8th class physics bridge program

CLASS-VIII MPC BRIDGE COURSE

NARAYANA GROUP OF SCHOOLS 44

PHYSICS

DAY-1 : SYNOPSISPhysical Quantity : The quantities whichare measurable in physics are calledphysical quantities.Ex : Length, mass, time, speed, etc.Unit : Unit is a standard which is usedfor the measurement of physical quantity.Ex : (i) unit of length is metre(ii) unit of time is secondNumerical value of physical quantity :The number of times a unit is present ina given physical quantity is calledNumerical value of physical quantity.Relation among Physical quantity,Numerical value and Unit :Physicalquantity = Numerical value × UnitEx : Let length of table = 3 metreHere 3 is the Numerical value and metreis the standard unit.Fundamental quantities : The physicalquantity which does not depend upon (orindependent of) other physical quantitiesare called fundamental quantities.Ex : Length, mass and time etc.Derived physical quantities : Thephysical quantities which depends onfundamental physical quantities arecalled derived physical quantities.Ex : Area, Volume, Speed, etc.Fundamental units : The units used formeasuring fundamental quantities arecalled Fundamental units. These areindependent of other units.Ex: The fundamental unit of length ismetre.Derived units : Derived units are theunits of derived physical quantities whichare expressed in terms of fundamentalunits.Ex: The derived unit of speed is ms–1 (read as metre per second).Systems of Units : There are four systemof units:

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 oftime is second.

ii. French or C.G.S system : In C.G.Ssystem, the unit of length is centimetre.The unit of mass is gram. The unit oftime is second.

iii. M.K.S system or metric system : InM.K.S system, the unit of length ismetre.The unit of mass is kilogram.The unit oftime is second.

iv. International system or S.I : S.I.system has seven basic units and twosupplementary units.

Quantity Unit Symbol

length metre mmass kilogram kgtime second stemperature kelvin Kluminous intensity candela cdelectric current ampere Aamount of mole molsubstancePlane angle Radian radsolid angle steradian sr

Multiple and sub multiple factors :

Multiplication factor Name Symbol

1012 tera T109

106

103

102

1010-1

10-2

10-3

gigamegakilohectodecadecicentimilli

GMkhdadcm

Multiple and sub multiple factors

10-6

10-9

10-12

micronanopico

np

10-15

10-18femtoatto

fa



DAY-1: WORKSHEET1. Pick the odd man out.

1] Length 2] Mass3] Time 4] Area

2. F.P.S stands for1] Foot, pound, second2] France, Paris, Spain3] Force, pressure, second4] Foot, Pace, Second

3. C.G.S stands for1] Centimetre, Gravitation, second2] Centisecond, gram, second3] Centimetre, gram, second4] None of these

4. Ampere is the unit of1] Length 2] Temperature3] Luminous intensity 4] Current

5. Number of fundamental physicalquantities in M.K.S system are1] Two 2] Three 3] Seven 4] Six

6. The temperature standard metre rodmade of platinum - Iridium alloy kept inthe archives of serves near Paris is1] 0°C 2] 27°C 3] 100° C 4] None

7. Statement I : Numerical value = Physicalquantity × unitStatement II : Current strength is afundamental physical quantity accordingto S.I system.1) Statement I & II is true2) 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 followinga) Length b) Massc) Current strength d) Temperaturepick the correct statement1] Length is the odd man out.2]All are Fundamental PhysicalQuantities according to M.K.S system.3]All are Fundamental PhysicalQuantities according to S.I system.4] Current strength is a derived quantity.

9. Match the following.List - A List - B(a) Temperature (e) Mole(b) Luminous intensity (f) Kelvin(c) Amount of substance (g) Candela1] a e; b f; c g2] a f; b g; c e3] a g; b f; c e4] 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) 1012. Match the following.

List A List B

i) 365 14 days a) 1 Century

ii) 10 decades b) 3600 sec

iii) 1

1440 the part of a mean solar dayc) 1 year

iv) 1 hour d) 1 minute1) i a, ii b, iii c, iv d2) i a, ii b, iii d, iv c3) i b, ii a, iii c, iv d4) i c, ii a, iii d, iv b

DAY-2 : SYNOPSISDensity : The density of a substance isdefined as the mass per unit volume ofthe substance.Density of a substance (d)

= mass of the subs tance(M)

volume of the subs tance(V) MdV

Unit : C.G.S unit : g cm–3

S.I. unit : kg m–3

Relationship between S.I. & C.G.S unitsof density : 1g/cm3 = 1000 kg/m3

Relative density : It is the ratio of thedensity of the substance to the densityof water at 40C.Thus, Relative density (R.D.)

0density thesubs tancedensityof water at 4 C

Units of relative density : As the relativedensity is the ratio of two similarquantities hence it is a pure numberand therefore has no units.



Relationship between density and relativedensity :

1) Density of a solid in S.I. Unit = R.D. ofthe 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. ofthe solid × Density of water (in C.G.S.unit) = R.D. of the solid × 1g/cm3

DAY-2: WORKSHEET1. 5 litres of alcohol has a mass of 4 kg. The

density of alcohol is ___

1) 0.8 3kg

m 2) 80 3kg

m

3) 800 3kg

m 4) 8000 3kg

m2. A 50kg mass of a body is immersed in

water then we observe the mark onmeasuring cylinder as 70cm3. If the bodyis taken out from the cylinder, the markobserved on cylinder is 50cm3. Then thedensity of that body is [The graduationsmarked 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 3g

cm and that

of wood is 800 3kg

m . what do you

understand by these statements ?1) The matter is only densely packed inlead than wood.2) The matter is only densely packed inwood than lead.3) The density of lead is greater than thedensity of wood.4) Both (1) and (3).

4. Take two identical 100cm3 beakers. Fillone beaker completely with water andthe other with kerosene oil. Place thebeakers in the scale pans of an ordinarybeam balance as shown in fig. It isobserved that the beaker filled with waterhave more mass than the beaker filledwith kerosene oil. Because

(1) the matter in water is more denselypacked than in kerosene oil.(2) the matter in kerosene oil is moredensely packed than in water.(3) Water and kerosene oil are occupiedthe same volume.(4) None of these.

5. Mass of liquid = 72 gInitial volume of water in measuringcylinder = 24 cm3

Final volume of water + solid inmeasuring cylinder = 42 cm2

From the above data the density of solidis ___________1) 4000 kg/m3 2) 3.0 g/cm3

3) 372 g

cm424) 4.0 kg/m3

6. If relative density of gold is 19.3. Thenthe density of gold is __________timesgreater 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

and the mass is 18 kg, then the relative

density of the wooden cube is __________

1) 19.53 × 10–7 2) 1

5123)

151200

4) 132

9. An iron cylinder of radius 1.4 cm andlength 8 cm is found to weigh 369.6 g.The relative density of iron cylinder is

Take volume of cylinder 2= r .1) 7.2 2) 7.5 3) 8 4) 8.2

10. Calculate the mass of a body whosevolume is 2m3 and relative density is 0.521) 1040 kg 2) 1000 kg 3) 950 kg 4) 750 kg



DAY-3 : SYNOPSISScalars:- The physical quantities which

have only magnitude but no direction, arecalled scalar quantities or simply calledas scalars.

Examples of scalar quantities:- 2 kg sugartells about the magnitude of its mass,but has no direction.Mass, length, time, distance covered,temperature, area, volume, density, tem-perature etc are a few examples of sca-lars. The scalars can be added, sub-tracted, multiplied and divided by ordi-nary laws of algebra.A scalar is specified by mere number andunit, where number represents its mag-nitude.A scalar may be positive or negative.A sca-lar can be represented by a single letter.

Vectors:- The physical quantities which areexpressed in magnitude as well as di-rection are called vector quantities orsimply called as a vectors. They shouldalso 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, andmultiplied by ordinary laws of algebra.A vector in writing, can be representedeither by a single letter in bold face or bya single letter with an arrow head onit.Diplacement = SGeometrically or graphically, a vector isrepresented by a straight line with anarrow head i.e. arrowed line. Here thelength of the arrowed the line drawn ona suitable scale represents the magni-tude and the arrow head represents thedirection 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 ob-ject is displaced from A to C.

DAY-3: WORKSHEET1.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 22. The physcial quantities which are ex-

pressed in magnitude as well as direc-tion are called

1)Scalars 2)Units 3)Vectors 4)both 1 and 23. Ordinary laws of algebra are applicable

for 1)Scalars 2)Vectors 3)both 1 and 2 4)neither 1 nor 24. _____ cannot be added, subtracted and

multiplied by ordinary laws of algebra. 1)Scalars 2)Numbers 3)algebraic equations 4)Vectors5. Graphically a vector is represented by a 1)line 2)curve 3)circle 4)directed line segment6. Directionless quantity is called ______ 1)vector 2)Scalar 3)both 1 and 2 4)neither 1 nor 27. The magnitude of a vector is 1)Scalar 2)vector 3)direction 4)can’t say8. Length of the directed line segment rep-

resents 1)Direction 2)Orientation 3)spin 4)Magnitude9. Arrow head of a vector represents 1)Magnitude 2)sense of direction 3)sense of rotation 4)spin10. Mass is a

1)Scalar quantity 2)vector quantity3)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 212. Mr.Satish travelled from Narayana

Olympiad school, Narayanaguda toDilsukhnagar 10km towards North. Dis-placement of Mr.Satish is a

1)Scalar 2)Vector 3)has only magnitude4)has only direction



13. Among the following the quantity whichis not a scalar?

1)20 kg 2)15m

3) 40 s 4)13m due north

14. Which of the following whcih is the ex-ample for vector?

1)15m due East 2)20m due west

3)10m due south 4)5m

15. Among the following which is the ex-ample for vector?

1)East direction 2)20m

3)20m due East 4)can’t say

16. Among the following the quantity whichis not a scalar?

1)8 kg 2)10m 3)20 sec 4)13m due west

17. In the following which set is a completescalar 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 whichone is scalar?

1)18m due west 2)20m due south

3)30m 4)23m due north

18. The set containing only vectorquantitites is

1)speed,velocity 2)time,displacement

3)energy,mass 4)displacement,velocity

19. A physical quantity which has only mag-nitude and no specific direction

1)weight 2)Mass

3)force 4)displacement

20. Which of the following pair are scalarquantities

1)Speed, velocity

2)relative density, volume

3)velocity,area4)density,displacement

DAY-4 : SYNOPSISBasic Terms related to Kinematics:Distance : It is defined as the actual pathfollowed by a body between the pointsbetween which its moves.Unit : C.G.S unit : cm S.I unit : mNote: The distance travelled by body isalways positive.Displacement : It is the shortest distancebetween between initial and final pointin a definite direction.Unit : C.G.S unit : cm S.I unit : mNote : (i) For a moving body displacementcan be positive, negative or zero. (ii) Ifinitial point and final points are samethen displacement is zero.Scalar quantities : A physical quantitywhich is described completely by itsmagnitude is called a scalar quantity. Ithas only magnitude and no specificdirection.Ex: Length, distance, area, volume, mass,time and energy are examples of scalarquantities.Vector quantities : A physical quantitywhich is described completely by itsmagnitude as well as specific directionis called vector quantity. It has bothmagnitude and direction.Ex : Displacement, velocity, acceleration,force and weight are examples of vectorquantities.Speed: The rate of change of motion iscalled speed. The speed can be found bydividing the distance covered by the timein which the distance is covered.

Formula : Dis tance travelledSpeedTime taken

Units : C.G.S unit : cm/sS.I unit : m/s Nature : ScalarKinds of Speed :

a) Uniform speed : When a body coversequal distances in equal intervals of time(however small the time intervals maybe), the body is said to be moving with auniform speed.Ex : A rotating fan, a rocket moving inspace, etc., have uniform speeds.



b) Non-Uniform Speed : When a body coversunequal distances in equal intervals oftime, the body is said to be moving witha nonuniform speed.Ex : A train starting from a station, a dogchasing a cat, have variable speeds.

c) Average Speed:- When a body is movingwith a variable speed, then the averagespeed of the body is defined as the ratioof total distance travelled by the body tothe total time taken i.e., Average speed

Total dis tance coveredTotal time taken to cover the dis tance

Velocity : Velocity is the rate of changeof motion in a specified direction.

Formula : Velocity = displacement

timeUnits : C.G.S. unit : cm/sS.I. unit : m/s Nature : VectorKinds of velocity :

a) Uniform velocity :- When a body coversequal distances in equal intervals of timein a specified direction, (howsoever short,the time intervals may be) the body issaid to be moving with a uniform velocity.Ex : Imagine a car is moving along astraight road towards east, such that inevery one second it covers a distance of5m then the car is said to be moving withuniform velocity.

b) Variable Velocity:- When a body coversunequal distances in equal intervals oftime in a specified direction or equaldistances in equal intervals of time, butits direction changes, then the body issaid to be moving with a variable velocity.Ex : Now imagine the car is moving alonga circular path, such that it is covering5m in every one second but as thedirection of the car is changing at everyinstant, we say the car is moving withvariable velocity.

c) Average velocity:- It is the ratio of totaldisplacement to total time taken.

Total displacementAveragevelocityTotal time taken

DAY-4: WORKSHEET1. Which is a vector quantity among these ?

1) My mass is 20 kg2) Himalayas are in the northern India3) The sun rises in the east.4) 200 m towards north is Ramoji film cityfrom my house.

2. A cyclist moves from a certain point X andgoes round a circle of radius ‘r’ andreaches Y, exactly at the other side of thepoint X, as shown in figure.

X Yr

O

The displacement of the cyclist would be__________

1) r 2) 2 r 3) 2r 4) 2r

3. In the above problem, the distancecovered by the cyclist would be

1) r 2) 2 r 3) 2r 4) 2r

4. A man walks 8m towards East and then6m towards north. His magnitude ofdisplacement is ___________1) 10 m 2) 14 m 3) 2 m 4) zero

5. A player completes a circular path ofradius ‘r’ in 40s. At the end of 2 minutes20 seconds, displacement will be1) 2r 2) 2 r 3) 7 r 4) Zero

6. A horse runs a distance of 1200m in 3min and 20 s. The speed of the horseis_________1) 60 ms–12) 65 ms–13) 40 ms–14) 6ms–1

7. When a body covers first one thirddistance with speed 1m/s, the secondone third distance with speed 2m/s andthe last one third distance with speed3m/s then average speed is __1) 2 m/s 2) 1.79 m/s3) 2.66 m/s 4) 1.64 m/s



8. An insect crawls a distance of 4m alongnorth in 10 seconds and then a distanceof 3m along east in 5 seconds. Theaverage velocity of the insect is _

1) 7 m/sec5

2) 1 m/sec5

3) 5 m/sec

154) None of these

9. A car travels half the distance withconstant velocity 50 km/h, and anotherhalf with a constant velocity of 40 km/halong a straight line. The average velocityof the car in km/h is _______

1) 45 2) 44.4 3) 0 4) 50 40

DAY-5 : SYNOPSISAcceleration : The rate of change ofvelocity of a body is called acceleration.

Formula : Acceleration = Change in VelocityTime taken

Units : C.G.S. unit : cm/s2

S.I. unit : m/s2 Nature : VectorUniform Acceleration: When a bodydescribes equal changes in velocity inequal intervals of time (however smallmay be the time intervals) it is said to bemoving with uniform acceleration.Equations of motion for a body movingwith uniform acceleration in a straightline:

a) First Equations of Motion : It gives thevelocity acquired by a body in time t whichis v = u + at where, v = Final velocity ofthe body [velocity after time (t) seconds]u = Initial velocity of the body [velocity attime (t) = 0 second] a = Acceleration(uniform) t = Time taken

b) Second Equation of Motion : It gives thedisplacement of the body in a time t,

which is 21s = ut+ at2

where, s =

displacement of the body in time tc) Third Equation of motion : It gives the

velocity acquired by a body indisplacement ‘s’ which is v2 – u2 = 2as

Points to remember :i) If a body starts from rest, its initial

velocity, u = 0ii) If a body comes to rest (it stops),its final

velocity, v = 0iii)If a body moves with uniform velocity, its

acceleration, a = 0Distance travelled in nth second : Itgives the distance travelled by the body

in nth second which is nas = u + 2n -12

where, sn = distance travelled by the bodyin nth second

DAY-5: WORKSHEET1. The velocity of car changes from 18 km/

h to 72 km/h in 30s. What will be itsacceleration 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 of72 km/h for 15s on a National Highway.Its acceleration and the distancetravelled are1) 0m/s2, 200m 2) 0m/s2, 300m3) 20m/s2, 300m 4) 2m/s2, 400m

3. A motor bike is moving with a velocity of5m/s. It is accelerated at a rate of 0.6m/s2 for 20s. Then the final velocity of motorbike is1) 16m/s 2) 18 m/s 3) 15m/s 4) 17m/s

4. If a bus starts from rest and attains aspeed of 36 km/h in 10 minutes whilemoving with uniform acceleration, thenthe acceleration of the bus is

1) 140

m/s2 2) 160

m/s2

3) 150

m/s2 4) 130

m/s2

5. A girl running in a race accelerates at2.5 m/s2 for the first 4s of the race. Howfar does she travel in this time? (Assumea girl has started from rest)1) 10m 2) 15m 3) 20m 4) 25m



6. A body starting from rest travels adistance of 200m in 10 seconds, then thevalue of acceleration is1) 2ms–2 2) 4ms–2 3) 5ms–2 4) 3ms2

7. A scooter moving at a speed of 10 m/s isstopped by applying brakes which producea uniform acceleration of, –0.5 m/s2. Howmuch distance will be covered by thescooter 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 uponthe application of brakes comes to restwithin a distance of 5m. The retardationproduced by the brakes is1) 90 m/s2 2) 80 m/s2

3) 95 m/s2 4) 85 m/s2

9. A body originally at rest is subjected touniform acceleration of 4ms –2. Thedistance travelled by it in 5th second is1) 10 m 2) 15 m 3) 18 m 4) 20 m

DAY-6 : SYNOPSISAcceleration due to gravity:Theacceleration of a freely falling body is calledacceleration due to gravity. The value ofacceleration due to gravity is 9.8m/s2.Equation of motion for freely fallingbody : For a freely falling body, with initialvelocity, u = 0, the velocity continuouslyincreases as it falls through a height.The direction of ‘g’ and the direction ofmotion of the body are same (i.e.downwards).Therefore g is taken aspositive under gravity. Here, g = +ve,u = 0, S = hThus the first equation of motion, v = gt

The second equation of motion, 21h gt2

The third equation of motion, v 2gh

The equation for the distance covered in

nth second, nthgh 2n 12

Equations of motion for a bodyprojected vertically upwards : When abody is projected vertically upwards, itsvelocity decreases continuously since itsmotion is against gravity, hence g is takenas –ve.

When a body is projected verticallyupwards equations of motion will be

v = u – gt, h = ut – 21 gt2

, v2 – u2 = –2gh

The equation for the distance covered in

nth second, nthgh u 2n 12

Maximum height reached by a body

thrown vertically up : 2uh

2g

Where, u = initial velocity h = maximumheight g = acceleration due to gravityTime of ascent (ta) : The time taken bybody thrown up to reach maximum height‘h’ is called its time of ascent. Let ta be

the time of ascent, autg

Time of descent : The time taken by afreely falling body to touch the ground is

called the time of descent, dutg

Note : Time of ascent is equal to the timeof descent in the case of bodies movingunder gravity (neglecting air resistance).Time of flight : It is defined as the totaltime for which the body stays in air whenprojected vertically up. It is equal to sumof time of ascent ta and time of descent td.

time of flight = ta + td u ug g

Time

of flight = 2ug

DAY-6: WORKSHEET1. If a stone is dropped from the top of a

building and is found to reach the groundin 1 second, then the height of thebuilding is1) 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 secondis1) 53.9 m 2) 35.9 m 3) 93.5 m 4) 59.3 m



3. A body dropped from a height h strikesthe ground with a velocity of 3 m/s.Another body of same mass is droppedfrom the same height h with an initialvelocity of 4 m/s. Final velocity of secondmass with which it strikes the groundwill be1) 3 m/s 2) 4 m/s 3) 8 m/s 4) 5 m/s

4. A body is falling under gravity. Thedistance covered in 1st, 2nd and 3rdsecond of its motion are 1) 1 : 3 : 5 2) 3 : 1 : 53) 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) maximumheight attained by it are(Take g = 10 ms–2).1) 15 ms–1, 11.25 m 2) 17 ms–1, 11.25 m3) 18 ms–1, 12.25 m 4) 16 ms–1, 11.25 m

6. A stone is thrown vertically up with aninitial velocity of 10 m/s. Then themaximum height reached and the time ofascent are (Take g = 10 m/s2)1) 6m, 2s 2) 3m, 1s 3) 4m, 2s 4) 5m, 1s

7. A ball is thrown vertically upwards. Itreturns 6 s later. Then (i) the greatestheight reached by the ball and (ii) theinitial velocity of the ball are (Take g = 10ms–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 withan initial velocity of 49 m s–1. Then (i) themaximum height attained, (ii) the timetaken by it before it reaches the groundagain are (g = 9.8 m s–2)1) 12.5 m, 10s 2) 122.5 m, 10s3) 122.5 m, 1s 4)122.5 m, 100s

DAY-7 : SYNOPSIS1. DISPLACEMENT – TIME GRAPHS

These graphs are very useful in studyingthe linear motion of the body. Thedisplacement is plotted on the Y – axisand the time on X – axis.These graphs are very helpful in findingthe velocity of body, as the slope of graph

Y axisX axis

is equal to Displacement

Time.

Following are various types of displacementtime graphs:(a) Displacement – time graph is parallel totime axis.If a displacement – time graph isparallel to time axis as shown in figure (a) itmeans that the body is not changing itsposition with respect to time. In otherwords, body is stationary.

0

2

46

8

1 2 3 4 5 6Time in (seconds)

Dis

plac

emen

t in

(m)

Fig. (a)(b)When displacement – time graph is astraight line, but is not parallel to time axis.If the displacement – time graph is a straightline, such that it starts from origin and movesaway from time and displacement axis, thebody is said to be moving with uniformvelocity. Figure (b) shows a displacement-time graph of a body. The values ofdisplacement and time are shown in the tablebelow:

Displacementin metres 0 5 10 15 20 25

Time in seconds 0 1 2 3 4 5

Since graph is a straight line, therefore itmeans that displacement is proportional totime. In other words, body is covering equaldistances in equal intervals of time i nspecified direction, and hence, is moving witha uniform velocity.The slope of this graph gives uniform velocity.

Thus, velocity of body x ABt BC

10m2s

= 5m/s



Where x is short distance and t is ashort interval of time.

0 1 2 3 4 5

5

10

15

20

25

x = 10m

A

C B

t = 2s

Time in (seconds)

Dis

plac

emen

t (x

) in

(m)

Fig. (b)

(c) When displacement – time graph isnot a straight line: If the displacement -time graph is a curve as shown infigure(c).It represents ‘variable velocity’of a moving body.

Slope of the graph BCDC

at any point

gives the velocity at that instant . Forexample, velocity at A is

8 2 mBC 6 m

DC 3 1 s 2 s

= 3m/s

Fig. (c)Conclusions from displacement–time

graph1. 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. Thevelocity can be found out by finding theslope of the graph.

3. The graph can never be parallel todisplacement axis, as it means thatdisplacement increases indefinitely,without any increase in time, which isimpossible.

4. If graph is a curve, it means the body ismoving with a variable velocity, andhence, has some acceleration.

1. VELOCITY – TIME GRAPHS(i) In these graphs generally, the velocityis plotted on Y-axis and time on X-axis.The slope of such graphs givesacceleration.(ii) As, velocity time = acceleration, theacceleration will be positive if the slopeis positive, and negative if the slope isnegative.(iii)The area of graph under velocity –time curve, gives displacement of body.Displacement = Velocity × Time.(a) When velocity – time graph isparallel to time axis(i) Figure(a) represents velocity – timegraph PQ, when a body is moving with auniform velocity of 20 ms–1.As the slope of graph is zero, thereforeits acceleration is zero.(ii) The distance covered by the body inspecified direction (displacement) can becalculated by finding the area ofrectangle PQRS. Thus, Displacement =PS × SR = 20ms–1 × 4s = 80m.

1 2 3 4

5

10

15

20

S

QP

Time in (seconds)

Vel

ocit

y in

(ms-

1 )

R

Fig. (a)

Conclusion : If velocity – time graph isparallel to time axis, then:a) Body is moving with uniform velocity.b) Its acceleration is zero.c) Its displacement can be found by findingthe area of the graph.(b) When velocity – time graph is astraight line, but not parallel to timeaxis.



C

Fig. (i) Fig. (ii)Figure (i) represents a velocity – timegraph when the body starts from rest,and its velocity increases at a uniformrate.

The slope of the graph AC, i.e., ABBC

gives

the acceleration of body.

Acceleration change in velocityTotal time

116 0 ms5s

= 3.2ms–2.

The area of triangle ABC, gives thedisplacement.Thus, displacement in 5s

1 1AB BC2 2

16ms–1 × 5s = 40m

Figure(ii) .represents velocity – timegraph, where the body is initially not atrest.

The slope of graph AC, i.e., ABBC

gives

the acceleration. Acceleration

125 5 msAB 20BC 4s 4

ms–2 = 5ms–2.

The distance covered by the body inspecified direction is area of trapeziumECAD.

Displacement 12

(CE + AD) × ED 12

(5 + 25)ms–1 × 4s = 60m.

Conclusions :(i) If velocity – time graph is a straight linebut moving away from velocity time axis,then:a) Body is moving with variable velocity.b) It has uniform acceleration, which canbe found by the slope of graph.c) Displacement can be found, by findingarea under the velocity - time graph.d) If slope is positive, then the body haspositive acceleration and vice – versa.(ii) If the velocity – time graph is a curve,then:a) The body has variable velocity andvariable acceleration.b) Area under the curve representsdisplacement.c) Acceleration at any instant can befound by finding slope at that point.

2. ACCELERATION – TIME GRAPHFigure (i) represents an acceleration –time graph, AB coinciding with time axis.From the figure it is clear thatacceleration of the body is zero, andhence, it is moving with a uniformvelocity.

Fig (i) Fig (ii) Fig (iii)Figure (ii) represents an acceleration –time graph, parallel to time axis. Fromfigure it is clear that as acceleration doesnot change, therefore body is moving witha uniform acceleration and variablevelocity. The area of graph, i.e.,Acceleration × Time gives change invelocity.Figure (iii) represents an acceleration –time graph moving away from time as wellas acceleration axis. From the graph itis clear that the body is moving withvariable velocity and variableacceleration. Area of the graph giveschange in velocity.



DAY-7: WORKSHEET1. The distance-time graph of an object

moving in a fixed direction is shown infigure. The object is

dist

ance

time O| | | | | |

| |

| |

| |

1) at rest 2) moving with constant speed3) moving with variable speed4) none of these

2. Figure shows the displacement (s) - time(t) graph of a particle moving on the X–axis.

S

tt0O

Which is correct given below?1) The particle is at rest2) The velocity of particle increases uptotime t0 and then increases3) The velocity of particle increases uptotime t0 and then becomes constant4) The particle moves at a constantvelocity up to a time t0, and then stops.

3. Which of the following displacement(x) –time(t) graph is not possible?

1) 2)

3) 4)

4. Refer Figure, the ratio of speed in firsttwo seconds to the speed in the next 4seconds is

S

0 1 2 3 4 5 6

s

t

1) 1 : 2 2) 2 : 1 3) 2 : 1 4) 3 :15. Figure shows the time acceleration graph

for a particle in rectilinear motion. Theaverage acceleration in first twentyseconds is

1) 45 m/s2 2) 40 m/s2

3) 15 m/s2 4) 20 m/s2

6. The graph below represents motion ofcar. The displacement of the car in 20sis

1) 160 m 2) 20 m 3) 90 m 4) 10 m7. Diagram shows a velocity time graph for

a car starting from rest, the graph hasthree sections AB, BC and CD.

The ratio of distance along AB and BC is

1) 1 : 2 2) 2 : 1 3) 1 : 2 4) 2 : 1



DAY-8 : SYNOPSISForce : Push or pull is called force. The

cause of motion is force.Effects of force :

A force can cause a motion in stationaryobjectA force can stop the moving objects orslow them downA force can make a moving object movefasterForce can change direction of movingobjectsForce can change the shape of objects.From the above examples, we are inposition to define force.Force is an external agent which changesor tends to change the state of rest oruniform motion of a body or changes itsdirection or shape.

Types of forces :a) Muscular forces : The force applied bythe muscles of our body is calledmuscular force or biological force.Ex : Lifting of heavy weight pulling ofwheel cart, pushing a lawn roller etc.involves muscular forces.b) Mechanical forces :The forcesgenerated by a machine are calledmechanical forces.Ex : The force used to run a motor carengine is produced by using the energy ofpetrol. The force used to run steamengine, is produced by using the energy ofcoal.c) Gravitational force : The force ofattraction exerted by the earth on all theobjects is called the force of gravity orgravitational force.Ex : A stone falls downwards due togravitational force.It is the gravitational force of the sun thatkeeps the planets in their orbits.d) Electrostatic force : The force exertedby electrostatic charge is calledelectrostatic force.Ex : Charged comb attracts small piecesof paper.

e ) Magnetic force : The force by which amagnet attracts or repels objects of iron,steel, nickel and cobalt is called magneticforce.

Newton’s first law of motion :A body at rest will remain at rest and abody in motion will remain in uniformmotion, unless it is compelled by anexternal force to change its state of restor of uniform motion.Ex : A book lying on a table is in the stateof rest w.r.t. the table. It will remain atrest unless some one picks it up or movesit from one position to another.Inertia : Inertia of a body may be definedas the tendency of a body to oppose anychange in its state of rest or uniformmotion.Ex : A book lying on a table will remainplaced at its place unless it is displaced.Measure of inertia : Mass is a measureof inertia.Ex : If a body has a mass of 1 kilogram andanother body has a mass of 20 kg, thenthe body having 20 kg mass will have moreinertia since its mass is more.

Types of inertia : Inertia can be dividedinto three types.1) Inertia of rest 2) Inertia of motion 3 )Inertia of directiona) Inertia of rest : The tendency of a bodyby virtue of which it cannot change itsstate of rest by itself is called inertia ofrest.Ex : A passenger in a bus tends to fallbackward when the bus starts suddenlyb) Inertia of motion : The tendency of abody by virtue of which it cannot changeits state of motion by itself is calledinertia of motion.Ex : A passenger in a moving bus tends tofall forward when the bus stops suddenlyc) Inertia of direction : The tendency ofa body to oppose any change in itsdirection of motion by itself is known asinertia of direction.Ex : A stone tied to a string and whirledalong a circular path flies off tangentiallydue to inertia of direction, if the stringbreaks.



DAY-8: WORKSHEET1. Identity the situations where a pull is

involved.a) Man sitting on a chairb) ball falling to the groundc) Woman drawing water from a welld) tube light fixed on the wall1) both ‘a’ and ‘b’ 2) both ‘b’ and ‘c’3) both ‘c’ and ‘d’ 4)both ‘a’ and ‘d’

2. A force cana) 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 true3) both ‘a’ and ‘b’ are ture4) ‘b’ and ‘c’ are true

3. Apple is falling from the tree towards theground due to1) Magnetic force 2) Gravitational force3) Mechanical force 4) Electrostatic force

4. A boy used ____ force to kick a football.1) Muscular force 2) Gravitational force3) Mechanical force 4) Electrostatic force

5. Magnetic force can cause1) only attraction 2) only repulsion3) both attraction and repulsion4) none of these

6. A tailor cuts a piece of cloth using a pairof scissors. The force involved here is1) electrostatic force2) mechanical force3) magnetic force 4) none of these

7. Inertia is that property of a body by virtueof which the body is1) unable to change by itself its state ofrest.2) unable to change by itself its state ofuniform motion.3) unable to change by itself its directionof motion. 4) all the above.

8. An athlete runs some distance beforetaking a long jump, because1) It helps him to gain energy

2) It helps to apply large force3) It gives himself large amount of inertia4) None of these

9. A rider on a horse back falls forward,when the horse suddenly stops, this isdue to1) inertia of the horse2) inertia of the rider3) large weight of the horse4) losing the balance

10. A passenger sitting in a bus gets abackward jerk when the bus startssuddenly due to the1) inertia of rest 2) inertia of motion3) inertia of direction 4) none of these

DAY-9 : SYNOPSISLinear momentum : The total quantityof motion contained in a body is calledlinear momentum.Formula : Momentum of a body is equalto the product of the mass (m) of the body

and the velocity v

of the body. It isdenoted by P

.Momentum = mass ×velocity (P

= m v

)

Units : C.G.S. unit: g cms–1

S.I. unit: kg ms–1

Nature : Vector , The direction ofmomentum of a body is same as that ofthe direction of the velocity of the body.Newton’s Second Law of motion :The magnitude of the resultant forceacting on a body is proportional to theproduct of the mass of the body and itsacceleration. The direction of the forceis the same as that of the acceleration.Newton’s second law gives thequantitative definition of force in otherwords it measures force. i.e. Fnet = ma,where, F = Force, m = Mass, a =AccelerationSo, Force acting on the body = mass ofthe body × acceleration produced in thebody.Newton’s 2nd Law in terms ofmomentum : The rate of change ofmomentum of an object is proportional tothe net force applied on the object. Thedirection of the change of momentum isthe same as the direction of the net force.Force Rate of change of momentum.



i.e., Change of momentumForceTime

Absolute Units of Force : C.G.S unit : gcm/s2 or dyne.Definition of dyne : The force is said tobe 1 dyne if it produces 1 cm/s2

acceleration in a body of 1g mass.S.I unit of force : kg m/s2 or newton(N).Definition of newton (N) : 1 newton isthat much force which produces anacceleration of 1 m/s2 in a body of mass1 kg.Relation between newton and dyne :1newton (N) = 105 dyne

DAY-9: WORKSHEET1. What will be the momentum of a toy car

of mass 200 g moving with a speed of5 m/s ?1) 1 kg m/s 2) 2 kg m/s3) 3 kg m/s 4) 4 kg m/s

2. A body of mass 25 kg has a momentum of125 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 movingwith velocity 25 m/s. Then themomentum of the ball is1) 7.5 kg.m/s 2) 3 kg.m/s3) 2 .5 kg.m/s 4) 4 kg.m/s

4. Two bodies A and B of same mass aremoving with velocities V and 3 Vrespectively, then the ratio of theirmomentum will be1) 1 : 2 2) 2 : 1 3) 3 : 1 4) 1 : 3

5. A cricket ball of mass 100g strikes thehand of a player with a velocity of 20 m/s and is brought to rest in 0.01 s, thenthe force applied by the hands of theplayer is1) 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 forcewill 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

7. A force of 200 dyne acts on a body of mass10 g for 5 sec. What will be the finalvelocity of body if it starts from rest ?1) 50 cm/s 2) 80 cm/s3) 100 cm/s 4) 200 cm/s

8. A force of 10 kg wt acting on a certainmass 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 : SYNOPSISImpulsive Force : A large force whichacts for a small interval of time is calledimpulsive force.Impulse : Impulse of a force is definedas the change in momentum producedby the given force and it is equal to theproduct of force and the time for which itacts.Formula : Impulse = Change inmomentum = Force × Time.Unit:C.G.S unit:dyne second (or) g cm/sS. I. unit: N s (or) kg m/s Nature : VectorMass : The quantity of matter containedin the body is called its mass.Unit : C.G.S unit : gram (g).S.I. unit : kilogram (kg). Nature : ScalarWeight : The weight of a body is the forcewith which it attracted towards thecentre of the earth.Formula : Weight = mass × accelerationdue to gravity.Unit : C.G.S unit : dyneS.I. unit : newton Nature : VectorNote: Mass remains constant whereasweight 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 equalin magnitude but opposite in direction.i.e. Action = – ReactionEx : Jet aeroplanes and rockets works onthe principle of Newton’s third law ofmotion.Note : Action and Reaction actsimultaneously but act on differentbodies. So they donot cancel with eachother.



Law of conservation of momentum :According to this law, the totalmomentum of a system remains constantif no net external force acts on thesystem.That is, momentum of a system.pr = constant, if net external force acting

on it is zero (i.e. externalF

= 0)

DAY-10: WORKSHEET1. A force of 50 N acts on a body for 10 s.

What will be the change inmomentum ?1) 200 Ns 2) 400 Ns 3) 500 Ns 4) 1000 Ns

2. A body of mass 100 kg moving straightline with a velocity of 30 m/s, moves inopposite direction with a velocity of 20m/s after hitting a wall. What is itsmagnitude of impulse ?1) 6000 Ns 2) 5000 Ns3) 4000 Ns 4) 3000 Ns

3. How much would a 70 kg man weigh onthe moon ? What will be his mass on theearth and on the moon ? [g on moon =1.7 m/s2]1) 119 N, 70 kg 2) 115 N, 68 kg3) 116 N, 65 kg 4) 114 N, 55 kg

4. If the weight of man on earth surface 30N. What will be his weight on moonsurface ? (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 oftable on book are equal and opposite andare inclined to vertical.2) action and reaction are equal andopposite and act perpendicular to thesurfaces of contact.3) action and reaction are equal but actin the same direction.4) action and reaction are not equal butare in opposite direction.

6. Whenever an object A exerts a force onanother object B, object B will exert areturn force back an object A. The twoforces are1) equal in magnitude and in samedirection

2) equal in magnitude but opposite indirection3) not equal and in opposite direction4) not equal and in the same direction

7. When two bodies of masses m1 and m2moving with velocities u1 and u2 in thesame direction collide with each otherand v1 and v2 are their velocities aftercollision in the same direction, then1) 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 m1 and m2moving with velocities u1 and u2 in theopposite direction coll ide with eachother and move together after collisionin the same direction with a commonvelocity v, then1) m1u1 + m2u2 = m1v – m2v2) m2u1 – m1u2 = (m1 + m2)v3) m1u1 – m2u2 = (m1 – m2)v

4) 1 1 2 2

1 2

m u m uvm m

9. The car A of mass 1500 kg travelling at25 m/s collides with another car B ofmass 1000 kg travelling at 15 m/s in thesame direction. After collision thevelocity of car A becomes 20 m/s. Thevelocity of car B after the collision is1) 12.2 m/s 2) 11.5 m/s3) 22.5 m/s 4) 5.22 m/s

DAY-11 : SYNOPSISFriction : The force which opposes the

relative motion of a body over another iscalled force of friction. The force offriction is always parallel to the twosurfaces.

Cause of friction : Friction is due to theirregularities (interlocking) of the twosurfaces in contact.

Factors on which frictional force depends:The nature of two surfaces in contactwith each other.Normal force with the surfaces are beingpressed together.



Note : The force of friction does notdepend upon the area of the surfaces incontact.

Types of friction :a) Static friction : (fs)Static friction is the force of friction actingon the body when it is rest position inspiteof the fact that some force is being appliedon it.Note : Static friction always equal toapplied force.Static frictional force is a self adjustingone. It can adjust not only in magnitudebut also in direction.b) Limiting friction : The maximum valueof the static friction is called limitingfriction. (or) The maximum frictional forcewhen the body is ready to start is calledlimiting frictional force.c) Kinetic friction (fk) : The force offriction which opposes when the body inmotion on the surface of another body.(or) When one body moves over theother, the force of fr ict ion actingbetween the two surfaces is cal l edkinetic friction.

Laws of limiting friction : The direction offorce of friction is always opposite to thedirection of motion.The force of limiting friction depends uponthe nature and state of polish of the twosurfaces in contact.The magnitude of limiting friction ‘F’ isdirectly proportional to the magnitude ofthe normal reaction R between the twosurfaces in contact, i.e, F R (or) F =

R where = coefficient of friction.The magnitude of the limiting frictionbetween two surfaces is independent ofthe area and shape of the surfaces incontact so long as the normal reactionremains the same.

Factors on which coefficient of frictiondepends : Coefficient of friction dependson Nature of the surfaces in contact.It depends on temperature.Note : Coefficient of friction have nounits.

DAY-11: WORKSHEET1. The maximum value of static friction is

called1) limiting friction 2) static friction3) kinetic friction 4)rolling friction

2. The friction acting on the body when thebody is in motion is called1) static friction 2) dynamic friction3) limiting friction 4) none of these

3. Choose the correct relation 1) f = R 2) f = /R3) f = + R 4) f = R/

4. Statement A: Frictional force increaseswith the increases external force, in caseof static friction.Statement B : Static friction always equalto applied force.1) Statement A is true2) Statement B is true3)Both the statements A and B are true 4) None of these

5. Force of limiting friction for a body and asurface1) increases if the surface is inclined2) decreases if the surface is inclined3) remains the same whether the surfaceis inclined on horizontal4) none of these

6. The coefficient of static friction betweentwo surfaces depends upon1) the normal reaction2) the shape of surfaces in contact3) the area of contact 4) all of the above

7. A force of 100 g wt. is required to pull abody weighing 1 kg over ice. What is theco-efficient of friction ? [g = 9.8m/s2]1) 0.01 2) 0.1 3) 1 4) 10

8. A body of mass 100 g is made just to slideon a rough surface by applying a force of0.8 N. Then the coefficient of friction is(Take g = 10 ms–2)1) 0.8 2) 0.08 3) 0.7 4) 0.6

9. What minimum force is required to movea body of mass 5 kg over a surface whosecoefficient of friction is 0.3 ? g = 10 ms–2.1) 15 N 2) 13 N 3) 12 N 4) 10 N



DAY-12 : SYNOPSISWork : Work is said to be done when a force

produces motion.Ex : When an engine moves a train alonga railway line, it is said to be doing work.Mathematical Expression for work :If a force F acts on a body and moves it adistance S in the direction of the forcethen work done. i.e., W = F × SUnits of work : C.G.S. unit : g cm2 s–2 orerg.S.I unit : kg m2 s–2 or joule(J).Relation between Joule and erg : 1 J = 1N × 1 m = 105 dyn × 100 cm = 107 dyn cm.(or) 1 J = 107 erg.Note : Work is a scalar quantity.

Types of Work : Work done can be positive,negative or zero depending upon thedirection of force and direction of motion.(displacement)Positive work done : Work done by a forceon a body (or an object) is said to bepositive work done when the body isdisplaced in the direction of applied force.Ex : The body falling freely under theaction of gravity has positive work done bythe gravitational force.Negative work done : The work done by aforce on a body is said to be negative workdone when the body is displaced in adirection opposite to the direction of theforce.Ex : Work done by frictional force as forceof friction and the displacement areopposite to each other.Zero work : Work done is zero if

i) The displacement is zero.Ex : When a person pushes a wall but failsto move the wall, then work done by theforce on the wall is zero.

ii) The force and the displacement areperpendicular to each other.Ex : When a person carrying a suitcase inhis hand or on his head is walkinghorizontally, the work done againstgravity is zero.No work is done on a body when the bodymoves along a circular path.

Work done against gravity :Work done in lifting a body = weight ofbody × vertical distance = W = mg × hwhere, m = mass of body g = accelerationdue to gravity at that place h = heightthrough which the body is lifted.

DAY-12: WORKSHEET1. When a stone tied to a string is whirled in

a circle, the work done on it by the string is1) positive 2) negative 3) zero 4) none

2. A man with a box on his head is climbingup a ladder. The work done by the man onthe box is1) positive 2) negative3) zero 4) undefined

3. Work is said to be done ifStatement A : a force is applied whichbrings about motionStatement B : a force is applied but nomotion is produced1) only statement A is true2) only statement B is true3) both the statement A and B are true4) none of these

4. Work done is zero1) When force and displacement of thebody are in the same direction2) When force and displacement of thebody are in the opposite directions3) When force acting on the body isperpendicular to the direction of thedisplacement of the body4) None of these

5. How much work is done by a force of 10N is moving a body through a distance of2 m in its own direction ?1) 20 J 2) 24 J 3) 26 J 4) 30 J

6. Calculate the work done by a passengerstanding on a platform holding a suitcaseof 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 ofheight 72 metre. The work done is [g = 9.8m/s2]1) 35280 J 2) 32580 J3) 52380 4) 58320 J



9. How much is the mass of a man if he hasto do 2500 joule of work in climbing a tree5m tall ? (g = 10 m/s2)1) 30 kg 2) 40 kg 3) 50 kg 4) 45 kg

10.An object of 100 kg is lifted to a height of10 m vertically. What will be the workdone? [g = 9.8 m/s2]1) 9800 J 2) 9008 J 3) 9.8 J 4) 8.9 J

DAY-13 : SYNOPSISEnergy : Energy is the ability to do workor the capacity to do work.Units : Unit of energy is same as that ofthe unit of work. As work is a form ofenergy.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 ofkineti c energy (K.E) and potentialenergy (P.E) of a body is known asmechanical energy.M.E = K.E + P.EKinetic energy (K.E) : The word kineticcomes from a Greek word which meansmotion.The energy possessed by a body by virtueof its motion is known as kinetic energy.Note : All moving bodies possess kineticenergy.Ex : A moving bus or a car or a train haskinetic energy.Formula of kinetic energy:Kinetic

energy, 21K.E mv2

where, m = Mass of the body, v = Velocityof the body.Potential energy : (P.E) : The energypossessed by a body by virtue of itsposition is called potential energy.Ex : Water stored in a dam has potentialenergy due to its position.Formula of potential energy :Potential energy (P.E) of a body at acertain height = P.E = mghwhere, m = mass is the body, g =acceleration due to gravity h = heightfrom the ground.

Examples of body possessing both thekinetic and potential energies at the sametime:i) A flying aeroplaneii) A bird flying in the skyRelation between kinetic energy andmomentum :We know , P = mv v = P/m and K.E. =

21 mv2

21 Pm

2 m

= 2P

2m

2P2m

K.E

Note : For a body having same

momentum, 1K.Em

For a body having

same kinetic energy, P m .

Law of conservation of energy : Accordingto this law “Energy can neither be creatednor be destroyed, but can be changedfrom one form to another form”.Ex :When a body falls from a certainheight, its P.E gradually changes intokinetic energy but the total sum of boththe energies remains the same.

Note : i) For a freely falling body, potentialenergy changes into kinetic energy.

Hence, Loss of P.E = Gain of K.Eii) For a body projected verticallyupwards, kinetic energy changes intopotential energy. Hence, Loss of K.E =Gain of P.E.

DAY-13: WORKSHEET1. What will be the K.E of a body of mass 2 kg

moving with a velocity of 0.1 metre persecond ?1) 0.1 J 2) 0.01 J 3) 0.001 J 4) 1 J

2. Two bodies of equal masses move withuniform 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 Joulewhen its velocity is1) 0.45 m/s 2) 1 m/s3) 1.4 m/s 4) 4.4 m/s

4. If acceleration due to gravity is 10 m/s2,what will be the potential energy of a bodyof mass 1 kg kept at a height of 5 m ?]1) 20 J 2) 30 J 3) 40 J 4) 50 J



5. An object of mass 1 kg has a potentialenergy of 1 J relative to the ground, whenit is at a height of [g=10 m/s2]1) 0.1 m 2) 1 m 3) 9.8 m 4) 32 m

6. A light and a heavy body have equalkinetic energy. Which one has greatermomentum ?1) The lighter body has greatermomentum2) The heavier body has greatermomentum3) both the bodies have same momentum4) none of these

7. What will be the momentum of a body ofmass 100 g having kinetic energy of20 J ?1) 2 kg m/s 2) 4 kg m/s3) 5 kg m/s 4) 6 kg m/s

8. Two bodies of mass 1 kg and 4 kgpossess equal momentum. The ratio oftheir kinetic energies is1) 4 : 1 2) 1 : 4 3) 2 : 1 4) 1 : 2

9. A body of mass 2kg moving up haspotential energy 400J and kinetic energy580J at a point ‘P’ in its path. Themaximum height reached by the body is(g = 10ms–2)1) 49m 2) 98m 3) 196m 4) 392m

10. A body is moving horizontally at aheight 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–14) 2.8ms–1

DAY-14 : SYNOPSISThrust : The total force exerted by thebody perpendicular to the surface isknown as thrust.Units of thrust : Since thrust is a type offorce its units are same as that of theforce.C.G.S. unit : dyne S.I. unit: newton (N)Pressure: Thrust acting over a unit areaof the surface is called pressure.Formula : Pressure

= normal force (or Thrust)

Area P = FA

Units of pressure : C.G.S. unit : 2dynecm

S.I. unit : 2N

m or pascal

Other units of pressure:Other units ofpressure are bar or atmosphere1 bar = 105 N/m2 or 105 pascals = 105 Pa1 milli bar = 102 N/m2 or 102 PaNote : Pressure is a scalar quantity.Atmospheric pressure :The force exerted on unit area of theEarth’s surface due to the atmosphere iscalled atmospheric pressure. We canexpress it in pascal as : 1 atm = 0.76 mHg = 105 PaMathematical expression for pressurein fluids :Pressure (P) exerted by liquid at depth(h)= P = h d g (neglecting atmosphericpressure)Here, h = height of the liquid column, d =density of the liquidg = acceleration due to gravityNote : Total pressure in a liquid at a depth‘h’ = Atmospheric pressure (pA) + Pressuredue to liquid column = pA + hdg

DAY-14: WORKSHEET1. 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 atable with different faces touching thetable. In different cases, the block exerts1) same thrust and same pressure2) same thrust and different pressure3) different thrust and same pressure4) different thrust and differen7t pressure

3. A force of 16N acts on an area of 50 cm2.What is the pressure in pascal?1) 3200 Pa 2) 4200 Pa3) 5200 Pa 4) 2200 Pa

4. What is the magnitude of force requiredin newton’s to produce a pressure of27500 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 areaA, produces a pressure of 1500 Pa. Whatis the magnitude of A in cm2 ?1) 1000 cm2 2) 3000 cm2

3) 4000 cm2 4) 2000 cm2



6. Pressure at any point inside a liquid is1) directly proportional to density of theliquid2) inversely proportional to density of theliquid3) directly proportional to square root ofdensity of the liquid4) inversely proportional to square ofdensity of liquid

7. What will be the pressure in dyne/cm2,due to a water column of height 10 cm ?[ take g = 980 cm /s2] (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 groundfloor of a building is 120000 Pa, where asthe pressure on the third floor is 30000Pa. What is the height of third floor ? [ Take g = 10 m /s2, density of water =1000 kg /m3]1) 11 m 2) 9 m 3) 10 m 4) 12 m

DAY-15 : SYNOPSISPascal’s Law : This law is also known as “the principle of transmission of fluid-pressure.”This law states that “ The pressureexerted at any point in an enclosed andincompressible liquid is transmittedequally in all direction.”

Buoyant Force and Buoyancy : Every liquidexerts an upward force on the objectsimmersed in it is called buoyant force.The tendency of a liquid to exert anupward force on an object placed in it iscalled buoyancy.Factors Affecting Buoyant Force: Themagnitude of buoyant force acting on anobject immersed in a liquid depends ontwo factors :i) volume of object immersed in theliquid, and ii) density of the liquid.

Statement of Archimedes’ Principle :When an object is wholly (or partially)immersed in a liquid, it experiences abuoyant force (or upthrust) which is equal

to the weight of liquid displaced by theimmersed part of the object. In otherwords, for a body inside the liquid.loss of weight or Buoyant force = Wt ofliquid displaced by the immersed part ofthe body.= mass of the liquid displaced × g= volume of liquid displaced × density ofliquid × g= volume of body × density of liquid × g =Vbody × dliquid × g(since for a body completely immersed inthe liquid , the volume of the liquiddisplaced is equal to the total volume ofthe body).Relative density of Solid: R.D of solid =

weight of solid in air

loss of weight of solid in waterair

air water

ww w

Relative density of Liquid :R.D of liquid =

air liquid

air water

loss of weight of the body in liquidloss of weight of the body in water

w ww w

DAY-15: WORKSHEET1. Archimedes’ principle states that when a

body is totally or partially immersed in afluid the upthrust is equal to1) the weight of the fluid displaced by it2) the weight of the body3) volume of the fluid displaced4) volume of the body

2. A body is lowered into a liquid. The bodyloses some of its weight. The loss ofweight depends upon1) volume of the body 2) density of liquid3) acceleration due to gravity4) all of these

3. Two balls, one of iron and the other ofaluminium experience the sameupthrust when dipped in water if1) both have same mass2) one has half the volume as that of theother3) both have equal volume4) one has one-fourth of the volume asthat of the other.



4. What buoyant force acts on a solid ofvolume 1m3, immersed in water of density1000kg/m3 ? (Take g = 10 m/s2)1) 104 N 2) 105 N 3) 102 N 4) 103 N

5. Determine the buoyant force acting on asolid of volume 1.6m3, immersed in seawater of density 1030 kg m–3. (Take g = 10m/s2)1) 16480 N 2) 164 N 3) 1648 N 4) 61840 N

6. A body whose volume is 200 cm3 weighs700gf 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 gfwhen completely immersed in water.Find (i) the apparent loss in the weight ofthe body.(ii) the upthrust on the body. iii. thevolume 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-sectionof 0.005 m2 and length 0.60m iscompletely immersed in water.Calculate thei. weight of the solid in S.I system ofunits.ii. upthrust acting on the solid in S.Isystem of units.iii. apparent weight of the solid in water.(Take g = 10 m/s2, density of water =1000 kg/m3, density of solid=1500 kg/m3)1) 20N, 25N, 11N 2) 45N, 30N, 15N3) 10N, 10N, 10N 4) 12N, 12N, 12N

9. The following observations were takenwhile determining the relative density ofa liquid.weight of the solid in air = 0.100 kgfweight of the solid in liquid = 0.080 kgfweight of the solid in water = 0.075 kgfCalculate (i) the apparent loss in weightof solid in liquid (ii) the apparent loss inweight of solid in water and (iii) therelative density of the liquid.1) 0.010 kgf, 0.02 kgf, 0.62) 0.2 kgf, 0.025 kgf, 0.93) 0.020 kgf, 0.025 kgf, 0.84) 0.030 kgf, 0.05 kgf, 1.8

10. A solid weighs 200 gf in air, 160 gf inwater and 170 gf in a liquid. Calculatethe relative density of the solid and thatof 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 electriccurrent.In other words, it is the amountof charge flowing per second.It is denotedby the letter I.If Q is the charge which is flowing througha conductor in time t, then current is

given by QIt

Unit Of Current: The S.I unit of current isampere and it is denoted by the letter ‘A’.The S.I unit of Q is coulomb and that of tis second.Thus, the S.I unit of electric current is

11seccoulomb

ond =1A

Definition of ampere: When a charge ofcoulomb flows through a conductor in onesecond, then the current flowing throughthe conductor is said to be one ampere.Thus, when 1 coulomb of charge flowsthrough a conductor in 1 second , thenthe current flowing through it is said to

be 1 ampere111seccoulombampere

ond

Smaller units of electric current:Some times smaller units of current arealso used.These are microampere and

milliampere. 61 1 10microampere A A 31 1 10milliampere mA A

Bigger unit of electric current:Sometimes the magnitude of the currentflowing in a conductor is very large.This large magnitude of current is ex-pressed in bigger units,such as kilo am-pere and mega ampere

31 ( ) 1000 10kiloampere KA A A 61 ( ) 1,000,000 10megaampere MA A A



Flow of current: In metals, the movingcharges are the electrons constitutingthe current, while in electrolytes andionized gases, electrons and positivelycharged ions are the ions moving chargeswhich constitute currentThe charge on an electron is negativeand is 191.6 10x coulomb(symbol C)Therefore, I C charge is carried by

1819

1 6.25 101.6 10

xx

electrons.Hence if I A

current flows through a conductor, itimplies that 186.25 10x electrons pass in 1second across the cross section of theconductorThe direction of current is convention-ally taken opposite to the direction ofmotion of electronsIf n electrons pass through a cross sec-tion of a conductor in time t, then totalcharge passed

Q nxe and current in conductor Q neIT t

Instrument by which current measured:Current is measured by an instrumentcalled ammeter

DAY-16: WORKSHEET1. A body is said to have 1 coulomb charge,

if it has excess or defict of: 1) 66.25 10x electrons 2) 166.25 10x electrons

3) 186.25 10x electrons 4) 96.25 10x electrons2. The charge on one electron is: 1) 91.6 10x C 2) 191.6 10x C

3) 121.6 10x C 4) 181.6 10x C

3. It a current of 25mA flows through acircuit for 1 hour, the charge flowingthrough conductor is:1)9C 2)900C 3)90C 4)60C

4. When one coulomb charge flows througha conductor in one second, the currentflowing through the conductor is1)1A 2)10A 3)2A 4)4A

5. The rate of flow of charge in a circuit iscalled1)Electric potential 2)Electric current3)Electric charge 4)All of these

6.If ‘Q’ is the charge which is flowing througha conductor in time ‘t’ then current ‘I’ isgiven by

1) I Qxt 2)QIt

3) I Q T 4)2QI T

7. If the electronic charge is 191.6 10x C, thenthe number of electrons passing througha section of wire per second, when thewire carries a current of 2ampere is1)1.2×10–12A 2) 1.25×1019A 3) 1.6×10–8A4) 1.6×104A

8. Current of 4.8 ampere is flowing througha conductor.The number of electronscrossing per second the cross-section ofconductor will be___1) 193 10x 2) 153 10x 3) 113 10x 4) 123 10x

9. A conductor carries a current of 2A.Howlong will it take for 1800C of electricityto 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 ofradius ‘r’ is ‘V’ units.Find the strength ofcurrent.

1)2Ve

r 2) 2Ve

r 3)2 rVe

4) 2r

Ve

11.Frequency of an electron in an orbit is‘f’.Find the strength of the current

1) fxe 2)fe 3) 2f xe 4) 2fxe

12.How many electrons pass through a wirein 1 minute if the current passing throughthe wire is 200mA?1) 167.5 10x 2) 117.5 10x 3) 197.5 10x 4) 192.5 10x

13.In metals moving charges constitutingcurrent are1)Electrons 2)Positively charged ions3)Protons 4)All of these

14.In ‘n’ electrons pass through a cross-sec-tion of a conductor in time ‘t’, then totalcharge passed is

1)nQe

2) Q nxe 3)2nQ

e 4) 2Q n e

15.In electrolytes and ionized gases mov-ing charges that constituent current aresame options1)Electrons 2)Positively charged ions3)Protons 4)Both (1) and (2)

16.One coloumb per second=_____ampere1)1 2)2 3)3 4)4



DAY-17 : SYNOPSISHot and Cold:Hold a piece of ice on your

palm.You feel cold on your palm.Now, dipyour finger in warm water. You feelwarm.Hold a glass of boiling tea in yourhand.You feel hot.Thus you feel cold,warm and hot. Try putting your index fin-ger in boiling tea. It may not be possibleto hold your finger in boiling tea or tryeven touching it. You may burn your fin-ger if you keep it in boiling tea even for asecond. You may say that tea is boilinghot and it should not be touched.Same is true about the weather.It is coolat night.It is warm in the shade duringday.It is hot in sun. We make use of oursense of touch to learn about cold, warmand hot.We are able to sense heat in anobject. Heat is something which pro-duces a sensation in our body by way ofwhich we make out whether a body iscold, warm or hot.

Heat is form of an energy:Energy is the ability to do work. Whenan object has the ability to do work, wesay that the object has energy. Heat hasalso the ability to do work.For example, the steam engine pulls atrain converting heat into mechanicalenergy.Heat can also be converted toother forms of energy.For example, when charcoal is heated,it gives light. Here heat produces light.In a hot air balloon the hot gases, beinglighter than the surrounding air, rise upin the air and are made to lift weights.Here heat is used to produce mechani-cal energy. The heat in a fire crackerproduces both sound and light.Other forms of energy can also be con-verted to heat energy. For example , youcan feel the heat produced from the me-chanical energy by rubbing your palmsvigorously against each other.When a candle burns in air. chemicalenergy is converted into heat. In an elec-tric blub, electrical energy is convertedinto light and heat.

So as in an electric heater the electricalenergy is converted into heat energy.When heat is given to a substance, wefind the following effects:1.Rise in temperature 2.Expansion3.Change in state

General effects of heat energy:a) Heat energy brings about change in tem-

peratureb) Heat energy brings about change in di-

mensionsc) Heat energy brings about change in state.d) Heat energy affects living beings

Flow of heat energy: Heat flows from hotbody to a cold body. A body which is los-ing heat is feeling the other body to becold. A body which is gaining heat is feel-ing the other body to be hot.Thus heatalways flows from a body of higher tem-perature to the body at lower tempera-ture.Concept of Heat:Heat is a form of energywhich always flows from a hot body to acold body. (or) Heat is a form of energywhich makes any object hot or cold. Heatenergy is also called thermal energy.

Unit of Heat:S.I unit of heat is joule(J).Another commonly used unit of heat iscalorie(cal).One calorie is the quantity of heat en-ergy required to raise the temperatureof 1g of water through 01 C.1 cal = 4.2J, 1k.cal = 1000calories.

Note: i) As heat is a form of energy.So itsunit 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 ourhand.We say that the object is at a highertemperature than our hand.If we touch a piece of ice our hand feelscold, because heat flows from our handto the ice.We say that ice is at a lowertemperature than our hand.This suggest that temperature tells ushow hot a body is.



It is the degree of hotness or coldness ofa body. Heat flows in the direction of fallof temperature. When an object is heated,its temperature rises. When it is cooled,its temperature decreases.If two bodies having unequal temperatureare brought together, heat flows from thebody at higher temperature to the bodyat lower temperature, till the two bodiesare at the sam temperature.Thus, the degree of hotness or coldnessof the body is called temperature.Mathematically, Temperature is heat perunit mass.

Unit of temperature:S.I unit of temperature is kelvin(K).Other unit of temperature is degree Cel-sius( 0 C ) and degree Fahrenheit( 0 F ).

Note: i)It is a scalar quantityii) Thermometry is the branch of heat deal-

ing with the measurement of tempera-ture.

Difference between heat and temperatureHeat Temperature

1.Heat is a form of energy.1. Temperature indicates

Hence it has the capaciythe thermal condition offor 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 theeffect

3.Two bodies of same 3.Two bodies ofsame

substance having different substancehaving

masses may have same differentmasses may

amount of heat but have samedifferent temperature4.Heat contents of a body 4.Temperature

of ado not decide the directio body decides

the-n of heat flow from the direction of

heat flow fr

body -om the body5.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 tempera-ture of a substance is called athermometer.(“thermo” is a Latin wordwhich means heat and ‘meter’ means ameasuring device).

Mercury Thermometer:We know that the substances expandwhen heated and contract when cooled.This principle is utilized for the construc-tion of a thermometer.

Uses of Mercury in thermometers:Mercury is used in thermometers becauseof the following advantages

i) It expands evenly as the temperaturerises.

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 involvesfixing of two points on it.One lower fixed point and other upperfixed point.Lower fixed point: The melting point ofpure ice at normal atmospheric pressureis taken as lower fixed point(L.F.P).Upper fixed point: The boiling point of purewater at normal atmospheric pressure istaken as upper fixed point(U.F.P).

Freezing and Boiling point:A thermometer has two standard mark-ing on its glass tube. These are called‘lower fixed point’ and ‘upper fixed point’.The lower fixed point of thermometerscale is the temperature of meltingice(ice point). It is given a value of

00 C .The upper fixed point of a thermom-eter scale is the temperature of boilingwater. It is given a value of 0100 C (Steampoint).



DAY-17: WORKSHEET1. Heat is the abilit to do

1)Work 2)Force 3)Time 4)Mass2. Heat is a form of

1)Matter 2)Energy 3)Fluid 4)None3. The most important naturally occuring

source of heat is 1)Earth 2)Sun 3)Moon 4)None4.The degree of hotness and coldness of the

body is called1)Heat 2)Temperature3)Pressure 4)Force

5. A Thermometer measures1)The quantity of heat 2)Density3)Temperature 4)Humidity

6. The thermometric liquid used in a mer-cury thermometer is1)alcohol 2)water3)Mercury 4)Benzene

7.Galileo’s thermometer was based on theproperty of expansion1)Gases 2)Liquids3)Solids 4)both (2) and (3)

8. S.I unit of temperature is1)Kelvin 2)Celsius3)Fahrenheit 4)Reaumer

9. Temperature is a _______quantity1)Vector 2)Scalar3)both (1) and (2)4)Neither a vector nor a scalar

10.When heat is given to a substance thefollowing effects observed are1)rise in temperature2)Expansion of the substance3)Change in scale of the substance4)All the above

11.Heat always flows from1)higher temperature to lower tempera-ture2)lower temperature to higher tempera-ture3)some times higher to lower to highertemperature4)none of these

12.1 cal=_______1)1.8J 2)2.4J 3)4.2J 4)5.1J

13.The device for measuring the tempera-ture of a substance is called a _______1)barometer 2)thermometer3)Voltmeter 4)Ammeter

14.Thermometer works on the principle of1)Substances expandson heating2)Substances contracts on heating3)Substances have no effect on heating4)None of these

15.The substance whose property is utilisedfor measuring temperature is called1)Thermometric substance2)Calorimetric Substance3)Solid substance 4)None of these

16.If a body is at a temperature higher thanthe room temperature the level of mer-cury in the thermometer’s stem1)Falls 2)remain at the same position3)rises 4)may rise or fail

17.The following one determines thedirection of flow of heat1)Temperature 2)Thermometer3)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 is1) 0180 C 2) 012 C 3) 040 C 4) 0100 C

20.The melting point of ice is1) 00 C 2) 0273 C 3) 040 C 4) 0100 C

21.Normal temperature of human body is1) 098.4 F 2) 0120 F 3) 080 F 4) 037 F

22.1 Kilo calorie=______calories1)10 2)100 3)1000 4)10000

23.Units(s) of heat energy is1)Joule 2)calorie 3)Kelvin 4)both 1&2

24.The lower fixed point of a thermometerscale is the temperature of melting of1)Ice 2)Water 3)Mercury4)Alcohol

25.The lower fixed point of a thermometerscale is also called1)Ice point 2)Water point3)Liquid Point 4)Steam Point

26.The boiling point of pure water at nor-mal atmospheric pressure is taken as1)Upper fixed point 2)Lower Fixed Point3)Both (1) & (2) 4)Neither (1) nor (2)

8th class physics bridge program

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