CHAPTER -2
NUMERICALS
1. The density of air is 1.28 g/litre⁻¹. Express it in:
(a) g/cm³ (b) kg/m³
Answer:
(a) Density in g/cm³:
Density in g/cm³ = 1.28 g/litre × (1 litre / 1000 cm³) = 0.00128 g/cm³
(b) Density in kg/m³:
Density in kg/m³ = 1.28 g/litre × (1 kg / 1000 g) × (1000 litre / 1 m³) = 1.28 kg/m³
2. The dimensions of a hall are 10 m × 7 m × 5 m. If the density of air is 1.1 kg/m³, find the
mass of air in the hall.
Answer:
Volume of hall = 10 m × 7 m × 5 m = 350 m³
Mass of air = Volume × Density = 350 m³ × 1.1 kg/m³ = 385 kg
3. The density of aluminium is 2.7 g/cm³. Express it in kg/m³.
Answer:
Density in kg/m³ = 2.7 g/cm³ × (1 kg / 1000 g) × (10^6 cm³ / 1 m³) = 2700 kg/m³
4. The density of alcohol is 600 kg/m³. Express it in g/cm³.
Answer:
Density in g/cm³ = 600 kg/m³ × (1000 g / 1 kg) × (1 m³ / 10^6 cm³) = 0.6 g/cm³
5. A piece of zinc of mass 438.6 g has a volume of 86 cm³. Calculate the density of zinc.
Answer:
Density = Mass / Volume = 438.6 g / 86 cm³ = 5.1 g/cm³
6. A piece of wood of mass 150 g has a volume of 200 cm³. Find the density of wood in:
(a) C.G.S. unit (b) S.I. unit.
Answer:
(a) Density in C.G.S. unit:
Density = Mass / Volume = 150 g / 200 cm³ = 0.75 g/cm³
(b) Density in S.I. unit:
Density = 0.75 g/cm³ × 1000 kg/m³ = 750 kg/m³
7. Calculate the volume of wood of mass 6000 kg if the density of wood is 0.8 g/cm³.
Answer:
Density in kg/m³ = 0.8 g/cm³ × 1000 = 800 kg/m³
Volume = Mass / Density = 6000 kg / 800 kg/m³ = 7.5 m³
8. Calculate the density of a solid from the following data:
(a) Mass of solid = 72 g
(b) Initial volume of water in measuring cylinder = 24 mL
(c) Final volume of water when solid is completely immersed in water = 42 mL
Answer:
Volume of solid = Final volume - Initial volume = 42 mL - 24 mL = 18 mL
Density = Mass / Volume = 72 g / 18 mL = 4 g/cm³
9. The mass of an empty density bottle is 21.8 g. When filled completely with water it is
41.8 g and when filled completely with liquid it is 40.6 g. Find:
(a) the volume of density bottle
(b) the density of liquid
(c) the relative density of liquid
Answer:
(a) Volume of density bottle:
Volume = Mass of water / Density of water
Mass of water = 41.8 g - 21.8 g = 20 g
Volume = 20 g / 1 g/cm³ = 20 cm³
(b) Density of liquid:
Mass of liquid = 40.6 g - 21.8 g = 18.8 g
Density = Mass / Volume = 18.8 g / 20 cm³ = 0.94 g/cm³
(c) Relative density of liquid:
Relative density = Density of liquid / Density of water = 0.94 g/cm³ / 1 g/cm³ = 0.94
10. From the following observations, calculate the density and relative density of a brine
solution:
Mass of empty density bottle = 22 g
Mass of bottle + water = 50 g
Mass of bottle + brine solution = 54 g
Answer:
(a) Density of brine solution:
Volume of bottle = Mass of water / Density of water
Mass of water = 50 g - 22 g = 28 g
Volume = 28 g / 1 g/cm³ = 28 cm³
Mass of brine solution = 54 g - 22 g = 32 g
Density = Mass / Volume = 32 g / 28 cm³ = 1.14 g/cm³
(b) Relative density of brine solution:
Relative density = Density of brine solution / Density of water = 1.14 g/cm³ / 1 g/cm³ =
1.14
11. The mass of an empty density bottle is 30 g. It is 75 g when filled completely with water
and 65 g when filled completely with a liquid. Find:
(a) Volume of density bottle
(b) Density of liquid
(c) Relative density of liquid
Answer:
(a) Volume of density bottle:
Volume = Mass of water / Density of water
Mass of water = 75 g - 30 g = 45 g
Volume = 45 g / 1 g/cm³ = 45 cm³
(b) Density of liquid:
Mass of liquid = 65 g - 30 g = 35 g
Density = Mass / Volume = 35 g / 45 cm³ = 0.77 g/cm³
(c) Relative density of liquid:
Relative density = Density of liquid / Density of water = 0.77 g/cm³ / 1 g/cm³ = 0.77
CHAPTER – 3
SHORT ANSWER TYPE:
1. Define the term moment of force.
Answer: Moment of force is the turning effect produced by a force applied at a distance
from a pivot point. It is calculated as the product of the force and the perpendicular
distance from the pivot.
2. State its S.I. unit.
Answer: The S.I. unit of moment of force is Newton-meter (Nm).
3. State two effects of a force when applied on a rigid body.
Answer: The two effects of a force when applied on a rigid body are:
1. Translational motion.
2. Rotational motion.
4. How does the effect of a force differ when it is applied on a (a) rigid body, (b) non-rigid
body?
Answer:
(a) On a rigid body, a force can cause translation or rotation without changing the shape of
the body.
(b) On a non-rigid body, a force can cause deformation along with translation and
rotation.
5. State the effect of force F in each of the following diagrams (a) and (b).
Answer:
(a) In diagram (a), force F causes a clockwise moment around the pivot.
(b) In diagram (b), force F causes an anticlockwise moment around the pivot.
6. Define the term thrust.
Answer: Thrust is a force applied perpendicular to the surface of an object.
7. State its unit.
Answer: The unit of thrust is Newton (N).
8. State the unit of pressure.
Answer: The unit of pressure is Pascal (Pa).
9. How does thrust differ from pressure?
Answer: Thrust is the total force acting perpendicular to a surface, whereas pressure is
the force per unit area exerted on a surface.
10. State two factors affecting the pressure exerted by a liquid.
Answer:
1. Depth of the liquid.
2. Density of the liquid.
11. Explain why it is easier to cut an apple with a sharp knife than with a blunt knife.
Answer: It is easier to cut an apple with a sharp knife because the sharp knife has a
smaller surface area at its edge, resulting in higher pressure for a given force, making it
easier to cut through the apple.
12. Explain why sleepers are used below the rails.
Answer: Sleepers are used below the rails to distribute the weight of the train over a larger
area, reducing the pressure on the ground and preventing the rails from sinking into the
ground.
13. Explain why cutting tools and machine parts are made of hard metals.
Answer: Cutting tools and machine parts are made of hard metals to withstand high
pressure and wear and tear, ensuring durability and effectiveness in cutting or machining
operations.
14. Describe a simple experiment to illustrate that liquid exerts pressure on the walls of the
container.
Answer: Fill a balloon with water and place it inside a jar with holes on the sides. As water
fills the balloon, it exerts pressure on the balloon walls, causing water to squirt out of the
holes, demonstrating that liquid exerts pressure on the container walls.
15. Explain why dams have broad walls at the bottom.
Answer: Dams have broad walls at the bottom to withstand the higher pressure exerted
by the water at greater depths, providing stability and preventing structural failure.
16. Explain why a wide steel belt is provided over the wheels of an army tank.
Answer: A wide steel belt is provided over the wheels of an army tank to distribute the
weight of the tank over a larger area, reducing the pressure on the ground and preventing
the tank from sinking into soft terrain.
17. Describe a simple experiment to illustrate that liquid pressure increases with depth.
Answer: Take a tall container filled with water and make three holes at different heights.
Observe that water squirts out with greater force from the hole at the lowest height,
illustrating that liquid pressure increases with depth.
18. How does pressure vary with depth in a liquid? Give reason.
Answer: Pressure in a liquid increases with depth due to the weight of the liquid above
exerting a force on the layers below. This is because deeper layers have to support the
weight of all the liquid above them.
19. Explain why the wall of a dam is made broader at the base than at the top.
Answer: The wall of a dam is made broader at the base to withstand the higher pressure
exerted by the water at greater depths, ensuring structural stability and preventing failure.
20. What do you mean by atmospheric pressure? Write the numerical value of the
atmospheric pressure on the earth's surface in Pascal.
Answer: Atmospheric pressure is the pressure exerted by the weight of the atmosphere.
The numerical value of atmospheric pressure on the earth's surface is approximately
101,325 Pa.
21. Explain why broad wooden sleepers are placed below railway tracks.
Answer: Broad wooden sleepers are placed below railway tracks to distribute the weight
of the train over a larger area, reducing the pressure on the ground and preventing the
tracks from sinking into the ground.
22. Explain why porters place a round piece of cloth on their heads when carrying heavy
loads.
Answer: Porters place a round piece of cloth on their heads to increase the area over
which the load is distributed, reducing the pressure on their heads and making it more
comfortable to carry heavy loads.
23. State two factors affecting atmospheric pressure.
Answer:
1. Altitude (height above sea level).
2. Temperature.
24. Why do mountaineers carry oxygen cylinders with them?
Answer: Mountaineers carry oxygen cylinders because atmospheric pressure and oxygen
levels decrease with altitude, making it difficult to breathe at higher elevations.
25. Why do your ears feel uncomfortable when you take off in an airplane?
Answer: Ears feel uncomfortable during airplane takeoff due to rapid changes in air
pressure, causing an imbalance between the external pressure and the pressure inside the
ear, leading to discomfort.
26. Write the numerical value of the atmospheric pressure on the earth's surface in Pascal.
Answer: The numerical value of atmospheric pressure on the earth's surface is
approximately 101,325 Pa.
LONG ANSWER TYPE:-
1. Explain the following:
(a) The spanner (or wrench) has a long handle.
Answer: A spanner has a long handle to increase the perpendicular distance from the
pivot point, which increases the moment of force, making it easier to turn a nut or bolt with
less effort.
(b) The steering wheel of a vehicle is of large diameter.
Answer: The steering wheel of a vehicle is of large diameter to increase the distance
from the center (pivot point), allowing the driver to apply less force to achieve the
necessary moment to turn the wheels of the vehicle.
(c) The hand flour grinder is provided with a handle near its rim.
Answer: The hand flour grinder is provided with a handle near its rim to increase the
distance from the axis of rotation, thereby increasing the moment of force and making it
easier to grind flour with less effort.
(d) It is easier to open a door by pushing it at its free end.
Answer: It is easier to open a door by pushing it at its free end because the distance
from the hinges (pivot) is maximized, increasing the moment of force and requiring less
effort to open the door.
(e) It is easier to turn the steering wheel by applying a force through the stick near the rim
of the wheel.
Answer: It is easier to turn the steering wheel by applying a force near the rim because
the increased distance from the center of rotation increases the moment of force, making it
easier to turn the wheel with less effort.
2. Explain the following:
(a) Sleepers are used below the rails.
Answer: Sleepers are used below the rails to distribute the weight of the train over a larger
area, reducing the pressure on the ground and preventing the rails from sinking.
(b) A tall building has wide foundations.
Answer: A tall building has wide foundations to spread the load of the building over a
larger area, reducing the pressure on the ground and ensuring the stability of the structure.
3. Describe an experiment to show that a liquid exerts pressure at the bottom of the
container in which it is kept.
Answer: Take a container with a small hole at the bottom, fill it with water, and observe
that water squirts out from the hole. This demonstrates that the liquid exerts pressure at
the bottom of the container.
4. Describe a suitable experiment to demonstrate that a liquid exerts pressure sideways
also.
Answer: Take a container with holes at different heights on the sides, fill it with water, and
observe that water squirts out from the side holes. This shows that liquid exerts pressure
sideways.
5. Describe an experiment to show that the liquid pressure at a point increases with the
increase in height of the liquid column above that point.
Answer: Take a tall container with holes at different heights, fill it with water, and observe
that water squirts out with more force from the lower holes. This demonstrates that liquid
pressure increases with depth.
6. Describe a simple experiment to show that at a given depth, a liquid exerts same
pressure in all directions.
Answer: Submerge a balloon in water at a certain depth and observe that it expands
uniformly. This indicates that the liquid exerts the same pressure in all directions at a given
depth.
7. Describe an experiment to show that liquid pressure depends on the density of liquid.
Answer: Take two containers with holes at the same height, fill one with water and the
other with oil, and observe the difference in the force of the liquid squirting out. The denser
liquid (water) will squirt out with more force, showing that liquid pressure depends on
density.
8. Describe a simple experiment to illustrate that air exerts pressure.
Answer: Inflate a balloon and observe that it expands. This shows that the air inside the
balloon exerts pressure on the balloon walls.
9. Describe the crushing tin can experiment. What do you conclude from this experiment?
Answer: Heat a small amount of water in a tin can until it produces steam, then quickly
seal the can and place it in cold water. The can will crush inward due to the higher
atmospheric pressure outside. This demonstrates that air exerts pressure.
10. Give reasons for the following:
(a) A balloon collapses when air is removed from it.
Answer: A balloon collapses when air is removed because the external atmospheric
pressure becomes greater than the pressure inside, causing it to collapse.
(b) Water does not run out of a dropper unless its rubber bulb is pressed.
Answer: Water does not run out of a dropper unless its rubber bulb is pressed because
pressing the bulb increases the pressure inside, forcing the water out.
(c) Two holes are made in a sealed oil tin to take out oil from it.
Answer: Two holes are made in a sealed oil tin to allow air to enter through one hole while
oil flows out through the other, preventing a vacuum and ensuring a smooth flow.
11. How does the atmospheric pressure change with altitude?
Answer: Atmospheric pressure decreases with increasing altitude because the density of
air decreases, resulting in fewer air molecules exerting pressure.
Think and answer
1. In Fig. 3.33 a force F is applied in a direction passing through the pivoted point O of the
body. Will the body rotate? Give reason to support your answer.
Answer: The body will not rotate because the force is applied along the line passing
through the pivot point, resulting in zero moment of force.
2. When does a man exert more pressure on the floor: while standing or while walking?
Answer: A man exerts more pressure on the floor while walking because, at certain points,
all his weight is concentrated on a smaller area (e.g., one foot), compared to standing
where the weight is more evenly distributed.
3. Why do camels or elephants have broad feet?
Answer: Camels and elephants have broad feet to distribute their body weight over a
larger area, reducing the pressure on the ground and preventing them from sinking into soft
surfaces like sand.
1. A sharp pin works better than a blunt pin. Explain the reason.
Answer: A sharp pin works better than a blunt pin because it has a smaller surface area at
the tip, leading to higher pressure when force is applied, making it easier to penetrate
surfaces.
2. Which fact about liquid pressure does the diagram in Fig. 3.34 illustrate?
Answer: The diagram in Fig. 3.34 illustrates that liquid pressure increases with depth.
3. A dam has broader walls at the bottom than at the top. Give a reason.
Answer: A dam has broader walls at the bottom to withstand the higher pressure exerted
by water at greater depths.
4. Sea-divers wear specially designed swim-suits. Why?
Answer: Sea-divers wear specially designed swim-suits to protect themselves from
increased pressure at greater depths, providing insulation and reducing pressure-related
injuries.
Numericals
1. Find the moment of force of 20 N about an axis of rotation at a distance of 0.5 m from the
force.
Answer: Moment of force = Force × Distance = 20 N × 0.5 m = 10 Nm
2. The moment of a force of 25 N about a point is 2.5 Nm. Find the perpendicular distance
of force from that point.
Answer: Distance = Moment of force / Force = 2.5 Nm / 25 N = 0.1 m
3. A spanner of length 10 cm is used to turn a nut by applying a minimum force of 50 N.
Calculate the moment of force required.
Answer: Distance in meters = 10 cm = 0.1 m
Moment of force = Force × Distance = 50 N × 0.1 m = 5 Nm
4. A wheel of diameter 2 m can be rotated about an axis passing through its center by a
moment of force equal to 2.0 Nm. What minimum force must be applied on its rim?
Answer: Radius = Diameter / 2 = 2 m / 2 = 1 m
Force = Moment of force / Radius = 2 Nm / 1 m = 2 N
5. A normal force of 200 N acts on an area of 0.02 m². Find the pressure in Pascal.
Answer: Pressure = Force / Area = 200 N / 0.02 m² = 10,000 Pa
6. Find the thrust required to exert a pressure of 50,000 Pascal on an area of 0.05 m².
Answer: Thrust = Pressure × Area = 50,000 Pa × 0.05 m² = 2,500 N
7. Find the area of a body which experiences a pressure of 50,000 Pa by a thrust of 100 N.
Answer: Area = Thrust / Pressure = 100 N / 50,000 Pa = 0.002 m²
8. Calculate the pressure in Pascal exerted by a force of 300 N acting normally on an area of
30 cm².
Answer: Area in m² = 30 cm² × 10^(-4) m²/cm² = 0.003 m²
Pressure = Force / Area = 300 N / 0.003 m² = 100,000 Pa
9. How much thrust will be required to exert a pressure of 200,000 Pa on an area of 1 cm²?
Answer: Area in m² = 1 cm² × 10^(-4) m²/cm² = 0.0001 m²
Thrust = Pressure × Area = 200,000 Pa × 0.0001 m² = 20 N
10. The base of a container measures 15 cm × 20 cm. It is placed on a table top. If the
weight of the container is 60 N, what will be the pressure exerted by the container on the
table top?
Answer: Area in m² = 15 cm × 20 cm × 10^(-4) m²/cm² = 0.03 m²
Pressure = Force / Area = 60 N / 0.03 m² = 2,000 Pa
11. Calculate the pressure exerted on an area of 0.5 m² by a thrust of 100 kgf.
Answer: 1 kgf = 9.8 N
Thrust in N = 100 kgf × 9.8 N/kgf = 980 N
Pressure = Thrust / Area = 980 N / 0.5 m² = 1,960 Pa
12. A boy weighing 60 kg stands on a platform of dimensions 2.5 cm × 5 cm. What will be
the pressure in Pascal does he exert on the platform?
Answer: Area in m² = 2.5 cm × 5 cm × 10^(-4) m²/cm² = 0.00125 m²
Weight in N = 60 kg × 9.8 N/kg = 588 N
Pressure = Force / Area = 588 N / 0.00125 m² = 470,400 Pa
