PROPERTIES OF MATER CHAPTER # 10 Physics 10th - Question Answers

 Physics 10th - Question Answers

PROPERTIES OF MATER
CHAPTER # 10

Q.1: Define matter?
Ans: MATTER:
Anything which occupies space and has mass is called matter.

Q.2: Write down the properties of solids?
Ans: PROPERTIES OF SOLIDS:

• The solid bodies have fixed volume and fixed shape.
• In solids, the molecules are closely packed.
• The force of attraction between molecules is very strong; it is because of strong attraction that the solid has fixed shape and fixed size.
• The motion of molecules in a solid body is very much restricted. They just vibrate about their mean position.
• In solids, molecules do not leave their mean position.
• The relative distance between the atoms or molecules of a solid body is fixed.

Q.3: Write down the properties of liquids?
Ans: PROPERTIES OF LIQUIDS:

• Liquids have fixed volume but no fixed shape.
• They assume the shape of the container.
• The force of attraction between the molecules is small as compared to that found in solids.
• The liquid molecules move freely inside the volume occupied by the liquid.
• Liquids are incompressible and maintain their level.

Q.4: Write down the properties of gases?
Ans: PROPERTIES OF GASES:

• Gases have neither fixed volume nor fixed shape.
• They can occupy any space available to them.
• The force of attraction between the molecules of a gas is negligibly small.
• This is the reason why the gas molecules move freely in all directions with all possible velocities.

Q.5: Why does a gas have neither a fixed shape nor a fixed volume?
Ans: The molecules of a gas keep moving very fast in all directions. They are far from one another. Thus, the cohesive forces are insignificant, and the gases do not have a fixed shape or a fixed volume. They occupy all the space available to them.

Q.6: Is it easy to compress air as compared to water? Why?
Ans: In gases, the molecules are far apart from each other. There are a lot of empty spaces between gas molecules. As air is a mixture of gases, it can be compressed quite easily. Whereas water is a liquid, and its molecules are closer than air. Liquids are incompressible. That is why it is easy to compress air as compared to water.

Q.7: Define the following terms?

• Elasticity
• Stress
• Strain
• Elastic limit
• Yield point

Ans: FOLLOWING TERMS:

Elasticity:
A phenomenon in which a material body comes back to its original shape after the removal of a deforming force is called elasticity.

Stress:
When a body is made to change in length, volume, or shape by the application of an external force to keep the body in equilibrium, the opposing force per unit area acts on the body, which resists such a change. This internal force arising from the deformation is called stress. It is denoted by $\sigma$.

Strain:
The change in the shape, volume, and length of an object caused due to stress is called strain. It is denoted by $E$.

Elastic Limit:
It is the maximum value of stress within which the body exhibits the property of elasticity. Below the elastic limit, the body regains its original state of shape and size after the removal of the deforming force.

Yield Point:
The point beyond the elastic limit of a material subjected to tensile stress at which an increase in load causes a proportionately larger increase in extension than predicted by elasticity.

Hook’s Law is called yield point. It arises from the internal changes in the structure of a material form which does not totally recover on removal of stress.

Q.8: Define stress? Write down its formula and unit?
Ans: STRESS:
When a body is made to change in length, volume, or shape by the application of an external force, to keep the body in equilibrium, the opposing force per unit area acts on the body, which resists such a change. This internal force arising from the deformation is called stress. It is denoted by $\sigma$.

Formula:

$$\text{Stress} = \frac{\text{Force}}{\text{Area}}$$ $$\sigma = \frac{F}{A}$$

Units:
The S.I. unit of stress is $\text{N/m}^2$ or $\text{Nm}^2$.

Q.9: Define longitudinal strain?
Ans: LONGITUDINAL STRAIN:
When the stress brings change in length, then the strain is called longitudinal strain.

$$\text{Longitudinal strain} = \frac{\text{Change in Length}}{\text{Original Length}} \Rightarrow e = \frac{\Delta L}{L}$$

Q.10: Define Hooke’s law? Give its mathematical derivation?
Ans: HOOKE’S LAW:
It states that for any elastic body, stress is proportional to strain. In other words, under the elastic limit, the deformation in a material is proportional to the force applied; then the material is said to obey Hooke’s Law.

Mathematically:

$$\text{Stress} \propto \text{strain}$$ $$\text{Stress} = \text{constant} \times \text{Strain}$$

Q.11: Define Young’s modulus of elasticity? Derive its equation?
Ans: YOUNG’S MODULUS OF ELASTICITY:
The ratio of the stress to the longitudinal strain is called Young’s modulus of elasticity.

Young’s Modulus=longitudinal strainstress​⟹Y=eσ​

We know that,

$$\sigma = \frac{F}{A}$$ $$\therefore Y = \frac{\frac{F}{A}}{\frac{\Delta L}{L}}$$

So,

$$Y = \frac{F}{A} \times \frac{L}{\Delta L}$$

Hence,

$$Y = \frac{FL}{A \Delta L}$$

Unit of Young’s Modulus:
The unit of Young’s modulus is $\text{N/m}^2$ or $\text{Nm}^2$.

Q.12: How can you show that steel is more elastic than rubber?
Ans: Steel is more elastic than rubber because more effort is needed to produce any permanent change in it than in rubber. In simple words, a body which easily deforms is less elastic than a body that undergoes deformation with difficulty when the applied stress is the same.

Q.13: How can you compare the elasticities of two substances?
Ans: A body is said to be more elastic if it retains its elasticity upon the application of a large external force, compared to a body that loses its elasticity at a lesser external force.

Consider two elastic substances $x$ and $y$ in the form of wires, which have the same length and the same area of cross-section.

• Let us suppose that the same stress is applied to both of them.
• We further suppose that $e_x$ and $e_y$ are the strains produced in $x$ and $y$, respectively.
• Let $E_x$ and $E_y$ be the moduli of elasticity of the two bodies, respectively.

Then,

$$E_x = \frac{\sigma}{e_x}, \quad E_y = \frac{\sigma}{e_y}$$

Where $\sigma$ = stress applied.

Ey​Ex​​=εx​εy​​ $$\therefore \frac{E_x}{E_y} = \frac{\sigma / \varepsilon_x}{\sigma / \varepsilon_y}$$ $$\frac{E_x}{E_y} = \frac{\sigma}{\varepsilon_x} \times \frac{\varepsilon_y}{\sigma}$$ $$\frac{E_x}{E_y} = \frac{\varepsilon_y}{\varepsilon_x}$$

Hence,

$$\frac{E_x}{E_y} = \frac{\varepsilon_y}{\varepsilon_x}$$ $$\therefore \text{if } \varepsilon_x > \varepsilon_y, \text{ then } E_x < E_y$$

Conclusion:

• A body with smaller strain is more elastic than the body having a larger strain when the applied stress is the same.
• In simple words, we can say that a body which is easily deformed is less elastic than the body which undergoes deformation with difficulty when the applied stress is the same.

Q.14: Apply Hooke’s law to a helical spring and derive its expression?
Ans: PROCEDURE:

• Let a helical spring be suspended vertically from a fixed support.
• A block is attached at the lower end.
• Due to the forces of gravity, the block is displaced downward.
• After a little while, the block comes to rest and attains equilibrium.
• Under this condition, the block is under the action of two forces which are equal and opposite:
  • The force of gravity acting downward.
  • The tension in the string acting upward.

ACCORDING TO HOOKE’S LAW:
Tension is proportional to extensions. If $F$ stands for tension and $x$ for extension, then:

$$F \propto x \implies F = -Kx$$

Where $K$ is called the spring constant, and the negative sign shows that tension and extension act in opposite directions.

Hooke’s law: $F \propto d$. If a force of 4N stretches a spring by a length of 2 cm, an 8N force will stretch it 4 cm.

Q.15: Define Pressure? Give one example and its unit?
Ans: PRESSURE:
Pressure is the effect of force which depends on the surface area on which the force is applied.

• The larger the area, the lesser will be the pressure.
• The smaller the area, the greater will be the pressure exerted by the force.

Example:
A woman putting on high heels sinks more into a soft ground than when the same woman puts on flat shoes.

In both cases, the force pulling down is the force of gravity (weight of the woman) acting on her. The force of gravity is the same at a place, so why the woman goes deeper in the first case (when wearing high heels) than she does with flat shoes?

Explanation:
No doubt, the gravitational pull is the same in both cases, but the pressure exerted by the force on the heels (having lesser area) is greater than that exerted on the flat shoes which have larger areas.

Definition:
It is the force acting normally per unit area of a surface. If $F$ stands for force and $A$ stands for area, then:

$$\text{Pressure} = \frac{\text{Force}}{\text{Area}}$$ $$P = \frac{F}{A}$$

Unit:
The unit of pressure is $\text{N/m}^2$ or $\text{Nm}^{-2}$ or Pascal (Pa).

Q.16: Compare the pressure exerted by a woman putting all her weight wearing one heel with that exerted by an elephant standing on one foot?
Ans: REASON:
A very surprising result is obtained by comparing the pressure exerted by a woman putting all her weight on one heel with that exerted by an elephant standing on one foot.

Suppose:

• Weight of woman = 500 N
• Area of heel = 1 cm $\times$ 1 cm $$= 0.01 m \times 0.01 m$$ $$= 10^{-4} m^2$$

So,

$$\text{Pressure} = \frac{\text{Force}}{\text{Area}}$$ $$P = \frac{500}{10^{-4}}$$ $$P = 500 \times 10^4$$ $$P = 500 \times 10000$$ $$\mathbf{P = 5000000 \, Pa \, \text{--- (i)}}$$

• Weight of the elephant = 54000 N
• Area of the foot = 30 cm $\times$ 30 cm $$A = 0.3 m \times 0.3 m$$

$$P = \frac{F}{A}$$ $$P = \frac{54000}{0.3 \times 0.3}$$ $$P = 600000 \, Pa$$ $$\mathbf{P = 600000 \, Pa \, \text{--- (ii)}}$$

Q.17: Define Atmospheric Pressure and Barometer?
Ans: ATMOSPHERIC PRESSURE:
The pressure exerted by the weight of the atmosphere on the surface of the earth is called atmospheric pressure.

Barometer:
The device used to measure the atmospheric pressure is called a Barometer.

At sea-level, the value of atmospheric pressure is about 1 Pa. This standard pressure is sometimes called one atmosphere or one bar.

Q.18: Explain the collapsing can experiment?
Ans: COLLAPSING CAN EXPERIMENT:
The effect of atmospheric pressure can be demonstrated by evacuating a tin using a vacuum pump.

• Before the air is pumped out, the pressure inside the tin is equal to the atmospheric pressure.
• When air is partially removed from the tin by the vacuum pump, the pressure of air inside the tin is less than the atmospheric pressure.

Hence, the collapsing of the tin takes place, as shown in the figure.

Diagram:
The diagram shows:

• A vacuum pump connected to a tin.
• Atmospheric pressure $A$ pushing in from all sides.
• Partial vacuum inside, causing the tin to collapse.

Q.19: Explain Magdeburg-hemisphere experiment?
Ans: MAGDEBURG HEMISPHERE EXPERIMENT:
The existence of atmospheric pressure was first demonstrated by a German scientist. His experiment is historically known as the Magdeburg hemisphere experiment.

Procedure:

• He took two hollow metallic hemispheres.
• The hemispheres were placed in contact.
• The air inside the hemispheres was pumped out by a vacuum pump.
• After the partial removal of the air from inside the hemispheres, it was almost impossible to separate them by pulling them apart. It is due to the fact that the pressure exerted by the atmosphere on the outer walls of the hemispheres is much greater than the pressure exerted by the air left inside the hemispheres.

Diagram:
The diagram shows the Magdeburg hemisphere apparatus with two hemispheres being held together by atmospheric pressure.

In the original experiment, with perfect vacuum inside the hemispheres, even sixteen horses (eight on each side) could not separate the hemispheres.

Q.20: Define the pressure of a liquid and derive the expression for it?
Ans: PRESSURE OF A LIQUID:
The force exerted per unit area by a liquid on the surface of a container or object is called the pressure of the liquid.

Calculation of Pressure of Liquid:
In order to find the pressure, we take a cylinder of height $h$ and area of cross-section $A$.

The volume of the cylinder is given by:

$$v = A \times h$$

If "d" is the density of the given liquid, then:

$$\text{Density} = \frac{\text{Mass of Liquid}}{\text{Volume}}$$ $$P = \frac{\text{Mass of the Liquid}}{A \times V}$$

As we know,

Weight of the liquid:
$W = mg = \rho Ahg$

But we know that:

$$F = W$$ $$\therefore \text{Force of liquid} = \rho Ahg$$

But

$$P = \frac{F}{A}$$

P=AρAhg​ $$P = \rho gh$$

Q.21: How can you show that pressure increases with depth?
Ans: INCREASE OF PRESSURE WITH DEPTH:
The pressure below the surface of a liquid depends on three things:

• Depth ($h$)
• Density ($\rho$)
• The pull of gravity ($g$)

The pressure that acts on a diver depends on the weight of the water above him. As he goes deeper, the weight on him increases, so the pressure also increases.

The pressure $P$ under a liquid is found using the formula:

$$P = \rho gh$$

Experiment:
To show that the pressure of a liquid increases with depth, fill a tank with three equal-sized outlets, with water.

Diagram Explanation:

• Spout A: Water shoots out from this spout with the least force, as the pressure $P_1$ is low due to the shallow depth $h_1$.
• Spout B: Water shoots out from this spout with medium force, as the pressure $P_2$ is greater than pressure $P_1$ since depth $h_2$ is greater than depth $h_1$.
• Spout C: Water shoots out with great force, as the pressure $P_3$ is greater than pressure $P_2$, which is greater than pressure $P_1$, since depth $h_3$ is greater than depth $h_2$, which is greater than depth $h_1$.

Conclusion:
The experiment demonstrates that water (hydrostatic) pressure increases with depth.

Water shoots out from the outlet A with the least force as the pressure $P_1$ is low due to the shallow depth $h_1$. Water shoots out from the outlet B with medium force as the pressure $P_2 > P_1$ since depth $h_2 > h_1$.

Water shoots out with greater force as the pressure $P_3 > P_2 > P_1$ since depth $h_3 > h_2 > h_1$.

This shows that pressure is greater at the deeper point in the liquid. The volume of water collected from the jets also increases with the depth.

Q.22: State and explain Pascal’s law?
Ans: STATEMENT:
When the pressure is applied to a liquid, it is transmitted equally in all directions.

Experiment:
Take a spherical vessel fitted with four water-tight pistons having the same area of cross-section as shown in the figure.

The vessel is filled completely with water. Force is applied on the position “A” to move it inward. Thus, a pressure is exerted on the water inside the vessel. From the experiment, it is found that the positions B, C, and D move outward through the same distance, showing that the pressure applied at “A” has been transmitted equally in all directions.

Q.23: Name some important applications of Pascal’s law?
Ans: APPLICATION OF PASCAL’S LAW:
A few applications of Pascal’s law are given below:

• Hydraulic lift or jack
• Hydraulic brakes
• Hydraulic press

Q.24: Write down the principle, construction, working, and use of the hydraulic lift or jack?
Ans: HYDRAULIC LIFT OR JACK:

Principle:
A hydraulic lift or jack works on the principle that a liquid transmits pressure equally in all directions.

Construction:
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It consists of two cylinders; one is of smaller area and the other of greater area. Both the cylinders are fitted with airtight pistons and connected with each other with the help of a tube. It contains an incompressible fluid.

Diagram:
The diagram shows a hydraulic lift with:

• A smaller piston with area $A_1$ and force $F_1$.
• A larger piston with area $A_2$ and force $F_2$.
• The fluid transfers pressure between the pistons.

Working:
The small force $F_1$ is applied to a piston of area $A_1$. Thus, the pressure on the piston is given by:

$$P = \frac{F_1}{A_1} \quad \text{--- (i)}$$

This pressure is transmitted equally throughout the liquid. Thus, an upward pressure of $P_2$ is applied to the larger piston of area $A_2$, which is greater than $A_1$. Therefore, the piston is capable of supporting a force:

$$F_2 = P \times A_2$$

As,

$$P = \frac{F_2}{A_2} \quad \text{--- (ii)}$$

By comparing equation (i) and equation (ii), we have:

$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$$ $$F_2 = \frac{F_1}{A_1} \times A_2$$

Thus, the applied force ($F_2$) has been increased by a factor of ($A_2$). It is clear from the above equation $F_2 > F_1$ since $A_2 > A_1$. Thus, a small force acting on a small area $A_1$ generates a large force $F_2$ acting on a large area $A_2$.

Uses:
It is used to lift heavy loads such as vehicles for service purposes.

Q.25: Write down the principle, construction, and working of the hydraulic brakes?
Ans: PRINCIPLE:
Its working is based on the principle of Pascal’s law. These brakes are used in automobiles. The foot exerts a small force on the brake pedal. The pressure created by this force is transmitted by the brake fluid to the brake pads. The brake pads have a large area and exert a large force on the wheel disc. The same pressure can be transmitted to all the four wheels just according to Pascal’s principle.

Construction:
A hydraulic brake consists of a tube which contains oil called brake oil. This tube is fitted with a piston working in a master cylinder. The master cylinder is linked with four other small cylinders, each containing a piston of larger area, for each wheel of the automobile.

Working:
Foot pressure on the brake pedal is transmitted via levers to the piston in the master cylinder and so increases the pressure in the brake fluid (oil). This pressure is transmitted equally and simultaneously to all four wheels. Modern cars often have disc brakes on the front wheels and drum brakes on the rear wheels, both being operated by hydraulic pressure.

Uses:

• It is used to stop the motion of vehicles.
• It is used to control the speed of the vehicles.

Q.26: Write down the principle, construction, and working of a hydraulic press?
Ans: PRINCIPLE:
A hydraulic press works on the principle that liquids transmit pressure equally in all directions.

Construction:
A hydraulic press consists of a narrow cylinder A connected to a wider cylinder B. Both cylinders are fitted with water-tight pistons. The piston B is provided with a rigid roof over it.

Working:

• The piston in the narrow cylinder A can be moved up and down by a lever.
• The pressure in A is transmitted equally.
• The piston B moves upward.
• The piston B compresses only the material placed between the piston and the roof.

Uses:
The hydraulic press is commonly used for compressing soft materials like cotton into compact bales.

Q.27: State Archimedes' principle with examples?
Ans: STATEMENT:
The principle states that when a body immersed completely or partially in a liquid will experience an upthrust (upward force) equal to the weight of the liquid (or fluid) displaced by the body.

Examples:

• A piece of cork keeps floating on the surface of water because its weight is less than that of the upthrust.
• Put the closed end of a test tube vertically on the surface of water and push it down in the water. A force is felt acting vertically upwards on the tube. The deeper you pull the tube into the water, the greater will be the upward force acting on it. This upward force of water acting on the test tube is called the upthrust.

Q.28: Derive an expression for Archimedes' principle?
Ans: VERIFICATION OF ARCHIMEDES' PRINCIPLE:

We consider a cylinder AB of length $l$ and area of cross-section $a$ be immersed in a liquid of density $\rho$. Let depth of A, the upper end of the cylinder below the liquid surface be $h_1$, and that of B, the lower end be $h_2$.

$$\therefore \text{Pressure on A} = \rho g h_1 \quad \text{--- (i)}$$

The cross-sectional area of the cylinder is $a$.

Diagram:
A cylindrical vessel full of liquid with a cylindrical object is shown.

The force on the cylinder, which is equal to the thrust, is given by:

$$\text{Pressure} = \frac{\text{Force}}{\text{Area}} \quad \text{Pressure} = \frac{\text{Thrust}}{\text{Area}}$$ $$\therefore \text{Thrust} = \text{Pressure} \times \text{Area}$$ $$\text{Thrust} = \rho g h_1 \times a \quad \text{--- (i)}$$

This thrust is in the downward direction. The pressure at B:

$$B = \rho g h_2$$

This thrust on B upward:

$$= \rho g h_2 \times a \quad \text{--- (ii)}$$

This thrust is greater than the first one. The resultant thrust in the upward direction on the cylinder is:

$$\implies \rho g h_2 - a \rho g h_1 - a = \rho g a (h_2 - h_1)$$

But:

$$h_2 - h_1 = l$$ $$\therefore \text{Thrust} = \rho g a l$$

But we know that $al$ is the volume of the cylinder “V”:

$$\therefore \text{Thrust} = \rho g V$$ $$\text{Thrust} = \rho g V \quad \text{--- (iv)}$$

If “m” is the mass of liquid displaced by this cylinder, then:

$$\text{Density of liquid} = \frac{\text{mass}}{\text{volume}} \implies \rho = \frac{m}{v}$$ $$\therefore m = \rho V$$

By putting the value of $\rho V$ in equation (iv), we have:

$$\text{Thrust} = m g$$

where $mg$ is the weight of the liquid displaced by the cylinder AB.

Upthrust = Weight of the liquid displaced by the body

Since an object immersed in a liquid experiences an upthrust, it has an apparent weight in the liquid given by:

$$\text{Apparent weight} = \text{Actual weight} - \text{Weight of the displaced liquid}$$

This means an object loses its weight in a liquid. This is the reason why it is easier to lift an object while in water than when it is above water.

Q.29: What is buoyancy? State the laws of floatation?
Ans: BUOYANCY:

It is the property of a fluid (liquid or gas) through which it exerts an upward force or upthrust on a body which is wholly or partially immersed in the liquid. This upward force is called buoyant force.

Explanation:
When an object is immersed partially or completely in a liquid, two forces act on it:

1. The weight of the object which acts downward.
2. The buoyant force which acts vertically upward.

It is the resultant of these two forces which decides whether the object will sink or float.

Laws of Floatation:

• If the weight of the body is more than the upthrust, it sinks.
• If the upthrust is more than the weight of the body, it floats.
• If the upthrust and the weight of the body become equal, it will neither sink nor float but remain in liquid.

Examples:

• Cork:
  A cork floats on the surface of water because the upthrust of water on it is greater than its weight.

• Ship:
  A ship is so designed that its volume is very large. Hence, the buoyant force, which is equal to the weight of equal volume of water displaced by the ship, is very large and greater than its weight. Hence, a ship floats on the surface of water in the sea.

• Needle:
  A needle made of iron sinks in water because the buoyant force acting on it is less than its weight. This is due to the fact that the volume of a needle is very small and hence the weight of the water displaced by the needle is less than the weight of the needle.

• Submarines:
  Submarines can float on the surface of water and, when needed, they can dive into water. They are fitted with large hollow ballast tanks.

• In order to dive in water, they fill the tanks with water. This increases the weight of the submarines and submerges them in water.
• To bring the submarine to the surface, extra load due to the water in the tanks is removed. This is done by forcing compressed air into the tanks, which expels water from the tanks.

Q.30: How can we find the density of copper from Archimedes' principle?
Ans: DENSITY OF COPPER:
Archimedes' principle can be used to find out the specific gravity or relative density of an insoluble solid, for example, a piece of copper.

$$\text{Specific gravity of copper} = \frac{\text{Density of Copper}}{\text{Density of Water}}$$ $$= \frac{\text{Mass of Copper Piece} / \text{Volume of Copper Piece}}{\text{Mass of Water} / \text{Volume of Water}}$$ $$= \frac{\text{Mass of Copper Piece} / \text{Volume of Copper Piece}}{\text{Mass of Water displaced by copper piece} / \text{Volume of displaced Water}}$$

Since:

The volume of displaced water = Volume of copper piece.

Hence:

$$\text{Specific gravity of copper} = \frac{\text{Mass of copper piece in air}}{\text{Mass of water displaced}}$$ $$= \frac{\text{Weight of copper piece in air}}{\text{Weight of water displaced}}$$

But the weight of water displaced is equal to the loss of weight of copper piece in water. Hence:

$$\text{Specific gravity of copper} = \frac{\text{Weight of copper in air}}{\text{Loss of weight of copper piece in water}}$$ $$\text{Specific gravity} = \frac{W_1}{W_1 - W_2}$$

Where:

• $W_1$ = Weight of copper piece in air
• $W_2$ = Weight of copper piece in water

From the above equation, we can find the density of copper.

$$\text{Specific gravity of copper} = \frac{\text{Density of copper}}{\text{Density of Water}}$$ $$\text{Density of copper} = \frac{W_1}{W_1 - W_2} \times \text{Density of Water}$$

If "d" is the density of water at 40°C, then:

$$\text{Density of copper} = \frac{W_1}{W_1 - W_2} \times d$$ $$\boxed{\text{Density of copper} = \frac{W_1}{W_1 - W_2} \times d}$$

Q.31: How can we find the density of oil from Archimedes’s principle?
Ans: DENSITY OF OIL (FOR EXAMPLE KEROSENE OIL):
We know that the density of copper using water is given by:

$$\text{Density of copper} = \frac{W_1}{W_1 - W_2} \times \text{Density of water} \quad \text{--- (i)}$$

Similarly, the density of copper using kerosene oil is given by:

$$\text{Density of copper} = \frac{W_1}{W_1 - W_3} \times \text{Density of kerosene oil} \quad \text{--- (ii)}$$

Where $W_3$ is the weight of copper in kerosene oil.

Comparing The Above Equations:

$$\frac{W_1}{W_1 - W_3} \times \text{Density of kerosene} = \frac{W_1}{W_1 - W_2} \times \text{Density of water}$$ $$\text{Density of kerosene oil} = \frac{W_1 - W_3}{W_1 - W_2} \times \text{Density of water}$$

If "d" is the density of water, then:

$$\text{Density of kerosene oil} = \frac{W_1 - W_3}{W_1 - W_2} \times d$$ $$\boxed{\text{Density of kerosene oil} = \frac{W_1 - W_3}{W_1 - W_2} \times d}$$

Where $W_3$ is the weight of copper piece in kerosene oil.

Q.32: How can we find the density of cork from Archimedes’ principle?
Ans: DENSITY OF CORK:
Let $W_1$ be the weight of cork in air. Since cork is lighter than water, its weight in water cannot be found directly. To find its weight in water, a sinker is used. Let the weight of cork in air be $W_1$, and that of sinker in water be $W_2$. Let the weight of sinker and cork both in water be $W_3$. Hence, the loss of weight of cork in water is $W_2 - W_3$.

$$\text{Density of cork} = \frac{\text{Weight of Cork in air}}{\text{Loss of weight of cork in water}} \times \text{Density of water}$$ $$\boxed{\text{Density of cork} = \frac{W_1}{W_2 - W_3}}$$

Where “d” is the density of water.

Q.33: Write down the main points of kinetic molecular theory?
Ans: KINETIC MOLECULAR THEORY:
The main points of kinetic molecular theory are as follows:

• Matter is composed of molecules.
• These molecules are always in motion; hence, the molecules could have all kinds of kinetic energy.
• The molecules of a substance attract each other with a force that depends on the distance between them.
• The velocity of molecules depends on the temperature of the substance. The higher the temperature, the greater the velocity of the molecules.
• The kinetic energy of a molecule increases with an increase in temperature.

Q.34: Define and explain surface tension?
Ans: SURFACE TENSION:
The surface tension of a liquid is a net inward pull that tends to draw the surface molecules into the body of the liquid.

OR
Surface tension is a force that acts along the surface of a liquid to keep it in a state of tension.

OR
Surface tension is the property of a liquid through which the free surface of the liquid behaves like a stretched membrane, tending to decrease the surface area.

Explanation:
Draw an imaginary line AB on the surface of a liquid. The molecules on either side of the line tend to pull away from the other side. Thus, the free surface of the liquid tends to decrease in area, producing a tension in it.

Mathematical Representation:
If “F” is the force per unit length acting on either side of the imaginary line, then the surface tension is given by:

$$\sigma = \frac{F}{L}$$

In the S.I. system, the unit of surface tension is N/m.

Explanation On The Basis Of Kinetic Molecular Theory:
Consider molecule “A” lying inside the liquid. It is attracted from all sides, and thus, this molecule has no resultant force acting on it. The molecule “B” lying on the surface has a resultant force acting downward. Due to the downward forces acting on the surface molecules, the free surface of the liquid behaves like a stretched membrane.

Molecule A lying in the interior has no resultant force acting on it. The resultant force acting on molecule B lying on the surface is downward.

In brief, liquids show surface tension because the attraction between molecules is strong enough to keep them clustered together.

Examples:

• The stable form of a mass of liquid is a sphere. It is shown by a free-falling drop of rain. A liquid tends to keep its surface area minimum due to surface tension. The surface area of a sphere is minimum for a given volume, so raindrops are spherical.
• A steel needle, if dropped in water, will sink because the density of steel is greater than that of water. However, if it is put on the surface of water horizontally, it does not sink into water because of the surface tension of water.

Q.35: Define and explain viscosity? Write down its formula and unit?
Ans: FLUID FRICTION (VISCOSITY):
Viscosity is the force that arises due to the force of friction between different layers of a fluid in flow.

OR
The property by virtue of which a fluid tends to oppose the relative motion between its different layers is called the viscosity of the liquid.

Explanation: Consider a liquid that flows on a smooth horizontal glass surface. We may suppose the liquid to be divided into different layers parallel to the fixed surface on which the liquid is flowing.

• It is found that different layers move with different velocities.
• For a streamline motion, the layer in contact with the fixed surface is stationary.
• The velocity of the layer increases with the distance from the fixed surface in a perpendicular direction.
• For any two layers, the upper layers move faster than the lower ones. Hence, the upper layer tends to accelerate the lower layer, which tends to retard the upper layer. Thus, the two layers together tend to destroy their relative motion as if there is a backward dragging tangential force acting between the two layers.

Formula: [Formula is not fully shown in the extracted content.]

Consider any layer lying at a distance $x$ from the stationary surface. Let $A$ be the area of the layer moving with a velocity $v$. It is found that the backward dragging force $F$ acting on any layer is found to vary as:

$$F \propto A \quad \text{--- (i)}$$ $$F \propto v \quad \text{--- (ii)}$$ $$F \propto \frac{1}{x} \quad \text{--- (iii)}$$

On combining equations (i), (ii), and (iii), we get:

$$F = -\eta \frac{Av}{x}$$

where $\eta$ is a constant depending on the nature of the liquid and is known as the coefficient of viscosity. The negative sign shows that the dragging force acts in a direction opposite to the flow of the liquid.

Coefficient Of Viscosity:
The coefficient of viscosity is numerically equal to the retarding force required to maintain a velocity of one m/s relative to the stationary layer.

Unit:
The unit of coefficient of viscosity is poise.

The coefficient of viscosity is said to be one poise if the backward dragging force required to maintain a relative velocity of 1 m/s between two layers, each of area $1 \, m^2$, separated by a distance of 1 m, is one Newton.

Q.36: Write down the uses of viscous fluid and their characteristics?
Ans: Characteristics Of Viscous Liquid:
Viscous fluids have slow motion because viscosity, which acts as a friction, opposes the motion. The viscosity of a liquid is usually much smaller compared to the friction between two solid surfaces.

Uses of Viscous Fluid:

• An oil of high viscosity may be used as a lubricant.
• In heavy machines where there is considerable pressure on the bearing, a viscous lubricant should be preferably used because light or thin oil is easily squeezed out.
• For light machinery like a sewing machine, comparatively less viscous oil may be used as a lubricant.