acceleration due to gravity = 9.8 m/s²

Required Data:

• Pressure = P = ?

Working Formula:

P = dρg

Calculation:

P = (1150) x (50) x (9.8)
P = 563500 pa

Ans: Therefore, the pressure on a body of 50 m below the surface of sea is 563500 pa.

Q.11: In a hydraulic press, a force of 100 N is applied on the pump of cross-sectional area 0.01 m². Find the force that compresses a cotton bale placed on larger piston of cross-sectional area 1 m².
Solution:
Data:

• F₁ = 100 N
• A₁ = 0.01 m²
• A₂ = 1 m²

Required Data:

• F₂ =?

Ans: Therefore, the force that compresses a cotton bale placed on larger piston of cross-sectional area 1 m² is 1 x 10⁴ N.

Forces And Matter - Physics For Class IX (Science Group) - Self Assessment Questions and Test book Exercise

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Physics For Class IX (Science Group)
UNIT 5: FORCES AND MATTER
Self Assessment Questions and Test book Exercise

Self Assessment Questions:

Q.1: An elastic spring is 70 cm long. When it is stretched by hanging some load its length increases to 100 cm. Calculate its extension?
Solution:
Data:

• Original length of the spring = 70 cm
• Length of the stretched spring = 100 cm
• Extension = ?

Formula:

Length of stretched spring = Original length + Extension

Calculation:

100 = 70 + Extension

Extension = 100 - 70 = 30 cm

Ans: Therefore, its extension is 30 cm.

Q.2: Following table shows the results of an activity to stretch an elastic spring. Complete the table and draw a graph to represent this data.

Table for load and extension

Load (N)	Length (cm)	Extension (cm)
0.0	00	0.0
1.0	30	1.0
2.0	32	2.0
3.0	34	3.0
8.0	36	4.0
4.0	38	5.0
5.0	40	6.0
6.0	41.5	8.0
7.0	42	10
8.0	43	10

Solution:
Given Data:

• Load (N) = 0.0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0
• Length (cm) = 30, 32, 34, 36, 38, 40, 41.5, 42, 43

Required Data:

• Extension (cm) = ?
• Graph = ?

Table for load and extension

Load (N)	Length (cm)	Extension (cm)
0.0	30	0.0
1.0	32	2.0
2.0	34	4.0
3.0	36	6.0
4.0	38	8.0
5.0	40	10
6.0	41.5	11.5
7.0	42	12
8.0	43	13

GRAPH:

Result: The graph shows that extension depends on the load, the extension increases in equal steps as the load increases.

Q.3: How much force is needed to pull a spring to a distance of 30cm, the spring constant is 15 Nm^-1?
Ans: For Solution of Numericals Click Here

Q.4: Write two properties of spring.
Ans: PROPERTIES OF SPRING:
An ideal spring material has two high strength properties, which are:

1. A high elastic limit:
   Because springs are resilient structures designed to undergo large deflection, spring materials must have property of extensive elastic range.
   
   
2. A low modulus:
   Modulus of elasticity is the measurement of stiffness and rigidity of spring material, or its elastic ability. The higher the value (modulus), the stiffer the material. Conversely, materials with lower values are more easily bent under load.

Use the idea of pressure to explain the following:

• Q.5: Sharks and crocodiles have sharp teeth.
  Ans: Sharks and crocodiles have sharp teeth because they have evolved successfully as carnivorous creatures. Most carnivores have long, sharp teeth adapted to ripping, tearing or cutting flesh. Sharp teeth are required to effectively grab and tear away chunks of tissue by applying high pressure.
  
  
• Q.6: Camels have wide, flatted feet.
  Ans: Camels have to walk on sand, and to walk fast, their feet should not sink in the sand. The feet area is broad to exert less pressure which makes the feet sink less in the sand.
  
  
• Q.7: If you walk on wooden floor wearing shoes with very narrow heels, you will damage the floor.
  Ans: If we walk on wooden floor wearing shoes with very narrow heels, we will damage the floor because of the smaller surface area of a heel there is a greater force per square inch that is produced against the floor.

Q.8: A wooden block of dimensions 0.5 m × 0.6 m × 1.0 m kept on the ground has a mass of 200 kg. Calculate the maximum pressure acting on the ground.
Ans: For Solution of Numericals Click Here

Q.9: If the density of sea water is 1150 kgm^-3, calculatethe pressure on a body of 50 m below the surface of sea?
Ans: For Solution of Numericals Click Here

Q.10: Dam holds water at high altitude. The walls of the dam are made wider at the base. Explain why?
Ans: The walls of dam is made much thicker at the bottom than it is at the top because pressure exerted by a liquid increases with Depth. Thus as depth increases more and more pressure is exerted by water on the walls of dam. A dam needs the thicker wall to withstand this large pressure, therefore, the wall of the dam is made wider and thickness increasing towards the base.
In other words,
A dam has broader walls at the bottom than at the top to withstand the great pressure of water which increases with an increase in depth.

Q.11: In a hydraulic press, a force of 100 N is applied on the pump of cross-sectional area 0.01 m². Find the force that compresses a cotton bale placed on larger piston of cross-sectional area 1 m².
Ans: For Solution of Numericals Click Here

Q.12: Write down the names of four machines that you have seen working on the principle of pascal’s law.
Ans: Machines Working On The Principle Of Pascal's Law:
A hydraulic machine works on this principle.

1. Hydraulic brakes
2. Car lifts
3. Hydraulic jacks
4. Forklifts
5. Hydraulic Pumps
6. Aircraft Hydraulic System

Text Book Exer cise

Section (B) Structured Questions
Stretching of Spring
1. Some students experimented to find out how a spring stretched when loads were added to it.

Table For Load and Extension

Load (N)	Extension (cm)
0	0
2	15
4	30
6	45
8	60
10	75
12	90
14	100

a) Use these results to plot a graph. (Plot x = load, y = extension in spring).
b) Use your graph to find
i) The extension when the load is 3 N;
ii) The load which produces an extension of 40 mm.
Ans: For Solution of Numericals Click Here

2. The variation in extension x of the force F for a spring is shown in Fig 5.14. The point L on the graph is the elastic limit of the spring.
a) Describe the meaning of elastic limit.
b) Calculate the force in extending the spring to its elastic limit ‘L’.

Ans 2 (a): ELASTIC LIMIT:
Hooke's law is applicable to all kinds of deformation and all types of matter i.e., solids, liquids or gases within certain limit. Such limit is called elastic limit. It can be define as:

"Elastic limit tells the maximum force or stress that can be safely applied on a body without causing permanent deformation in its length, volume or shape."

In other words,

"Elastic limit is a limit within which a body recovers its original length, volume or shape after the deforming force is removed. Beyond this limit spring deforms permanently."

Ans 2 (b): For Solution of Numericals Click Here

Hooke's Law
3. State Hooke's law.
Ans: HOOKE'S LAW:
Robert Hooke, an English scientist first described the mathematical pattern of stretching a spring. He observed the dependence of displacement or size of the deformation upon the deforming force or load.
Hooke's law states that:

"Within elastic limit, the displacement produced in the spring is directly proportional to the force applied."

In other words,
Hooke’s Law state that the extension in spring is proportional to the load applied to it, provided the limit of elasticity is not exceeded.

Mathematically:
If ‘F’ is the applied force and ‘x’ is the displacement (extension) in the spring then the equation for Hooke's law may be written as:

F ∝ x

or F = kx ....... (i)

Where, k is spring constant (stiffness of spring).

4. Calculate the spring constant for a spring which extends by a distance of 3.5 cm when a load of 14 N is hung from its end.
Ans: For Solution of Numericals Click Here

5. Table 5.4 shows the results of an experiment to stretch a spring.

Table 5.4 For load and extension in spring

Load (N)	Extension (cm)
0.0	0.0
2.0	80.0
4.0	83.0
6.0	86.0
8.0	89.0
10.0	92.0
12.0	93.0

a) Use the result to plot an extension against load graph.
b) On the graph mark the limit of proportionality and state the value of the load at this point.
c) Calculate the spring constant k.
d) Show maximum force at which Hooke's law is applicable.
Ans: For Solution of Numericals Click Here

Pressure
6. a) Define the term pressure.
Ans: PRESSURE:
Pressure is defined as:

"The force acting normally per unit area on the surface of a body is called pressure."

OR

"The quantity that depends upon the force and increases with decrease in the area on which force is acting is called pressure."

Examples:

• Press a pencil from its ends between the palms. The palm pressing the tip feels much more pain than the palm pressing its blunt end.
• We can push a drawing pin into a wooden board by pressing it by our thumb. It is because the force we apply on the drawing pin is confined just at a very small area under its sharp tip. A drawing pin with a blunt tip would be very difficult to push into the board due to the large area of its tip.
• When force is exerted on area, this is known as pressure, e.g. A hammer hits a nail, driving the nail downwards into a piece of wood.

b) Write down the S.I unit of pressure.
Unit:
PASCAL (S.I Unit):
As the force is measured in newtons (N) and area in square meters (m²).
Therefore in SI system unit of pressure is newton per square meter (Nm^-2). It is also known as Pascal (Pa).

1 Pa = 1 Nm^-2

ATMOSPHERE:
The pressure exerted by the air molecule at sea level is measured in atmosphere.
Where,

1 atmosphere = 1.013 x 10^-2) Pascal

1 atm = 1.013 x 10^-2) Pa

7. Why does pressure increases as you dig deeper; Explain in detail.
Ans: Pressure Increases As We Dig Deeper:
Pressure is force per unit area and pressure and depth have a directly proportional relationship.
In this case, the force is primarily the weight of the overlying rock. So, in very simple terms, the deeper we go the more rock must be supported so the more force is required and the pressure goes up. It is the same thing that happens in the ocean and the atmosphere. The deeper we go in the ocean the higher the pressure. And in the atmosphere the density of air becomes heavier near the surface of the earth (due to gravity). Thus, the atmospheric pressure is high at lower altitudes, the density being higher.

8. A boy is pressing a thumbtack into a piece of wood with a force of 20 N. The surface area of head of thumbtack is 1 cm² and the cross-section area of the tip of the thumbtack is 0.01 cm². Calculate:
a) The pressure exerted by boy's thumb on the head of thumbtack.
b) The pressure of the tip of the thumbtack on the wood.
c) What conclusion can be drawn from answers of part (a) and (b)?
Ans: For Solution of Numericals Click Here

9. The Fig 5.15 shows a basic hydraulic system that has small and large pistons of cross section area of 0.005 m² and 0.1 m² respectively. A force of 20 N is applied to small piston. Calculate:
a) The pressure transmitted into hydraulic fluid.
b) The force at large piston.
c) Discuss the distance travelled by small and large pistons.
Ans: For Solution of Numericals Click Here