A Crunk Book * Main Points A . CoM CCF ibook Q/ABy Coa) ICAL SCIENCE HANDBOOK Chapter - 1 poms Mf 1. Describe the position of an object to specify a reference point called the origin. 2. The total path length covered by the object is called distance. 3. The shortest distance measured from the initial to the final position of an object is known as the displacement. 4, The object covers equal distances in equal intervals of time, it is said to be in uniform motion. 5. The object covers unequal distances in equal intervals of time, it is said to be in non-uniform motion. 6. The distance travelled by the object in unit time is called as speed. The SI unit of speed is mvs or ms"! Distance 1. Speed =~ Tine 8. The displacement travelled by the object in unit time is called as velocity. The SI unit of velocity is m/s or ms* Displacement Time | | 10. The average speed of an object is obtai Avge psd = ate ete 11. The average velocity of an object is obtained by dividing the total displacement travelled by the total time taken Average velocity = 242i diplacementtravelet 9. Velocity = xd by dividing the total distance travelled by the total time taken. Total time taken Initial velocitys Final vel uty Average velocity = 4 12. During uniform motion of an object along a straight line, the velocity remains constant with time (the change in velocity of the object for any time interval is zero). 13, During non-uniform motion of an object along a straight line, the velocity varies velocity of the object for any time interval is not zero). 14, The change in the velocity of an object per unit time is called as acceleration. The SI unit is m/s? or ms” sceletati Change in velocity _ vu Acceleration (a) = Mtge rete . 15, The acceleration is taken to be positive if itis in the direction of velocity and negative when it is opposite to the irection of velocity. 16, Distance-Time graph: It is the graph between the time t and distance s of a particle relative to a fixed origin. Its slope at any point gives the instantaneous velocity at that point. 17. Velocity-Time graph: It is the graph of time vs velocity. Its slope at any point gives the acceleration at the corresponding instant. Distance covered in time t equals area under the velocity-time graph. 18, Equations of motion ut it + Yeat” vu? =2as 19, If an object moves in a circular path with uniform speed, its motion is called uniform circular motion. 20. Ifthe athlete takes t seconds to go once around the circular path of radius r, the speed v is given by v=" T Exercises / 1. An athlete completes one round of a circular track of diameter 200 m in 40 s, What will be the distance covered and the displacement at the end of 2 minutes 20 s? Ans: Given Diameter = 200 m Radius (r) = 200/2= 100 m ‘Time taken = 2 minutes 20s =2 x 60+ 20= 140s 140 1 No.of rounds =~" = 3.5 = 35 i) Distance covered = 2ar x 3.5 =2x = x 100 x 3.5 = 2200m ii) Atend of his motion, the athlete will be in the diametrically opposite position. Displacement = Diameter = 200 m e (the change inCay goat} PHYSICAL SCIENCE HANDBOOK 2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C? Ans: (a) Distance covered from A to B = 300 m Displacement covered = 300 m ‘Time taken = 2 minutes 30 seconds = 2 x 60 + 30= 150s Distance coverea _ 300 Distance 7B-300m in Zin 30sec Avergespecd =: Time taken 150 Average velocity = 2sbitement coverét ‘one taken "nein (b) Distance covered (AB + BC) = 300 + 100 Displacement covered (AB- BC) = 300 ~ 100 = 200 m Time taken = 150s + 60s = 210s Distance covered _ 400 1 ‘Average speed = “Fine taken — zag — 1-9 MS Average velocity = 2spieement covered _ 200 _ 9.959 mye! 3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h, On his return trip along the same route, there is less traffic and the average speed is 30 km h-!. What is the average speed for Abdul's trip? Ans: Let the distance from home to school Average speet Time taken from home to school (ti) = km 20 km b! Distance _ x ‘Speed 20 Let the distance from home to school km Average speed = 30 km h! Distance _ x ‘Time taken from school to home (ta) = "St = XX actan sx Total time taken =t += += SA = Total distance covered = X +X =2X Totaldistance 2X _ 2X x60 Totaltime ~ 5x/60 8X 4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s~* for 8.0 s, How far does the boat travel during this time? Ans: Given u=0, 0 ms, s=ut+ Yat? =0x8+%x3x8°=4x3x64=96m 5. A driver of a car travelling at 52 km h applies the brakes. (a) Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car? Ans: (a) The shaded region of the graph represents the distance travelled by the car during the period. Average speed = =24kmh" (b) The slope of the velocity-time graph is the same throughout. So, the whole graph represents the uniform motion of the ca 6. Fig 1.10 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:By Coa) PHYSICAL SCIENCE HANDBOOK (a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (©) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C? Ans: (a) Slope = = = PSOne = speed The slope of the object B is greater than object A and object C. So, object B is travelling fastest. (b) No, all three do not meet at any point on the road. (©)7 square box = 4 km | square box = = km Cis 4 blocks away from origin posi Initial distance of C from origin = * km Distance of C from origin when B passes A =8 km 16 _ (56-16) _ 40 ion, Distance travelled by C when B passes A = 8 - 4 = 2719-9 5.714 km (d) Distance travelled by B by the time it passes C = 9 square boxes 36 = 9x5 = = 5.143 km 7. A ball is gently dropped from a height of 20 m. If its veloci with what velocity will it strike the ground? After what time wi eases uniformly at the rate of 10 ms strike the ground? Ans: Given u=0ms! h=20m g=10ms? v=? Final velocity v= /2gh =Vv2x 10x20 = 20x20 =20ms Time, = s a 0 8. The speed-time graph for a car is shown is Fig. 1.11. (a) Find how far does the car trayel in the first 4 seconds. i * mia Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car? Ans: (a) Speed (m= 7 6 8 0 ——> Time) ‘The distance travelled by the car in the first 4 s = The shaded area =ix4x6 =12m (b) After 6 s, the car has a uniform motion 9. State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero veloci (b) an object moving with an acceleration but with uniform speed. (©) an object moving in a certain direction with an acceleration in the perpendicular direction. Ans: (a) Yes. When a body is thrown up, at maximum height its velocity is zero but it has constant acceleration. (b) Yes. In uniform circular motion, an object moves with a uniform speed but it has an acceleration towards the centre of the circle. (©) Yes. When the object is moving along a circular motion its path is along the tangent. The direction of the body is always perpendicular to its acceleration,By Coa) ICAL SCIENCE HANDBOOK 10. An artificial satellite is moving in a circular orbit of radius 42250 km, Calculate its speed if it takes 24 hours to revolve around the earth, Ans: Given r= 42250 km = 42250 x 1000 m T=24h=24x 60x 608 = 2m _ 2x3.14.x 42250 x 1000 i Speed, VS ag = 3070.9 ms In Textbook Questions 1, An object has moved through a distance, Can it have zero displacement? If yes, support your answer with an example. (Page No:08) Ans: Yes, an object which has moved through a distance can have zero displacement. If it comes its initial position. Example: If person starts at one initial point on circular path and complete one round, the final point and tial points are same. 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position? (Page No: ) Ans: Side of square field = 10 m Perimeter = 4 x 10=40 m ‘Time taken = 40 Given time = 2 minutes 20 second = 2 x 60 + 20 = 140s No. of rounds = Total time taken/ Time of one complete round = “ =3.5=3% So, Farmer moves diagonal opposite point on the square field Displacement = V10? + 10? = \200=14.14m 3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object. Ans: (a) False, because displacement can be zero. (b) False, because displacement is less than or equal to the distance travelled by the object. 1. Distinguish between speed and velocity. (Page No:12) Ans; Speed Velocity 1. Speed is the distance travelled by an object ina | 1. Velocity is the displacement of an object in a given given interval of time. interval of time. 2, Speed = distance / time 2, Velocity = displacement / time 3. SI unit is m/s 3. SLunit is m/s 4. Speed is scalar quantity 4. Velocity is vector quantity 2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed? Ans: Only when an object is moving in a straight line. 3. What does the odometer of an automobile measure? Ans: The distance covered by an automobile. 4. What does the path of an object look like when it is in uniform motion? Ans: Straight line path 5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 ms Ans: Speed=3 x 10° ms! Time=5 min = 5 x 60 = 300 secs. Distance = Speed x Time 3 x 108 x 300=9 x 10! m 1, When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration? (Page No:14) Ans: (i) A body is said to be in uniform acceleration if the rate of change of its velocity is constant. i) A body is said to be in nonuniform acceleration if the rate of change of its velocity is not constant, 2, A bus decreases its speed from 80 km h~ to 60 km h” in 5 s, Find the acceleration of the bus. ‘Ans: Change in speed (v-u) = 60 km h! - 80 km h ‘Time ()=5sBy Coa) PHYSICAL SCIENCE HANDBOOK Acceleration (a) = (v-u)t=-5.55 ms"/5 = -1.11 m 3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h-! in 10 minutes. Find its acceleration, 100 et Ans: Given u=Oms' — y=40 kmh! =40x 2 =" ms t= 10 min = 10 x 60 = 600s 100 ~~ 600 ~ xeon ~ 548 1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object? (Page No:22) Ans: The nature of the distance ~ time graph for uniform motion: Straight line The nature of the distance ~ time graph for non-uniform motion: Curve 2. What can you say about the motion of an object whose distance-time graph is a straight line paral time axis? Ans: The object is at rest 3. What can you say about the motion of an object if its speed time graph axis? Ans: The object is moving uniformly. 4. What is the quantity which is measured by the area occupied below the velocity-time graph? Ans: The distance covered by the object. 1. A bus starting from rest moves with a uniform acceleration of 0.1 ms? for 2 minutes. Find (a) the speed acquired, (b) the distance travelled. (Page No:24&26) Ans: Given u=0 m: a=0.1 ms? (a) The speed acquired, v = u + at = 0 +0.1x120 () Distance, s = ut + % at? =0 x 120 + x 0.1 x (120) = = 720 m 2. A train is travelling at a speed of 90 km h~!, Brakes are applied so as to produce a uniform acceleration of —0.5 ms* id how far the train will go before it is brought to rest. Ans: Here u=90kmh'=90x == 25ms" v=0m Acceleration, @ = to the a straight line parallel to the time minutes = 2 x 60 = 120s 12ms* 1440 -0.5 ms? ©. Distance = 625 m 3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s®. What will be its velocity 3s after the start? Ans: Given u=Ocms! a=2cms? t veutat =042x3=6cems" 4. A racing car has a uniform acceleration of 4 ms, What distance will it cover in 10 s after start? Ans: Given u=0 ms? a=4ms* =10s distance, s = ut + at? =0x 10+%x 4x (10)? = 200 m 5. A stone is thrown in a vertically upward direction with a velocity of 5 ms“. If the acceleration of the stone during its motion is 10 ms in the downward direction, what will be the height attained by the stone and how much time will it take to reach there? Ans: Here u=Sms-1 a=10ms2 v=0ms1 we x Height, h = == 5*5 125m 2g 2x10 20 ‘Time taken, 2=05s toBy Coa) Ney Maple Ney PHYSICAL SCIENCE HANDBOOK 1. What is motion? a) The change of position of an object with time b) The colour change of an object ) The size change of an object d) The shape change of an object Ans:a) The change of position of an object with time 2. Which of the following is the unit of speed? a)m/s b)kgim —c) mis? admis Ans: a) m/s 3. What does uniform motion mean? a) Changing speed _b) Changing direction ¢) Constant speed) Constant acceleration Ans: c) Constant speed 4. What is the formula for speed? a) Speed = Distance x Time b) Speed = Distance / Time ) Speed = Time / Distance d) Speed = Distance + Time ‘Ans: b) Speed = Distance / Time 5, What is the unit of distance? a) Second b) Meter c) Kilogram d) Newton Ans: b) Meter 6. If an object moves 100 meters in 20 seconds, what is its speed? a)Sm/s b)2m/s c)10m/s_— d) 20 m/s ‘Ans: a) 5 m/s 7. What is velocity? a) Speed in a particular direction b) Speed in all directions ©) Speed without direction d) Distance per unit time Ans: a) Speed in a particular direction 8. Which of the following is a vector quantity? a) Speed —b) Distance c) Velocity d) Mass Ans: c) Velocity 9. What is acceleration? a) Change in speed c) Change in direction Ans:b) Change in velocity 10. Which of the following is the unit of acceleration? a) m/s b) m/s? Ans: b) m/s? 11. If an object accelerates from 0 to 20 m/s in 5 seconds, what is its acceleration? ) Change in velocity 4) Change in distance om d) mis a) Speed b) Distance c) Acceleration d) Time Ans: c) Acceleration 14. What is the acceleration of an object moving at constant velocity? a) Zero_b) Positive c) Negative d) Infinity Ans: a) Zero 15. If'a car is traveling at 30 m/s and comes to a stop in 10 seconds, what is its acceleration? a)-3 m/s? b)-S m/s? c)-10 m/s? d) -2 m/s? Ans: b) -3 m/s? 16. What is the difference between speed and velocity? a) Speed is a scalar, veloc’ b) Speed is a vector, velocity is a scalar c) Speed includes direction, velocity does not d) There is no difference Ans: a) Speed is a scalar, velocity is a vector 17. Ifan object covers equal distances in equal intervals of time, it is said to be in a) Non-uniform motion b) Uniform motion c) Accelerated motion —_) Circular motion Ans: b) Uniform motion 18. What is the formula for acceleration? a) Acceleration = Velocity / Time b) Acceleration = Time / Velocity ¢) Acceleration = Change in velocity / Time 4) Acceleration = Distance / Time Ans: c) Acceleration = Change in velocity / Time 19. What does a horizontal line on a distance- time graph indicate? a) Constant speed —_b) Increasing speed ¢) Stationary object d) Decreasing speed Ans: c) Stationary object 20.1f an object is moving with uniform velocity, what will its acceleration be? a) Zero. b) Positive c) Negative d) Variable Ans: a) Zero 21. Which physical quantity is described by the area under the velocity-time graph? a) Speed b) Acceleration ¢) Distance 4d) Displacement Ans: d) Displacement 22. If'a cyclist is moving at a constant speed of 10 as, how far will they travel in 5 seconds? a)4m/s —b)S mV/s? c) Ons? d) 3 m/s? a)50m_ b)100m ce) 10m_—d) Sm Ans:a) 4 m/s? Ans: a) 50m 12. What does the slope of a distance-time graph | | 23. What kind of motion is described by an object represent? moving in a circular path at constant speed? a) Speed b) Distance) Time d) Acceleration | | a) Uniform motion _ b) Non-uniform motion Ans:a) Speed ¢) Uniform circular motion 13. What does the slope of a velocity-time graph 4d) Non-uniform circular motion represent? Ans: c) Uniform circular motion29:00) t--) PHYSICAL SCIENCE HANDBOOK Chapter - 2 fh Points Me 1. Everything in this universe is made up of material which scientists have named “matter”. 2. Early Indian philosophers classified matter in the form of five basic elements — the “Panch Tatva” — air, earth, fire, sky and water. 3. Matter is made up of small particles. 4. The particles of matter are very small they are small beyond our imagination. 5. Characteristics of Particles of Matter: - i) Particles of matter have spaces between them. ii) Particles of matter are continuously moving. iii) Particles of matter attract each other. 6. The matter around us exists in three states—solid, liquid and gas. 7. The forces of attraction between the particles are maximum in solids, intermediate in liquids and minimum in gases. 8. Solids have a definite shape, distinct boundaries, fixed volumes and negligible compressi 9. Liquids have no fixed shape but have a fixed volume. They take up the shape of the container in which they are kept. 10. The rate of diffusion of liquids is higher than that of solids and less than that of gases. 11, Gases are highly compressible as compared to solids and liquids. 12. Liquefied Petroleum Gas (LPG) cylinder that we get in our home for cooking or the oxygen supplied to hospitals in cylinders is compressed gas. 13. Compressed Natural Gas (CNG) is used as fuel these days in vehicles 14. The states of matter are inter-convertible. The state of matter can be changed by changing temperature or pressure. 15. On increasing the temperature of solids, the kinetic energy of the particles increases. 16. Solid carbon dioxide is also known as dry ice. 17. The minimum temperature at which a solid melts to become a liquid at the atmospheric pressure is called its melting point. 18. The temperature at which a liquid starts boiling at the atmospheric pressure is known as its boiling point. Boiling isa bulk phenomenon. 19. The melting point of ice is 273.15 K, Boiling point of water is 373 K (100°C). 20. Sublimation is the change of solid state directly to gaseous state without going through liquid state. 21. Deposition is the change of gaseous state directly to solid state without going through liquid state. 22. The phenomenon of change of liquid into vapours at any temperature below its boiling point is called Evaporation. 23. The rate of evaporation depends upon the surface area exposed to the atmosphere, the temperature, the humidity and the wind speed. 24. Evaporation is a surface phenomenon. Evaporation causes cooling. 25. Latent heat of fusion is the amount of heat energy required to change I kg of point. 26. Latent heat of vaporisation is the heat energy required to change 1 kg of a liquid to gas at atmospheric pressure at its boiling point. id into Tiquid at its melting / ercises_/ 1. Convert the following temperatures to the Celsius scale. (a) 293 K (b) 470 K Ans: °C=K-273 (a) 293 K= 293 20°C (b) 470 K = 470 - 273 = 197°C 2. Convert the following temperatures to the kelvin scale. (a) 25°C (b) 373°C Ans: K=°C+273 (a) 25°C = 20 + 273 = 298 KAy Co) poet (b) 373°C = 373 +273 = 646 K 3. Give reason for the following observations. (a) Naphthalene balls disappear with time without leaving any solid. (b) We can get the smell of perfume sitting several metres away. Ans: (a) Naphthalene is a substance which directly changes from solid to gas on heating by the process of Sublimation. So, Naphthalene balls disappear with time without leaving any solid. (b) Perfumes have high degree of vaporisation and its vapour diffuses into the air easily. Therefore, we can get the smell of perfume sitting several metres away. 4, Arrange the following substances in increasing order of forces of attraction between the particles— water, sugar, oxygen. Ans: oxygen < water < sugar 5. What is the physical state of water at— (a) 25°C orc (©) 100°C 2 Ans: (a) Liquid state (b) Solid or/and liquid state (Transition state) (©) Liquid or/and gaseous state (Transition state) 6. Give two reasons to just (a) water at room temperature is a liquid. (b) an iron almirah is a solid at room temperature. Ans: (a) Water at room temperature is a liquid because it has fluidity also it has no shape but has a fixed volume that is, it occupies the shape of the container in which it is kept. (b) An iron almirah is a solid at room temperature it has rigid and fixed shape. 7. Why is ice at 273 K more effective in cooling than water at the same temperature? Ans: Ice at 273 K has less energy than water (although both are at the same temperature). Water possesses the additional latent heat of fusion. Hence, at 273 K, ice is more effective in cooling than water. 8. What produces more severe burns, boiling water or steam? Ans: Steam has more energy than boiling water. It possesses the additional latent heat of vaporisation, Therefore, burns produced by steam are more severe than those produced by boiling water. 9, Name A,B,C,D,E and F in the following diagram showing change in its state. ICAL SCIENCE HANDBOOK Ans: ‘Melting or fusion. B = Evaporation or vaporization. C = Condensation or liquification. D =Freezing or solidification. E = Sublimation. 1. Which of the following are matter? (Page No:38) Chair, air, love, smell, hate, almonds, thought, cold, lemon water, smell of perfume. ‘Ans: Chair, air, almonds and cold drink 2. Give reasons for the following observation: The smell of hot sizzling food reaches you several metres away, but to get the smell from cold food you have to go close. Ans: Evaporation is directly proportional to temperature, means hot food evaporates easily. Diffusion of hot food vapor with air becomes very fast and can reach to a distant place within a very short time.Cay goat} poet ICAL SCIENCE HANDBOOK 3. A diver is able to cut through water in a swimming pool. Which property of matter does this observation show? Ans: The phenomena of cutting the water by the diver show that matter has space between its particles. 4. What are the characteristics of the particles of matter? Ans: i) Particles of matter have spaces between them. ii) Particles of matter are continuously moving. iii) Particles of matter attract each other. 1. The mass per unit volume of a substance is called density. (density = mass/volume). Arrange the following in order of increasing density — air, exhaust from chimneys, honey, water, chalk, cotton and iron. (Page No:44) Ans: Air, Exhaust from chimneys, water, honey, cotton, chalk and iron. 2. (a) Tabulate the differences in the characteristics of states of matter. (b) Comment upon the following: rigidity, compressibility, fluidity, filling a gas container, shape, kinetic energy and density. Ans: Rigidity: It is the property of matter to resist the change of its shape Compressibility: It is the property of matter in which its volume is decreased by applying force. Fluidity: It is the ability of matter to flow. Filling a gas container: On filling a gas takes the shape of the container. Shape: Having definite boundaries. Kinetic Energy: It is the energy possessed by the particles of matter due to its motion. Density: The mass per unit volume of a substance is called density. 3. Give reasons (a) A gas fills completely the vessel in which it is kept. (b) A gas exerts pressure on the walls of the container. (c) A wooden table should be called a solid. (@) We can easily move our hand in air but to do the same through a soli expert. Ans: (a) The gas particles have negligible intermolecular forces. These are move freely and occupy all the space available to them, (b) The gas particles collide with the container walls due to their random motion in all directions, exerting a force on the surface per unit area. (©) A wooden table has fixed shape and fixed volume. It is very rigid and cannot be compressed. (@) In air, there is a less force of attraction between the particles. So, very less amount of external force can break it. But in the case of solid, the force of attraction is very strong and the molecular space is so high. Hence a large amount of force is required to break it 4. Liquids generally have lower density as compared to solids. But you must have observed that ice floats on water. Find out w Ans: Ice is solid but its density is lower than that of water due to its network structure and forms a cage-like structure with a lot of vacant space. So that's why ice floats on water. 1. Convert the following temperature to Celsius scale: (Page No:50) a. 300K b. 573K Ans: °C = K -273 a) 300 K = 300 - 273 = 27°C b) 573 K = 573 - 273 = 300°C 2, What is the physical state of water at: a. 250°C —_b. 100°C? ‘Ans: a, Gaseous State b. Can exist in both liquid and gaseous form 3. For any substance, why does the temperature remain constant during the change of state? Ans: The heat energy being supplied is used to break the bonds between molecules, not to increase their energy and raise the temperature. 4, Suggest a method to liquefy atmospheric gases. Ans: High pressure is applied at low temperature. 1, Why does a desert cooler cool better on a hot dry day? (Page No:52) Ans: Evaporation increases on temperature and humidity. On a hot dry day, there is high temperature and low block of wood we need a karateBy Coa) poet ICAL SCIENCE HANDBOOK humidity. This faster evaporation absorbs heat from the surroundings and a desert cooler cool. 2. How does the water kept in an earthen pot (matka) become cool during summer? Ans: An earthen pot is porous in nature. Water gets evaporated through the pores of pot by absorbing heat from the water present inside the pot leading to cooling. 3. Why does our palm feel cold when we put some acetone or petrol or perfume on it? ‘Ans: Our palm feels cold when we put some acetone/petrol/perfume on it because evaporation. During the evaporation, liquid absorb energy from the palm. Evaporation is a cooling process. 4. Why are we able to sip hot tea or milk faster from a saucer rather than a cup? Ans: When the surface area increases, evaporation also increase. Saucer has larger surface area than the cup. So, we are able to sip hot tea or milk faster than a saucer rather than a cup. 5. What type of clothes should we wear in summer? Ans: We should wear cotton cloths during summer season. 1, Explain with the help of an activity which shows that particles of matter are very small? Aim: To shows that particles of matter are very small. Required materials: Potassium permanganate crystals, Beakers, water Procedure: 1, Take 2-3 crystals of potassium permanganate and dissolve them in 100 mL of water. ‘Take out approximately 10 mL of this solution and put it into 90 mL. of clear water. Take out 10 mL of this solution and put it into another 90 mL of clear water. Keep diluting the solution like these 5 to 8 times. ‘The water remains coloured till the last dilution. frac SUEEE Observation: The particles of matter are very small. 2. Explain with the help of an activity which shows that particles of matter are continuously moving. ‘Aim: Shows that particles of matter are continuously moving asses! beakers, Blue or Red ink, Honey weun 1. Take two glasses/beakers filled with water. 2. Puta drop of blue or red ink slowly and carefully along the sides of the first beaker and honey in the same way in the second beaker. Leave them undisturbed in your house or in a comer of the class. Record your observations. The colour of ink starts spreading soon. The honey drops slowly disappear. The colour of ink spreads evenly throughout the water within a few minutes. Observation: Particles of matter are continuously moving. 3. Explain with the help of an activity which shows that gases can be compressed easily. Aim: Shows that gases can be compressed easily. Required materials: Syringes, Rubber corks, Water, Pieces of chalk, Vaseline Procedure: Noawew 1. Take three 100 mL syringes and close their nozzles by rubber corks. 2. Remove the pistons from all the syringes. 3. Leaving one syringe untouched, fill water in the second and pieces of chalk in the third. 4. Insert the pistons back into the syringes. 5. Apply some vaseline on the pistons before inserting them into the syringes for their | - smooth movement. 6. Now, try to compress the content by pushing the piston in each syringe. BaerBy Coa) poet ICAL SCIENCE HANDBOOK 7. The piston was pushed in easily in the syringe in the which nothing except air was filled. Observation: Gases can be compressed easily. 4, Explain with the help of an activity which shows that effect of changes of temperature on substance. im: Effect of changes of temperature on substance. Required materials: Ice, Beaker, Thermometer, Iron stand, Burner. Procedure: 1. Take about 150 g of ice in a beaker and suspend a laboratory thermometer so that its bulb is in contact with the ice. Start heating the beaker on a low flame. Note the temperature when the ice starts melting. Note the temperature when all the ice has converted into water. Record your observations for this conversion of solid to liquid state. Now, put a glass rod in the beaker and heat while stirring till the water starts boiling. Keep a careful eye on the thermometer reading till most of the water has vaporised. Record your observations for the conversion of water in the liquid state to the gaseous state. eI Awa chef ae ee ear Beaker Explain with the help of an activity which shows that sublimation of solid. ‘To study the changes in state of Sublimate solids on heating. Ree "amphor/Ammonium chloride, China dish, Funnel. bw tert seemm | 1. Take some camphor or ammonium chloride. Crush it and put itin a china dish. Sus" \ = 2. Putan inverted funnel over the china dish. e = 3. Puta cotton plug on the stem of the funnel, as shown in figure. ZA os tn 4. Now, heat slowly and observe. 5. Vapour of ammonium chloride get deposited on the inner stem of funnel. ue Observation: Solid ammonium chloride gets converted into vapours directly. \ 6. Explain with the help of an activity which shows that factors affecting evaporation. : Factors affecting evaporation. Required materials: Test tube, China dish, Water. Procedure: 1, Take 5 mL of water in a test tube and keep it near a window or under a fan. 2. Take 5 mL of water in an open china dish and keep it near a window or under a fan. 3. Take 5 mL of water in an open china dish and keep it inside a cupboard or on a shelf in your class. 4, Repeat the above three steps of activity on a rainy day and record your observations. Observation: 1. Evaporation will be fastest from the china dish kept near the window. 2. The evaporation increases with temperature, surface area of the container and speed of wind.By Coa) Ney Maple Ney PHYSICAL SCIENCE HANDBOOK 1. Which of the following cannot be considered a form of matter? a) Atom — b) Water Ans: c) Humidity 2. Which of the following causes the temperature of a substance to remain constant while it is undergoing ‘a change in its state? a)Latentheat —_b) Lattice energy ¢) Loss of heat) None of these Ans: a) Latent heat 3. What is the physical state of matter in which particles have the least kinetic energy? a) Solid b) Liquid c) Gas) Plasma Ans: a) Solid 4. Which of the following is not characteristic of solid a) High Rigidity b) Regular Shape ©) High Density d) High compressibility Ans: d) High compressibility 5. Which of the following is a characteristic property of solids? a) Definite shape and volume b) Indefinite shape and volume c) Indefinite shape but definite volume 4) Definite shape but indefinite volume Ans: a) Definite shape and volume 6. What happens to the arrangement of particles when a substance changes from solid to liquid state? a) Particles come closer together b) Particles move further apart c) Particles become arranged in a regular pattern 4) None of the above Ans: b) Particles move further apart 7. Which of the following statements about gases istrue? a) Gases have definite shape and volume b) Gases have definite shape but indefinite volume c) Gases have indefinite shape and volume 4d) Gases have indefinite shape but definite volume Ans: c) Gases have indefinite shape and volume 8. The process of conversion of a solid into a gas without passing through the liquid state is called a) Evaporation b) Condensation ) Sublimation 4) Fusion Ans: c) Sublimation 9. Which of the following substances does not show the property of sublimation? a) lodine _b) Camphor c) Naphthalene d) Copper Ans: d) Copper 10. Which of the following is a physical change? a) Rusting of iron _b) Burning of paper c) Melting of ice. —_d) Cooking of food Ans: c) Melting of ice ©) Humidity —d) Electron LL. Which of the following has the highest intermolecular force? a) Solid _b) Liquid Ans: a) Solid 12.The process of changing a liquid into a gas is called a) Condensation b) Evaporation ) Sublimation 4d) Fusion Ans: b) Evaporation 13. Which of the following is not a characteristic property of matter? a) Mass b) Volume Ans: c) Colour 14, A form of matter has no fixed shape but it has a fixed volume. An example of this matter is a) Krypton b) Kerosene ¢) Carbon steel d) Carbon dioxide Ans: b) Kerosene 15. When converting 308 K, 329 K, and 391 K to the Celsius scale, what is the correct sequence of temperatures? a) 33°C, 56°C and 118°C b) 35°C, 56°C and 119°C ©) 35°C, 56°C and 118°C d) 56°, 119°C and 35°C Ans: c) 35°C, 56°C and 118°C 16. Which of the following decreases the rate of evaporation? a) Surface area b) Humidity c) Temperature d) Wind Ans: b) Humidity 17. Large volume of CNG is available in small cylinders to us due to its property of a) High inflammability _b) Easy availability ¢) High Compressibility d) Low density Ans: c) High Compressibility 18, Which of the following phenomena always results in the cooling effect? a) Condensation ) Sublimation Ans; b) Evaporation 19. During evaporation particles of a liquid change into vapours a) From the surface ) From the bottom Ans: a) From the surface 20. The boiling point of water is a) 100 °C at atmospheric pressure b) 273 K at atmospheric pressure ¢) 0 °C at atmospheric pressure 4) 0 K at atmospheric pressure Ans: a) 100 atmospheric pressure °C at 21. At higher altitudes the boiling points of liquids a) Decreases b) Increases ) Increases then decreases d) Remains all same Ans: a) Decreases ©)Gas — d) Plasma ©) Colour d) Density b) Evaporation ) None of these b) From the bulk ) From all over the liquidBy Coa) ICAL SCIENCE HANDBOOK Chapter - 3 Mainpoints Mf 1. A force can be used to change the magnitude of velocity or change size of objec 2. If the resultant of several forces acting on a body is zero, the forces are said to be Balanced forces. 3. If the resultant of several forces acting on a body is not zero, the forces are said to be Unbalanced forces. 4. An object moves with a uniform velocity when the forces acting on the object are balanced and there is no net external force on it. 5. If an unbalanced force is applied on the object, there will be a change either in its speed or in the direction of its motion 6. First law of motion: An object continues to be in a state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force. 7. The first law of motion is also known as the law of inertia. 8. The natural tendency of objects to resist a change in their state of rest or of uniform motion is called inertia 9, Inertia: a) Inertia of rest b) Inertia of motion _c) Inertia of direction 10. The mass of an object is a measure of its inertia. Its ST unit is kilogram (kg). 11, Heavier or more massive objects offer larger inertia. 12. The momentum of an object is the product of its mass and velocity and has the same direction as that of the velocity. Its SI unit is kg ms"! 13, Second law of motion: The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force. 's direction of motion or change the shape and 14, Mathematical formula of Newton’s second law of motion is F=ma (or) F=m 15, Force of friction always opposes motion of objects. 16, The unit of force is kg m $* or newton, which has the symbol N. 17.One Newton: A force of one newton produces an acceleration of | ms on an object of mass 1 kg. 18, Third law of motion: To every action, there is an equal and opposite reaction and they act on two different bodies. T Exercises Ff 1, An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with anon-zero velocity? If yes, state the conditions that must be placed on Ans: Yes, an object may travel with a non-zero velocity even when the net external force on it is zero. A rain drop falls down with a constant velocity. The weight of the drop is balanced by the upthrust and the viscosity of ai The net force on the drop is zero. 2. When a carpet is beaten with a stick, dust comes out of it, Explain. Ans: When a carpet is beaten with a stick, then the carpet comes to motion. But, the dust particles try to resist their state of rest. According to Newton’s first law of motion, the dust particles stay in a state of rest, while the carpet moves. Hence, the dust particles come out of the carpet. 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope? Ans: When the bus accelerates and moves forward, it acquires a state of motion. The luggage kept on the roof of a bus to its inertia, tends to remain in its state of rest. Hence, with the forward movement of the bus, the luggage tends to remain at its original position and ultimately falls from the roof of the bus. To avoid this, itis advised to tie any luggage kept on the roof of a bus with a rope. 4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because (a) the batsman did not hit the ball hard enough. (b) velocity is proportional to the force exerted on the ball. (c) there is a force on the ball opposing the motion. (@) there is no unbalanced force on the ball, so the ball would want to come to rest. Ans: (c)Cay goat} PHYSICAL SCIENCE HANDBOOK 5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.) Ans: Given _m=7 tonnes = 7 x 1000 kg = 7000 kg u=Oms! — s=400m 20s From 2" equation of motion, s = ut +14 at” 400 = 0x 20+ ¥xax 207 400 = 200 a a=2ms? Force, F= ma = 7000 x 2 = 14000 N 6. A stone of 1 kg is thrown with a velocity of 20 m s* across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice? Ans: Given m=1kg u=20ms" v=Oms' = s=50m From 3" equation of motion, v? coy az 2a(S0) 4 ms? Friction force, F= ma = | x (-4) 7. 4.8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force and (b) the acceleration of the train. Ans: Mass of engine = 8000 kg Mass of 5 wagons = 5 x 2000 = 10000 kg Total mass = 8000 + 10000 = 18000 kg (a) The net accelerating force, F = Engine force ~ Friction force = 40000 — 5000 = 35000 N (b) The acceleration of train, a == = 350° — 35 — 1.94 ms? mm ~ e000 ~ i 8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms? Ans:Given m= 1500kg =-1.7ms? The force between the vehicle and road, F = ma = 1500 x (-1.7) = - 2550 N 9. What is the momentum of an object of mass m, moving with a velocity v? (a) mv (b) my? (© mv (@) my Ans: (d) 10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet? Ans: The cabinet will move with constant velocity only when the net force on it is zero. From Newton's third law of motion, an equal amount of friction force will act in the opposite direction. :. The frictional force exerted on the cabinet = 200 N. 1. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. ‘Comment on this logic and explain why the truck does not move. Ans: Action and reaction always act on different bodies, so they cannot cancel each other. When we push a massive truck, the friction force between its tyres and the road is very large and so the truck does not move. 12. A hockey ball of mass 200 g travelling at 10 m s~is struck by a hockey stick so as to return it along its original path with a yelocity at S ms“, Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick. Ans:Given m=200g=02kg 9 u=10ms? Change in momentum = m (v ~ u) = 0.2 (-5 -10) 13. A bullet of mass 10 g travelling horizontally with a velocity of 150 ms“ strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet. Ans: Given m= 10g = 0.01 kg s0ms' v=O0ms' — t=0.03s Acceleration, a = * =5000 ms“?TH CLASS [CAL SCIENCE HANDBOOK The distance of penetration of the bullet, s = ut + 4 at” = 150 x 0.03 + % x (-5000) x (0.03? 5 =2.25m The magnitude of the force, F = ma = 0.01 x 5000 = 50 N 14, An object of mass 1 kg travelling in a straight line with a velocity of 10 ms“! collides with, and sticks to, a stationary wooden block of mass § kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object. Ans: Mass of object (m:) Initial velocity (uy Ikg Mass of wooden block (m2) 10 ms Initial velocity (uz) =0 m/s Let the velocity of combined after the collusion = v Total momentum just before the impact = my wu + m2 U2 = 1x 10+5x0= 10 kg mvs ‘Total momentum just after the impact = my v + m2 v= (m; + ma) v=(1 +5) v = 6v kg m/s From conservation of momentum, Total momentum before collusion = Total momentum after collusion 6v=10 v = 1.67 mis 15, An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s~! to 8 ms“ in 6 s, Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object. Ans: Given m=100kg 9 u=Sms" v=8ms" t=6s Initial momentum (pi) = mu = 100 x 5 = 500 kg ms Final momentum (p2) = my = 100 x 8 = 800 kg m ‘The magnitude of the Force exerted on the object, F = P==P = = = = =50N 16. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions. Ans: Both the motorcar and insect experience the equal force and same change in their momentum. So, I agree with Rahul. 17. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 ms, Ans: Given m=10kg, u=0m/s, h=80cm=0.8m, g=10ms* Final velocity (v) = /2gh = V2x 10x 0.8 = V16=4 mis Momentum of a dumb-bell (p) =m v = 10.x 4 = 40 kg mvs eco ge ost ela 1. Which of the following has more inertia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a train? (c) a five rupees coin and a one-rupee coin? (Page No:68) Ans: We know that, heavier or more massive objects offer larger inertia. (a) Stone (b) Train (©) A five-rupee coin 2. In the following example, try to identify the number of times the velocity of the ball changes: “A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”. Also identify the agent supplying the force in each case. Ans: The velocity of ball changes 3 times. kgBy Coa) PHYSICAL SCIENCE HANDBOOK 1 time, the velocity changes when football player of one team kicks the ball. 2" time, the velocity changes when another player of sane team kicks the football 3% time, the velocity changes when goalkeeper of the opposite team kicks the football 3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch. Ans: When a tree is vigorously shaken, the branches of the tree come in motion but the leaves tend to continue in their state of rest due to inertia of rest. As result, leaves get detached from a tree. 4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest? Ans: When a moving bus brakes to a stop, we fall in the forward direction due to inertia of motion. When the bus accelerates from rest, we fall in the backward direction due to inertia of rest. 1. Describe an activity of a body will remain at rest unless acted upon by an unbalanced force. ‘Aim: A body will remain at rest unless acted upon by an unbalanced force. (Inertia of rest) Materials required: Carom board, Carom coins. Procedure: 1, Make a pile of similar carom coins on a table, as shown in Figure. 2. Attempt a sharp horizontal hit at the bottom of the pile using another carom coin or the striker. 3. If the hit is strong enough, the bottom coin moves out quickly. Once the lowest coin is removed, the inertia of the other coins makes them “fall” vertically on the table. - Observation/Conclusion: A body will remain at rest unless acted upon by an unbalanced force. 2. Describe an activity of a body will remain at rest unless acted upon by an unbalanced force. Aim: A body will remain at rest unless acted upon by an unbalanced force. (Inertia of rest) ‘Materials required: Five-rupee coin, Glass tumbler, Stiff card. Procedure: . Seta five-rupee coin om a stiff card covering an empty glass tumbler standing on a table as shown in Figure. 2. Give the card a sharp horizontal flick with a finger. 3. If we do it fast then the card shoots away, allowing the coin to fall vertically into the glass tumbler due to its inertia. 4. The ine: of the coin tries to maintain its state of rest even when the card flows off. Observation/Conclusion: A body will remain at rest unless acted upon by an unbalanced force.29:00) t--) PHYSICAL SCIENCE HANDBOOK Sa Soy Mote i, 1. What is the ultimate outcome when balanced forces 11. Which of the following is a unit of force in the ST act upon an object? system? a) The object remains stationary. a) kg b)m/s oN d) Pa b) The object accelerates c) The object decelerates Ans: c) N d) The object moves at a constant velocity Ans: a) The object remains stationary 2. According to the fundamental law of motion, an object persists in its state of rest or uniform motion unless influenced by what? a) Unbalanced forces) Balanced forces ¢) Frictional forces d) Gravitational forces Ans: a) Unbalanced forces. 3. Inertia embodies the resistance of an object to alterations in which of the following? a) Direction b) Velocity c) Acceleration d) All the mentioned Ans: d) All the mentioned. 4. As per the second law of motion, what parameter is irectly linked to the acceleration of an object? a) Its mass b) Its inertia c) The applied force d) None Ans: c) The applied force. 5. What is the precise mathematical expression representing the second law of motion? a)F=mxa b)F=m/a c)F=m+a d)F=m-a Ans: a) F=m x a. 6. Ifa person exerts a force of 500 N against a wall, what is the corresponding reaction force applied by the wall? a)500N Ans: b) ON. 7. Which of the following scenarios best illustrates Newton’s first law of motion? a) A ball rolling down a hill. b) A book sliding on a table coming to a stop. ©) A person riding a bicycle at a steady pace. d) A car accelerating after being at rest. Ans: c) A person riding a bicycle at a steady pace. 8. When a force of 10 N is applied to an object weighing 2 kg, what would be the resultant ‘acceleration of the object? a) 20 m/s*_b) Sm/s* ‘Ans: b) 5 m/s” 9, Which of the following is an example of Newton's, third law of motion? a) A rocket launching into space b) A car accelerating on a straight road ©) A person pushing a box across the floor d) A book resting on a table Ans: a) A rocket launching into space 10, An object of mass 10 kg is subjected to a force of 20 N, What is the acceleration of the object? a) 0.5 m/s? b) 2 m/s? c) 200 m/s? d) 10 m/s? Ans: b) 2 nvs? b)ON c)250N_ d) 100N. c)2m/s*— d) 0.2 ms* 12, A force of 50 N is applied to an object of mass 10 kg. What is the net force acting on the “object if it is moving with a constant velocity? a)5N b)ION 6)50N ON Ans: d) ON 13. Which of the following is an example of inertia? a) A ball rolling down a hill b) A car tuming around a curve c) A book at rest on a table remaining stationary d) A person pushing a heavy box Ans:c)A book at rest on a table remaining stationary 14, If the mass of an object is doubled while the force acting on it remains constant, what happens to its acceleration? a) It doubles b) It remains the same ©) Ithalves 4) It quadruples Ans: c) It halves 15. Which of Newton’s laws states that an object will remain at rest or in uniform motion unless acted upon by an external force? a) Newton’s First Law _b) Newton’s Second Law ¢) Newton’s Third Law d) Law of Gravitation Ans: a) Newton's First Law 16, When a car accelerates on a straight road, the force responsible for its acceleration is primarily due to a) Frictional force b) Gravitational force ©) Applied forced) Normal force Ans: c) Applied force According to Newton’s second law of motion, the acceleration of an object is directly proportional to the a) Net force acting on the object b) Mass of the object) Velocity of the object d) Weight of the object Ans: a) Net force acting on the object 18. When a person jumps from a boat to then shore, the boat moves backward due to: a) Action-reaction force b) Gravitational force ) Frictional force 4) Buoyant force Ans: a) Action-reaction force 19, Inertia is a measure of a) Mass b) Velocity c) Force Ans: a) Mass 20.What is the SI unit of linear momentum. d) Acceleration a)kgms' b)kgms — c)kgms?_—d) kg 21.Which is more inertia a)Bicycle b) Bike ¢)Car_—d) Bus Ans: d) BusPHYSICAL SCIENCE HANDBOOK Chapter — 4 i Main Points Me 1. Pure substances have the same colour, composition, taste and texture at a given temperature and pressure. 2. A mixture contains more than one substance mixed in any proportion. Ifthe mixture has a uniform composition throughout, then itis called homogeneous mixtures or solutions. If the mixture has physically distinct parts and has non-uniform compositions, then it is called heterogeneous mixture, 5. Mixtures can be separated into pure substances using appropriate separation techniques. 6. A solution is a homogeneous mixture of two or more substances. 7. The component of the solution that dissolves the other component in it (usually the component present in larger amount) is called the solvent. The component of the solution that is dissolved in the solvent (usually present in lesser quantity) is called the solute. 8, They do not scatter a beam of light passing through the solution. So, the path of light is not visible in a solution. 9. Depending upon the amount of solute present in a solution, it can be called dilute, concentrated or saturated solution 10. The amount of the solute present in the saturated solution at this temperature is called its solubility. 1. The concentration of a solution is the amount of solute present in a given amount of solution. 12. There are various ways of expressing the concentration of a solution. Mass of solute Mass of solution * Mass of solute Vonume of solution * Volume of solute Volume of solution 13. A suspension is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of the medium, Particles of a suspension are visible to the naked eye and unstable. 14. Colloids are heterogeneous mixtures in which the particle size is too small to be seen with the naked eye. ‘The particles are called the dispersed phase and the medium in which they are distributed is called the dispersion medium. 15, Scattering of a beam of light is called the Tyndall effect 16, Colloids are classified according to the state (solid, liquid or gas) of the dispersing medium and the dispersed phase 17. A physical change brings change in the state of matter, without a change in composition and no change in the chemical nature of the substance. 18, Chemical change brings change in the chemical properties of matter and we get new substances. 19. Pure substances can be classified either as elements or compounds. 20. An element as a basic form of matter that cannot be broken down into simpler substances by chemical reactions. 21, Elements can be normally divided into metals, non-metals and metalloids. 22. A compound is a substance composed of two or more different types of elements, chemically combined in a fixed proportion. Mass by mass percentage of a solution Mass by volume percentage of a solution Volume by volume percentage of a solution x 100 Exercises 1. Which separation techniques will you apply for the separation of the following? (a)Sodium chloride from its solution in water (b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride (c) Small pieces of metal in the engine oil of a car (@) Different pigments from an extract of flower petals (e) Butter from curd (0) Oil from waterCay goat} PHYSICAL SCIENCE HANDBOOK (g) Tea leaves from tea (h) Iron pins from sand (i) Wheat grains from husk (j) Fine mud particles suspended in water ‘Ans: (a) Crystallization or evaporation (b) Sublimation (©) Centrifugation or sedimentation (d) Chromatography (e) Centrifugation (f) Separating funnel (g) Filtration (h) Magnetic separation (i) Winnowing @) Centrifugation 2.Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue. Ans: Take one cup of water as solvent and boil it for few minutes. When solvent boils, add solutes, i.e., milk, tea leaves, sugar. Boil the solution for a few minutes. Milk is soluble in water. Sugar also dissolves in the solution but tea leaves remain as filtrate. The insoluble residue of tea leaves is filtered using a strainer and the tea is ready. 3. Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution). ~ - Solubility Potassium nitrate 2 167 Sodium chloride 36 37 Potassium chloride 35 54 ‘Ammonium chloride: (a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K? (b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain. (c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature? (d) What is the effect of change of temperature on the solubility of a salt? Ans: (a) At 313 K, 62 gm of potassium is added in 100 gm of water to form a saturated solution. So, 50 gm of water requires 31 gm potassium nitrate (b) As solution cools, potassium chloride gets crystallized or precipitated. The solubility of potassium chloride is 54 gm and at room temperature is 35 g. So, excess of potassium chloride dissolved gets precipitated (©) Solubility of potassium nitrate at 293 K = 32 g Solubility of sodium chloride at 293 K = 36 ¢ Solubility of potassium chloride at 293 K = 35 g Solubility of ammonium chloride at 293 37g Ammonium chloride has the highest solubility at 293 K. (d) The solubility of the salt increases with an increase in temperature. Explain the following giving examples. (a) Saturated solution (b) Pure substance (©) Colloi (@ Suspension Ans: (a) Saturated solution: At any particular temperature, a solution that has dissolved as much solute as it is capable of dissolving, is said to be a saturated solution.By Coa) PHYSICAL SCIENCE HANDBOOK Example: When no more solute (sugar) can be dissolved in a solution (water sugar solution) at a given ‘Temperature. (b) Pure substance: Pure substances can be elements or compounds. An element is a form of matter that cannot be broken down by chemical reactions into simpler substances. A compound is a substance composed of two or more different types of elements, chemically combined in a fixed proportion Example: Gold, Iron, Water, Table salt (c) Colloid: Colloids are heterogeneous mixtures in which the particle size is too small to be seen with the naked eye. The particles are called the dispersed phase and the medium in which they are distributed is called the dispersion medium. It shows Tyndall effect. Example: Milk, fog, blood. (d) Suspension: A suspension is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of the medium, Particles of a suspension are visible to the naked eye and unstable. It shows Tyndall effect. Example: Muddy water, Chalk powder in water. 5. Classify each of the following as a homogeneous or heterogeneous mixture. Soda water, wood, air, soil, vinegar, filtered tea, Ans: Homogeneous mixture: Soda water, air, vinegar, filtered tea Heterogeneous mixture: Wood, soil 6. How would you confirm that a colourless liquid given to you is pure water? Ans: Every liquid has a characteristic boiling point at 1 atmospheric pressure. If the given colourless liquid boils exactly at 100°C or 373 K), then it is pure water. 7. Which of the following materials fall in the category of a “pure substance”? (a) Tee (b) Milk (c) Tron (@) Hydrochloric acid (e) Calcium oxide (0 Mercury (g) Brick (h) Wood @ Air Ans: Ice, Iron, Hydrochloric acid, Calcium oxide, Mercury 8, Identify the solutions among the following mixtures. (a) Soil (b) Sea water (Air (a) Coal (e) Soda water Ans: Sea water, Air, Soda water 9. Which of the following will show “Tyndall effect”? (a) Salt solution (b) Milk (C) Copper sulphate solution (@) Starch solution Ans: Milk and Starch solution are colloids and will show Tyndall effect. 10. Classify the following into elements, compounds and mixtures. (a) Sodium (b) Soil (c) Sugar solution (a) Silver (e) Calcium carbonate (® Tin (g) Si (h) Coal @ AirTH CLASS (i) Soap (k) Methane (D Carbon dioxide (m) Blood Ans: Elements: Sodium, Silver, Tin, Silicon ICAL SCIENCE Compounds: Calcium carbonate, Methane, Carbon dioxide Mixtures: Soil, Sugar, Coal, Air, Soap, Blood 1. Which of the following are chemical changes? (a) Growth of a plant (b) Rusting of iron (©) Mixing of iron filings and sand (a) Cooking of food (e) Digestion of food (0) Freezing of water (g) Burning of a candle Ans: (a) Growth of a plant (b) Rusting of iron (d) Cooking of food (©) Digestion of food (g) Burning of a candle Rocca ee os ay 1, What is meant by a substance? (Page No:88) Ans: A substance is a specific type of matter that ha type of particles or more than one type of particles. uniform and definite composition. It can be made up of one 2. List the points of differences between homogeneous and heterogeneous mixtures. Ans: ‘Homogeneous mixture Heterogeneous mixture If the mixture has a uniform composition throughout, then it is called homogeneous mixtures If the mixture has physically distinet parts and has non-uniform compositions, then it is called heterogeneous mixture. It does not show Tyndall effect Itshows Tyndall effect Tthas no visible boundary or boundaries oF separation between its constituents. Ithas visible boundary or boundaries of separation between its constituents. Ans: Differentiate between homogeneous and heterogeneous mixtures with examples. (Page Ne 394) Homogeneous mixture ‘Heterogeneous mixture If the mixture has a uniform composition throughout, then it is called homogeneous mixtures Ifthe mixture has physically distinct parts and has non-uniform compositions, then itis called heterogeneous mixture. It does not show Tyndall effect Itshows Tyndall effect Ithas no visible Boundary or boundaries of separation between its constituents. thas visible boundary or boundaries of separation between its constituents. Ex: a mixture of alcohol and water Ex: a mixture of sodium chloride and sand 2. How are sol, solution and suspension different from each other? Ans: A sol is a type of colloid with solid particles dispersed in a liquid. Solutions are homogeneous mixtures where a solute is mpletely dissolved in a solvent. ‘Suspensions are heterogeneous mixtures where larger particles remain suspended but settle out over time. 3. To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature. Ans: Mass of sodium chloride = 36 g HANDBOOKBy Coa) PHYSICAL SCIENCE HANDBOOK Mass of solution = 36 + 100 = 136 g = Mass of solute 8 - Concentration of a solution = oF x 100 = 35> x 100 = 26.47% 6) 1. Classify the following as chemical or physical changes: (Page No a) cutting of trees, b) melting of butter in a pan, ) rusting of almirah, 4) boiling of water to form steam, ©) passing of electric current, through water and the water breaking down into hydrogen and oxygen gases, £) dissolving common salt in water, g) making a fruit salad with raw fruits, and h) burning of paper and wood. Ans: a) Physical change b) Physical change ©) Chemical change 4) Physical change e) Chemical change ) Physical change g) Physical change h) Chemical change 2. Try segregating the things around you as pure substances or mixtures. Ans: Pure substances: Copper wire, Distilled water, Diamond, Sugar, Salt, Pure gold, Ice, Baking soda etc. Mixtures: Air, Paper, Milk, Curd, Cooking oils, Steel, Ice ream, Ink, LPG, Lemon juice etc. 1, Explain with the help of an activity which shows that homogeneous and heterogeneous mixtures. : Identification of homogeneous and heterogeneous mixtures. Materials required: Beakers, Water, Spatula, Copper sulphate powder, Potassium permanganate/ Common salt Procedure: 1. Let us divide the class into groups A, B, C and D. 2. Group A takes a beaker containing 50 mL of water and one spatula full of copper sulphate powder. Group B takes 50 mL of water and two spatula full of copper sulphate powder in a beaker. 3. Groups C and D can take different amounts of copper sulphate and potassium permanganate or common salt (sodium chloride) and mix the given components to form a mixture 4, Report the observations on the uniformity in colour and texture, 5. Groups A and B have obtained a mixture which has a uniform composition throughout. Compare the colour of the solutions of the two groups. Though both the groups have obtained copper sulphate solution but the intensity of colour of the solutions is different. This shows that a homogeneous mixture can have a variable composition, 6. Groups C and D have obtained mixtures, which contain physically distinct parts and have non-uniform compositions. Such mixtures are called heterogeneous mixtures. Observation/Conclusion: Identification of homogeneous and heterogeneous mixtures. 2, Explain with the help of an activity which shows that solution, suspension and colloidal solution. Aim: Preparation of solution, suspension and colloidal solution. ‘Materials required: Crystal of copper sulphate, Chalk powder/Wheat flour, Milk/Ink, Test tubes, Torch light, Filter paper, Funnel, Procedure: 1. Let us again divide the class into four groups— A, B, C and D. 2. Distribute the following samples to each group: - Few crystals of copper sulphate to group A. - One spatula full of copper sulphate to group B. - Chalk powder or wheat flour to group C. - Few drops of milk or ink to group D.By Coa) PHYSICAL SCIENCE HANDBOOK Each group should add the given sample in water and stir properly using a glass rod. Particles of mixture are visible only in case of group C. Direct a beam of light from a torch through the beaker containing the mixture and observe from the front. The path of beam of light was visible in the case of group C and D. Leave the mixtures undisturbed for a few minutes (and set up the filtration apparatus in the meantime). The Particles settle down after sometime in the case of group C. Filter the mixture, Residue will be left in case of group C. 10, Groups A and B have got a solution, Group C has got a suspension. Group D has got a colloidal solution. — sero Per awaw ty & Pier paper Cone of Mlded a her paper Observation/Conclusion: Identification and preparation of solution, suspension and colloidal solution,29:00) t--) PHYSICAL SCIENCE HANDBOOK & jultiple Soy Mote a, L.Which of the following is a homogeneous mixture? 11, Cloud is an example of a) Saltwater 'b) Sand and iron filings a) Solid dispersed in a gas c) Oil and water d) Milk and water b) Liquid dispersed in a gas Ans: a) Saltwater 2. The process of separating a mixture of li different boiling points is called? a) Filtration b) Evaporation ©) Distillation d) Condensation Ans: c) Distillation 3. Which of the following is a pure substance? a) Air b) Sugar solution c) Salt solution d) Gold Ans: d) Gold 4. The process of conversion of a solid directly into gas, without passing through the liquid state is called? a) Melting ) Evaporation ©) Sublimation d) Condensation Ans: c) Sublimation 5. A solution that cannot dissolve any more solute at a given temperature is said to be? a) Unsaturated b) Saturated ©) Supersaturated ——_) Dilute Ans: b) Saturated 6. Which method is used to separate a mixture of salt and water? a) Filtration ©) Decantation Ans: b) Evaporation 7. Which of the following is not a physical change? a) Boiling of water b) Rusting of iron ) Freezing of water 4) Melting of wax Ans; b) Rusting of iron 8. Which of the following statements about solutions is true? a) Solutions are always homogeneous mixtures. b) Solutions can be separated into their components by filtration, ©) The solute is present in larger quantity than the solvent in a solution, 4) Solutions cannot be formed with gases. Ans: a) Solutions are always homogeneous mixtures. 9. Which of the following is a chemical change? a) Freezing water to form ice b) Cutting a piece of wood into smaller pieces c) Digesting food in the stomach 4) Melting chocolate to form a liquid Ans: c) Digesting food in the stomach 10. The smallest particle of an element that retains all the properties of that element is called a) Atom b)Molecule c) Compound d) Mixture Ans: a) Atom uids with b) Evaporation 4) Sublimation €) Liquid dispersed in a solid 12. A colloidal solution is a) Homogeneous mixture b) Heterogeneous and transparent mixture ©) Heterogeneous mixture, Particles can be seen with naked eyes 4d) Heterogeneous mixture, Particles cannot be seen with naked ey: Ans: d) Heterogeneous mixture, Particles cannot be seen with naked eyes 13. What is the primary difference between a compound and a mixture? a) Mixtures have a fixed composition b) Compounds are chemically bonded ) Mixtures are always pure substances d)Compounds can be separated by physical methods Ans: b) Compounds are chemically bonded 14, Name the mixture whose particles are large enough to scatter light. a) Colloid c) Homogeneous solution Ans: a) Colloid 15. Which method is used to separate cream from milk? a) Centrifugation ©) Distillation Ans: a) Centrifugation 16. Ifa solution contains 60g of common salt in 340g of water, the mass by mass percentage will be a)25% — b)15% — c)20% =) 17.6% Ans: b) 15 % 17. Which of the following is a true solution? a) Mi b) Chalk powder in water ©)Salt solution) Blood Ans: c) Salt solution 18. Which of the following are homogeneous in nature? a) Ice, Soil b) Wood, Air c) Wood, Soil d) Ice, Air Ans: d) Ice, Air 19. Which of the following parameters of a substance does not alter during a physical change? a) State b)Mass c) Volume —d) Size ‘Ans: b) Mass 20. An example of a liquid metal and a liquid non- metal is a) gallium, mercury ) mercury, bromine Ans: c) mercury, bromine b) True solutions d) All of the above b) Adsorption 4) Crystallization b) mercury, chlorine 4d) bromine, sulphur29:00) t--) PHYSICAL SCIENCE HANDBOOK Chapter - 5 1. Gravitation: It is the force of attract-on between any two bodies in the universe. 2. The force of attraction between any two objects in the universe is called the gravitational force. 3. Universal law of gravitation: Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. where G is Universal gravitation constant 4. The SI unit of G is Nm* kg? and G = 6.67 x 10"! Nm? kg? 5. The value of G was found out by Henry Cavendish by using a sensitive balance. 6. Whenever an object falls towards the earth, an acceleration is involved. This acceleration is due to the earth’s gravitational force. Therefore, this acceleration is called the acceleration due to gravity. 7. The value of acceleration due to gravity of the earth, g = 9.8 ms. 8. The force of gravity decreases with altitude. It also varies on the surface of the earth, decreasing from poles to the equator. 9. Relation between g and Gis g = s 10, Acceleration due to gravity independent of the mass of the object. 11. The motion of a body under the influence of force of gravity alone is called a free fall. 12. The mass of a body is the quantity of matter contained in it. ST unit is kg 13. The weight of an object is the force with which it is attracted towards the earth. SI unit is newton (N). W=mxeg. 14, The weight may vary from place to place but the mass stays constant, 15, Weight of the object on the moon =~ x Its weight on the earth 16. The force acting on an object perpendicular to the surface is called thrust. 17. The thrust on unit area is called pressure. The SI unit is Nm? Thrust ‘Area Pressure 18, The SI unit of pressure is called pascal, denoted as Pa. 19. Pressure exerted in any confined mass of fluid is transmitted undiminished in all directions, 20. The upward force acting on a body immersed in a fluid is called upthrust or buoyant force. 21. All objects experience a force of buoyancy when they are immersed in a fluid. 22. Objects having density less than that of the liquid in which they are immersed, float on the surface of the li 23. Object having density more than the density of the liquid in which it is immersed then it sinks in the liquid. 24. The mass of the object per unit volume is called as density. The SI unit is kg m* ‘Mass Density = [oper 25. Archimedes’ principle: When a body is immersed fully or partially in a fluid, it experiences an upward force that is equal to the weight of the fluid displaced by it. 26. Archimedes’ principle is used in designing ships, submarines, Lactometers and hydrometers. T Exercises / 1. How does the force of gravitation between two objects change when the distance between them is reduced to half? Ans: According to universal law of gravitation, F = G ss ‘When the distance is reduced to half (d/2) vag ¢ mm m, Mz "dy ea d?By Coa) poet ICAL SCIENCE HANDBOOK ‘Thus, the gravitational force becomes 4 times increased. 2. Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object? Ans: Gravitational force acts on the object is directly proportional to mass. But acceleration due to gravity does not depends on mass of the objects. Hence, heavy objects do not fall faster than light objects. 3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6x10 kg and radius of the earth is 6.4x10° m.) Ans: Mass of the earth (M) = 6 x 10" kg Mass of the object (m) = 1 kg Radius of the earth © = 6.4 x 10°m G=6.7x 10"! Nm’ kg? According to universal law of gravitation, F = G = F 6.7x 107"? x6x10"* x1 (ax 108)? 4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why? Ans: According to universal law of gravitation, two objects with masses attract each other with equal gravitational force, but in opposite directions. So, the earth attracts the moon with the same force with which the moon attracts the earth. 5. If the moon attracts the earth, why does the earth not move towards the moon? Ans: According to universal law of gravitation, the Earth and the moon experience equal gravitational forces from each other. The mass of the Earth is much larger than the mass of the moon. Hence, it accelerates at a rate lesser than the acceleration rate of the moon towards the Earth. So, the Earth does not move towards the moon. 6. What happens to the force between two objects, if (i) the mass of one object is doubled? (ii) the distance between the objects is doubled and tripled? (ii) the masses of both objects are doubled? =98N law of gravitation, Fo mim2 and F cc + Ans: According to unive1 z ( Fis directly proportional to the masses of the objects. If the mass of one object is doubled, then the gravitational force will also get doubled. (ii) F is inversely proportional to the square of the distances between the objects. If the distance is doubled, then the gravitational force becomes one-fourth of its original value. If the distance is tripled, then the gravitational force becomes one-ninth of its original value. (iii) F is directly proportional to the product of masses of the objects. If the masses of both the objects are doubled, then the gravitational force becomes four times the original value. 7. What is the importance of universal law of gravitation? Ans: (i) the force that binds us to the earth (ii) the motion of the moon around the earth ii) the motion of planets around the Sun (iv) the tides due to the moon and the Sun 8. What is the acceleration of free fall? Ans: Whenever an object falls towards the earth, an acceleration is involved. This acceleration is due to the earth’s gravitational force. Therefore, this acceleration is called the acceleration due to gravity. 9. What do we call the gravitational force between the earth and an object? Ans: Weight of the object. 10. Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.] Ans: No. The value of g at the equator is less than that at the poles, The few grams of gold at poles will measure less when taken to the equator. Hence, Amit’s friend will not agree with the weight of the gold bought. 11. Why will a sheet of paper fall slower than one that is crumpled into a ball? Ans: A sheet of paper has a larger surface area in comparison to a crumpled paper ball. The larger surface area will experiences a larger air resistance. Hence, a sheet of paper falls slower than the crumpled ball. 12. Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. ‘Whaat is the weight in newtons of a 10 kg object on the moon and on the earth? Ans: Mass of the object (m) = 10 kg Weight of the object on the earth (W) = mg = 10x 9.8 =98.NBy Coa) PHYSICAL SCIENCE HANDBOOK Weight of the abject on the moon (W) == x Weight of the object on the earth = = x 98 = 16.3 N 13. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (i the maximum height to which it rises, i) the total time it takes to return to the surface of the earth. Ans: Initial velocity, u=49 m/s Acceleration due to gravity (g) = 9.8 ms? () Maximum height (h) ze = B28 = BL 1225m 8 =2x5=10s (ii) Total time taken = 14, A ston a 98 released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground. Ans: Initial velocity (u)=0 ms! Height (h) = 19.6 m Acceleration due to gravity (g) = 9.8 ms? Final velocity (v) = /2gh = V2x98x 196 = V19.6 x 19.6 = 19.6 m/s 15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s?, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone? Ans: Initial velocity (u)=40 m/s Acceleration due to gravity (g) = 10 mvs? we 40x40 80m Maximum height (h) == S28 Net displacement = 80 — 80 = Total distance = h +h =80 + 80= 160m 16. Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6x 104 kg and of the Sun = 2 x10" kg. The average distance between the two is 1.5 x 10!! m, Ans: Mass of the earth (mi) =6 x 10™kg Mass of the sun (m2)=2x 10%kg d= 1.5x 10" m Gravitational force, F = G ™7# p= 867x107" x6x107x2x10" _ 3 56 gy (415x102)? 17. A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet. Ans: Let the two stones meet after a time = t Let the displacement of the stone in time t from the top of the tower = Si Let the displacement of the stone in time t from the ground = S> Height of the tower = Si + S2 = 100m When the stone dropped from the tower: Si =" gt When the stone thrown upwards Sp=ut- Yet? Si +5: Two stones will meet (0) = = P= 45 18. A ball thrown up vertically returns to the thrower after 6 s. Find (a) the velocity with which it was thrown up, (b) the maximum height it reaches, and (c) its position after 4 s. Ans: Total time of flight = 6 s, So time of ascent or time of decent = 3 s (a) Initial velocity (u) = g x t=9.8x3=29.4ms" (b) Maximum height (h) == = 2242224 _ 2048 gg _2xo8 2 © Time for downward fall =4—3= 1s h=%gr="x98x Position after 4 s = 44.1—4.9 19. In what direction does the buoyant force on an object immersed in a liquid act? Ans: Vertically upward direction. 20. Why does a block of plastic released under water come up to the surface of water? Ans: The upthrust or buoyant force exerted by water brings the plastic block to the surface of water. 21. The volume of 50 g of a substance is 20 cm*. If the density of water is 1 g cm”, will the substance float or sink? Ans: Mass (m) = 50 g Volume (v) = 20 g/em* =44.1m5-00 ON) Bt [CAL SCIENCE HANDBOOK Density of given substance as = 2=25 g/cm’ Density of water = | g/em* Density of substance is greater than that of water, so the substance will sink in water. 22. The volume of a 500 g sealed packet is 350 cm, Will the packet float or sink in water if the density of water is 1 g em-*? What will be the mass of the water displaced by this packet? Ans: Mass of sealed packet (m) = 500 g Volume (v) = 350 g/em* Density of sealed packet = one = fn = 1.43 g/cm? Density of water = 1 g/em* Density of sealed packet is greater than that of water, so the substance will sink in water. Mass of water displaced = Volume x Density = 350 x 1 = 350 g. In Textbook Questions 1, State the universal law of gravitation. (Page No:114) Ans: Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centres of two objects. 2. Write the formula to find the magnitude of the gravi surface of the earth, Ans: F = G St 1, What do you mean by free fall? (Page Ni Ans: The earth attracts objects towards it. This is due to the gravitational force. Whenever objects fall towards the earth under this force alone, we say that the objects are in free fall, 2. What do you mean by acceleration due to gravity? Ans: The acceleration produced in an object under the influence of the gravitational force alone is called acceleration due to gravity. 1, What are the differences between the mass of an object and its weight? (Page No:122) Ans: jonal force between the earth and an object on the 18) ‘Mass Weight ‘The mass of a body is the quantity of matter ‘The weight of an object is the force with which it contained in it is attracted towards the earth. M=— W=mg STunit is kg STunit is newton (N). “The mass stays constant at all places. ‘The weight may vary from place to place. ‘Scalar quantity Vector quantit} Itis measured by common balance Itis measured by spring balance 2. Why is the weight of an object on the moon 1/6" its weight on the earth? Ans: Weight of an object depends on Acceleration due to gravity. Acceleration due to gravity on the moon is one~ sixth of that on the earth. Hence, the weight of an object on the moon 1/6" its weight on the earth. 1, Why is it difficult to hold a school bag having a strap made of a thin and strong string? (Page No:128) Ans: The thin string has smaller area and exerts a large pressure on our hand. So, it is difficult to hold a school bag having thin string, 2. What do you mean by buoyancy? Ans: The upward force acting on a body immersed in a fluid is called upthrust or buoyant force and this phenomenon is called buoyancy. 3. Why does an object float or sink when placed on the surface of water? Ans: An object sinks in water if its density is greater than that of water. An object floats on water if its density is less than that of water. 1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg? (Page No:130) ‘Ans: More than 42 kg. The weighing machine reads slightly less value due to the upthrust of air acting on our body. 2. You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?Cay goat} PHYSICAL SCIENCE HANDBOOK Ans: We know that 100 kg of cotton bag has a much more volume that the 100 kg of iron bar. 100 kg of cotton bag will displace much more weight of air than 100 kg of iron bar. The cotton bag experiences larger upthrust of air than the iron bar. So, reality 100 kg of cotton bag weighted in air will be heavier than 100 kg of iron weighed in air. 1, Explain with help of an activity which shows that motion of objects under the influence of gra‘ force of earth. To understand whether all objects hollow or solid, big or small, will fall from a height at the same rate. Materials required: Paper sheet, Stone. Procedure: Take a sheet of paper and a stone. Drop them simultaneously from the first floor of a building. 2. Observe whether both of them reach the ground simultaneously. We see that paper reaches the ground little later than the stone. 3. This happens because of air resistance. The air offers resistance due to friction to the motion of the falling objects. 4. The resistance offered by air to the paper is more than the resistance offered to the stone. 5. If we do the experiment in a glass jar from which air has been sucked out, the paper and the stone would fall at the same rate. Observations/ Conclusion: All objects hollow or solid, big or small, will fall from a height at the same rate. 2. Explain with help of an activity which shows that the concept of buoyance. Aim: To understand the concept of buoyance Materials required: Empty plastic bottle, Airtight stopper, Bucket, Water. Procedure: ‘Take an empty plastic bottle. Close the mouth of the bottle with an airtight stopper. Put it in a bucket filled with water. You see that the bottle floats. Push the bottle into the water. You feel an upward push. Try to push it further down. ‘You will find it difficult to push deeper and deeper. Thi the upward direction. 5. The upward force exerted by the water goes on increasing as the bottle is pushed deeper till it is completely immersed, 6. Now, release the bottle. It bounces back to the surface, 7. The gravitational attraction of the earth act on this bottle in a downward direction. The upward buoyant force exerted by the water is greater than the downward force of gravity acting on the bottle. Observations/ Conclusion: 1. When the bottle is immersed, the upward force exerted by the water on the bottle is greater than its weight. Therefore, it rises up when released. 2. The upward force exerted by the water on the bottle is known as upthrust or buoyant force. 3. Explain with help of an activity which shows that the concept of force of buoyancy. Aim: To understand the concept of force of buoyancy. Materials required: Stone, Rubber string/Spring balance, Bucket. Procedure: ‘Take a piece of stone and tie it to one end of a rubber string or a spring balance. Suspend the stone by holding the balance or the string as shown in Figure (a). Note the elongation of the string or the reading on the spring balance due to the weight of the stone, Now, slowly dip the stone in the water in a container as shown in Figure(b). Observe what happens to elongation of the string or the reading on the balance. ‘The elongation of the string or the reading on the balance will decrease. tional eye indicates that water exerts a force on the bottle in auaune