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CLASS-VIII
MPC BRIDGE COURSE
PHYSICS
DAY-1 : SYNOPSIS
Physical Quantity : The quantities which
are measurable in physics are called
physical quantities.
Ex : Length, mass, time, speed, etc.
Unit : Unit is a standard which is used
for the measurement of physical quantity.
Ex : (i) unit of length is metre
(ii) unit of time is second
Numerical value of physical quantity :
The number of times a unit is present in
a given physical quantity is called
Numerical value of physical quantity.
Relation among Physical quantity,
Numerical value and Unit :Physical
quantity = Numerical value × Unit
Ex : Let length of table = 3 metre
Here 3 is the Numerical value and metre
is the standard unit.
Fundamental quantities : The physical
quantity which does not depend upon (or
independent of) other physical quantities
are called fundamental quantities.
Ex : Length, mass and time etc.
Derived physica l quantities : The
physical quantities which depends on
fundamental physical quantities are
called derived physical quantities.
Ex : Area, Volume, Speed, etc.
Fundamental units : The units used for
measuring fundamental quantities are
called Fundamental units. These are
independent of other units.
Ex: The fundamental unit of length is
metre.
Derived units : Derived units are the
units of derived physical quantities which
are expressed in terms of fundamental
units.
Ex: The derived unit of speed is ms –1 (
read as metre per second).
Systems of Units : There are four system
of units:
NARAYANA GROUP OF SCHOOLS
i. British or F.P.S. system : In F.P.S.
system, the unit of length is foot.
The unit of mass is pound. The unit of
time is second.
ii. French or C.G.S system : In C.G.S
system, the unit of length is centimetre.
The unit of mass is gram. The unit of
time is second.
iii. M.K.S system or metric system : In
M.K.S system, the unit of length is
metre.
The unit of mass is kilogram.The unit of
time is second.
iv. International system or S.I : S.I.
system has seven basic units and two
supplementary units.
Quantity
length
mass
time
temperature
luminous intensity
electric current
amount of
substance
Plane angle
solid angle
Unit
Symbol
metre
kilogram
second
kelvin
candela
ampere
mole
m
kg
s
K
cd
A
mol
Radian
steradian
rad
sr
Multiple and sub multiple factors :
Multiple and sub multiple factors
Multiplication factor Name
1012
109
106
103
102
10
10-1
10-2
10-3
10-6
10-9
10-12
10-15
10-18
tera
giga
mega
kilo
hecto
deca
deci
centi
milli
micro
nano
pico
femto
atto
Symbol
T
G
M
k
h
da
d
c
m

n
p
f
a
44
CLASS-VIII
DAY-1: WORKSHEET
1. Pick the odd man out.
1] Length
2] Mass
3] Time
4] Area
2. F.P.S stands for
1] Foot, pound, second
2] France, Paris, Spain
3] Force, pressure, second
4] Foot, Pace, Second
3. C.G.S stands for
1] Centimetre, Gravitation, second
2] Centisecond, gram, second
3] Centimetre, gram, second
4] None of these
4. Ampere is the unit of
1] Length
2] Temperature
3] Luminous intensity
4] Current
5. Number of fundamental physical
quantities in M.K.S system are
1] Two
2] Three
3] Seven 4] Six
6. The temperature standard metre rod
made of platinum - Iridium alloy kept in
the archives of serves near Paris is
1] 0°C
2] 27°C 3] 100° C 4] None
7. Statement I : Numerical value = Physical
quantity × unit
Statement II : Current strength is a
fundamental physical quantity according
to S.I system.
1) Statement I & II is true
2) Statement-I is true; Statement II is false.
3) Statement I is false ; Statement II is true.
4) Statement I is false ; Statement II is false.
8. Observe the following
a) Length
b) Mass
c) Current strength d) Temperature
pick the correct statement
1] Length is the odd man out.
2]All are Fundamental Physical
Quantities according to M.K.S system.
3]All are Fundamental Physical
Quantities according to S.I system.
4] Current strength is a derived quantity.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
9. Match the following.
List - A
List - B
(a) Temperature
(e) Mole
(b) Luminous intensity (f) Kelvin
(c) Amount of substance (g) Candela
1] a  e; b  f; c  g
2] a  f; b  g; c  e
3] a  g; b  f; c  e
4] a  g; b  e; c  f
10.1cm2 = _________m2
1) 104
2) 10–2
3) 10–3
4) 10–4
11. 1litre = ____cm3
1) 100
2) 1000 3) 500
4) 10
12. Match the following.
List A
List B
1
i) 365
days
a) 1 Century
4
ii) 10 decades
b) 3600 sec
1
iii)
the part of a mean solar day
1440
c) 1 year
iv) 1 hour
d) 1 minute
1) i  a, ii  b, iii  c, iv  d
2) i  a, ii  b, iii  d, iv  c
3) i  b, ii  a, iii  c, iv  d
4) i  c, ii  a, iii  d, iv  b
DAY-2 : SYNOPSIS
Density : The density of a substance is
defined as the mass per unit volume of
the substance.
Density of a substance (d)
mass of the subs tance(M)
M
= volume of the subs tance(V)  d  V
Unit : C.G.S unit : g cm–3
S.I. unit : kg m–3
Relationship between S.I. & C.G.S units
of density : 1g/cm3 = 1000 kg/m3
Relative density : It is the ratio of the
density of the substance to the density
of water at 40C.
Thus, Relative density (R.D.)

density the subs tance
density of water at 40 C
Units of relative density : As the relative
density is the ratio of two similar
quantities hence it is a pure number
and therefore has no units.
45
CLASS-VIII
MPC BRIDGE COURSE
Relationship between density and relative
density :
1) Density of a solid in S.I. Unit = R.D. of
the solid × Density of water (in S.I. Unit)=
R.D. of the solid × 1000 kg/m3
2) Density of a solid in C.G.S.unit = R.D. of
the solid × Density of water (in C.G.S.
unit) = R.D. of the solid × 1g/cm3
DAY-2: WORKSHEET
1. 5 litres of alcohol has a mass of 4 kg. The
density of alcohol is ___
kg
kg
1) 0.8
2) 80
m3
m3
kg
kg
3) 800
4) 8000
m3
m3
2. A 50kg mass of a body is immersed in
water then we observe the mark on
measuring cylinder as 70cm3. If the body
is taken out from the cylinder, the mark
observed on cylinder is 50cm3. Then the
density of that body is [The graduations
marked in cm3]
1) 2500 g/cm3
2) 250 g/cm3
3) 2.5 g/cm3
4)0.025×103g/cm3
3. The density of lead is 11.6
of wood is 800
kg
g
cm3
and that
. what do you
m3
understand by these statements ?
1) The matter is only densely packed in
lead than wood.
2) The matter is only densely packed in
wood than lead.
3) The density of lead is greater than the
density of wood.
4) Both (1) and (3).
4. Take two identical 100cm3 beakers. Fill
one beaker completely with water and
the other with kerosene oil. Place the
beakers in the scale pans of an ordinary
beam balance as shown in fig. It is
observed that the beaker filled with water
have more mass than the beaker filled
with kerosene oil. Because
NARAYANA GROUP OF SCHOOLS
(1) the matter in water is more densely
packed than in kerosene oil.
(2) the matter in kerosene oil is more
densely packed than in water.
(3) Water and kerosene oil are occupied
the same volume.
(4) None of these.
5. Mass of liquid = 72 g
Initial volume of water in measuring
cylinder = 24 cm3
Final volume of water + solid in
measuring cylinder = 42 cm2
From the above data the density of solid
is ___________
1) 4000 kg/m3
2) 3.0 g/cm3
72 g
4) 4.0 kg/m3
3
42 cm
6. If relative density of gold is 19.3. Then
the density of gold is __________times
greater than the density of water.
1) 8.9
2) 19.3
3) 1.29
4) 0.8
7. The density of copper is 8.9 × 103 kg/m3.
The relative density of copper is____
1) 8.9 × 103
2) 8.9 × 102
3) 8.9 × 101
4) 8.9 × 100
8. If the length of the wooden cube is 4m
3)
and the mass is
1
kg, then the relative
8
density of the wooden cube is __________
1
1
1
3)
4)
512
51200
32
9. An iron cylinder of radius 1.4 cm and
length 8 cm is found to weigh 369.6 g.
The relative density of iron cylinder is
1) 19.53 × 10–7 2)
Take volume of cylinder = r 2 .
1) 7.2
2) 7.5
3) 8
4) 8.2
10. Calculate the mass of a body whose
volume is 2m3 and relative density is 0.52
1) 1040 kg 2) 1000 kg 3) 950 kg 4) 750 kg
46
CLASS-VIII
MPC BRIDGE COURSE
DAY-3 : SYNOPSIS
DAY-3: WORKSHEET
Scalars:- The physical quantities which
have only magnitude but no direction, are
called scalar quantities or simply called
as scalars.
1.The physcial quantities which have only
magni tude but no direction are called--1)Vectors
2)Scalars
3)both (1) and (2) 4)neither 1 nor 2
2. The physcial quantities which are expressed in magnitude as well as direction are called
1)Scalars
2)Units
3)Vectors
4)both 1 and 2
3. Ordinary laws of algebra are applicable
for
1)Scalars
2)Vectors
3)both 1 and 2
4)neither 1 nor 2
4. _____ cannot be added, subtracted and
multiplied by ordinary laws of algebra.
1)Scalars
2)Numbers
3)algebraic equations 4)Vectors
5. Graphically a vector is represented by a
1)line
2)curve
3)circle
4)directed line segment
6. Directionless quantity is called ______
1)vector
2)Scalar
3)both 1 and 2
4)neither 1 nor 2
7. The magnitude of a vector is
1)Scalar 2)vector 3)direction 4)can’t say
8. Length of the directed line segment represents
1)Direction
2)Orientation
3)spin
4)Magnitude
9. Arrow head of a vector represents
1)Magnitude
2)sense of direction
3)sense of rotation 4)spin
10. Mass is a
1)Scalar quantity 2)vector quantity
3)both 1 and 2
4)Neither 1 nor 2
11. Density is a ______
1)scalar quantity 2)vector quantity
3)both 1 and 2
4)neither 1 nor 2
12. Mr.Satish travelled from Narayana
Olympiad school, Narayanaguda to
Dilsukhnagar 10km towards North. Displacement of Mr.Satish is a
1)Scalar
2)Vector
3)has only magnitude4)has only direction
Examples of scalar quantities:- 2 kg sugar
tells about the magnitude of its mass,
but has no direction.
Mass, length, time, distance covered,
temperature, area, volume, density, temperature etc are a few examples of scalars. The scalars can be added, subtracted, multiplied and divided by ordinary laws of algebra.
A scalar is specified by mere number and
unit, where number represents its magnitude.
A scalar may be positive or negative.A scalar can be represented by a single letter.
Vectors:- The physical quantities which are
expressed in magnitude as well as direction are called vector quantities or
simply called as a vectors. They should
also obey law of vector addition.
Examples of vector quantities:Displacement, velocity, acceleration,
force etc. are a few examples of vectors.
Vectors cannot be added, subtracted, and
multiplied by ordinary laws of algebra.
A vector in writing, can be represented
either by a single letter in bold face or by
a single letter with an arrow head on
it.Diplacement = S
Geometrically or graphically, a vector is
represented by a straight line with an
arrow head i.e. arrowed line. Here the
length of the arrowed the line drawn on
a suitable scale represents the magnitude and the arrow head represents the
direction of the given vector. For example,
when an object goes on the path ABC,
then the displacement of the object is
.The arrow head at C shows that the object is displaced from A to C.
NARAYANA GROUP OF SCHOOLS
47
CLASS-VIII
MPC BRIDGE COURSE
13. Among the following the quantity which
is not a scalar?
1)20 kg
2)15m
3) 40 s
4)13m due north
14. Which of the following whcih is the example for vector?
1)15m due East
2)20m due west
3)10m due south
4)5m
15. Among the following which is the example for vector?
1)East direction
2)20m
3)20m due East
4)can’t say
16. Among the following the quantity which
is not a scalar?
1)8 kg
2)10m 3)20 sec 4)13m due west
17. In the following which set is a complete
scalar set
1)Length,time and displacement
2)Area, velocity and volume
3)Mass, tempeature and volume
4)tempeature,density and force
17. Among the following the quantity which
one is scalar?
1)18m due west
2)20m due south
3)30m
4)23m due north
18. The set containing
quantitites is
only
vector
1)speed,velocity 2)time,displacement
3)energy,mass
4)displacement,velocity
19. A physical quantity which has only magnitude and no specific direction
1)weight
2)Mass
3)force
4)displacement
20. Which of the following pair are scalar
quantities
1)Speed, velocity
2)relative density, volume
3)velocity,area
4)density,displacement
NARAYANA GROUP OF SCHOOLS
DAY-4 : SYNOPSIS
Basic Terms related to Kinematics:
Distance : It is defined as the actual path
followed by a body between the points
between which its moves.
Unit : C.G.S unit : cm S.I unit : m
Note: The distance travelled by body is
always positive.
Displacement : It is the shortest distance
between between initial and final point
in a definite direction.
Unit : C.G.S unit : cm S.I unit : m
Note : (i) For a moving body displacement
can be positive, negative or zero. (ii) If
initial point and final points are same
then displacement is zero.
Scalar quantities : A physical quantity
which is described completely by its
magnitude is called a scalar quantity. It
has only magnitude and no specific
direction.
Ex: Length, distance, area, volume, mass,
time and energy are examples of scalar
quantities.
Vector quantities : A physical quantity
which is described completely by its
magnitude as well as specific direction
is called vector quantity. It has both
magnitude and direction.
Ex : Displacement, velocity, acceleration,
force and weight are examples of vector
quantities.
Speed: The rate of change of motion is
called speed. The speed can be found by
dividing the distance covered by the time
in which the distance is covered.
Formula : Speed 
Dis tance travelled
Time taken
Units : C.G.S unit : cm/s
S.I unit : m/s Nature : Scalar
Kinds of Speed :
a) Uniform speed : When a body covers
equal distances in equal intervals of time
(however small the time intervals may
be), the body is said to be moving with a
uniform speed.
Ex : A rotating fan, a rocket moving in
space, etc., have uniform speeds.
48
CLASS-VIII
MPC BRIDGE COURSE
b) Non-Uniform Speed : When a body covers
unequal distances in equal intervals of
time, the body is said to be moving with
a nonuniform speed.
Ex : A train starting from a station, a dog
chasing a cat, have variable speeds.
c) Average Speed:- When a body is moving
with a variable speed, then the average
speed of the body is defined as the ratio
of total distance travelled by the body to
the total time taken i.e.,
 Average speed

DAY-4: WORKSHEET
1. Which is a vector quantity among these ?
1) My mass is 20 kg
2) Himalayas are in the northern India
3) The sun rises in the east.
4) 200 m towards north is Ramoji film city
from my house.
2. A cyclist moves from a certain point X and
goes round a circle of radius ‘r’ and
reaches Y, exactly at the other side of the
point X, as shown in figure.
Total dis tan ce cov ered
Total time taken to cov er the dis tan ce
Velocity : Velocity is the rate of change
of motion in a specified direction.
displacement
time
Units : C.G.S. unit : cm/s
S.I. unit : m/s Nature : Vector
Kinds of velocity :
a) Uniform velocity :- When a body covers
equal distances in equal intervals of time
in a specified direction, (howsoever short,
the time intervals may be) the body is
said to be moving with a uniform velocity.
Ex : Imagine a car is moving along a
straight road towards east, such that in
every one second it covers a distance of
5m then the car is said to be moving with
uniform velocity.
b) Variable Velocity:- When a body covers
unequal distances in equal intervals of
time in a specified direction or equal
distances in equal intervals of time, but
its direction changes, then the body is
said to be moving with a variable velocity.
Ex : Now imagine the car is moving along
a circular path, such that it is covering
5m in every one second but as the
direction of the car is changing at every
instant, we say the car is moving with
variable velocity.
c) Average velocity:- It is the ratio of total
displacement to total time taken.
r
X
Y
O
Formula : Velocity =
Average velocity 
Total displacement
Total time taken
NARAYANA GROUP OF SCHOOLS
The displacement of the cyclist would be
__________
2
r
3. In the above problem, the distance
covered by the cyclist would be
1)  r
2) 2 r
3) 2r
4)
2
r
4. A man walks 8m towards East and then
6m towards north. His magnitude of
displacement is ___________
1) 10 m 2) 14 m 3) 2 m
4) zero
5. A player completes a circular path of
radius ‘r’ in 40s. At the end of 2 minutes
20 seconds, displacement will be
1)  r
2) 2 r
1) 2r
2) 2 r
3) 2r
4)
3) 7r
4) Zero
6. A horse runs a distance of 1200m in 3
min and 20 s. The speed of the horse
is_________
1) 60 ms–1 2) 65 ms–1 3) 40 ms–14) 6ms–1
7. When a body covers first one third
distance with speed 1m/s, the second
one third distance with speed 2m/s and
the last one third distance with speed
3m/s then average speed is __
1) 2 m/s
2) 1.79 m/s
3) 2.66 m/s
4) 1.64 m/s
49
CLASS-VIII
MPC BRIDGE COURSE
8. An insect crawls a distance of 4m along
north in 10 seconds and then a distance
of 3m along east in 5 seconds. The
average velocity of the insect is _
1)
7
m/sec
5
2)
1
m/sec
5
5
m/sec
4) None of these
15
9. A car travels half the distance with
constant velocity 50 km/h, and another
half with a constant velocity of 40 km/h
along a straight line. The average velocity
of the car in km/h is _______
3)
1) 45
2) 44.4
3) 0
4)
Acceleration : The rate of change of
velocity of a body is called acceleration.
Change in Velocity
Time taken
Units : C.G.S. unit : cm/s2
S.I. unit : m/s2
Nature : Vector
Uniform Acceleration: When a body
describes equal changes in velocity in
equal intervals of time (however small
may be the time intervals) it is said to be
moving with uniform acceleration.
Equations of motion for a body moving
with uniform acceleration in a straight
line:
a) First Equations of Motion : It gives the
velocity acquired by a body in time t which
is v = u + at where, v = Final velocity of
the body [velocity after time (t) seconds]
u = Initial velocity of the body [velocity at
time (t) = 0 second] a = Acceleration
(uniform) t = Time taken
b) Second Equation of Motion : It gives the
displacement of the body in a time t,
1 2
at where, s =
2
displacement of the body in time t
c) Third Equation of motion : It gives the
velocity acquired by a body in
displacement ‘s’ which is v2 – u2 = 2as
which is
s = ut +
NARAYANA GROUP OF SCHOOLS
a
2n - 1
2
where, sn = distance travelled by the body
in nth second
in n th second which is
 50  40 
DAY-5 : SYNOPSIS
Formula : Acceleration =
Points to remember :
i) If a body starts from rest, its initial
velocity, u = 0
ii) If a body comes to rest (it stops),its final
velocity, v = 0
iii)If a body moves with uniform velocity, its
acceleration, a = 0
Distance travelled in nth second : It
gives the distance travelled by the body
sn = u +
DAY-5: WORKSHEET
1. The velocity of car changes from 18 km/
h to 72 km/h in 30s. What will be its
acceleration in km/h2 and in m/s2
1) 6480 km/h2, 0.50 m/s2
2) 6450 km/h2, 0.40 m/s2
3) 6580 km/h2, 0.30 m/s2
4) 6840 km/h2, 0.50 m/s2
2. A car is moving with uniform velocity of
72 km/h for 15s on a National Highway.
Its acceleration and the distance
travelled are
1) 0m/s2, 200m
2) 0m/s2, 300m
3) 20m/s2, 300m
4) 2m/s2, 400m
3. A motor bike is moving with a velocity of
5m/s. It is accelerated at a rate of 0.6m/
s2 for 20s. Then the final velocity of motor
bike is
1) 16m/s 2) 18 m/s 3) 15m/s 4) 17m/s
4. If a bus starts from rest and attains a
speed of 36 km/h in 10 minutes while
moving with uniform acceleration, then
the acceleration of the bus is
1
1
m/s2
2)
m/s2
40
60
1
1
3)
m/s2
4)
m/s2
50
30
5. A girl running in a race accelerates at
2.5 m/s2 for the first 4s of the race. How
far does she travel in this time? (Assume
a girl has started from rest)
1) 10m
2) 15m
3) 20m
4) 25m
1)
50
CLASS-VIII
6. A body starting from rest travels a
distance of 200m in 10 seconds, then the
value of acceleration is
1) 2ms–2 2) 4ms–2 3) 5ms–2 4) 3ms2
7. A scooter moving at a speed of 10 m/s is
stopped by applying brakes which produce
a uniform acceleration of, –0.5 m/s2. How
much distance will be covered by the
scooter before it stops?
1) 200 m 2) 300 m 3) 100 m 4) 250 m
8. A car moving with a speed of 30ms–1 upon
the application of brakes comes to rest
within a distance of 5m. The retardation
produced by the brakes is
1) 90 m/s2
2) 80 m/s2
3) 95 m/s2
4) 85 m/s2
9. A body originally at rest is subjected to
uniform acceleration of 4ms –2 . The
distance travelled by it in 5th second is
1) 10 m 2) 15 m 3) 18 m 4) 20 m
DAY-6 : SYNOPSIS
Acceleration due to gravity:The
acceleration of a freely falling body is called
acceleration due to gravity. The value of
acceleration due to gravity is 9.8m/s2.
Equation of motion for freely falling
body : For a freely falling body, with initial
velocity, u = 0, the velocity continuously
increases as it falls through a height.
The direction of ‘g’ and the direction of
motion of the body are same (i.e.
downwards).Therefore g is taken as
positive under gravity. Here, g = +ve,
u = 0, S = h
Thus the first equation of motion, v = gt
The second equation of motion, h  1 gt 2
2
The third equation of motion, v  2gh
The equation for the distance covered in
nth second, hnth 
g
2n  1
2
Equa tions o f motion for a body
projected vertically upwards : When a
body is projected vertically upwards, its
velocity decreases continuously since its
motion is against gravity, hence g is taken
as –ve.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
When a body is projected vertically
upwards equations of motion will be
v = u – gt, h = ut –
1 2
gt , v2 – u2 = –2gh
2
The equation for the distance covered in
nth second, hnth  u 
g
2n  1
2
Maximum height reached by a body
thrown vertically up : h 
u2
2g
Where, u = initial velocity h = maximum
height g = acceleration due to gravity
Time of ascent (t a) : The time taken by
body thrown up to reach maximum height
‘h’ is called its time of ascent. Let t a be
the time of ascent, t a 
u
g
Time of descent : The time taken by a
freely falling body to touch the ground is
called the time of descent, t d 
u
g
Note : Time of ascent is equal to the time
of descent in the case of bodies moving
under gravity (neglecting air resistance).
Time of flight : It is defined as the total
time for which the body stays in air when
projected vertically up. It is equal to sum
of time of ascent ta and time of descent td.
 time of flight = ta + td 
of flight =
u u

 Time
g
g
2u
g
DAY-6: WORKSHEET
1. If a stone is dropped from the top of a
building and is found to reach the ground
in 1 second, then the height of the
building is
1) 2.9m 2) 3.9m 3) 4.9 m 4) 5.9m
2. A ball is dropped freely from a height,
the distance travelled in the sixth second
is
1) 53.9 m 2) 35.9 m 3) 93.5 m 4) 59.3 m
51
CLASS-VIII
DAY-7 : SYNOPSIS
1. DISPLACEMENT – TIME GRAPHS
These graphs are very useful in studying
the linear motion of the body. The
displacement is plotted on the Y – axis
and the time on X – axis.
These graphs are very helpful in finding
the velocity of body, as the slope of graph
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Displacement
 Y  axis 
.

 is equal to
Time
 X  axis 
Following are various types of displacement
time graphs:
(a) Displacement – time graph is parallel to
time axis.If a displacement – time graph is
parallel to time axis as shown in figure (a) it
means that the body is not changing its
position with respect to time. In other
words, body is stationary.
Displacement in (m)
3. A body dropped from a height h strikes
the ground with a velocity of 3 m/s.
Another body of same mass is dropped
from the same height h with an initial
velocity of 4 m/s. Final velocity of second
mass with which it strikes the ground
will be
1) 3 m/s 2) 4 m/s 3) 8 m/s 4) 5 m/s
4. A body is falling under gravity. The
distance covered in 1st, 2nd and 3rd
second of its motion are
1) 1 : 3 : 5
2) 3 : 1 : 5
3) 5 : 1 : 3
4) 3 : 2 : 4
5. A stone projected vertically upwards,
takes 1.5s to reach highest point. Then (i)
initial velocity of stone and (ii) maximum
height attained by it are
(Take g = 10 ms –2).
1) 15 ms–1, 11.25 m 2) 17 ms–1, 11.25 m
3) 18 ms–1, 12.25 m 4) 16 ms–1, 11.25 m
6. A stone is thrown vertically up with an
initial velocity of 10 m/s. Then the
maximum height reached and the time of
ascent are (Take g = 10 m/s 2)
1) 6m, 2s 2) 3m, 1s 3) 4m, 2s 4) 5m, 1s
7. A ball is thrown vertically upwards. It
returns 6 s later. Then (i) the greatest
height reached by the ball and (ii) the
initial velocity of the ball are (Take g = 10
ms –2 )
1) 45 m, 30 ms –1
2) 40 m, 35 ms –1
3) 40 m, 30 ms –1
4) 45 m,35 ms–1
8. A ball is thrown vertically upwards with
an initial velocity of 49 m s –1. Then (i) the
maximum height attained, (ii) the time
taken by it before it reaches the ground
again are (g = 9.8 m s –2)
1) 12.5 m, 10s
2) 122.5 m, 10s
3) 122.5 m, 1s
4)122.5 m, 100s
MPC BRIDGE COURSE
8
6
4
2
0
1
2
3
4 5 6
Time in (seconds)
Fig. (a)
(b)When displacement – time graph is a
straight line, but is not parallel to time axis.
If the displacement – time graph is a straight
line, such that it starts from origin and moves
away from time and displacement axis, the
body is said to be moving with uniform
velocity. Figure (b) shows a displacementtime graph of a body. The values of
displacement and time are shown in the table
below:
Displacement
inmetres
0
5
10
15
20
25
Timeinseconds
0
1
2
3
4
5
Since graph is a straight line, therefore it
means that displacement is proportional to
time. In other words, body is covering equal
distances in equal intervals of time i
n
specified direction, and hence, is moving with
a uniform velocity.
The slope of this graph gives uniform velocity.

x AB
Thus, velocity of body 

t BC

10m
= 5m/s
2s
52
CLASS-VIII
Where  x is short distance and  t is a
short interval of time.
Displacement (x) in (m)
25
A
20
x = 10m
15
C
10
B
t = 2s
5
0
1
2
3
4
5
Time in (seconds)
Fig. (b)
(c) When displacement – time graph is
not a straight line: If the displacement time graph is a curve as shown in
figure(c).It represents ‘variable velocity’
of a moving body.
  
 BC 
Slope of the graph  DC  at any point


gives the velocity at that instant . For
example, velocity at A is

Velocity in (ms-1)
BC  8  2  m 6 m


= 3m/s
DC
 3  1 s 2 s
MPC BRIDGE COURSE
4. If graph is a curve, it means the body is
moving with a variable velocity, and
hence, has some acceleration.
1. VELOCITY – TIME GRAPHS
(i) In these graphs generally, the velocity
is plotted on Y-axis and time on X-axis.
The slope of such graphs gives
acceleration.
(ii) As, velocity  time = acceleration, the
acceleration will be positive if the slope
is positive, and negative if the slope is
negative.
(iii)The area of graph under velocity –
time curve, gives displacement of body.
Displacement = Velocity × Time.
(a) W hen velocity – time graph is
parallel to time axis
(i) Figure(a) represents velocity – time
graph PQ, when a body is moving with a
uniform velocity of 20 ms–1.
As the slope of graph is zero, therefore
its acceleration is zero.
(ii) The distance covered by the body in
specified direction (displacement) can be
calculated by finding the area of
rectangle PQRS. Thus, Displacement =
PS × SR = 20ms–1 × 4s = 80m.
P
20 
Q
15 
10 
5
S
1
2
3
4R
Time in (seconds)
Fig. (c)
Conclusions from displacement–time
graph
1. If the graph is parallel to time axis, then
body is stationary.
2. If graph is a straight line, then body is
moving with a uniform velocity. The
velocity can be found out by finding the
slope of the graph.
3. The graph can never be parallel to
displacement axis, as it means that
displacement increases indefinitely,
without any increase in time, which is
impossible.
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Fig. (a)
Conclusion : If velocity – time graph is
parallel to time axis, then:
a) Body is moving with uniform velocity.
b) Its acceleration is zero.
c) Its displacement can be found by finding
the area of the graph.
(b) When velocity – time graph is a
straight line, but not parallel to time
axis.
53
CLASS-VIII
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C
Fig. (i)
Fig. (ii)
Figure (i) represents a velocity – time
graph when the body starts from rest,
and
its velocity increases at a uniform
rate.
The slope of the graph AC, i.e.,
AB
gives
BC
the acceleration of body.

Acceleration

change in velocity
Total time
16  0  ms 1


= 3.2ms–2.
5s
The area of triangle ABC, gives the
displacement.
Thus,
displacement
in
5s
1
1
 AB  BC   16ms–1 × 5s = 40m
2
2
Figure(ii) .represents velocity – time
graph, where the body is initially not at
rest.
Conclusions :
(i) If velocity – time graph is a straight line
but moving away from velocity time axis,
then:
a) Body is moving with variable velocity.
b) It has uniform acceleration, which can
be found by the slope of graph.
c) Displacement can be found, by finding
area under the velocity - time graph.
d) If slope is positive, then the body has
positive acceleration and vice – versa.
(ii) If the velocity – time graph is a curve,
then:
a) The body has variable velocity and
variable acceleration.
b) Area under the curve represents
displacement.
c) Acceleration at any instant can be
found by finding slope at that point.
2. ACCELERATION – TIME GRAPH
Figure (i) represents an acceleration –
time graph, AB coinciding with time axis.
From the figure it is clear that
acceleration of the body is zero, and
hence, it is moving with a uniform
velocity.

 AB 
The slope of graph AC, i.e., 
 gives
 BC 
the acceleration.
Acceleration

1
AB  25  5  ms
20


ms–2 = 5ms–2.
BC
4s
4
The distance covered by the body in
specified direction is area of trapezium
ECAD.

1
1
(CE + AD) × ED 
2
2
× 4s = 60m.
 Displacement 
(5 + 25)ms–1
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Fig (i)
Fig (ii)
Fig (iii)
Figure (ii) represents an acceleration –
time graph, parallel to time axis. From
figure it is clear that as acceleration does
not change, therefore body is moving with
a uniform acceleration and variable
velocity. The area of graph, i.e.,
Acceleration × Time gives change in
velocity.
Figure (iii) represents an acceleration –
time graph moving away from time as well
as acceleration axis. From the graph it
is clear that the body is moving with
variable
velocity
and
variable
acceleration. Area of the graph gives
change in velocity.
54
CLASS-VIII
DAY-7: WORKSHEET
S
distance 
||||||
1. The distance-time graph of an object
moving in a fixed direction is shown in
figure. The object is
MPC BRIDGE COURSE
4. Refer Figure, the ratio of speed in first
two seconds to the speed in the next 4
seconds is
O
s
0
||||||
time 
1) at rest 2) moving with constant speed
3) moving with variable speed
4) none of these
2. Figure shows the displacement (s) - time
(t) graph of a particle moving on the X–
axis.
1
2
3
4
5
6
t
1) 1 : 2 2) 2 : 1 3)
2 : 1 4) 3 :1
5. Figure shows the time acceleration graph
for a particle in rectilinear motion. The
average acceleration in first twenty
seconds is

S
O
t0
t
Which is correct given below?
1) The particle is at rest
2) The velocity of particle increases upto
time t0 and then increases
3) The velocity of particle increases upto
time t0 and then becomes constant
4) The particle moves at a constant
velocity up to a time t0, and then stops.
3. Which of the following displacement(x) –
time(t) graph is not possible?
1)
1) 45 m/s2
2) 40 m/s2
3) 15 m/s2
4) 20 m/s2
6. The graph below represents motion of
car. The displacement of the car in 20s
is
1) 160 m 2) 20 m 3) 90 m 4) 10 m
7. Diagram shows a velocity time graph for
a car starting from rest, the graph has
three sections AB, BC and CD.
2)
3)
4)
The ratio of distance along AB and BC is
1) 1 : 2
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2) 2 : 1
3) 1 :
2 4)
2 :1
55
CLASS-VIII
DAY-8 : SYNOPSIS
Force : Push or pull is called force. The
cause of motion is force.
Effects of force :
A force can cause a motion in stationary
object
A force can stop the moving objects or
slow them down
A force can make a moving object move
faster
Force can change direction of moving
objects
Force can change the shape of objects.
From the above examples, we are in
position to define force.
Force is an external agent which changes
or tends to change the state of rest or
uniform motion of a body or changes its
direction or shape.
Types of forces :
a) Muscular forces : The force applied by
the muscles of our body is called
muscular force or biological force.
Ex : Lifting of heavy weight pulling of
wheel cart, pushing a lawn roller etc.
involves muscular forces.
b) Mech anical force s : The forces
generated by a machine are called
mechanical forces.
Ex : The force used to run a motor car
engine is produced by using the energy of
petrol. The force used to run steam
engine, is produced by using the energy of
coal.
c) Gravitational force : The force of
attraction exerted by the earth on all the
objects is called the force of gravity or
gravitational force.
Ex : A stone falls downwards due to
gravitational force.
It is the gravitational force of the sun that
keeps the planets in their orbits.
d) Electrostatic force : The force exerted
by electrostatic charge is called
electrostatic force.
Ex : Charged comb attracts small pieces
of paper.
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e ) Magnetic force : The force by which a
magnet attracts or repels objects of iron,
steel, nickel and cobalt is called magnetic
force.
Newton’s first law of motion :
A body at rest will remain at rest and a
body in motion will remain in uniform
motion, unless it is compelled by an
external force to change its state of rest
or of uniform motion.
Ex : A book lying on a table is in the state
of rest w.r.t. the table. It will remain at
rest unless some one picks it up or moves
it from one position to another.
Inertia : Inertia of a body may be defined
as the tendency of a body to oppose any
change in its state of rest or uniform
motion.
Ex : A book lying on a table will remain
placed at its place unless it is displaced.
Measure of inertia : Mass is a measure
of inertia.
Ex : If a body has a mass of 1 kilogram and
another body has a mass of 20 kg, then
the body having 20 kg mass will have more
inertia since its mass is more.
Types of inertia : Inertia can be divided
into three types.
1) Inertia of rest 2) Inertia of motion 3 )
Inertia of direction
a) Inertia of rest : The tendency of a body
by virtue of which it cannot change its
state of rest by itself is called inertia of
rest.
Ex : A passenger in a bus tends to fall
backward when the bus starts suddenly
b) Inertia of motion : The tendency of a
body by virtue of which it cannot change
its state of motion by itself is called
inertia of motion.
Ex : A passenger in a moving bus tends to
fall forward when the bus stops suddenly
c) Inertia of direction : The tendency of
a body to oppose any change in its
direction of motion by itself is known as
inertia of direction.
Ex : A stone tied to a string and whirled
along a circular path flies off tangentially
due to inertia of direction, if the string
breaks.
56
CLASS-VIII
DAY-8: WORKSHEET
1. Identity the situations where a pull is
involved.
a) Man sitting on a chair
b) ball falling to the ground
c) Woman drawing water from a well
d) tube light fixed on the wall
1) both ‘a’ and ‘b’
2) both ‘b’ and ‘c’
3) both ‘c’ and ‘d’
4)both ‘a’ and ‘d’
2. A force can
a) move a body from rest.
b) change the direction of a moving body.
c) increases the mass of a body.
1) only ‘a’ is true
2) only ‘b’ is true
3) both ‘a’ and ‘b’ are ture
4) ‘b’ and ‘c’ are true
3. Apple is falling from the tree towards the
ground due to
1) Magnetic force 2) Gravitational force
3) Mechanical force 4) Electrostatic force
4. A boy used ____ force to kick a football.
1) Muscular force 2) Gravitational force
3) Mechanical force 4) Electrostatic force
5. Magnetic force can cause
1) only attraction 2) only repulsion
3) both attraction and repulsion
4) none of these
6. A tailor cuts a piece of cloth using a pair
of scissors. The force involved here is
1) electrostatic force
2) mechanical force
3) magnetic force
4) none of these
7. Inertia is that property of a body by virtue
of which the body is
1) unable to change by itself its state of
rest.
2) unable to change by itself its state of
uniform motion.
3) unable to change by itself its direction
of motion.
4) all the above.
8. An athlete runs some distance before
taking a long jump, because
1) It helps him to gain energy
2) It helps to apply large force
3) It gives himself large amount of inertia
4) None of these
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9. A rider on a horse back falls forward,
when the horse suddenly stops, this is
due to
1) inertia of the horse
2) inertia of the rider
3) large weight of the horse
4) losing the balance
10. A passenger sitting in a bus gets a
backward jerk when the bus starts
suddenly due to the
1) inertia of rest
2) inertia of motion
3) inertia of direction 4) none of these
DAY-9 : SYNOPSIS
Linear momentum : The total quantity
of motion contained in a body is called
linear momentum.
Formula : Momentum of a body is equal
to the product of the mass (m) of the body

and the velocity v of the body. It is

denoted by P .Momentum = mass ×


velocity ( P = m v )
Units : C.G.S. unit: g cms–1
S.I. unit: kg ms–1
Natur e : Vector , The direction of
momentum of a body is same as that of
the direction of the velocity of the body.
Newton’s Second Law of motion :
The magnitude of the resultant force
acting on a body is proportional to the
product of the mass of the body and its
acceleration. The direction of the force
is the same as that of the acceleration.
Newton’s second law gives the
quantitative definition of force in other
words it measures force. i.e. F net = ma,
where, F = Force, m = Mass, a =
Acceleration
So, Force acting on the body = mass of
the body × acceleration produced in the
body.
Newton ’s 2 nd Law in t erms of
momentum : The rate of change of
momentum of an object is proportional to
the net force applied on the object. The
direction of the change of momentum is
the same as the direction of the net force.
Force  Rate of change of momentum.
 
57
CLASS-VIII
Change of momentum
Time
Absolute Units of Force : C.G.S unit : g
cm/s2 or dyne.
Definition of dyne : The force is said to
be 1 dyne if it produces 1 cm/s 2
acceleration in a body of 1g mass.
S.I unit of force : kg m/s2 or newton(N).
Definition of newton (N) : 1 newton is
that much force which produces an
acceleration of 1 m/s2 in a body of mass
1 kg.
Relation between newton and dyne :1
newton (N) = 105 dyne
i.e., Force 
DAY-9: WORKSHEET
1. What will be the momentum of a toy car
of mass 200 g moving with a speed of
5 m/s ?
1) 1 kg m/s
2) 2 kg m/s
3) 3 kg m/s
4) 4 kg m/s
2. A body of mass 25 kg has a momentum of
125 kg m/s what is its velocity?
1) 6 m/s 2) 5 m/s 3) 4 m/s 4) 3 m/s
3. A cricket ball of mass 100 g is moving
with velocity 25 m/s. Then the
momentum of the ball is
1) 7.5 kg.m/s
2) 3 kg.m/s
3) 2 .5 kg.m/s
4) 4 kg.m/s
4. Two bodies A and B of same mass are
moving with velocities V and 3 V
respectively, then the ratio of their
momentum will be
1) 1 : 2
2) 2 : 1
3) 3 : 1
4) 1 : 3
5. A cricket ball of mass 100g strikes the
hand of a player with a velocity of 20 m/
s and is brought to rest in 0.01 s, then
the force applied by the hands of the
player is
1) 200 N 2) 300 N 3) 400 N 4) 500 N
6. Two bodies have masses in the ratio 3 :
4. When a force is applied on first body,
it moves with an acceleration of 6 m/s2.
How much acceleration the same force
will produce in the other body ?
1) 5.5 m/s2
2) 3.5 m/s2
3) 4.5 m/s2
4) 2.5 m/s2
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7. A force of 200 dyne acts on a body of mass
10 g for 5 sec. What will be the final
velocity of body if it starts from rest ?
1) 50 cm/s
2) 80 cm/s
3) 100 cm/s
4) 200 cm/s
8. A force of 10 kg wt acting on a certain
mass for 2 second gave it a velocity 10m/s.
What is the mass in kg ? (g = 9.8 m/s2)
1) 19.6
2) 9.8
3) 15
4) 5
DAY-10 : SYNOPSIS
Impulsive Force : A large force which
acts for a small interval of time is called
impulsive force.
Impulse : Impulse of a force is defined
as the change in momentum produced
by the given force and it is equal to the
product of force and the time for which it
acts.
Formula :
Impulse = Change in
momentum = Force × Time.
Unit:C.G.S unit:dyne second (or) g cm/s
S. I. unit: N s (or) kg m/s Nature : Vector
Mass : The quantity of matter contained
in the body is called its mass.
Unit : C.G.S unit : gram (g).
S.I. unit : kilogram (kg). Nature : Scalar
Weight : The weight of a body is the force
with which it attracted towards the
centre of the earth.
Formula : Weight = mass × acceleration
due to gravity.
Unit : C.G.S unit : dyne
S.I. unit : newton
Nature : Vector
Note: Mass remains constant whereas
weight changes from place to place.
Newton’s Third Law : ‘To every action,
there is an equal and opposite reaction’
Note : Action and reaction force are equal
in magnitude but opposite in direction.
i.e. Action = – Reaction
Ex : Jet aeroplanes and rockets works on
the principle of Newton’s third law of
motion.
Note : Action and Reaction act
simultaneously but act on different
bodies. So they donot cancel with each
other.
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CLASS-VIII
Law of conservation of momentum :
According to this law, the total
momentum of a system remains constant
if no net external force acts on the
system.
That is, momentum of a system.
r
p = constant, if net external force acting
on it is zero

(i.e. Fexternal = 0)
DAY-10: WORKSHEET
1. A force of 50 N acts on a body for 10 s.
What
will
be
the
change
in
momentum ?
1) 200 Ns 2) 400 Ns 3) 500 Ns 4) 1000 Ns
2. A body of mass 100 kg moving straight
line with a velocity of 30 m/s, moves in
opposite direction with a velocity of 20
m/s after hitting a wall. What is its
magnitude of impulse ?
1) 6000 Ns
2) 5000 Ns
3) 4000 Ns
4) 3000 Ns
3. How much would a 70 kg man weigh on
the moon ? What will be his mass on the
earth and on the moon ? [g on moon =
1.7 m/s2]
1) 119 N, 70 kg
2) 115 N, 68 kg
3) 116 N, 65 kg
4) 114 N, 55 kg
4. If the weight of man on earth surface 30
N. What will be his weight on moon
surface ? (g moon = (1/6) gearth)
1) 5 N
2) 4 N
3) 3 N
4) 2 N
5. In case of a book lying on a table.
1) action of book on table and reaction of
table on book are equal and opposite and
are inclined to vertical.
2) action and reaction are equal and
opposite and act perpendicular to the
surfaces of contact.
3) action and reaction are equal but act
in the same direction.
4) action and reaction are not equal but
are in opposite direction.
6. Whenever an object A exerts a force on
another object B, object B will exert a
return force back an object A. The two
forces are
1) equal in magnitude and in same
direction
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2) equal in magnitude but opposite in
direction
3) not equal and in opposite direction
4) not equal and in the same direction
7. When two bodies of masses m 1 and m 2
moving with velocities u 1 and u 2 in the
same direction collide with each other
and v 1 and v 2 are their velocities after
collision in the same direction, then
1) m1v1 + m2v2 = m2u2 – m1u1
2) m1v1 + m2v2 = m1u1 – m2u2
3) m2u2 + m2u1 = m2v1 + m1v2
4) m1u1 + m2u2 = m1v1 + m2v2
8. When two bodies of masses m 1 and m 2
moving with velocities u 1 and u 2 in the
op po si te di re cti on c oll id e with ea ch
other and move together after collision
in the same direction with a common
velocity v, then
1) m1u1 + m2u2 = m1v – m2v
2) m2u1 – m1u2 = (m1 + m2)v
3) m1u1 – m2u2 = (m1 – m2)v
4) v 
m1u1  m2u2
m1  m2
9. The car A of mass 1500 kg travelling at
25 m/s collides with another car B of
mass 1000 kg travelling at 15 m/s in the
same direction. After collision the
velocity of car A becomes 20 m/s. The
velocity of car B after the collision is
1) 12.2 m/s
2) 11.5 m/s
3) 22.5 m/s
4) 5.22 m/s
DAY-11 : SYNOPSIS
Friction : The force which opposes the
relative motion of a body over another is
called force of friction. The force of
friction is always parallel to the two
surfaces.
Cause of friction : Friction is due to the
irregularities (interlocking) of the two
surfaces in contact.
Factors on which frictional force depends:
The nature of two surfaces in contact
with each other.
Normal force with the surfaces are being
pressed together.
59
CLASS-VIII
Note : The force of friction does not
depend upon the area of the surfaces in
contact.
Types of friction :
a) Static friction : (fs)
Static friction is the force of friction acting
on the body when it is rest position inspite
of the fact that some force is being applied
on it.
Note : Static friction always equal to
applied force.
Static frictional force is a self adjusting
one. It can adjust not only in magnitude
but also in direction.
b) Limiting friction : The maximum value
of the static friction is called limiting
friction. (or) The maximum frictional force
when the body is ready to start is called
limiting frictional force.
c) Kinetic friction (f k ) : The force of
friction which opposes when the body in
motion on the surface of another body.
(or) W h e n o n e b o d y m o v e s o v e r th e
o t h e r , th e f o r c e o f f r i c t i o n a c ti n g
b e tw e e n th e t w o s u r f a c e s i s c a l l e d
kinetic friction.
Laws of limiting friction : The direction of
force of friction is always opposite to the
direction of motion.
The force of limiting friction depends upon
the nature and state of polish of the two
surfaces in contact.
The magnitude of limiting friction ‘F’ is
directly proportional to the magnitude of
the normal reaction R between the two
surfaces in contact, i.e, F  R (or) F =
 R where  = coefficient of friction.
The magnitude of the limiting friction
between two surfaces is independent of
the area and shape of the surfaces in
contact so long as the normal reaction
remains the same.
Factors on which coefficient of friction
depends : Coefficient of friction depends
on Nature of the surfaces in contact.
It depends on temperature.
Note : Coefficient of friction have no
units.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
DAY-11: WORKSHEET
1. The maximum value of static friction is
called
1) limiting friction 2) static friction
3) kinetic friction 4)rolling friction
2. The friction acting on the body when the
body is in motion is called
1) static friction
2) dynamic friction
3) limiting friction 4) none of these
3. Choose the correct relation
1) f =  R
2) f =  /R
4.
5.
6.
7.
8.
9.
3) f =  + R
4) f = R/ 
Statement A: Frictional force increases
with the increases external force, in case
of static friction.
Statement B : Static friction always equal
to applied force.
1) Statement A is true
2) Statement B is true
3)Both the statements A and B are true
4) None of these
Force of limiting friction for a body and a
surface
1) increases if the surface is inclined
2) decreases if the surface is inclined
3) remains the same whether the surface
is inclined on horizontal
4) none of these
The coefficient of static friction between
two surfaces depends upon
1) the normal reaction
2) the shape of surfaces in contact
3) the area of contact 4) all of the above
A force of 100 g wt. is required to pull a
body weighing 1 kg over ice. What is the
co-efficient of friction ? [g = 9.8m/s2]
1) 0.01
2) 0.1
3) 1
4) 10
A body of mass 100 g is made just to slide
on a rough surface by applying a force of
0.8 N. Then the coefficient of friction is
(Take g = 10 ms –2)
1) 0.8
2) 0.08
3) 0.7
4) 0.6
What minimum force is required to move
a body of mass 5 kg over a surface whose
coefficient of friction is 0.3 ? g = 10 ms –2.
1) 15 N
2) 13 N
3) 12 N
4) 10 N
60
CLASS-VIII
DAY-12 : SYNOPSIS
Work : Work is said to be done when a force
produces motion.
Ex : When an engine moves a train along
a railway line, it is said to be doing work.
Mathematical Expression for work :
If a force F acts on a body and moves it a
distance S in the direction of the force
then work done. i.e., W = F × S
Units of work : C.G.S. unit : g cm2 s–2 or
erg.
S.I unit : kg m2 s–2 or joule(J).
Relation between Joule and erg : 1 J = 1
N × 1 m = 105 dyn × 100 cm = 107 dyn cm.
(or) 1 J = 10 7 erg.
Note : Work is a scalar quantity.
Types of Work : Work done can be positive,
negative or zero depending upon the
direction of force and direction of motion.
(displacement)
Positive work done : Work done by a force
on a body (or an object) is said to be
positive work done when the body is
displaced in the direction of applied force.
Ex : The body falling freely under the
action of gravity has positive work done by
the gravitational force.
Negative work done : The work done by a
force on a body is said to be negative work
done when the body is displaced in a
direction opposite to the direction of the
force.
Ex : Work done by frictional force as force
of friction and the displacement are
opposite to each other.
Zero work : Work done is zero if
i) The displacement is zero.
Ex : When a person pushes a wall but fails
to move the wall, then work done by the
force on the wall is zero.
ii) The force and the displacement are
perpendicular to each other.
Ex : When a person carrying a suitcase in
his hand or on his head is walking
horizontally, the work done against
gravity is zero.
No work is done on a body when the body
moves along a circular path.
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
Work done against gravity :
Work done in lifting a body = weight of
body × vertical distance = W = mg × h
where, m = mass of body g = acceleration
due to gravity at that place h = height
through which the body is lifted.
DAY-12: WORKSHEET
1. When a stone tied to a string is whirled in
a circle, the work done on it by the string is
1) positive 2) negative 3) zero 4) none
2. A man with a box on his head is climbing
up a ladder. The work done by the man on
the box is
1) positive
2) negative
3) zero
4) undefined
3. Work is said to be done if
Statement A : a force is applied which
brings about motion
Statement B : a force is applied but no
motion is produced
1) only statement A is true
2) only statement B is true
3) both the statement A and B are true
4) none of these
4. Work done is zero
1) When force and displacement of the
body are in the same direction
2) When force and displacement of the
body are in the opposite directions
3) When force acting on the body is
perpendicular to the direction of the
displacement of the body
4) None of these
5. How much work is done by a force of 10
N is moving a body through a distance of
2 m in its own direction ?
1) 20 J
2) 24 J
3) 26 J
4) 30 J
6. Calculate the work done by a passenger
standing on a platform holding a suitcase
of weight 10 kgwt.
1) 15
2) 10
3) 0
4) 5
7. Which of the following represents joule ?
1) Nm
2) dyn m 3) N/m 4) m/N
8. A person of mass 50 kg climbs a tower of
height 72 metre. The work done is [g = 9.8
m/s2]
1) 35280 J
2) 32580 J
3) 52380
4) 58320 J
61
CLASS-VIII
9. How much is the mass of a man if he has
to do 2500 joule of work in climbing a tree
5m tall ? (g = 10 m/s 2)
1) 30 kg 2) 40 kg 3) 50 kg 4) 45 kg
10.An object of 100 kg is lifted to a height of
10 m vertically. What will be the work
done? [g = 9.8 m/s 2]
1) 9800 J 2) 9008 J 3) 9.8 J 4) 8.9 J
MPC BRIDGE COURSE
Examples of body possessing both the
kinetic and potential energies at the same
time:
i) A flying aeroplane
ii) A bird flying in the sky
Relation between kinetic energy and
momentum :
We know , P = mv  v = P/m and K.E. =
2
DAY-13 : SYNOPSIS
Energy : Energy is the ability to do work
or the capacity to do work.
Units : Unit of energy is same as that of
the unit of work. As work is a form of
energy.
C.G.S unit of energy is erg.
S.I. unit of energy is Joule (J).
Nature : Energy is a scalar quantity.
Mechanical energy (M.E) : The sum of
k i n e ti c e n e r g y (K.E) a n d p o te n ti a l
e n e r g y (P.E) o f a b o d y i s k n o w n a s
mechanical energy.  M.E = K.E + P.E
Kinetic energy (K.E) : The word kinetic
comes from a Greek word which means
motion.
The energy possessed by a body by virtue
of its motion is known as kinetic energy.
Note : All moving bodies possess kinetic
energy.
Ex : A moving bus or a car or a train has
kinetic energy.
Form ula of kineti c ener gy: Kinetic
energy, K.E  1 mv 2
2
where, m = Mass of the body, v = Velocity
of the body.
Potential energy : (P.E) : The energy
possessed by a body by virtue of its
position is called potential energy.
Ex : Water stored in a dam has potential
energy due to its position.
Formula of potential energy :
Potential energy (P.E) of a body at a
certain height = P.E = mgh
where, m = mass is the body, g =
acceleration due to gravity h = height
from the ground.
NARAYANA GROUP OF SCHOOLS
1 P
1
P2
mv 2  m   =
2
2 m
2m
Note
:
K.E 
For a body
momentum, K.E 
P2
2m
having
same
1
For a body having
m
same kinetic energy, P  m .
Law of conservation of energy : According
to this law “Energy can neither be created
nor be destroyed, but can be changed
from one form to another form”.
Ex :When a body falls from a certain
height, its P.E gradually changes into
kinetic energy but the total sum of both
the energies remains the same.
Note : i) For a freely falling body, potential
energy changes into kinetic energy.
Hence, Loss of P.E = Gain of K.E
ii) For a body
projected vertically
upwards, kinetic energy changes into
potential energy. Hence, Loss of K.E =
Gain of P.E.
DAY-13: WORKSHEET
1. What will be the K.E of a body of mass 2 kg
moving with a velocity of 0.1 metre per
second ?
1) 0.1 J 2) 0.01 J 3) 0.001 J 4) 1 J
2. Two bodies of equal masses move with
uniform velocities v and 3v respectively.
Find the ratio of their kinetic energies.
1) 9 :1
2) 2 : 9
3) 1 : 9
4) 1 : 1
3. A 1kg mass has a kinetic energy of 1 Joule
when its velocity is
1) 0.45 m/s
2) 1 m/s
3) 1.4 m/s
4) 4.4 m/s
4. If acceleration due to gravity is 10 m/s 2,
what will be the potential energy of a body
of mass 1 kg kept at a height of 5 m ?]
1) 20 J
2) 30 J
3) 40 J
4) 50 J
62
CLASS-VIII
MPC BRIDGE COURSE
5. An object of mass 1 kg has a potential
energy of 1 J relative to the ground, when
it is at a height of [g=10 m/s 2]
1) 0.1 m 2) 1 m
3) 9.8 m 4) 32 m
6. A light and a heavy body have equal
kinetic energy. Which one has greater
momentum ?
1) The lighter body has greater
momentum
2) The heavier body has greater
momentum
3) both the bodies have same momentum
4) none of these
7. What will be the momentum of a body of
mass 100 g having kinetic energy of
20 J ?
1) 2 kg m/s
2) 4 kg m/s
3) 5 kg m/s
4) 6 kg m/s
8. Tw o b o d i e s o f m a s s 1 k g a n d 4 k g
possess equal momentum. The ratio of
their kinetic energies is
1) 4 : 1
2) 1 : 4
3) 2 : 1
4) 1 : 2
9. A body of mass 2kg moving up has
potential energy 400J and kinetic energy
580J at a point ‘P’ in its path. The
maximum height reached by the body is
(g = 10ms–2)
1) 49m
2) 98m
3) 196m 4) 392m
10. A body is moving horizontally at a
height of 10m has its P.E equal to K.E.
Then velocity of that body is (g=9.8 m/s2)
1) 7ms–1 2) 14ms–1 3) 3.5ms–1 4) 2.8ms–1
DAY-14 : SYNOPSIS
Thrust : The total force exerted by the
body perpendicular to the surface is
known as thrust.
Units of thrust : Since thrust is a type of
force its units are same as that of the
force.
C.G.S. unit : dyne S.I. unit: newton (N)
Pressure: Thrust acting over a unit area
of the surface is called pressure.
Formula : Pressure
=
normal force (or Thrust)
F
 P=
Area
A
Units of pressure : C.G.S. unit :
NARAYANA GROUP OF SCHOOLS
dyne
cm2
S.I. unit :
N
m2
or
pascal
Other units of pressure:Other units of
pressure are bar or atmosphere
1 bar = 105 N/m2 or 105 pascals = 105 Pa
1 milli bar = 10 2 N/m2 or 10 2 Pa
Note : Pressure is a scalar quantity.
Atmospheric pressure :
The force exerted on unit area of the
Earth’s surface due to the atmosphere is
called atmospheric pressure. We can
express it in pascal as : 1 atm = 0.76 m
Hg = 105 Pa
Mathematical expression for pressure
in fluids :
Pressure (P) exerted by liquid at depth
(h)= P = h d g (neglecting atmospheric
pressure)
Here, h = height of the liquid column, d =
density of the liquid
g = acceleration due to gravity
Note : Total pressure in a liquid at a depth
‘h’ = Atmospheric pressure (pA) + Pressure
due to liquid column = pA + hdg
DAY-14: WORKSHEET
1. One pascal is the pressure generated by
1) force of 1N on 1 m2
2) force of 1 kg on 1 m2
3) force of on 1N an 1000cm2
4) force of 1N on 1cm2
2. A rectangular iron block is kept over a
table with different faces touching the
table. In different cases, the block exerts
1) same thrust and same pressure
2) same thrust and different pressure
3) different thrust and same pressure
4) different thrust and differen7t pressure
3. A force of 16N acts on an area of 50 cm 2.
What is the pressure in pascal?
1) 3200 Pa
2) 4200 Pa
3) 5200 Pa
4) 2200 Pa
4. What is the magnitude of force required
in newton’s to produce a pressure of
27500 Pa on an area of 200 cm2 ?
1) 650 N 2) 750 N 3) 550 N 4) 450 N
5. A force of 300 N, while acting on an area
A, produces a pressure of 1500 Pa. What
is the magnitude of A in cm 2 ?
1) 1000 cm2
2) 3000 cm2
3) 4000 cm2
4) 2000 cm2
63
CLASS-VIII
6. Pressure at any point inside a liquid is
1) directly proportional to density of the
liquid
2) inversely proportional to density of the
liquid
3) directly proportional to square root of
density of the liquid
4) inversely proportional to square of
density of liquid
7. What will be the pressure in dyne/cm 2,
due to a water column of height 10 cm ?
[ take g = 980 cm /s 2] (density of water =
103 kg/m3)
1) 9.8 × 103 dyne/cm2
2) 9.8 × 104 dyne/cm2
3) 9.8 × 105 dyne/cm2
4) 9.8 × 106 dyne/cm2
8. The pressure in water pipe at the ground
floor of a building is 120000 Pa, where as
the pressure on the third floor is 30000
Pa. What is the height of third floor ?
[ Take g = 10 m /s 2, density of water =
1000 kg /m3]
1) 11 m 2) 9 m
3) 10 m 4) 12 m
DAY-15 : SYNOPSIS
Pascal’s Law : This law is also known as “
the principle of transmission of fluidpressure.”
This law states that “ The pressure
exerted at any point in an enclosed and
incompressible liquid is transmitted
equally in all direction.”
Buoyant Force and Buoyancy : Every liquid
exerts an upward force on the objects
immersed in it is called buoyant force.
The tendency of a liquid to exert an
upward force on an object placed in it is
called buoyancy.
Factors Affecting Buoyant Force: The
magnitude of buoyant force acting on an
object immersed in a liquid depends on
two factors :
i) volume of object immersed in the
liquid, and ii) density of the liquid.
Statement of Archimedes’ Principle :
When an object is wholly (or partially)
immersed in a liquid, it experiences a
buoyant force (or upthrust) which is equal
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
to the weight of liquid displaced by the
immersed part of the object. In other
words, for a body inside the liquid.
loss of weight or Buoyant force = Wt of
liquid displaced by the immersed part of
the body.
= mass of the liquid displaced × g
= volume of liquid displaced × density of
liquid × g
= volume of body × density of liquid × g =
Vbody × d liquid × g
(since for a body completely immersed in
the liquid , the volume of the liquid
displaced is equal to the total volume of
the body).
Relative density of Solid: R.D of solid =
weight of solid in air
loss of weight of solid in water

w air
w air  w water
Relative density of Liquid :
R.D of liquid =
loss of weight of the body in liquid
loss of weight of the body in water

w air  w liquid
w air  w water
DAY-15: WORKSHEET
1. Archimedes’ principle states that when a
body is totally or partially immersed in a
fluid the upthrust is equal to
1) the weight of the fluid displaced by it
2) the weight of the body
3) volume of the fluid displaced
4) volume of the body
2. A body is lowered into a liquid. The body
loses some of its weight. The loss of
weight depends upon
1) volume of the body 2) density of liquid
3) acceleration due to gravity
4) all of these
3. Two balls, one of iron and the other of
aluminium experience the same
upthrust when dipped in water if
1) both have same mass
2) one has half the volume as that of the
other
3) both have equal volume
4) one has one-fourth of the volume as
that of the other.
64
CLASS-VIII
4. What buoyant force acts on a solid of
volume 1m3, immersed in water of density
1000kg/m3 ? (Take g = 10 m/s 2)
1) 104 N 2) 105 N 3) 102 N 4) 103 N
5. Determine the buoyant force acting on a
solid of volume 1.6m 3, immersed in sea
water of density 1030 kg m–3. (Take g = 10
m/s2)
1) 16480 N 2) 164 N 3) 1648 N 4) 61840 N
6. A body whose volume is 200 cm 3 weighs
700gf in air. What is its weight in water?
1) 500 gf 2) 300 gf 3) 200 gf 4) 100 gf
7. A body weighs 500gf in air and 300 gf
when completely immersed in water.
Find (i) the apparent loss in the weight of
the body.
(ii) the upthrust on the body. iii.
the
volume of the body.
1) 300 gf, 300 gf, 300 cm3
2) 200 gf, 200 gf, 200 cm3
3) 100 gf, 100 gf, 100 cm3
4) 50 gf, 50 gf, 50 cm3
8. A cylindrical solid of area of cross-section
of 0.005 m 2 and length 0.60m is
completely immersed in water.
Calculate the
i. weight of the solid in S.I system of
units.
ii. upthrust acting on the solid in S.I
system of units.
iii. apparent weight of the solid in water.
(Take g = 10 m/s 2 , density of water =
1000 kg/m3, density of solid=1500 kg/m3)
1) 20N, 25N, 11N
2) 45N, 30N, 15N
3) 10N, 10N, 10N
4) 12N, 12N, 12N
9. The following observations were taken
while determining the relative density of
a liquid.
weight of the solid in air = 0.100 kgf
weight of the solid in liquid = 0.080 kgf
weight of the solid in water = 0.075 kgf
Calculate (i) the apparent loss in weight
of solid in liquid (ii) the apparent loss in
weight of solid in water and (iii) the
relative density of the liquid.
1) 0.010 kgf, 0.02 kgf, 0.6
2) 0.2 kgf, 0.025 kgf, 0.9
3) 0.020 kgf, 0.025 kgf, 0.8
4) 0.030 kgf, 0.05 kgf, 1.8
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
10. A solid weighs 200 gf in air, 160 gf in
water and 170 gf in a liquid. Calculate
the relative density of the solid and that
of the liquid.
1) 5, 0.75 2) 6, 1.7 3) 3, 2.75 4) 4, 1.25
DAY-16 : SYNOPSIS
Electric current: The rate of flow of
charge in a circuit is called electric
current.In other words, it is the amount
of charge flowing per second.It is denoted
by the letter I.
If Q is the charge which is flowing through
a conductor in time t, then current is
given by I 
Q
t
Unit Of Current: The S.I unit of current is
ampere and it is denoted by the letter ‘A’.
The S.I unit of Q is coulomb and that of t
is second.
Thus, the S.I unit of electric current is
1coulomb
=1A
1sec ond
Definition of ampere: When a charge of
coulomb flows through a conductor in one
second, then the current flowing through
the conductor is said to be one ampere.
Thus, when 1 coulomb of charge flows
through a conductor in 1 second , then
the current flowing through it is said to
be 1 ampere
1ampere 
1coulomb
1sec ond
Smaller units of electric current:
Some times smaller units of current are
also used.These are microampere and
milliampere. 1microampere  1 A  106 A
1milliampere  1mA  103 A
Bigger unit of electric current:
Sometimes the magnitude of the current
flowing in a conductor is very large.
This large magnitude of current is expressed in bigger units,such as kilo ampere and mega ampere
1kiloampere( KA)  1000 A  103 A
1megaampere(MA)  1,000, 000 A  106 A
65
CLASS-VIII
Flow of current: In metals, the moving
charges are the electrons constituting
the current, while in electrolytes and
ionized gases, electrons and positively
charged ions are the ions moving charges
which constitute current
The charge on an electron is negative
and is 1.6 x1019 coulomb(symbol C)
Therefore, I C charge is carried by
1
 6.25 x1018 electrons.Hence if I A
19
1.6 x10
current flows through a conductor, it
implies that 6.25 x1018 electrons pass in 1
second across the cross section of the
conductor
The direction of current is conventionally taken opposite to the direction of
motion of electrons
If n electrons pass through a cross section of a conductor in time t, then total
charge passed
Q  nxe and current in conductor I 
Q ne

T
t
Instrument by which current measured:
Current is measured by an instrument
called ammeter
DAY-16: WORKSHEET
1. A body is said to have 1 coulomb charge,
if it has excess or defict of:
1) 6.25 x106 electrons 2) 6.25 x1016 electrons
3) 6.25 x1018 electrons 4) 6.25 x109 electrons
2. The charge on one electron is:
1) 1.6 x109 C
2) 1.6 x1019 C
3) 1.6 x1012 C
4) 1.6 x1018 C
3. It a current of 25mA flows through a
circuit for 1 hour, the charge flowing
through conductor is:
1)9C
2)900C 3)90C
4)60C
4. When one coulomb charge flows through
a conductor in one second, the current
flowing through the conductor is
1)1A
2)10A
3)2A
4)4A
5. The rate of flow of charge in a circuit is
called
1)Electric potential
2)Electric current
3)Electric charge
4)All of these
NARAYANA GROUP OF SCHOOLS
MPC BRIDGE COURSE
6.If ‘Q’ is the charge which is flowing through
a conductor in time ‘t’ then current ‘I’ is
given by
2
Q
3) I  Q  T 4) I  Q
T
t

19
7. If the electronic charge is 1.6 x10 C, then
1) I  Qxt 2) I 
the number of electrons passing through
a section of wire per second, when the
wire carries a current of 2ampere is
1)1.2×10–12A 2) 1.25×1019A 3) 1.6×10–8A
4) 1.6×104A
8. Current of 4.8 ampere is flowing through
a conductor.The number of electrons
crossing per second the cross-section of
conductor will be___
1) 3 x1019 2) 3x1015 3) 3 x1011 4) 3x1012
9. A conductor carries a current of 2A.How
long will it take for 1800C of electricity
to flow through a given cross-section?
1)15min 2)10 min 3)5 min 4)1 min
10.The speed of an electron in an orbit of
radius ‘r’ is ‘V’ units.Find the strength of
current.
1)
2Ve
r
2)
Ve
2 r
3)
2 r
Ve
4)
r
2Ve
11.Frequency of an electron in an orbit is
‘f’.Find the strength of the current
1) fxe
2)
f
e
3) f 2 xe
4) fxe 2
12.How many electrons pass through a wire
in 1 minute if the current passing through
the wire is 200mA?
1) 7.5 x1016 2) 7.5 x1011 3) 7.5 x1019 4) 2.5 x1019
13.In metals moving charges constituting
current are
1)Electrons
2)Positively charged ions
3)Protons
4)All of these
14.In ‘n’ electrons pass through a cross-section of a conductor in time ‘t’, then total
charge passed is
1) Q 
n
e
2) Q  nxe 3) Q 
n2
4) Q  n 2  e
e
15.In electrolytes and ionized gases moving charges that constituent current are
same options
1)Electrons
2)Positively charged ions
3)Protons
4)Both (1) and (2)
16.One coloumb per second=_____ampere
1)1
2)2
3)3
4)4
66
CLASS-VIII
DAY-17 : SYNOPSIS
Hot and Cold:Hold a piece of ice on your
palm.You feel cold on your palm.Now, dip
your finger in warm water. You feel
warm.Hold a glass of boiling tea in your
hand.You feel hot.Thus you feel cold,
warm and hot. Try putting your index finger in boiling tea. It may not be possible
to hold your finger in boiling tea or try
even touching it. You may burn your finger if you keep it in boiling tea even for a
second. You may say that tea is boiling
hot and it should not be touched.
Same is true about the weather.It is cool
at night.It is warm in the shade during
day.It is hot in sun. We make use of our
sense of touch to learn about cold, warm
and hot.We are able to sense heat in an
object. Heat is something which produces a sensation in our body by way of
which we make out whether a body is
cold, warm or hot.
Heat is form of an energy:
Energy is the ability to do work. When
an object has the ability to do work, we
say that the object has energy. Heat has
also the ability to do work.
For example, the steam engine pulls a
train converting heat into mechanical
energy.Heat can also be converted to
other forms of energy.
For example, when charcoal is heated,
it gives light. Here heat produces light.
In a hot air balloon the hot gases, being
lighter than the surrounding air, rise up
in the air and are made to lift weights.
Here heat is used to produce mechanical energy. The heat in a fire cracker
produces both sound and light.
Other forms of energy can also be converted to heat energy. For example , you
can feel the heat produced from the mechanical energy by rubbing your palms
vigorously against each other.
When a candle burns in air. chemical
energy is converted into heat. In an electric blub, electrical energy is converted
into light and heat.
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So as in an electric heater the electrical
energy is converted into heat energy.
When heat is given to a substance, we
find the following effects:
1.Rise in temperature 2.Expansion
3.Change in state
General effects of heat energy:
a) Heat energy brings about change in temperature
b) Heat energy brings about change in dimensions
c) Heat energy brings about change in state.
d) Heat energy affects living beings
Flow of heat energy: Heat flows from hot
body to a cold body. A body which is losing heat is feeling the other body to be
cold. A body which is gaining heat is feeling the other body to be hot.Thus heat
always flows from a body of higher temperature to the body at lower temperature.
Concept of Heat:Heat is a form of energy
which always flows from a hot body to a
cold body. (or) Heat is a form of energy
which makes any object hot or cold. Heat
energy is also called thermal energy.
Unit of Heat:
S.I unit of heat is joule(J).
Another commonly used unit of heat is
calorie(cal).
One calorie is the quantity of heat energy required to raise the temperature
of 1g of water through 10 C.
1 cal = 4.2J, 1k.cal = 1000calories.
Note: i) As heat is a form of energy.So its
unit is same as energy.
ii) It is a scalar quantity.
Concept of Temperature: When we touch
a hot object our hand becomes warm,
because heat flows from the object to our
hand.We say that the object is at a higher
temperature than our hand.
If we touch a piece of ice our hand feels
cold, because heat flows from our hand
to the ice.We say that ice is at a lower
temperature than our hand.
This suggest that temperature tells us
how hot a body is.
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CLASS-VIII
It is the degree of hotness or coldness of
a body. Heat flows in the direction of fall
of temperature. When an object is heated,
its temperature rises. When it is cooled,
its temperature decreases.
If two bodies having unequal temperature
are brought together, heat flows from the
body at higher temperature to the body
at lower temperature, till the two bodies
are at the sam temperature.
Thus, the degree of hotness or coldness
of the body is called temperature.
Mathematically, Temperature is heat per
unit mass.
Unit of temperature:
S.I unit of temperature is kelvin(K).
Other unit of temperature is degree Celsius( 0 C ) and degree Fahrenheit( 0 F ).
Note: i)It is a scalar quantity
ii) Thermometry is the branch of heat dealing with the measurement of temperature.
Difference between heat and temperature
Heat
Temperature
1.Heat is a form of energy.
1. Temperature indicates
Hence it has the capaciy
the thermal condition of
for doing work
a body which may be
stated as how much hot
or how much cold the
2.Heat is the cause
2.Temperature is the
effect
3.Two bodies of same
3.Two bodies of
same
substance having different
substance
having
masses may have same
different
masses may
amount of heat but
have same
different temperature
4.Heat contents of a body
4.Temperature
of a
do not decide the directio
body decides
the
-n of heat flow from the
direction of
heat flow fr
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body
-om the body
5.S.I unit of heat if joule
5.S.I unit of
temperature
(J) (energy unit)
is Kelvin(K)
Thermometer:
The device for measuring the temperature of a substance is called a
thermometer.(“thermo” is a Latin word
which means heat and ‘meter’ means a
measuring device).
Mercury Thermometer:
We know that the substances expand
when heated and contract when cooled.
This principle is utilized for the construction of a thermometer.
Uses of Mercury in thermometers:
Mercury is used in thermometers because
of the following advantages
i) It expands evenly as the temperature
rises.
ii) It is a good conductor of heat.
iii) Its density is higher.
iv) Its very sensitive in expansion.
v) It does not stick on the wall of a glass
tube.
vi) It has very low freezing point and a very
high boiling point.
The calibration of thermometer involves
fixing of two points on it.
One lower fixed point and other upper
fixed point.
Lower fixed point: The melting point of
pure ice at normal atmospheric pressure
is taken as lower fixed point(L.F.P).
Upper fixed point: The boiling point of pure
water at normal atmospheric pressure is
taken as upper fixed point(U.F.P).
Freezing and Boiling point:
A thermometer has two standard marking on its glass tube. These are called
‘lower fixed point’ and ‘upper fixed point’.
The lower fixed point of thermometer
scale is the temperature of melting
ice(ice point). It is given a value of
00 C .The upper fixed point of a thermometer scale is the temperature of boiling
water. It is given a value of 1000 C (Steam
point).
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CLASS-VIII
DAY-17: WORKSHEET
1. Heat is the abilit to do
1)Work
2)Force 3)Time
4)Mass
2. Heat is a form of
1)Matter 2)Energy 3)Fluid
4)None
3. The most important naturally occuring
source of heat is
1)Earth
2)Sun
3)Moon
4)None
4.The degree of hotness and coldness of the
body is called
1)Heat
2)Temperature
3)Pressure
4)Force
5. A Thermometer measures
1)The quantity of heat 2)Density
3)Temperature
4)Humidity
6. The thermometric liquid used in a mercury thermometer is
1)alcohol
2)water
3)Mercury
4)Benzene
7.Galileo’s thermometer was based on the
property of expansion
1)Gases
2)Liquids
3)Solids
4)both (2) and (3)
8. S.I unit of temperature is
1)Kelvin
2)Celsius
3)Fahrenheit 4)Reaumer
9. Temperature is a _______quantity
1)Vector
2)Scalar
3)both (1) and (2)
4)Neither a vector nor a scalar
10.When heat is given to a substance the
following effects observed are
1)rise in temperature
2)Expansion of the substance
3)Change in scale of the substance
4)All the above
11.Heat always flows from
1)higher temperature to lower temperature
2)lower temperature to higher temperature
3)some times higher to lower to higher
temperature
4)none of these
12.1 cal=_______
1)1.8J 2)2.4J
3)4.2J 4)5.1J
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13.The device for measuring the temperature of a substance is called a _______
1)barometer
2)thermometer
3)Voltmeter
4)Ammeter
14.Thermometer works on the principle of
1)Substances expandson heating
2)Substances contracts on heating
3)Substances have no effect on heating
4)None of these
15.The substance whose property is utilised
for measuring temperature is called
1)Thermometric substance
2)Calorimetric Substance
3)Solid substance
4)None of these
16.If a body is at a temperature higher than
the room temperature the level of mercury in the thermometer’s stem
1)Falls
2)remain at the same position
3)rises
4)may rise or fail
17.The following one determines the
direction of flow of heat
1)Temperature 2)Thermometer
3)Altimeter
4)Ammeter
18.If 1 cal=4.2J, then 1 K cal=_______J.
1)420
2)4200
3)42000 4)42
19.The boiling point of water is
1) 1800 C
2) 120 C
3) 400 C
4) 1000 C
20.The melting point of ice is
1) 00 C
2) 2730 C
3) 400 C
4) 1000 C
21.Normal temperature of human body is
1) 98.40 F 2) 1200 F 3) 800 F
4) 37 0 F
22.1 Kilo calorie=______calories
1)10
2)100
3)1000
4)10000
23.Units(s) of heat energy is
1)Joule 2)calorie 3)Kelvin 4)both 1&2
24.The lower fixed point of a thermometer
scale is the temperature of melting of
1)Ice
2)Water 3)Mercury4)Alcohol
25.The lower fixed point of a thermometer
scale is also called
1)Ice point
2)Water point
3)Liquid Point
4)Steam Point
26.The boiling point of pure water at normal atmospheric pressure is taken as
1)Upper fixed point 2)Lower Fixed Point
3)Both (1) & (2)
4)Neither (1) nor (2)
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