Solution (Based on the provided table):

To solve for total extension and spring constant k.

Arrangement 1 (Springs in Series):
Total Extension(x):

In series, the total extension is the sum of the extensions for each spring.

Since both springs are identical.

Xtotal = x + x = 2x

Spring Constant (Karrangement):

For springs in series, the effective spring constant is given by.

 1/Karrangement = 1/k + 1/k = 2/k

Karrangement = k/2

Arrangement 2 (Springs in Parallel):
Total Extension(x):

In parallel, the extension remains the same as for one spring:

Xtotal = x

Spring constant (Karrangement):

For springs in parallel, the effective spring constant is the sum of individual spring constants:

Karrangement = k + k

Karrangement= 2k.

Springs are made of steel because steel has greater elasticity, strength and resistance to deformation compared to iron. 

This ensures better durability and performance.

(a) Iron or Rubber:

Iron is more elastic because it regains its shape faster after deformation.

b) Air or Water: 

Water is more elastic as it is less compressible compared to air.

Water pressure, one meter below the surface of both the swimming pool and the lake will be the same. Pressure depends on depth, not the size or volume of the water body.

The pressure in all parts of a confined liquid increases equally. This principle is applied in hydraulic brakes where pressure on the brake pedal is transmitted to the brake pads.

Trapped air in the mercury barometer would reduce the height of the mercury column because the trapped air exerts additional pressure, counteracting the atmospheric pressure.

 Giraffes have a unique blood vessel system with valves that regulate blood pressure, preventing damage to the brain when they suddenly raise or lower their necks.

 A leak in the glass tube would allow air to enter, reducing the vacuum. This would decrease the height of the mercury column, making pressure readings inaccurate.

Density is a property of a material, independent of the size or shape of an object made from that material. For example: a small gold nugget and a gold bar have the same density.

Engineers estimate the load of a structure using calculations based on the material's density, cross-sectional area, and the force it must support. They use principles of stress, strain and elasticity.

Hooke's Law:

If a force F is applied on a spring to stretch or compress it, the extension or compression x has been found directly proportional to the applied force within the elastic limit. Thus,

F ∝ x

or  

F = kx        

or  k = F/x

Where k is the constant of proportionality and is known as spring constant. In fact, it is a measure of stiffness of the spring. The greater the value of spring constant, the greater will be the stiffness or strength of the spring.

Its unit is Nm⁻¹.

Applications of Hooke's law:
1. Spring scales:

Spring scales the extension or compression of a spring to determine the weight of objects. In a common spring balance the extension or elongation produced is a measure of the weight. In compression balance, the spring is compressed by the load (force) and the compression produced is measured by means of a pointer moving over a scale. Weighing machine usually use this type of balance.

2. Balance wheel of mechanical clocks:

The balance wheel in mechanical clocks use spring to control the back and forth motion that regulates the speed of the hands of a clock.

3. Galvanometer:

Galvanometer is a current detecting device. It makes use of a tiny spring called hair spring. Which provides electrical connection to the galvanometer coil and also restores the pointer back to zero position. The deflection of the pointer is proportional to the current flowing through it with in the range

Simple Mercury Barometer:

Working: 

A simple mercury barometer is used to measure atmospheric pressure. It consists of a long glass tube (about 1 meter long) filled with mercury and inverted into a dish containing mercury.

1. The tube is completely filled with mercury and then inverted in the dish.

2. Some mercury flows out, creating a vacuum (Torricelli's vacuum) at the top of the tube.

3. The mercury level inside the tube stabilizes at a height where the pressure exerted by the mercury column equals the atmospheric pressure

4. The height of the mercury column (about 76cm at sea level) directly represents atmospheric pressure.

Application:
• Used in weather forecasting to monitor atmospheric pressure changes
• Helps in studying altitude variations (as pressure decreases with height).
• Used in laboratories for calibration and scientific research.

Manometer

Working:

A manometer measures the pressure of gases or liquids in a closed system. It consists of a U-shaped tube partially filled with liquid. (e.g, mercury, water, or oil). There are two main types:

1. Open tube Manometer:
        • One end is open to the atmosphere, and the other is connected to the gas.
        • The difference in liquid levels in the two arms indicates the gas pressure.

2. Closed tube Manometer:
        • One end is sealed (vacuum), and the other is connected to the gas.
        • The height difference represents absolute pressure.

Application:
        • Used in laboratories to measure gas pressures.

•  Helps in HVAC (heating, ventilation, and air conditioning) systems to monitor air pressure.
• Applied in industrial processes to control gas and liquid flow.

Pascal's law:

Statement:

"In a confined fluid, any change in pressure is transmitted equally in all directions without being diminished."

This means that when pressure is applied to an enclosed fluid, it is distributed uniformly throughout the fluid.

Mathematically Expression:

P = F / A

        - P = Pressure (N/m² or Pa)
        - F = Force applied (N)
        - A = Area of the surface (m²).

Applications of Pascal's law.

1. Hydraulic Brakes:
        - Used in vehicles to transmit braking force from the brake pedal to the brake pads.
        - When the driver presses the brake pedal, the fluid transmits pressure equally to all wheels, stopping the vehicle efficiently.
2. Hydraulic lift:
        - Used in car service stations and elevators.
        - A small force applied to a small piston generates a larger force on a bigger piston, lifting heavy objects easily.

Factors Affecting the pressure of a liquid in a Container:

 The pressure of a liquid at a given depth depends on the following factors:

1. Depth (Height) of the liquid (h):
        - Pressure increases with depth.
        - The deeper you go, the higher the pressure.

2. Density of the liquid (ρ):
        - Heavier (denser) liquid exert more pressure than lighter liquids at the same depth.
        - Example: Mercury exerts more pressure than water at the same height.

3. Acceleration due to Gravity (g):
        - The pressure of a liquid is higher in places with stronger gravitational force.
        - Gravity varies slightly with location (e.g., it is slightly weaker at higher altitudes).

Determination of liquid Pressure:

The pressure exerted by a liquid at a depth is given by the formula:

        P = ρgh

        Where,

        . P = Pressure (Pa)
        . ρ = Density of the liquid (kg/m³)
        . g = Acceleration due to gravity (9.8 m/s²).
        . h = Depth of the liquid (m).

Atmospheric Pressure and Its Applications.

Explanation:
The atmosphere exerts pressure due to the weight of air above us. Although we do not feel it, atmospheric pressure is significant- about 101.3kPa (1 atm) at sea level- This pressure can be demonstrated through various experiments and real-life phenomena.

Applications of Atmospheric Pressure:

1. Drinking with a straw:
        - When you suck on a straw, air is removed from it.
        - The external atmospheric pressure pushes the liquid up into the straw.

2. Syringe Working Mechanism:
        - When the plunger is pulled, air pressure inside the syringe decreases.
        - Higher atmospheric pressure outside pushes the liquid into the syringe.

3. Vacuum Sealing of food:
        - Air is removed from food packaging creating a vacuum.
        - The external atmospheric pressure keeps the packaging tightly sealed.

Atmospheric pressure plays a crucial role in everyday life, engineering and meteorology.

Given data:
        F = 40N
        x = 20mm = 0.02m
        New extension x = 16mm = 0.016m

To Find:
        Spring constant (K) = ?
        New force (F) = ?

Solution:
        According to Hooke's law F = kx
        => k = F / x

        k = 40 / 0.02
        k = 2000N/m = 2kN/m

Again,
        F = kx
        F = 2000 x 0.016
        F = 32N

Given data:
        m = 4.5 kg
        V = 5 liters = 0.005 m³

To Find:
        Density (ρ) = ?

Solution:
        We know that ρ = m / V

        ρ = 4.5 / 0.005
        ρ = 900 kg/m³
        ρ = 0.9 x 10³ kg/m³

Given data:
        mass (m) = 60g = 0.06kg
       Vᵢ = 40cm³
        Vᶠ = 44cm³
        Volume of solid (V) = Vᶠ - Vᵢ
                              = 44cm³ - 40cm³
                              = 4cm³
        V = 4 x 10⁻⁶ m³

To Find:
        Density (ρ) = ?

Solution:
        We know that ρ = m / V

        ρ = 0.06 / 4 x 10⁻⁶
        ρ = 15000 kg/m³
        ρ = 15 x 10³ kg/m³

Given data:
        Density (ρ) = 8 x 10³ kg/m³
        V = 60cm³ = 6 x 10⁻⁵ m³

To Find:
        mass (m) = ?

Solution:
        We know that ρ = m / V
        => m = ρ x V

        m = 8 x 10³ x 6 x 10⁻⁵
        m = 48 x 10⁻²
        m = 0.48 kg

Given data:
        m = 5kg
        g = 9.8 m/s²
        dimension of the brick = 5cm x 10cm x 20cm

To Find:
        Pmax and Pmin = ?

Solution:
        Force = m x g
                = 5 x 10
        F = 50N

Surfaces: Areas of the brick faces:

• Face 1 (5cm x 10cm)
• Area₁ = 5 x 10 = 50cm² = 5 x 10⁻³ m²
• Face 2 (10cm x 20cm)
• Area₂ = 10 x 20 = 200cm² = 2 x 10⁻² m²
• Face 3 (5cm x 20cm)
• Area₃ = 5 x 20 = 100cm² = 1 x 10⁻² m²

Maximum Pressure:
        Pmax = F / Area₁
             = 50 / 5 x 10⁻³
        Pmax = 1 x 10⁴ Pa

Minimum Pressure:
        Pmin = F / Area₂
             = 50 / 2 x 10⁻²
        Pmin = 25 x 10² Pa

Given data:

density of mercury (ρHg) = 13.6 x 10³ kg/m³

Density of Water (ρwater) = 1000 kg/m³
Height of mercury (hHg) = 0.76 m   

To Find:
        hwater = ?

Solution:
        ρHg . g . hHg = ρwater . g . hwater
        ρHg . hHg = ρwater . hwater

        => hwater = (ρHg . hHg )/ ρwater

        putting values;
        hwater = (13.6 x 10³) x 0.76 / 1000
         hwater = 10.3 m

Given data:
        F₁ = 500N
        A₁ = 5cm² = 5 x 10⁻⁴ m²
        A₂ = 20cm² = 20 x 10⁻⁴ m²

To Find:
        P = ?
        F₂ = ?

Solution:
        We know that

        P = F₁ / A₁

        P = 500 / 5 x 10⁻⁴
        P = 1.0 x 10⁶ N/m²

        and
        F₂ = P . A₂
        F₂ = (1.0 x 10⁶) . (20 x 10⁻⁴)
        F₂ = 2000N

Given data:
        Depth (h) = 10m
        ρ = 1030 kg/m³
        g = 9.8 m/s²

To Find:
        Water Pressure (P) = ?

Solution:
        We know that P = ρgh
        P = 1030 x 9.8 x 10
        P = 103000 N/m²
        P = 1.03 x 10⁵ N/m²

Given data:
        A₁ = 10cm² = 10 x 10⁻⁴ m²
        A₂ = 100cm² = 100 x 10⁻⁴ m²
        F₂ = 4000N

To Find:
        F₁ = ?

Solution:
        Apply Pascal's law F₁ / A₁ = F₂ / A₂
        => F₁ = F₂ . A₁ / A₂

        Putting values;
        F₁ = 4000 x 10 x 10⁻⁴ / 100 x 10⁻⁴
        F₁ = 400 N

Given data:
       

Solution:

The x-axis will represent the altitude and y-axis represent the pressure.

(a) We know that at sea level (altitude = 0m) the pressure is typically around 101.3 kPa. It appears that air pressure at sea level would be approximately 1.01 x 10⁵ Pa.

(b) We can estimate the altitude at which the air pressure would be 90 kPa.  It appears that the altitude at which the pressure is 90 kPa is approximately 1.02 km.

Given data:
        P = 10 N/cm²
        A = 50cm²

To Find:
        F = ?

Solution:
        We know that P = F / A
        => F = P x A
        F = 10 x 50
        F = 500N

Given data:
        F₁ = 500N
        A₁ = 5cm² = 5 x 10⁻⁴ m²
        A₂ = 20cm² = 20 x 10⁻⁴ m²

(a) To Find:
        P = ?

        F₂ = ?

Solution:
        P = F₁ / A₁
        P = 500 / 5 x 10⁻⁴
        P = 1.0 x 10⁶ N/m²

(b) We know that

         P = F₂ / A₂
        => F₂ = P x A₂
        F₂ = (1.0 x 10⁶) x (20 x 10⁻⁴)
        F₂ = 2000N