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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) 7r 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 NARAYANA GROUP OF SCHOOLS 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. NARAYANA GROUP OF SCHOOLS 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 MPC BRIDGE COURSE 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 NARAYANA GROUP OF SCHOOLS 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 NARAYANA GROUP OF SCHOOLS 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. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE 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 NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE 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 NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE 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. 58 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 NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE 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 106 A 1milliampere 1mA 103 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 x1019 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 x109 C 2) 1.6 x1019 C 3) 1.6 x1012 C 4) 1.6 x1018 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. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE 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. 67 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 NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE 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). 68 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 NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE 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) 69