Properties of Matter MCQs Qs and Numerical

Properties of Matter MCQs Qs and Numerical (School)

For more Physics Quizzes Notes and Numerical:

MCQs

1. In which of the following state molecules do not leave their position?

1. Solid
2. Liquid
3. Gas
4. Plasma

2. Which of the substance is the lightest one?

1. copper
2. mercury
3. aluminum
4. lead

3. SI unit of pressure is pascal, which is equal to:

1. 10⁴ Nm^-2
2. 1 Nm^-2
3. 10² Nm^-2
4. 10³ Nm^-2

4. What should be the approximate length of a glass tube to construct a water barometer?

1. 0.5 m
2. 1 m
3. 2.5 m
4. 11 m

5. According to Archimedes, up-thrust is equal to:

1. weight of displaced liquid
2. volume of displaced liquid
3. mass of displaced liquid
4. none of these

6. The density of a substance can be found with the help of:

1. Pascal's law
2. Hooke's law
3. Archimedes principle
4. Principle of flotation

7. According to Hooke's law:

1. Stress x strain = constant
2. Stress / strain = constant
3. Strain / stress = constant
4. Stress = strain

8. Tensile strain produces change in:

1. length
2. volume
3. area
4. density

9. Unit of strain is:

1. Nm^-2
2. Nm
3. N
4. No unit

10. Hydraulic press is a machine which works on:

1. Hook's law
2. Pascal's law
3. Boyle's law
4. Archimedes

11. In SI system, unit of density is:

1. kgm^-1
2. kgm^-3
3. kgm^-2
4. kgm

12. Which of the following works on Pascal's law?

1. Screw gauge
2. Vernier Calipers
3. Hydraulic press
4. Wedge

13. How many times Mercury is denser than water?

1. 12.5
2. 13.6 ANS
3. 14.7
4. 15.8

14. Atmospheric pressure is measured by an instrument called.

1. Barometer
2. Altimeter
3. Meter rod
4. Hydrometer

15. At sea level, the atmospheric pressure is about:

1. 101325Pa
2. 101300Pa
3. 10110Pa
4. 10112Pa

16. Density of water is:

1. 5500 kgm^-3
2. 10000 kgm^-3
3. 1000 kgm^-3
4. 330 kgm^-3

17. Most of the matter that fills the universe is in state:

1. solid
2. plasma
3. liquid
4. gas

Question Answers

SQs

Q1. How kinetic molecular model of matter is helpful in differentiating various states of matter?

Ans. By finding density, solubility, motion of molecules we can differentiate between solids liquids and gases. With the help of kinetic molecular theory we add properties of solid, liquid, gases and of plasma.

Q2. Does there exist a fourth state of matter? What is that?

Ans. Yes! there is fourth state of matter which is called plasma. Plasma consists of ions and electrons which exists on Sun and in florescent tubes.

Q3. What is meant by density? What is its SI unit?

Ans. Density of a substance is defined as its mass per unit volume.

Density = mass of substance / volume of that substance

Density = m /V

SI unit = kg/ m³  = kgm^-3

Q4. Can we use a hydrometer to measure the density of milk?

Ans. Yes! we can find the density of milk by hydrometer. But for this we use a special hydrometer called lactometer.

Q5. Define the term pressure.

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

P = Force / Area

P = F/ A

Q6. Show that atmosphere exerts pressure.

Ans. Atmosphere Pressure: The atmospheric pressure is the weight of gases above a certain point on the Earth. The Earth is surrounded by a cover of air called atmosphere. It extends to few hundred kilometers above sea level. Just as certain sea creatures live at the bottom of ocean. We live at the bottom of a huge ocean of air. Air is a mixture of gases. The density of air in the atmosphere is not uniform. It decreases continuously as we go up. The fact that atmosphere exerts pressure can be explained by experiment.

Experiment:

Take an empty tin can with a lid. Open its cap and put some water in it. Place it over flame. Wail till water begins to boil and the steam expels the air out of the can. Remove it from the flame. Close the can firmly by its cap. Now place the can under tap water.

When the can is cooled by tap water, the steam in it condenses. As the steam changes into water, it leaves an empty space behind it. This lowers  the pressure inside the can as compared to the atmospheric pressure outside the can. This will cause the can to collapse from all directions. This experiment shows that atmosphere exerts pressure in all directions.

Q7. It is easy to fill air in a balloon but it is very difficult to remove air from a glass bottle. Why?

Ans. A balloon has elasticity and it is easy to fill air inside a balloon. An air filled balloon has greater air pressure than surrounding air pressure. The glass bottle is rigid. So it is easy to fill air in a balloon but it is difficult to remove air from glass bottle because air pressure inside the glass bottle is less than atmospheric pressure.

Q8. What is a barometer?

Ans. It is an instrument which is used to measure atmospheric pressure.

Q9. Why water is not suitable to be used in a barometer?

Ans. Since the density of water is much less than mercury so we cannot use water in barometer. If we use water in barometer longer glass tube is required.

Q10. What makes a sucker pressed on a smooth wall sticks to it? Ans. The sucker is dish shaped, when pressed against a smooth surface the air is forced from beneath the sucker. The rubber makes an air tight seal and the air pressure outside is greater than the air pressure beneath the sucker, thus forcing the rubber sucker to stick.

Q11. Why does the atmospheric pressure vary with height?

Ans. With the increase of height quantity of air began to decrease due to which atmospheric pressure also becomes low. Density of air and value of' 'g' decreases with height so atmospheric pressure vary with height.

Q12. What does it mean when the atmospheric pressure at a place fall suddenly?

Ans. When the atmospheric pressure fall suddenly at a place, it may followed the storm or rain at that place.

Q13. What changes are expected in weather if the barometer reading shows a sudden increase?

Ans. When the atmospheric pressure increases suddenly this means that there is poor weather ahead.

Q14. State Pascal's law.

Ans. Pressure applied at any point of a liquid enclosed in a container is transmitted without loss to all other parts of the liquid.

Q15. Explain the working of hydraulic press.

Ans. Hydraulic press is a machine which works on Pascal's law. It consists of two cylinders of different cross sectional areas. They are filled with pistons of cross sectional areas a and A. 

Q16. What is meant by elasticity?

Ans. The property of a body to restore its original size and shape as the deforming force ceases to act is called elasticity.

Q17. State Archimedes principle.

Ans. When an object is totally or partially immersed in a liquid, an upthrust acts on equal to the weight of the liquid it displaces.

Q18. What is upthrust? Explain the principle of floatation.

Ans. Upthrust: Upthrust is the force of liquid which acts on the floating object in the upward direction, which are immersed in liquid and is equal to the weight of displaced water.

Principle of Floatation: A floating object displace a fluid having weight equal to the weight of object.

Q19. Explain how a submarine moves up the water surface and down into water.

Ans. A Submarine can travel over as well as under water. It also works on the principle of floatation. It floats over water when the weight of water equal to its volume is greater than its weight. Under this condition, it is similar to ship and remains partially above water level. it has a system of tanks which can be filled with and emptied from ca water. When these tanks are filled with seawater, he weight of the submarine increases. As soon as its weight becomes greater than the upthrust, it dives into water and remains under water. To come up on the surface, the tanks are emptied from sea water.

Q20. Why does piece of stone sink in water but a ship with a huge weight floats?

Ans. According to principle of floatation "A floating object displaces a fluid having weight equal to the weight of object."

A stone sink in water because upthrust force is less than the displaced liquid weight but a huge ship floats because it displaces water equal to the weight of body when it is partially or completely immersed in water. In this case upthrust of liquid is equal or greater than the weight of displaced liquid. So a piece of stone sink and a ship with huge weight floats.

Q21. What is Hooke's Law? What is meant by elastic limit?

Ans. HOOKE'S LAW:

The strain produced in a body by the stress applied to t s directly proportional to the stress within the elastic limit of the body".

Thus stress proportional to strain

Or stress = constant x strain

Or stress / strain = constant

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 This limit tells the maximum stress that can be safely applied on a body without causing permanent deformation in its length, volume or shape. In other words, it is a limit within which a body recover its original length, volume or shape after the deforming force is removed. When a stress crosses this limit, called the elastic limit, a body is permanently deformed and is unable to restore its original state after the stress is removed. 

Q22. Take a rubber band. Construct a balance of your own using a rubber band. Check its accuracy by weighing various objects.

Ans. Rubber band is the elastic body and it changes its length if the deforming force is acting if we hang an object with Rubber band with scale marked at one end of box. The length of rubber band increases on stretching it. The pointer of balance is lowered when body is suspended from it. It can measure the weight of an object.

Conceptual Qs

Q1. If you climbed a mountain carrying a mercury barometer, would the level of mercury in the glass tube of the barometer increases or decreases as you climb the mountain? Explain.

Answer: As we climb up the mountain the level of mercury in the glass tube of the barometer decreases.

The height of the mercury column depends on the atmospheric pressure. A higher atmospheric pressure results in an increase of the height of the mercury column in the barometer and vice versa.

The atmospheric pressure decreases as we go towards the higher altitude, hence, the level of the barometer also decreases.

Q2. Walnuts can be broken in the hand by squeezing two together but not one, why? 

Answer: When two walnuts are placed in the hand, the area of the palm of the hand is greater than the area of surfaces of walnuts in contact.

As we know: Pressure = $\frac{Force}{Area}$

When a force is applied to squeeze these walnuts together, the pressure between the surfaces in contact increases as the contact area is small. Hence, by the above equation, a decrease in the area results in a higher pressure which causes the walnuts to break at the surface in contact.

Q3. Why is the cutting edge of the knife made very thin?

Answer: The cutting edges of the knife are made very thin because the smaller the area more will be the pressure that is exerted on the area and vice versa.

Mathematically, we know the relation between pressure and applied force where ‘P’ is pressure, ‘F’ is the applied force and ‘A’ is area. Hence, a small area with large pressure is easy for things like cutting, slicing or stabbing etc.

Q4. Why water tanks are constructed at the highest level in our houses? 

Answer: As we know that: P = ρgh

The water tanks are constructed at the highest level in our houses because the pressure depends upon the height.

From the above equation, the pressure of water increases with height. So as the pressure applied by the water on pipes increases it results in an easy flow of water in a pipe system.

Q5. Why a small needle sinks in water and huge ships travel easily in water without sinking?

Answer: A small needle sinks in water while a huge ship travels easily in water without sinking because the density of needle is greater than water.

F (buoyant force of fluid) = w (displaced fluid) = m (mass of fluid) x g

                                                       ρ  = $\frac{m}{V}$

The mass of needle is more than the mass of water having same volume. Hence, the density of the needle is greater than the density of water displaced by it. As a result the buoyant force of fluid is less than the weight of the needle, so the needle sinks in water.

In the case of a huge ship, the weight of water displaced is greater than the weight of the ship due to which the ship does not sink and floats on water.

Q6. Explain how and why camels have adapted to allow them to walk more easily in desert conditions?

Answer: Camels can walk easily in the desert because of their wide feet. Their wide feet allow the weight of their body to act on a larger surface of land reducing the pressure exerted on the land by the camel. we know that pressure exerted and area are inversely proportional.

Q7. You would have probably experienced your ears ‘popping’ while driving in mountains. Why ears’ pop’?

Answer: In our ears a fluid maintains the internal and external pressure normally. While driving to an elevation the change in the atmospheric pressure results in changing the pressure on the ear drums, so a net pressure arise on ear drums which cause popping.

It can be relieved in a few minutes or by swallowing, chewing or blowing out nose.

Q8. If you filled an airtight balloon at the top of a mountain, would the balloon expand or contact as you descend the mountain? Explain.

Answer: The air balloon will contract as the atmospheric pressure increase going down the mountain. So the external pressure is more as compared to the pressure on the top of the mountain.

Q9. A rowboat is floating in a swimming pool when the anchor is dropped over the side. When the anchor is dropped, will the water level in the swimming pool increase, decrease, or remain the same? Explain.

Answer:  When the anchor is dropped, the water level in the swimming pool increases as the anchor displaces a volume of water resulting in the water level to rise. In this case, the swimming pool attains more volume of the anchor as it is dropped in it.

Q10. Which material is more elastic, steel or rubber and why?

Answer: Elasticity is defined as the ability to resist a change.

Steel is more elastic than rubber because steel produces more resistance against the deformation than rubber. So, more deforming force is required to deform steel while rubber can be easily deformed.

Descriptive Qs

Q1. Using kinetic molecular model of matter, explain three states of matter. Answer: Matter has three states solids, liquid and gases. These three sates of matter are explained not he basis of kinetic molecular theory. It is assumed that:

Solids:

1. Solids are made up of molecules which are arranged closely in a fixed pattern.
2. Molecules in solids vibrate about their mean position.
3. The attractive forces between the molecules are strong. 

  Liquids:

1. Liquids are also made up of molecules which are close together.
2. The pattern of molecules is not fixed and does not extend far. The molecules in a pattern keep changing their position.
3. Molecules are able to move about, which means that a liquid is able to change its shape and can adopt the shape of the container.
4. The attractive forces between the molecules of liquid is less than the solid.

Gases:

1. A gas is made up of molecules which are in constant random motion.
2. The distance between molecules id larger as compared to the size of molecules.
3. The molecules are constantly colliding elastically with each other and with the walls of the container.
4. Forces between molecules are negligible, except during collisions.

Q2. Define and explain density and pressure.

Answer: Density: “Mass per unit volume of a body is called density”. It is a scalar quantity.

Density = $\frac{Mass}{Volume}$

ρ = $\frac{m}{V}$

The SI unit of density kgm^-3

Explanation:

Let’s take the example of iron and wood. The same volume of iron is heavier than wood due to the reason that the density of iron is more than that of wood. In other words, the mass of iron for 1 m³ is more than the mass of wood of 1 m³ volume. Molecules of iron are held closely together, whereas the molecules of wood are arranged in a different manner and have voids in the structure so the number of molecules in the same volume is more in case of iron than the wood.

Pressure:

“The normal force applied per unit area is called pressure”. It is denoted by “P”.

Pressure = $\frac{Force}{Area}$

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

Mathematically:

The SI unit of pressure is pascal (Pa).

Pascal:

It can be defined as “when a one newton force acts on a body of area one meter square, then the pressure is one pascal”. Mathematically

Explanation:

Let’s take an example of a drawing pin. Hold the pin between a finger and the thumb, and the pointed part directed towards the thumb. By applying pressure we feel pain at the contact point at the thumb but not at the finger. This is because the pressure at the thumb is much higher than the pressure at the finger.

The reason is that the pressure is inversely proportional to the area. The pressure increase as the area decreases keeping the force constant. The area of the pointed part is very small as compared to the head of the drawing pin, so a pain is felt on the thumb.

Q3. What is atmospheric pressure? How is it measured by using a mercury barometer? Also, describe how weather changes with atmospheric pressure? Answer:

Atmospheric Pressure:

“The pressure exerted by the gases in the atmosphere is known as atmospheric pressure”.

Barometer: Atmospheric pressure is measured by using mercury barometer.

Construction:

A long tube opened at one end is filled with mercury and inverted in a dish of mercury. The mercury does not empty into the bowl. Instead, the atmospheric pressure pushes the mercury in the tube to some height ‘h’ above the bowl. In this way the force exerted on the bowl of mercury by the atmosphere is equal to the weight of the column of mercury in the tube. A change in the height means a change in the atmospheric pressure.

 Weather changes:

When we keep a barometer at some height at sea-level, it shows some variation in atmospheric pressure from day to day. The winds move from higher to lower pressure regions and the strength of the wind depends upon the pressure gradient. There will be strong winds if the pressure gradient is high and vice versa in this way weather changes with atmospheric pressure.

Q4. State Pascal’s principle and explain with example?

Answer: Pascal’s principle

This principle states that ″whenever an external pressure is applied in a liquid, the pressure is transmitted equally to every point of the liquid in all directions.”

Or

“All liquids exert the same pressure in all directions.”

Explanation:

There are so many systems which work on the principle of Pascal’s law. For example, hydraulic press, hydraulic brakes in vehicles and the hydraulic lifts etc. Here, we take a simple example of a hydraulic press to understand it. Consider two interconnected cylinders having different diameters. The chamber is filled with incompressible fluid, and their top ends fitted with pistons.

Suppose a F₁ is applied to a small piston of A₁. The pressure P₁ is exerted by this piston is.

P₁ = $\frac{F_1}{A_1}$

Similarly, the pressure on the large piston P₂ is

P₂ = $\frac{F_2}{A_2}$

Where A₂ is being of larger piston and F₂ is the force on the plunger. Since the pressure is the same on both sides, So

P₁ = P₂

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

${F_2}$ = $(\frac{F_1}{A_1}){A_2}$

${F_2}$ = $(\frac{A_2}{A_1}){F_1}$

Therefore, the Force F₂ is larger than F₁ by multiplying factor $\frac{A_2}{A_2}$. Hence, a small force can lift a much larger load. Hydraulic brakes, car lifts, hydraulic jacks, and forklifts all make use of this principle.

Q5. How pressure varies with depth in liquids? Explain.

Answer: The pressure in liquid increases with depth. Because the further down you go, the greater the weight of the liquid above. It can be observed in the figure below that the water spurts out fastest and furthest from the lowest hole. A simple set-up in the following figure shows how water pressure increases with depth.

As we know P = ρgh

Consider a container is filled with water and having three opening 'P₁', 'P₂' and 'P₃' as shown in figure.

The water form opening 'P₁', shoots out with the least force because the pressure at 'P₁' is very small due to smaller depth 'h₁'.

At opening 'P₂', the water shoots out with medium force because, the pressure at 'P₂' is greater due to greater depth 'h₂' as compared to 'h₁'.

At opening 'P₃', the water shoots out with maximum force because, the pressure at 'P₃' is greatest due to maximum depth 'h₃'.

Hence, it is proved that pressure varies directly with depth.

Q6. What is meant by buoyant force or upthrust in fluids?

Answer: Buoyant force: The force exerted by liquids in opposite direction against the weight of the body is known as upward thrust.

Explanation:

When an object is immersed in a fluid, the pressure on the lower surface of the object is higher than the pressure on the upper surface. The difference in pressure leads to an upward net force acting on the object due to the fluid pressure called up-thrust or buoyant force and phenomenon is called buoyancy.

If you try to push a piece of cork underwater, you feel the buoyant force pushing the cork back up.

The buoyant force arise because the fluid pressure at the bottom of the cylinder is larger than at the top.

Net force F_net of the fluid on the cork is the buoyant force F_B.

F_up > F_down because the pressure is greater at bottom of the beaker, hence the fluid exerts a net upward force.

Q7. State and explain Archimedes principle .

Answer: Archimedes principle: “The buoyant force acting on an object fully or partially submerged in a fluid is equal to the weight of the fluid displaced by the object”.

Explanation:

According to Archimedes principle every object experiences a buoyant force. The buoyant force will decrease the weight of the object which termed as apparent lose of weight. Whether it floats or sinks depends on the object’s density relative to the fluid. When we push a wooden block underwater and you will feel the upward buoyant force as in figure.By Archimedes principle, the buoyant force equals the weight of water displaced; since water is denser than wood, the buoyant force is greater than the wood’s weight, and that is why wood floats. Now when submerged a coin. It’s denser than water, so the coin’s weight is greater than the weight of the displaced water. Hence the coin’s weight is greater than the upward buoyant force, and it sinks.

Whether an object will float or sink depends on the net force acting on it. This net force can be calculated as follows: F_net = F_B – W (object)

Now we can apply Archimedes principle, using ‘m_o‘ to represent the mass of the submerges object and ‘m_f‘ as mass of fluid ‘g’ is acceleration due to gravity.

F_net = m_fg – m_og

Remember that ‘m = ρV’, so the expression can be rewritten as follows:

F_net = ρ_fV_fg – ρ_oV_og or F_net = (ρ_fV_f – ρ_oV_o)g

A simple relationship between the weight ‘W’ of submerged object of mass mo and density ρo can be found by considering their ratio as follows:

$\frac{W}{F_B} = \frac{m_o g}{m_B g}$

$\frac{W}{F_B} = \frac{m_o}{m_B}$ --- (1)

For submerged object equal volume is displaced therefore

V_o = V_B = V

therefore m_o = ρ_oV and m_B = ρ_BV

$\frac{W}{F_B} = \frac{ρ_o g}{ρ_B g}$

$\frac{W}{F_B} = \frac{ρ_o}{ρ_B}$

putting these values in eq. 1, we get

$\frac{m_o}{m_B}$ = $\frac{ρ_o g}{ρ_B g}$

Q8. What is elasticity? Explain

Answer: Elasticity: “The property of Solid materials to return to their original shape and size after removal of deforming force is called elasticity”.

For example, when we stretch a rubber band, its size increases, but after the removal of the force, the rubber band gets their original shape.

Similarly, when a racket hits a tennis ball, the shape of the ball is distorted or deformed, but it regains its original shape when it bounces off the tennis racket.

But the materials are elastic up to a certain limit known as the elastic limit. Beyond this limit, a material deforms and do not attain its original position.

Q9. State and explain Hooke’s law.

Answer: Hook’s law: This law states that “within the elastic limit, the extension (or compression) is directly proportional to the restoring force”.

Explanation

Consider that a spring is connected with a firm support at its one end. And a force ‘F’ is applied on the object the other end of the spring to produce an extension ‘x’ Then release it. A restoring force F restoring of spring act on the object to restore its original position. According to Hook’s law, we have

F_res ∝ -x

F_res = k x

Where ‘k’ is constant of proportionality and is known as spring constant or modulus of elasticity. Its unit is Nm^-1. The value of ‘k’ depends upon the nature of the spring and system of units.

Q10. Define and explain, Stress, strain and Young’s modulus

Answer: Stress: “The force applied per unit area of cross-section to produce deformation”.

Mathematically:

Stress = $\frac{Force}{Area}$

σ = $\frac{F}{A}$

The SI unit of stress is Nm^-2 or pascal.

Strain: The deformation produced in a body due to stress is called strain.

Consider a wire has an initial length “L”. After the applied deforming force, the length of the wire change to an amount “ΔL”. The linear strain can be defined as the change in length per unit original length is called linear strain.

It is denoted by a Greek letter Epsilon “ε”.

Mathematically:

Strain = $\frac{Change in length}{Original length}$

ε = $\frac{ΔL}{L}$

The strain has no unit because it is the ratio of two similar quantities.

Young’s modulus: The ratio of the tensile stress to the strain is called Young’s modulus or modulus of elasticity. It is represented by ‘Y’.

Mathematically:

Young's Modulus = $\frac{Stress}{Tensile Strain}$

Mathematically:

Y = $\frac{\frac{F}{A}}{\frac{ΔL}{L}}$

Y = $\frac{\frac{F}{L}}{\frac{ΔL}{A}}$

The SI unit of Young’s modulus is Nm^-2.

Numerical Problems

1 A wooden block measuring 40cmx10cmx5cm has a mass of 850g. Find the density of wood.

Given Data: Mass = m = 850g = 0.85kg

Volume = V= $\frac{40}{100} × \frac{10}{100} × \frac{5}{100}$ = 2 x 10^-3 m³

To find: Density = ρ =?

Solution: ρ = $\frac{mass}{volume} = \frac{0.85}{2×10^{-3}}$

ρ = 425kgm^-3 ANS

2 What would be the volume of ice formed by freezing 1 liter of water?

Given Data: Volume (water) = V_w = 1 liter = 1x10^-3m³

Density (water) = ρ_w = 1000kgm^-3

Density (ice) = ρ_i = 920kgm^-3

To find: Volume (ice) = V_i =?

Solution: We know that

m_w = ρ_wV_w ⇒ ρ_w = $\frac{m_w}{V_w}$

Also

m_i = ρ_iV_i ⇒ ρ_i = $\frac{m_i}{V_i}$

m_w = m_i

ρ_wV_w = ρ_iV_i

⇒ ρ_i = $\frac{ρ_w × V_w}{ρ_i}$

ρ_i = 1.09x10^-3m³

ρ_i = 1.09l ANS

3 Calculate the volume of the following objects:

(i) An iron sphere of mass 5kg, the density of iron is 8200kgm^-3.

(ii) 200g of lead shot having density 113000kgm^-3.

(iii) A gold bar of mass 0.2kg. The density of gold is 19300 kgm^-3.

Given Data: (a) m = 5kg          ρ = 8400kgm^-3

To find: V =?

Solution: V = $\frac{m}{ρ} = \frac{5}{8400}$

V = 6.1 × 10^-4 m³ ANS

(b) m = 200g = 0.2kg        ρ = 11300kgm^-3

To find: V =?

Solution: V = $\frac{m}{ρ} = \frac{0.2}{11300}$

V = 1.77 × 10^-5 m³ ANS

(c) m = 200g = 0.2kg        ρ = 19300kgm^-3

To find: V =?

Solution: V = $\frac{m}{ρ} = \frac{0.2}{19300}$

V = 1.04 × 10^-5 m³ ANS

4 The density of air is 1.3kgm^-3. Find the mass of air in a room measuring 8mx5mx4m.

Given Data: ρ = 1.3kgm^-3              V = 8 x 5 x 4 = 160m³

To find: m =?

Solution: m = ρV = 1.3 x 160

m = 208kg ANS

5 A student presses her palm by her thumb with a force of 75N. What would be the pressure under her thumb having contact area 1.5cm²?

Given Data:  F = 75N       A = 1.5cm² = 1.5x10^-4m²

To find: P =?

Solution: P= $\frac{75}{1.5 × 10^{-4}}$

⇒  P = 5x10⁵ Nm^-2 ANS

6 The head of pin is square of side 10mm. Find the pressure on it due to a force of 20N.

Given Data: F = 20N        L = $\frac{10}{1000}$ = 1 × 10^-2 m

A = L² = (1x10^-2)² = 1x10^-4m²

To find: P =?

Solution: P= $\frac{F}{A} = \frac{20}{1 × 10^{-4}}$

⇒P = 2x10⁵Nm^-2 ANS

7 A uniform rectangular block of wood 20cmx7.5cmx7.5cm and of mass 1000g stands on a horizontal surface with its longest edge vertical. Find (i) the pressure exerted by the block on the surface (ii) density of the wood.

Given Data: Volume=V= V= $\frac{20}{100} × \frac{7.5}{100} × \frac{7.5}{100}$ =1.125 x 10^-3m³

Mass = m = 1kg

To find: (i) P =?                 (ii) ρ of wood

Solution: F= w = mg = 1 x 10 = 10N

A = 0.075 X 0.075 = 5.625 x 10^-3m²

P= $\frac{F}{A} = \frac{10}{5.625 × 10^{-3}}$

⇒ (i) P = 1778Nm^-2 ANS

Density= $\frac{m}{V} = \frac{1}{1.125 × 10^{-3}}$

⇒ (ii)density= ρ =889kgm^-3ANS

8 A cube  of glass of 5cm side and mass of 306g, has a cavity inside it. If the density of glass is 2.55gcm^-3. Find the volume of the cavity.

Given Data: Side of cube = x = 5cm = 0.05m

Mass of cube = m = 306g = 0.306kg

To find: V_cavity =?

Solution: V₁ = x³ = (0.05)³ = 1.25x10^-4 m³

Density = 2.55 gcm^-3 = 2550 kgm^-3

Volume = $\frac{m}{ρ} = \frac{0.306}{2550}$

⇒ V = 1.2x10^-4 m³

V_Cavity = V – V₁ = 1.25 x 10^-4 – 1.2 x 10^-4 = 0.05 x 10^-4 m³

V_Cavity = 5x10⁶ m³ = 5 cm³  ANS

9 An object has weight 18N in air. Its weight is found to be 11.4N when immersed in water. Calculate its density. Can you guess the material of the object?

Given Data:  Weight of the object in air = w₁ = 18N

Weight of the object immersed in water = w₂ = 11N

ρ = 1000 kgm^-3

To find: D =?      Material Name =?

Solution: D = $\frac{w_1}{w_1 - w_2} x ρ = \frac{18}{18 - 11}$ x 1000

D = 2727kgm^-3 Material in Aluminum ANS

10 A solid block of wood of density 0.6gcm^-3 weighs 3.06N in air. Determine (a) volume of the block (b) the volume of the block immersed when placed freely in a liquid of density 0.9gcm^-3?

Given Data: Density of solid block = D = 0.6 gcm^-3 = 600 kgm^-3

Weight of block in air = 3.06 N

To find: (a) Volume of block        (b) Volume of block immersed in liquid of density = 900kgm^-3

Solution: (a) w = mg

⇒ m = $\frac{w}{g} =\frac{3.06}{10}$ = 0.306 kg

V =$\frac{m}{D} = \frac{0.306}{600}$ = 5.1 x 10^-4 m³

V = 510 cm³ ANS

(b) As solid replaces equal liquid so weight is equal then the mass = 0.306 kg density = 900kgm^-3

V =$\frac{m}{D} = \frac{0.306}{900}$ = 3.4 x 10^-4 m³

V = 340 cm³ ANS

11 The diameter of the piston of hydraulic press is 30cm. How much force is required to lift a car weighing 20000N on its piston if the diameter of the piston of the pump is 3cm?

Given Data: Diameter of piston of hydraulic press = d₁=30cm= 0.3m

Radius of piston of hydraulic press = r₁ = 0.15m

Cross-sectional areas of piston of hydraulic press = A = πr₁² = π(0.15)² = 0.0225π

Force on piston of hydraulic press = weight of car = F₁ = 20000N

Diameter of piston of pump= d₂ = 3cm = 0.03m

Radius of piston of pump = r₂ = 0.015m

Cross-sectional areas of piston of pump = a= πr₂² = π(0.015)² = 0.000225π

To find: Force on small piston = Force required to lift a car = F₂ =?

Solution: According to Pascal Law the pressure on both pistons will be same ⇒ P₁ = P₂

$\frac{F_1}{a} = \frac{F_2}{A}$

⇒ F₂ = $\frac{F_1 x A}{a}$

F₂ = $\frac{20000 × 0.000225π}{0.0225π}$

F₂ = 20000 x 0.01

F₂ = 200N ANS

12 A steel wire of cross-sectional area 2x10^-5m² is stretched through 2mm by a force of 4000N. Find the Young’s modulus of het wire. The length of the wire is 2m.

Given Data: Cross sectional area of wire = A = 2.5x10^-5 m²

Change in its length = ∆L = 2mm = 0.002m

Force applied = F = 4000N

Actual length of wire = L_o = 2m

To find: Young’s Modulus = Y =?

Solution: Y =$\frac{FL_o}{∆LA}$

Y = $\frac{4000 × 2}{0.002 × 2 × 10^{-5}}$

Y = 2x10¹¹Nm^-2 ANS