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Subject- Physics. Class-VII

PHYSICAL QUANTITIES AND MEASUREMENT. Chapter -1

Stop and check

1. area 2. cubic metre (m3) 3. capacity 4. displaces 5. 25 cm3

6. measuring jars and measuring cylinders

Stop and check

1. closely or densely 2. mass 3. greater 4. 0.001 g/cm3 5. distance travelled

6. m/s (metre/s)

Checkpoint

A. 1. c, 4 cm 2. d, 1/100,000 m3 3. b, mass / volume 4. a, 2500 kg/m3

5. b, 600 kg/m3

6. d, ρ 7. b, kg/m3

B. 1. area 2. displacement 3. 1000 4. capacity 5. mass,

volume 6. kg/m3

7. lesser 8. km/h

C. 1. iv 2. i 3. ii 4. vi 5. iii 6. v

D. 1. False 2. False 3. True 4. True

E. 1. Keep the object on a graph sheet and outline its boundary using a pencil. Remove the object and

count the number of squares fully covered by the outline. Now count all the squares for which more

than half lie inside the outline and ignore the others. Add the two numbers. This will be

approximately equal to the area of the object measured in sq.cm. (cm2)

2. The volume of an irregular object can be measured by using displacement method.

3. In the laboratory, the volumes of liquids are measured using measuring flasks, measuring

cylinders

and measuring beakers.

4. The correct way of taking readings using a measuring cylinder is without parallax error. For this

our

eye level must be at the same level as the liquid meniscus.

5. Matter is packed differently in different substances. In some materials, matter is packed very

densely. Such substances weigh more than other substances in which matter is packed more

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loosely.

6. Since the density of steel is large, 100 kg of steel will occupy much less space when compared to

100 kg of cotton. In cotton, matter is packed loosely. Hence it is easy to store steel than cotton.

7. The SI unit of density is kg/m3 while the smaller, commonly used unit is g/cm3.

1 kg/m3 = 0.001 g/cm3 and 1 g/cm3 = 1000 kg/m3.

8. When we say that the density of copper is 8.9 g/cm3, we understand that one cubic centimetre

of

copper weighs 8.9 g.

9. Speed is defined as the distance travelled by an object in unit time. The SI unit for speed is

metre/

second (m/s).

F. 1. Take sufficient water in a measuring cylinder such that the stone will be fully immersed without

spilling water. Note the water level in the cylinder. Slowly immerse the stone without splaying

water. Water level in the cylinder will rise. Note the reading corresponding to the new level. The

difference in reading will give the volume of the stone in cubic centimetre.

2. The commonly used devices to measure volumes of liquids in the laboratory are measuring

flasks,

measuring cylinders and measuring beakers.

i. A measuring flask is a glass jar with a narrow neck and a mark on the neck indicating

the level

up to which liquids must be filled. Flasks of different capacities are used to measure different

volumes in millilitre.

ii. The measuring cylinder is a tall cylinder made of glass or plastic. It has markings in

millilitre for

different volumes. This is also available in different capacities.

iii. Measuring beakers are wider and shorter versions of measuring cylinders and

graduated in

steps of 25 ml. The beaker is used when precise measurements are not required.

3. Take an empty beaker and weigh it. Let its mass be m . Pour the liquid into it and weigh it again.

Let 1

the mass be m2 this time. The mass of the liquid can be obtained as m = m2 −

m1. Now pour the

liquid

into a measuring cylinder and note down its volume as V ml. The density of the liquid is given by

m

= g/cm3.

V

distance (d) d

ρ

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4. Speed is the rate at which a body covers distance. Thus it is defined as = = . On cross

time (t) t

d

multiplication, we get, the distance travelled in time t with speed v is d = v × t. Similarly, t = .

These are the different relations between d, v and t v

G. 1. Length of the box l = 80 cm = 0.8 m

Breadth of the box b = 50 cm = 0.5 m

Depth of the box d = 40 cm = 0.4 m

The volume of the box V = l × b × d = 0.8 × 0.5 × 0.4 m3 = 0.16 m3

2. The initial level of water h = 50 ml 1

The final level of water h = 54.5 ml 2

The difference in water level = volume of water displaced V = (54.5 – 50) ml = 4.5 ml

Since the volume of the stopper is equal to the volume of water displaced. Hence the volume of the

stopper is equal to 4.5 ml / 4.5 cm3.

3. Mass of the slab m = 48 kg

Volume of the slab V = 0.016 m3

m 48 kg

Hence the density ρ = = = 3000 kg/m3

V 0.016 m3

4. Density of aluminium ρ = 2710 kg/m3

The volume of the block V = 2 m × 1.5 m × 1 m = 3 m3

The mass of the block m = ρ × V = 2710 × 3 = 8130 kg

5. Mass of the brass block m = 340 kg Density of brass ρ = 8500 kg/m3

340

Volume of

the brass block V =

= 0.04

m3 8500

g

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6. Density of copper ρ = 8.96 = 8960 kg/m3

cm3

7. Capacity of beaker (volume of glycerin) V = 50 ml

Mass of empty beaker m = 27 g 1

Mass of beaker filled with glycerin m2 = 90 g

90 – 27 63 g

Density of glycerin ρ = = = 1.26 = 1260 kg/m3

50 50 cm3

8. Mass of iron sphere m = 390 g

Volume of displaced water = volume of iron sphere V = 5 cm3

390 g

Density of iron ρ = = 78 g/cm3

5 cm3

9. Distance covered d = 27 km = 27000 m

Time taken t = 90 minutes = 5400 s = 1.5 h.

27000 m m 27 km

Speed v = = 5 = = 18 km/h

5400 s s 1.5 h

10. Speed of motor cycle v = 20 m/s

Time of travel t = 4 h = 4 × 60 × 60 s = 14400 s

Distance travelled d = v × t = 20 × 14400 = 288000 m = 288 km

11. Convert the following

190

i. 190 cm3 = m3 = 0.00019 m3

1000000

ii. 10l = 10 × 1000 ml = 10000 ml

100

iii. 100 m = l = 0.1 l

1000

iv. 750 ml = 750 cm3

l

63

65

v. 65 = m3 = 0.065 m3

1000

750

vi. 750 m = 750 cm3 = m3 = 0.00075 m3

1000000

g vii. 600

= 600000 kg/m3

cm3

kg 60000 g

viii. 600 = = 0.6 g/cm3 m3 1000000 cm3

km 72000 m

ix. 72 = = 20 m/s

h 3600 s

m 45 km

x. 45 = ×

3600 = 162 km/m3 s 1000

m3

l

l

Think and answer

1. The volume of the cuboid V = 10 cm × 12 cm × 12 cm = 1440 cm3

The volume of the cubical cavity v = 5 cm × 5 cm × 5 cm = 125 cm3

The volume of wax in the block is V – v = 1440 – 125 = 1315 cm3

2. Yes. First put the sand in the measuring jar, shake the jar well for the sand to evenly fill the jar. Note

the reading corresponding to the surface of sand in the jar to get the volume of sand. Transfer

most of the sand to a paper leaving some in the jar. Now place the stone on the sand in the jar and

press it down well. Put back the sand on the paper to cover the stone. Once again shake mildly so

that sand fills the space in the jar evenly. Now measure the volume of sand and the stone. The

difference in the two readings will give the volume of the stone.

3. Distance of the school from Malini’s home d = 5 km Speed of the bus v = 20 km/h

d 5 1

Time taken to travel 5 km, t = = h = × 60 min = 15 min

v 20 4

If Malini starts at 7.40 a.m., she will reach school at 7.40 + 15 min = 7.55 a.m. So she will

reach school on time.

4. Since density of freshwater is 1 kg/m3, and that of seawater is 1.025 kg/m3, seawater is more denser

than fresh water. For this reason seawater can support heavier objects than freshwater and thus

floating is easier in seawater.

FORCE AND PRESSURE: MOTION. Chapter -2

Stop and check

1. position, fixed 2. rectilinear, curvilinear 3. revolution 4. periodic 5. random

Checkpoint

A. 1. b, a boy in a giant wheel 2. b, oscillatory 3. c, multiple 4. c, 75

m/s

B. 1. rest 2. curvilinear 3. curvilinear 4. stationary (or at rest) 5.

unequal 6. average

7. speed 8. circular / revolution 9. average speed 10. equal, equal

C. 1. False 2. False 3. False 4. False 5. False

D. 1. i. translatory ii. rotatory iii. vibratory iv. random v. rotatory vi. random vii. translatory

2. When an object moves to and fro along about a mean position, it is called vibratory motion.

The

motion of the pendulum of a pendulum clock is vibratory motion.

3. Irregular motion of objects which change direction of motion and speed frequently is known as

random motion. The motion of a bee flying from flower to flower in a garden is an example of

random motion.

4. When an object possesses more than one type of motion, it is said to have multiple motion. An

example is a ball rolling on the floor. It has both rotatory and translatory motion.

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5. Periodic motion is a repeated motion in which time taken for one complete cycle is fixed. Hence,

periodic motion must be repetitive. But repetitive motion need not happen at regular intervals.

For example consider a person swimming back and forth between the two ends of a pool. Since

he can

change his speed, the motion need not be periodic though repetitive.

6. Average speed is the ratio of the total distance travelled and the total time taken. Since speed

varies during non-uniform motion, the motion is represented by an average value such that the

total distance travelled can be expressed as product of average speed and the total time taken.

E. 1. Consider two people sitting on a moving train. They do not move with respect to each other and

hence are at rest with respect to each other. But for a person standing on the platform, they are

moving with the speed of train. From this we can see that rest and motion are relative.

2. When an object undergoes circular motion about an axis passing through its body, the motion is

called rotational motion or spin motion. An example is the rotation of the Earth around its axis. On

the other hand, if the circular motion is about an external axis, then it is called revolution. Example

of revolution is motion of the Earth around the Sun.

3. Periodic motion is a repetitive motion that repeats itself in definite intervals of time. The

rotation and revolution of the Earth are examples of periodic motion. A non-periodic motion is

also a repetitive motion, but does not repeat itself in definite interval of time. A ball bouncing of

the floor

and the motion of a swing are not periodic though repetitive.

4. When an object travels equal distances in equal intervals of time, it has a constant velocity. Then

its

motion is said to be uniform motion. A cyclist (vehicle) travelling along a straight path with constant

speed is in uniform motion. If the cyclist/vehicle changes his (its) direction or speed during the

motion, the motion is said to be non-uniform motion.

F. Numerical problems

1. The total distance travelled by the train d = 144 km

The total time taken t = 2h

total distance travelled 144

The average speed v = = = 72 km/h

total time taken 2

km 72 × 1000 m

72 = = 20 m/s

h 60 × 60s

2. The total distance travelled by the sprinter d = 50 + 25 + 25 = 100 m The total time

taken t = 8 + 3 + 2 = 13s

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total distance travelled 100

The average speed v = = = 7.69 m/s

total time taken 13

3. The one way distance d = 900 m

km 6000 m

Speed of onward journey v = 6 = = 1.67 m/s

h 3600 s

900

Time taken for onward journey t = = 539s = 9 min 1

1.67

km 9000 m

Speed of return journey v = 9 = = 2.5 m/s

h 3600 s

900

Time taken for return journey t = = 360s = 6 min 2

2.5

The total time taken for the complete trip t = 9 + 6 = 15 min

4. The total distance travelled by the car d = 35 + 25 + 10 = 70 km

Let the time taken for the last part of the journey is t min 3

The total time taken to travel 70 km t = (30 + 20 + t3) min

total distance travelled km 70 km

The average speed v = = 70 =

total time taken h (30 + 20 + t ) min 3

So (30 + 20 + t3) min = 1 h. Hence t3 = 10 min

20 1

5. The speed of the cyclist for 20 min = h = h = 30 km/h

60 3

1 1

So distance travelled in h, d = × 30 km = 10 km 1

3 3

40 2

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The speed of the cyclist for the next 40 min = h = h = 15 km/h

60 3

2 2

So distance travelled in h, d = × 15 km = 10 km 2

3 3

The total distance travelled by the cyclist is d = d1 + d2 = 10 + 10 = 20 km

total distance travelled 20 km

The average speed v = = = 20 km/h

1 2

h

3 3

6. The speed of the car during onward journey v = 50 km/h 1

If d is the distance between Chennai and Pondicherry, the time taken for the onward journey

d

t1 = h

50

The speed of the car during return journey v = 60 km/h 2

d

Since the distance travelled the same both ways, the time taken for return journey t = h

2 60

d d

So the total distance travelled is 2d and the total time taken t = t1 + t2 = + h

50 60

total distance travelled

The average speed v =

total time taken

2d 2d 2d × 3000 6000

= = = =

= 54.55 km/h

d d 60d + 50d

50 60 60 × 50

total time taken +

+

110 d 110

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WEIGHT

Checkpoint

A. 1. b, weight only 2. d, newton

B. 1. weight 2. zero 3. 4 kg 4. 1 kg

C. 1. False 2. False

D. 1. Weight of an object on the Earth is the force with which the Earth attracts the object towards its

centre. The unit of weight is newton.

2. A can of water weighing 42 newton on the Earth weighs only 7 newton on the Moon because

the

1 1

gravitational force of the moon is only th of that of the Earth. th of 42 newton is 7 newton.

6 6

3. Beam balance measures weight of an object by comparing it with standard, known weights. For

balancing the beam balance, equal weights are to be placed on both the pans. Hence the

standard

weight to be used to balance the beam is 1 kg.

E. 1. Mass is the amount of substance contained in a body. Its unit in SI system is kg. It does not have any

direction. On the other hand, weight is the force with which the Earth attracts objects near it. Its

unit is newton or kgf. Since the force always acts towards the centre of the Earth, it has a direction.

While mass is an intrinsic property of a substance and hence a constant, weight varies from place to

place as it depends on the strength of the gravitational force.

2. A spring expands when it is pulled. The expansion is larger when we pull with a larger force. The

spring balance works on this principle. It has a spring with an attached pointer that moves on a

scale and a hook to hang weights to be weighed. Depending on the weight, the spring stretches

moving the pointer over the scale. The scale reading corresponding to the pointer will give the

weight of the

object.

F. Numerical problems

1. The weight of the 30 kg mass on earth is W = 30 × 10 N = 300 N earth

1

On the moon, the weight of the object is only th of its weight on the earth. Hence, the weight of

6

1

the object on the moon is W = × 300 = 50 N moon

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6

2. Weight of the 9 kg mass on moon W = 15 N moon

Weight of the body on earth W = 9 × 10 N = 90 N earth

It is also 6 times the weight on the moon. Thus Wearth = 6 × 15 N = 90 N

Its mass remains unchanged in earth and moon. Hence its mass is 9 kg on earth.

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Think and answer

1. Displacement of an object is the straight line distance between the starting point and end point.

Since the particle comes back to the starting point, the displacement is zero.

2. Random motion can neither be periodic nor uniform. By definition, an object is said to be in random

motion when it changes direction and speed frequently in different intervals of time. Thus the

speed is neither constant nor the motion is repeated.

3. Circular motion can be uniform motion because it is possible for the object to move around a circle

with constant speed. But it can be non-uniform also in the general case.

4. Using the meter scale measure the length of one of its side as L cm. From this the volume of the

cube can be found as V = L3cm3. From the data table, the density of iron, ρ = 7.7 g/cm3. But,

density

mass g is ρ = and hence mass = ρ × volume = 7.7 × L3 cm3 =

7.7 L3g.

volume cm3

ENERGY. Chapter -3

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Stop and check

1. Kinetic energy 2. Potential energy 3. Magnetic energy 4. Chemical energy

5. Nuclear energy

Checkpoint

A. 1. c, joule 2. a, kinetic, potential 3. b, light 4. d, chemical 5.

b, lowest

B. 1. motion 2. energy 3. kinetic 4. mass and velocity 5. double

6. kinetic

C. 1. False 2. False 3. False 4. False 5. True 6. True 7. True

D. 1. Work is the product of the force acting on an object and the displacement of the object along the

direction of the force, while energy is the ability to do work. Both work and energy are

interconvertible and has the same unit, joule (J)

2. Potential energy and kinetic energy are two different forms of mechanical energy. Potential energy

of an object is due to its position and state (configuration), while kinetic energy is due to its motion.

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E. 1. In scientific terms, work is said to be done if the force produces motion in the direction of the force.

No work is done if there is no motion.

2. Energy is the ability of an object to do work.

3. Mechanical work is measured as the product of the force acting on an object and the

displacement

of the object along the direction of the force.

4. The principle of conservation of energy states that energy can neither be created nor be destroyed.

It can undergo a change from one form of energy to another.

5. Example for energy transformation:

i. Fan does mechanical work in rotating using electrical energy.

ii. In a hydel power station, turbines are rotated using the potential energy of stored water to

turn

turbines to generate electrical energy, thus changing mechanical energy to electrical energy.

iii. In a loudspeaker, electrical energy is transformed into sound energy.

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iv. While burning a fuel, the fuel undergoes chemical change during burning. The energy

released is

converted to heat energy.

v. In a solar cell (or in a photoelectric cell), light is absorbed and its energy is converted to electrical

energy.

F. 1. The SI system unit for energy is joule (J). A lower unit of joule is erg (1 erg = 10–7 J) and a higher unit

is the calorie (1 cal = 4.2 J)

2. i. Mechanical energy: It can replace human labour. For example, instead of drawing water from a

well, we can make use of a water pump to pull water for us. Similarly, the potential energy

stored in a wound spring is used in moving the needles of a clock.

ii. Light energy: This is the source of all energy on the Earth and plants convert this into chemical

energy stored in the food. When the food we eat is digested, this chemical energy is released

and converted to muscle energy.

iii. Thermal energy: This form of energy is stored in all fuels. When the fuel burns, thermal

energy

is released. This energy is used to generate electricity.

iv. Electrical energy: This form of energy is due to the motion of charged particles. It has multitude

of application like running the fan, AC, refrigerator and similar appliances.

3. Potential energy is the energy possessed by an object due to its position of a particular state (configuration). An example is the gravitational potential energy. It is due to the position of an

object and depends on its height from the surface of the Earth. The larger the height, the

larger the potential energy. A second example is the potential energy stored in a wound

spring. It is due to the

particular state (wound state) of the spring.

4. Kinetic energy is the energy possessed by a body by virtue of its motion. It depends on the

mass and

velocity of the object. A cricket ball bowled with a high speed has a large amount of kinetic energy.

If it hits anyone, the person may get hurt.

5. i. Mechanical energy → magnetic energy → light energy ii. Potential energy of

stored water → mechanical energy of the turbine → electrical energy →

mechanical energy

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iii. Thermal energy → mechanical energy

iv. Muscle energy → chemical energy → light energy

v. Muscle energy → mechanical energy

6. The law of conservation of energy states that energy can neither be created nor destroyed.

Energy can only be changed from one form to another. For example, a swinging pendulum has zero

potential energy at its equilibrium position and all its energy is kinetic. On the other hand, at the

end points (turning points), its kinetic energy is zero and the potential energy is a maximum. The

maximum potential energy at the end points is equal to the maximum kinetic energy at the

equilibrium point. At any intermediate position, it will have some kinetic energy and some potential

energy such that their sum is equal to the maximum kinetic energy or potential energy.

G. Numerical problems

1. The potential energy of 24 kg mass at a height of 12 m is mgh = 24 × 10 × 12 = 2880 J

The potential energy of 18 kg mass at a height of 22 m is mgh = 18 × 10 × 22 =

3960 J Hence the mass of 18 kg has a higher potential energy.

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1 1

2. The kinetic energy of the ball of mass 160 g is mv2 = × 0.16 × 8 × 8 = 5.12 J. Since total energy is

2 2

conserved, the maximum potential energy at the highest point will be equal to the maximum

kinetic

5.12

energy with which it was thrown up. Hence, mgh = 0.16 × 10 × h = 5.12. ∴ h = = 3.2 m

0.16 × 10

Think and answer

1. When we do physical work heat energy is produced in the body through metabolism of food to provide

energy to the muscles. We sweat to keep our body cool for the smooth functioning of the internal

organs by evaporation of water. We breathe faster because we require more oxygen to

carry out the increased metabolism rate.

2. When we bring the similar poles of two magnets we have to do mechanical work against the repulsive

force between similar poles. This work done will be stored as the magnetic potential energy. Thus when

the two poles are closer, the magnetic potential energy is more. When we release them they move

away from each other to minimise the magnetic potential energy. (Optional: The work we did to bring

them closer will be spent on moving away from each other,

conserving the total energy of the system.)

3. When the iron nail is rubbed against a rough surface, the mechanical work we do is converted to

the kinetic energy of the nail which in turn is converted to heat energy due to friction between the

two surfaces. The principle operating behind this process is conservation of energy.