202110244-TRIUMPH-STUDENT-WORKBOOK-PHYSICAL_SCIENCE-G09-PART2

Physical Science Workbook_9_P_2.pdf 1 11/5/19 6:23 PM Name: ___________________________________ Section: ________________ Roll No.: _________ School: __________________________________

Table of Contents 1 37 10 WORK AND ENERGY 69 11 HEAT 106 12 SOUND PROJECT BASED QUESTIONS TABLE OF CONTENTS

10. WORK AND ENERGY SESSION 1 WORK 1.1 Mind Map 1.2 Terminology i. Work – Product of force and displacement. 1.3 Solved Examples Q1. Example 1: A boy pushes a book kept on a table by applying a force of 4.5 N. Find the work done by the force if the book is displaced through 30 cm along the direction of push. [Refer to TB page 170] A. Force applied on the book (F) = 4.5 N SESSION 1. WORK 1

Displacement (s) = 30 cm = = 30 m = 0.3 m 100 Work done, W = F s = 4.5 × 0.3 = 1.35 J Q2. Example 2: Calculate the work done by a student in lifting a 0.5 kg book from the ground and keeping it on a shelf of 1.5 m height. (g = 9.8m/s2 ) [ Refer to TB page 170] A. The mass of the book = 0.5 kg The force of gravity acting on the book is equal to ‘mg’ That is mg = 0.5 x 9.8 = 4.9 N The student has to apply a force equal to that of force of gravity acting on the book in order to lift it. Thus force applied by the student on the book, F = 4.9 N Displacement in the direction of force, s = 1.5 m Work done, W = F s = 4.9 x 1.5 = 7.35 J Q3. Example 3: A box is pushed through a distance of 4 m across a floor offering 100 N resistance. How much work is done by the resisting force?[ Refer to TB page 171] A. The force of friction acting on the box, F = 100 N The displacement of the box, s = 4 m The force and displacement are in opposite directions. Hence work done on the box is negative. That is, W = – F s = – 100 x 4 = – 400 J Q4. Example 4: A ball of mass 0.5 kg thrown upwards reaches a maximum height of 5 m. Calculate the work done by the force of gravity during this vertical displacement considering the value of g = 10 m/s2. [ Refer to TB page 172] SESSION 1. WORK 2

A. Force of gravity acting on the ball, F = mg = 0.5 x 10 = 5 N Displacement of the ball, s = 5 m The force and displacement are in opposite directions. Hence work done by the force of gravity on the ball is negative. W = – F s = – 5 x 5 = – 25 J 1.4 Key Concepts i. If a force acting on a body (F) causes a displacement (s) or produces change in the position of body, then work (W) is said to be done. ii. The amount of work done, W = F x s iii. The formula W = F x s can be used only in case only in translatory motion. iv. Work is a scalar because it has only magnitude but no direction. v. The work done by a force is considered as negative if the force acting on the object and displacement are in opposite directions. vi. If the work has positive value, the body on which work has been done would gain energy. If work has negative value, the body on which work has been done loses energy. 1.5 Reflection on Concepts Q1. [AS1] What is work according to science and write its units. [Refer to TB page 189 Q1] A. Work: The work done is equal to the product of the force and the displacement. Work done = Force x Displacement W=Fxs Unit of work = newton–metre (Nm) or Joule (J) Q2. [AS1] Give a few examples where displacement of an object is in the direction opposite to the force acting on the object. [Refer to TB page 189 Q2] SESSION 1. WORK 3

A. Examples where displacement of an object is in the direction opposite to the force acting on the object: i. When a ball is thrown up, the motion is in the upward direction, whereas the force due to earth’s gravity is in the downward direction. ii. A rubber ball moving on a plane ground will come to stop after sometime due to frictional force acting on it in opposite direction. 1.6 Application of Concepts Q1. [AS1] A cycle together with its rider weighs 100 kg. How much work is needed to set it moving at 3 m/s. [Refer to TB page 189 Q6] A. Mass of cycle together with cycle (m) = 100 kg Velocity (v) = 3 m/s Kinetic energy, K.E = 1 mv2 2 1 (3)2 = 2 × 100 × = 450J Work done by the rider of cycle = 450 J Objective Questions (1) A person is climbing a ladder with a suitcase on his head. Then the work done by that person on that suitcase is (Pg 190;Q 4) (A) Positive (B) Negative (C) Zero (D)Cannot be defined Correct Answer: A (2) If you have lifted a suitcase and kept it on a table, then the work done by you will depend on (Pg 190;Q 5) (A) The path of the motion of the suitcase (B) The time taken by you to do the work (C)Weight of the suitcase (D)Your weight SESSION 1. WORK 4

Correct Answer: C (B) Kg–m (3) S.I. unit of work (Pg 190; Q 1) (D) N-m2 (A) N–m (C) N/m Correct Answer: A SESSION 1. WORK 5

SESSION 2 ENERGY 2.1 Mind Map 2.2 Terminology i. Transfer of energy – Energy transfer is the conveyance of energy from one item to another or the conversion of one form of energy into another; the transfer occurs among different scales and motions. ii. Sources of energy – Sources from which energy can be obtained to provide heat, light, and power. iii. Kinetic energy – The energy possessed by a body by virtue of its motion is called kinetic energy. iv. Potential energy – The energy possessed by a body by virtue of its position or state is called potential energy. SESSION 2. ENERGY 6

v. Mechanical energy – Mechanical energy is the sum of kinetic energy and potential energy. 2.3 Solved Examples Q1. Example 5: Find the kinetic energy of a ball of 250 g mass, moving at a velocity of 40 cm/s. [Refer to TB page 176] A. Mass of the ball, m = 250 g = 0.25 kg Speed of the ball, v = 40 cm/s = 0.4 m/s Kinetic Energy, K.E. = 1 (0.25) x (0.4)2 = 0.02 J 2 Q2. Example 6: The mass of a cyclist together with the bicycle is 90 kg. Calculate the work done by cyclist if the speed increases from 6 km/h to 12 km/h. [Refer to TB page 176] A. Mass of cyclist together with bike, m = 90 kg Initial velocity, u = 6 km/h = 6 x 5 = 5 m/s 18 3 Final velocity, v = 12 km /h = 12 x 5 = 10 m/s 18 3 Initial kinetic energy, K.E.i = 1 mu2 2 = 1 (90) 5 2 2 3 = 125 J Final kinetic energy, K.E. f = 1 mv2 2 = 1 (90) 10 2 2 3 = 1 × (90) × 10 × 10 2 3 3 = 500 J The work done by the cyclist = Change in kinetic energy = K.E. f – K.E.i = 500 J – 125J = 375 J SESSION 2. ENERGY 7

Q3. Example 7: A block of 2 kg is lifted up through 2 m from the ground. Calculate the potential energy of the block at that point. [Take g = 9.8m/s2. [Refer to TB page 179] A. Mass of the block, m = 2 kg Height raised, h = 2 m Acceleration due to gravity, g = 9.8 m/s2 Potential energy of the block, P.E. = m g h = 2 × 9.8 × 2 = 39.2 J Q4. Example 8: A book of mass 1 kg is raised through a height ‘h’. If the potential energy increased by 49 J, find the height raised. [Refer to TB page 179] A. The increase in potential energy = mgh That is, mgh = 49 J i.e., 1 × 9.8 × h = 49 J The height raised, h = 49 =5m 1×9.8 2.4 Key Concepts i. Energy is defined as the capacity to do work. The energy depends on position and state of the object which is doing work. ii. Whenever work has been done on an object its energy either increases or decreases. iii. The energy possessed by a body by virtue of its motion is called kinetic energy. iv. The energy possessed by a body by virtue of its position or state is called potential energy. v. Mechanical energy is the sum of kinetic energy and potential energy. SESSION 2. ENERGY 8

2.5 Application of Concepts Q1. [AS1] A 10 kg ball is dropped from a height of 10 m. Find (a) The initial potential energy of the ball. (b) The kinetic energy just before it reaches the ground and (c) The speed just before it reaches the ground. [Refer to TB page 189 Q3] A. (a) Potential energy of the body: m = mass of ball = 10 kg ; g = 9.8 m/s2, height from which dropped (h) = 10 m. but P.E. = mgh Potential energy = 10 × 9.8 × 10 = 980 J (b) The potential energy is converted into kinetic energy as the body reaches the ground. K.E.= P.E. Kinetic energy before it reaches ground = 980 J (c) If ’v’ is the speed on reaching the ground, then Kinetic energy = ½ m v2 = 980 J m v2 = 1960 m = 10 kg, v2 = 1960/10 = 196 √ v = 196 = 14 SESSION 2. ENERGY 9

Speed just before it reaches the ground = 14 m/s. Q2. [AS1] Find the mass of a body which has 5 J of kinetic energy while moving at a speed of 2 m/s. [Refer to TB page 189 Q4] A. Kinetic energy of a body K.E.= ½ m v2 K.E. = 5 J; speed = 2 m/s mass of body (m) = ? K.E. = 5 = ½ m (2)2 m = 10 / 4 = 2.5 kg Mass of the body = 2.5 kg 2.6 Higher Order Thinking Skills Q1. [AS1] What is potential energy? Derive equation for gravitational potential energy of a body of mass ‘m’ at a height ‘h’. [Refer to TB page 189 Q1] A. Potential energy: The energy possessed by an object because of its position or shape is called its potential energy. i. Consider an object of mass ‘m’. ii. Let it be raised through a height, ‘h’ from the ground. iii. The minimum force required to raise the object, is equal to the weight of the object, viz., mg. iv. The object gains energy equal to the work done on the object against gravity i.e.; work done w = force × displacement = mg × h Therefore, Potential Energy = mgh Q2. [AS1] What is kinetic energy? Derive the equation for the kinetic energy of a body of mass ‘m’ moving at a speed ‘v’. [Refer to TB page 191 Q2] A. Kinetic Energy: The energy possessed by a body due to its motion is called kinetic energy. Consider an object of mass ‘m’ moving at a speed ‘v’. Let it be displaced through a distance ‘s’ when a constant force ‘F’ acts on it in the direction of its displacement. SESSION 2. ENERGY 10

Now the work done W = F × s − − − − − (1) The work done on the object will cause a change in its speed from ‘u’ to say ‘v’. Let ‘a’ be the acceleration produced. From the equation of motion : v2 – u2 = 2 a s We have s = (v2–u2) − − − −(2) 2a We know F = ma - - - - - - - - (3) From equations (1), (2) and (3) So the work done by the force F is w = ma × (v2–u2) 2a If the object is starting from rest, u = 0 Then w = 1 mv2 2 It is clear that the work done is equal to the change in the kinetic energy in the object. So, the kinetic energy possessed by an object of mass ‘m’ and moving at a speed ‘v’ is K.E = 1 mv2. 2 Q3. [AS7] When an apple falls down from a tree what happens to its gravitational potential energy just as it reaches the ground? After it strikes the ground. [Refer to TB page 189 Q4] A. i. When an apple falls down from a tree, its gravitational potential energy is gradually converted into kinetic energy required for its motion. ii. After it strikes the ground, the whole gravitational potential energy is converted into kinetic energy such that the total mechanical energy is conserved. Objective Questions (1) The energy possessed by a body by virtue of its motion is called as (Pg 190; Q2) (A) Potential energy (B) Kinetic energy (C)Attractive energy (D) Gravitational energy Correct Answer: B SESSION 2. ENERGY 11

(2) Two objects with same masses have been dropped from same height at same time. Which of the following will remain same in case of these objects? (Pg 190; Q3) (A) Speed (B) Gravitational force (C) Potential energy (D)Kinetic energy Correct Answer: C SESSION 2. ENERGY 12

SESSION 3 CONVERSION AND CONSERVATION OF ENERGY 3.1 Mind Map 3.2 Terminology i. Conservation of energy – ‘Energy can neither be created nor be destroyed. The total energy in this universe is always conserved’. This is called law of conservation of energy. ii. Gravitational energy – Energy due to the action of gravity. iii. Power – Rate of work done. 3.3 Solved Examples Q1. Example 9: A person performs 420 J of work in 5 minutes. Calculate the power delivered by him. [Refer to TB 183] A. Work done by the person, W = 420 J Time taken to complete the work, t = 5 min = 5 x 60 s = 300 s Power delivered, P = W = 400 = 1.4 W t 300 Q2. Example 10: A woman does 250 J of work in 10 seconds and a boy does 100 J of work in 4 seconds. Who delivers more power? [Refer to TB 183] SESSION 3. CONVERSION AND CONSERVATION OF ENERGY 13

A. Power, P = W t Power delivered by woman = 250 = 25 W 10 Power delivered by boy = 100 = 25 W 4 Both woman and boy deliver same power, i.e., rate of doing the work by woman and boy is equal. 3.4 Key Concepts i. ‘Energy can neither be created nor destroyed. The total energy in this universe is always conserved’. This is called law of conservation of energy. ii. Power is the rate at which work is done. iii. Power (P) = w/t, where w is the work done in time t. 3.5 Reflection on Concepts Q1. [AS1] Write few daily life examples in which you observe conservation of energy. [Refer to TB page 189 Q3] A. conversion of electrical energy into heat energy by Iron box, chemical energy into light energy by a torch etc. Q2. [AS1] Give some examples for renewable sources of energy. [Refer to TB page 189 Q5] A. Solar energy, Bio mass energy, Bio gas, Energy from sea (Tidal energy and Ocean thermal energy), Geo thermal energy , Wind energy and Atomic energy etc. Q3. [AS5] Draw a diagram to show conservation of mechanical energy in case of free falling body [Refer to TB page 189 Q4] SESSION 3. CONVERSION AND CONSERVATION OF ENERGY 14

A. 3.6 Application of Concepts Q1. [AS1] A man carrying a bag of total mass 25 kg climbs up to a height of 10 m in 50 seconds. Calculate the power delivered by him on the bag. [Refer to TB page 189 Q1] A. The work done by the man, W = F × s Force acting on the bag, F = mg m = mass of bag = 25 kg; g = 9.8 m /s2 F = 25 × 9.8N S = height = 10 m W =25 × 9.8 × 10 t = time taken by the man to carry the bag = 50 s Power delivered = w/t = 2450/50 =49 W SESSION 3. CONVERSION AND CONSERVATION OF ENERGY 15

The power delivered by him on the bag = 49 W Q2. [AS1] Calculate the work done by a person in lifting a load of 20 kg from the ground and placing it 1 m high on a table. [Refer to TB page 189 Q3] A. Mass of load, m = 20 kg Displacement s = 1 m Work done, w = F × s = mg × s = 20 kg × 9.8 m/s2 × 1 m = 196 kg m × m/s2 = 196 Nm = 196 J Work done by the person = 196 J Q3. [AS7] Which of the renewable sources of energy would you think suitable to produced in you native place. Why? [Refer to TB page 189 Q6] A. Students’ response. Hint –Solar energy, Biomass energy , Biogas and Wind energy etc can be produced in native place. 3.7 Higher Order Thinking Skills Q1. [AS7] When you lift a box from the floor and put it on an almirah potential energy of the box increases but there is no change in its kinetic energy. Is it violation of conservation of energy? Explain. [Refer to TB page 189 Q3] A. i. No. It is not a violation of conservation of energy. ii. Actually when we put the box at a higher level from ground, the force of friction offered by the plank of the almirah also increases and so the increased P.E. is thus balanced. As such, there is no increase or decrease in the energy of the system and it is still conserved. 3.8 Suggested Experiments Q1. [AS3] Conduct an experiment to prove the conservation of mechanical energy and write a report on it. [Refer to TB page 190 Q1] SESSION 3. CONVERSION AND CONSERVATION OF ENERGY 16

A. i. Take a long thread say 50 –60 cm long. ii. Attach a small heavy object like a metal ball at one end. iii. Tie other end to a nail fixed to the wall as shown in figure. iv. Now pull the bob of the pendulum to one side to the position A1 and release it. v. The bob swings towards opposite side and reaches the point A2. It repeats the motion over and over again. vi. The potential energy of the bob is minimum at ‘A’ and reaches maximum at A1 because the height of the bob is maximum at that position. vii. When the bob is released from this position (A1), its potential energy decreases and kinetic energy starts increasing slowly. viii. When the bob reaches the position A, its kinetic energy reaches maximum and potential energy becomes minimum. ix. As the bob proceeds from A to A1, its potential energy increases slowly and becomes maximum at A2 . x. The total potential energy and kinetic energy remains constant at any point of path of motion during the oscillation of the pendulum. xi. Thus the total mechanical energy in the system of pendulum remains constant. Q2. [AS3] Conduct an experiment to calculate the total energy of a freely falling body at different heights. [Refer to TB page 190 Q2] A. Drop an object of mass 20 kg from a height of 4 m. Observation: a) In a freely falling body, the total energy of system (i.e. the sum of potential energy and kinetic energy) is same at any instance of its travel. SESSION 3. CONVERSION AND CONSERVATION OF ENERGY 17

b) The sum of potential energy and kinetic energy at any instance of its travel is same. Hence we can say that the mechanical energy is conserved in the system. Height at Velocity of ePnoetergnytial Kenineergtiyc eTontearlgy which object at Ep = mgh Ek =½ mv2 P.E. + object different [in Joules K.E. located heights (in [in Joules (J)] [in Joules in metres (m) m/s) (J)] J] 4.0 0 800 0 800 3.55 3 710 90 800 3.0 √ 20 600 200 800 2.35 √ 33 470 330 800 0.8 8 160 640 800 3.9 Suggested Projects Q1. [AS4] Collect information about different sources of energy and write a report on the advantages and disadvantages in harnessing energy from these sources. [Refer to TB page 190 Q2] A. Students’ activity. Q2. [AS4] Make different models showing the harnessing energy from different energy sources. [Refer to TB page 190 Q3] A. Student’s activity. Q3. [AS7] How will the increasing energy needs and conservation of energy influence the international peace, co–operation and security? Collect information on this and write a report. [Refer to TB page 190 Q1] SESSION 3. CONVERSION AND CONSERVATION OF ENERGY 18

A. i. As the world’s population is increasing, the consumption of energy sources both nat- ural and fossil are increasing at a high rate. ii. The forest wealth biodiversity are seriously dwindling and this scarcity has a serious effect on economically backward countries. iii. In the struggle for existence the well developed countries with their deadly weapons and armies may declare war against poor countries to plunder their natural resources. iv. These would seriously influence the international peace, co–operation and security to a larger extent. v. Then the world will be in a hot tea pot and finally we may face a third world war. —— CCE Based Practice Questions —— AS1-Conceptual Understanding Very Short Answer Type Questions 1. State true or false. [Refer to Session 10.1 ] (i) The work done by a force is positive when the direction of force is opposite to the direction of displacement. [] (ii) If the position of a bag is changed by a person, then his muscular force is doing the work. [] (iii) If a ball is thrown upwards, the work done by gravity is negative. [ ] (iv) It is difficult to lift a heavy bag because friction acts against it. [ ] (v) W = –F s represents; force and displacement are in opposite directions. ] [ CHAPTER 10. WORK AND ENERGY 19

2. Match the following. [] Column B [(Session 10.1)] a. joule b. ms-2 Column A i. Force c. newton d. watt ii. Displacement [] e. metre iii. Work [] iv. Acceleration due to gravity [ ] v. Power [] 3. State true or false. [Refer to Session 10.3 ] (i) Potential energy can be converted into gravitational energy. [ ] ] (ii) Solar energy can only be converted into heat energy. ] ] [ ] (iii) Petroleum and coal are produced by dead plants and animals. [ (iv) When potential energy decreases kinetic energy also decreases. [ (v) In a pendulum, sum of potential energy and kinetic energy remains constant. [ CHAPTER 10. WORK AND ENERGY 20

4. Match the following. [] Column B [(Session 10.3)] [] a. Electrical energy to mechanical energy Column A [] b. Solar energy to heat energy [] c. Solar energy to chemical energy i. Photosynthesis [] d. Chemical energy to electrical energy e. Potential energy to electrical energy ii. Hydroelectric plants iii. Batteries iv. Solar cooker v. Fan 5. Fill in the blanks. [Refer to Session 10.3 ] (i) Solar batteries can be used to convert solar energy to energy. (ii) In a windmill, wind energy is converted into energy when the blades of the windmill rotate. (iii) Water heaters generate heat by converting energy to heat energy. (iv) Dead plants and animals decompose to form petroleum and coal which have energy stored in them. (v) Total energy of free fall is calculated using the formula . 6. State true or false. [Refer to Session 10.2 ] (i) Energy can neither be created nor be destroyed but can be changed from one form to another. [] CHAPTER 10. WORK AND ENERGY 21

(ii) Object which does work gains energy and the object on which work is done loses energy. [] (iii) Energy possessed by a moving object is known as kinetic energy. [ ] (iv) Energy possessed by an object due to its position or shape is known as mechanical energy. [] (v) The formula for potential energy is P.E = 12mv2. [ ] 7. Match the following. [] Column B [(Session 10.2)] a. Mechanical energy Column A i. Energy stored in water in dams ii. Energy possessed by a moving cycle [ ] b. Gravitation iii. K.E + P.E [ ] c. Friction iv. Force acting against moving objects [ ] d. Kinetic energy v. Force acting on a falling object [ ] e. Potential energy 8. Fill in the blanks. [Refer to Session 10.2 ] (i) K.E = . (ii) Sum of kinetic energy and potential energy is called __________________ ____________ . (iii) Friction does work on the moving object. (iv) A moving object can do . (v) As the height of an object increases, its energy increases. CHAPTER 10. WORK AND ENERGY 22

Short Answer Type Questions 9. Answer the following questions in 3-4 sentences. (i) [(Session 10.1)] Define work. Give the expression for work done. (ii) [(Session 10.1)] Calculate the work done by a person in lifting the load of 25 kg from the ground and placing it on 1 m high table. 10. Answer the following questions in 3-4 sentences. (i) [(Session 10.2)] Define kinetic energy. Give equation for kinetic energy. CHAPTER 10. WORK AND ENERGY 23

(ii) [(Session 10.2)] Derive the expression of kinetic energy. CHAPTER 10. WORK AND ENERGY 24

Long Answer Type Questions 11. Answer the following questions in 6-8 sentences. (i) [(Session 10.1)] A ball of mass 0.5 kg thrown upward reaches a maximum height of 5 m. Calculate the work done by the force of gravity during this vertical displacement considering the value of g = 10 m/s2. (ii) [(Session 10.1)] Calculate the work done by a student in lifting a 4.7 kg wooden block from the ground and keeping it on a shelf at a height of 2.9 m. Take g = 9.8 ms-2. 12. Answer the following questions in 6-8 sentences. (i) [(Session 10.3)] State the principle of law of conservation of energy. A woman does 600 J of work in 3 minutes and a boy does 400 J of work in 2 minutes. Who delivers more power? CHAPTER 10. WORK AND ENERGY 25

13. Answer the following questions in 6-8 sentences. (i) [(Session 10.2)] What will happen to the kinectic energy of a body, if the mass of the body is doubled (when its velocity remains the same)? (ii) [(Session 10.2)] Define potential energy. A wood of 4 kg is lifted up through 3 m from ground. Calculate the potential energy of the block at that point. CHAPTER 10. WORK AND ENERGY 26

AS2-Asking questions and making hypothesis Short Answer Type Questions 14. Answer the following questions in 3-4 sentences. (i) [(Session 10.2)] Rajeev said that it is easier to stop a car than a bus moving with the same speed. Why did he say that? AS3-Experimentation and field investigation 27 Short Answer Type Questions CHAPTER 10. WORK AND ENERGY

15. Answer the following questions in 3-4 sentences. (i) [(Session 10.2)] Write an activity to show a moving object can do work. CHAPTER 10. WORK AND ENERGY 28

AS4-Information skills and projects Short Answer Type Questions 16. Answer the following questions in 3-4 sentences. (i) [(Session 10.2)] Complete the table by writing the formula. S.No. Quantity Formula 1 Kinetic energy 2 Potential energy 3 Work done 4 Force CHAPTER 10. WORK AND ENERGY 29

AS5-Communication through drawing and model making Long Answer Type Questions 17. Answer the following questions in 6-8 sentences. (i) [(Session 10.3)] Explain what you notice from the diagram in terms of potential energy and kinetic energy. AS6-Appreciation and aesthetic sense, Values 30 Long Answer Type Questions CHAPTER 10. WORK AND ENERGY

18. Answer the following questions in 6-8 sentences. (i) [(Session 10.3)] How do you appreciate the law of conservation of energy? AS7-Application to daily life, concern to bio diversity Short Answer Type Questions 19. Answer the following questions in 3-4 sentences. (i) [(Session 10.1)] Do you think a person standing on a railway platform carrying luggage is doing any work? Justify your answer. CHAPTER 10. WORK AND ENERGY 31

(ii) [(Session 10.1)] Work done by a boy to push a wooden block by 30 cm is 1.35 J. Calculate the force applied by the boy. Long Answer Type Questions 20. Answer the following questions in 6-8 sentences. (i) [(Session 10.1)] A person holding a suitcase is at rest and a person is lifting a suit- case from the ground. Between these two people who is doing work? Explain. 21. Answer the following questions in 6-8 sentences. (i) [(Session 10.3)] By giving an example from your daily life, explain how one form of energy is converted to a different form. Also using the same example explain how energy is conserved. CHAPTER 10. WORK AND ENERGY 32

22. Answer the following questions in 6-8 sentences. (i) [(Session 10.2)] A ball is dropped from a height. Explain how energy is getting con- served in this example. CHAPTER 10. WORK AND ENERGY 33

Objective Questions AS1-Conceptual Understanding 23. Choose the correct answer. (i) Work is said to be done when (A) force acts upon a body, but the body does not move (B) force acts upon a body and moves it in the direction of force (C)force acts upon a body, but does not move it in the direction of applied force (D)none of these (B) newton (ii) Force is measured in (A) joule (C) kelvin (D) mole (iii) Unit of work is (B) joule (A) newton metre (C) kelvin (D)both A and B (iv) Potential Energy, P.E = (B) m/gh (A) mg/h (C) mgh (D) g/mh (v) In equation, W=F s, if F=1 and s=1 then work done by force is equal to (A) 0 J (B) 1 J (C)3 J (D)2 J (vi) The correct formula for work done is (B) Force x Acceleration (A) Force x Speed CHAPTER 10. WORK AND ENERGY 34

(C)Force x Velocity (D)Force x Displacement (vii) An object raised to a height from the ground possesses (A) kinetic energy (B) potential energy (C)chemical energy (D)mechanical energy (viii) If work has positive value, the body on which work has been done would (A) lose energy (B) either gain or lose energy (C)gain energy (D)neither gain nor lose energy (ix) Energy can neither be created nor be destroyed. It can only be changed from one form to another. This is the (A) Law of constant energy (B) Law of gaining energy (C)Law of losing energy (D)Law of conservation of energy (x) If force acting on an object and displacement are in opposite directions, then the work done by the force is taken as (A) positive (B) negative (C) neutral (D)either positive or negative (xi) What is the unit of power? (B) joule per second (A) kilojoule (C) joule (D) watt CHAPTER 10. WORK AND ENERGY 35

AS2-Asking questions and making hypothesis 24. Choose the correct answer. (i) What is the main source of energy on earth? (A) Sun (B) Moon (C) Wind (D) Fire (ii) Which type of energy conversion takes place in an electric fan? (A) Mechanical energy to electrical energy (B) Electrical energy to chemical energy (C)Electrical energy to mechanical energy (D)Chemical energy to electrical energy AS3-Experimentation and field investigation 25. Choose the correct answer. (i) A spring is compressed. The potential energy of the compressed spring (A) increases (B) decreases (C)remains unchanged (D)becomes zero AS7-Application to daily life, concern to bio diversity 26. Choose the correct answer. (i) A vehicle is accelerated on a levelled road and attains a velocity 4 times its initial velocity. In this process the potential energy of the car (A) becomes twice that of initial (B) becomes 4 times of that of initial (C)does not change (D)becomes 16 times that of initial CHAPTER 10. WORK AND ENERGY 36

11. HEAT SESSION 1 INTRODUCTION TO HEAT, THERMAL EQUILIBRIUM, TEMPERATURE AND KINETIC ENERGY 1.1 Mind Map 1.2 Terminology 37 SESSION 1. INTRODUCTION TO HEAT, THERMAL EQUILIBRIUM, TEMP...

i. Temperature – The degree or intensity of heat present in a substance or object. ii. Heat – Heat is a form of energy in transit, that flows from a body at higher temperature to a body at lower temperature. iii. Thermal Equilibrium – Two objects are said to be in thermal equilibrium if there is no flow of heat between them. 1.3 Key Concepts i. Heat flows from a body of higher temperature to a body of lower temperature. ii. There is no flow of heat when two bodies are in thermal equilibrium with each other. iii. The temperature of water rises with rise in temperature only until 100 degree celsius. iv. The rise in temperature of any substance in inversely proportional to the mass of the substance. 1.4 Application of Concepts Q1. [AS1] Convert following into kelvin scale. [Refer to TB page 208 Q4] (i) 20°C (ii) 27°C (iii) –273°C A. i. 20°C t = 20°C (t + 273) K (20 + 273) K = 293 K ii. 27 °C t = 27°C (t + 273) K (27 + 273) K = 300 K iii. –273 °C SESSION 1. INTRODUCTION TO HEAT, THERMAL EQUILIBRIUM, TEMP... 38

t = –273°C (t + 273) K (–273 + 273) K =0 K 1.5 Suggested Experiments Q1. [AS3] Take different metal pieces of same size and heat them to same temperature, dip them immediately in the beakers containing water of same level. Observe the tempera- ture differences in the water. Write your observations. [Refer to TB page 209 Q3] A. Students’ activity. Objective Questions (1) Three bodies A, B and C are in thermal equilibrium. The temperature of B is 45o C. then the temperature of C is .(Pg 208;TB Q 3) (A) 45°C (B) 50°C (C) 40°C (D)Any temperature Correct Answer: A (2) The temperature of a steel rod is 330K. Its temperature in °C is .(Pg 209;TB Q 4) (A) 55°C (B) 57°C (C) 59°C (D) 53°C Correct Answer: B SESSION 1. INTRODUCTION TO HEAT, THERMAL EQUILIBRIUM, TEMP... 39

SESSION 2 SPECIFIC HEAT AND ITS APPLICATION, DETERMINATION OF SPECIFIC HEAT OF A SOLID 2.1 Mind Map 2.2 Terminology i. Specific Heat – The specific heat of a substance is the amount of heat required to raise the temperature of unit mass of the substance by one unit. 2.3 Key Concepts i. The specific heat of a substance is the amount of heat required to raise the temperature of unit mass of the substance by one unit. S=Q/m∆t. ii. Rate of rise in temperature depends on the nature of the substance. iii. CGS unit of specific heat is cal/g– ˚ C and SI unit of it is J / kg –K. iv. Rise in temperature depends on the nature of the substance; hence the specific heat of a substance depends on its nature. v. Net heat loss = Net heat gain. This is known as principle of method of mixtures. SESSION 2. SPECIFIC HEAT AND ITS APPLICATION, DETERMINATIO... 40

2.4 Reflection on Concepts Q1. [AS1] Why specific heat is different for different substances? Explain. [Refer to TB page 208 Q6] A. We know that the temperature of a body is directly proportional to the average kinetic energy of particles of the body. The molecules of the system (body or substance) have different forms of energies such as linear kinetic energy, rotational kinetic energy, vibra- tional energy and potential energy between molecules. The total energy of the system is called internal energy of the system. When we supply heat energy to the system, the heat energy given to it will be shared by the molecules among the various forms of energy. This sharing will vary from substance to substance. The rise in temperature is high for a substance, if the maximum share of energy is utilized for increasing its linear kinetic energy. This sharing of heat energy with the system also varies with temperature. That is why the specific heat is different for different substances. Q2. [AS7] What role does specific heat play in keeping a watermelon cool for a long time after removing it from a fridge on a hot day? [Refer to TB page 207 Q4] A. If specific heat is high, the rate of rise or fall in temperature is low. So watermelon takes long time to rise in its temperature. Hence watermelon brought out from a fridge retains its coolness for a long time than other fruits. The watermelon contains more water and water has greater specific heat value. 2.5 Application of Concepts Q1. [AS1] If 50g of water at 20° temperature and 50 g of water 40°C temperature are mixed, what is the final temperature of the mixture of ? [Refer to TB page 208 Q2] A. 1st water sample data : Mass (m1) = 50g, Temperature (T1) = 20°C. 2nd water sample data : Mass (m2) = 50g, Temperature (T2) = 40°C. According to method of mixture, T = (m1)(T1)+(m2)(T2) (m1+m2) SESSION 2. SPECIFIC HEAT AND ITS APPLICATION, DETERMINATIO... 41

T = (50×20+50×40) (50+50) T = (1000+2000) 100 T = 3000 100 T = 30oC 2.6 Higher Order Thinking Skills Q1. [AS6] How do you appreciate the role of the higher specific heat of water in stabilizing atmospheric temperature during winter and summer seasons? [Refer to TB page 208 Q1] A. The sun delivers a large amount of energy to the earth every day. The water sources on the earth, particularly the oceans, absorb energy for maintaining a relatively constant temperature. They can absorb large amounts of heat at the equator without any rise in temperature due to a high specific heat of water. At night time, water cools releasing energy into the atmosphere. The wind transports the heat energy away from water maintaining a relatively constant temperature. This transportation of heat energy helps in moderating the earth’s temperature. Q2. [AS7] Suppose that 1 L of water is heated for a certain time to rise its temperature for 2°C. If 2 L of water is heated for the same time, how much of its temperature would rise? [Refer to TB page 208 Q3] A. Heat is given by Q = m.s.∆T where m = mass, s = specific heat, ∆T = change in temperature We can see from the above formula that, if T is the subject of the equation, then the equation will be ∆T = Q/m.s which indicates that T is inversely proportional to mass, which means the temperature increases as the mass decreases or the temperature decreases as the mass increases. ∆T1 = m2 ∆T2 m1 2 = 2L ∆T2 1L SESSION 2. SPECIFIC HEAT AND ITS APPLICATION, DETERMINATIO... 42

= 1 °C Therefore, the temperature rises by 1°C 2.7 Suggested Experiments Q1. [AS3] Find the specific heat of solids experimentally. Write a report. [Refer to TB page 209 Q1] A. Aim: To find the specific heat of given solid. Material required: Calorimeter, thermometer, stirrer, water, steam heater, wooden box and lead shots. Procedure: Measure the mass of the calorimeter along with stirrer. Mass of the calorimeter, m1 = ................ Now fill one third of the volume of calorimeter with water. Measure its mass and its temperature. Mass of the calorimeter plus water, m2 = ................ Mass of the water, m2–m1 = Temperature of water in calorimeter, T1 = ........... Note: Calorimeter and water are at same temperature. Take a few lead shots and place them in hot water or steam heater. Heat them upto a temperature 100°C. Let this temperature be T2 . Transfer the hot lead shots quickly into the calorimeter (with minimum loss of heat). We will notice that the mixture settles to a certain temperature after some time. Measure this temperature T3 and mass of the calorimeter along with contents (water and lead shots). Mass of the calorimeter along with contents, m3= ......... Mass of the lead shots, m –m2 = ............ 3 Since there is no loss of heat to surroundings, we can assume that the entire heat lost by the solid (lead shots) is transferred to the calorimeter and water to reach the final temperature. Let the specific heats of the calorimeter, lead shots and water be Sc , Sl and Sw re- spectively. SESSION 2. SPECIFIC HEAT AND ITS APPLICATION, DETERMINATIO... 43

According to the method of mixtures, we know; Heat lost by the solid = Heat gain by the calorimeter + Heat gain by the water (m3 − m2)S l(T2 − T3) = m1S c(T3 − T1) + (m2 − m1)S w(T3 − T1) Sl = [m1S c+(m2−m1)S w](T3−T1) (m3−m2)(T2−T3) Knowing the specific heats of calorimeter and water, we can calculate the specific heat of the solid (lead shots). Objective Questions (1) Specific heat S = (Pg 209;TB Q 5) (A) Q (B) Q T T (C) Q (D) mT mT Q Correct Answer: C SESSION 2. SPECIFIC HEAT AND ITS APPLICATION, DETERMINATIO... 44

SESSION 3 EVAPORATION, CONDENSATION, HUMIDITY AND DEW AND FOG 3.1 Mind Map 3.2 Terminology i. Evaporation – The process of molecules escaping from the surface of a liquid at any temperature is called evaporation. ii. Condensation – The change of phase from gas to liquid is called condensation. iii. Humidity – The amount of water vapour present in air is called humidity. iv. Dew – The water vapour that condenses into droplets is called as dew. v.Fog – Fog is a thick cloud of condensed water droplets floating in the air that makes visibility difficult. 3.3 Key Concepts i. Temperature of a system falls during evaporation. 45 SESSION 3. EVAPORATION, CONDENSATION, HUMIDITY AND DEW AND...

ii. Evaporation is a surface phenomenon. iii. Rate of evaporation of a liquid depends on its surface area, temperature and amount of vapour already present in the surrounding air. iv. Condensation can also be defined as “the phase change from gas to liquid”. It is a warming process. v. The amount of water vapour present in air is called humidity. vi. The process of escaping of molecules from the surface of a liquid at any temperature is called evaporation and it is a cooling process. vii. Condensation is the reverse process of evaporation. 3.4 Reflection on Concepts Q1. [AS1] Why do we get dew on the surface of a cold soft drink bottle kept in open air? [Refer to TB page 207 Q1] A. The temperature of the cold soft drink bottle is less than the temperature of atmosphere. The water molecules present in air touches the surface of the bottle and lose their kinetic energy. As a result the temperature of water molecules decrease and condenses on the surface of the bottle. These water droplets are seen as dew. SESSION 3. EVAPORATION, CONDENSATION, HUMIDITY AND DEW AND... 46

Q2. [AS2] Your friend is asked to differentiate between evaporation and boiling. What ques- tions could you ask to make him to know the differences between evaporation and boil- ing? [Refer to TB page 207 Q2] A. The questions one could ask are: i. At what temperature does evaporation takes place? ii. What is the temperature at which the water boils? iii. Does the water get converted to vapour faster during evaporation or boiling? Q3. [AS3] Water can evaporate at any temperature. Explain with an example? [Refer to TB page 207 Q3] A. Since evaporation is a surface phenomenon, the water molecules at the surface will be in random movement which will come in contact with air molecules which are also in random movement. This contact between water molecules on the surface and the air molecules is a collision because of which the water molecules acquire energy and fly off the surface thereby effecting evaporation at any temperature. (Room temperature is sufficient enough for the water molecules to be in random movement which is the main criterion for evaporation.) Q4. [AS3] Equal amounts of water are kept in a cap and in a dish. Which will evaporate faster? Why? [Refer to TB page 207 Q5] A. The rate of evaporation increases with increase in surface area (evaporation is a surface phenomenon). The surface area of water in dish is more than the surface area of water in the cup. So, water in the dish will evaporate quickly. Q5. [AS7] If you are chilly outside the shower stall, why do you feel warm after the bath if you stay in the bathroom? [Refer to TB page 208 Q7] A. After bath, the number of vapour molecules per unit volume in the bath room is greater than the number of vapour molecules per unit volume outside the bath room. When we try to dry ourselves with a towel, the vapour molecules surrounding us condenses on our body. Hence we feel warm after bath if we continue to stay in the bathroom. SESSION 3. EVAPORATION, CONDENSATION, HUMIDITY AND DEW AND... 47

3.5 Application of Concepts Q1. [AS1] Using the concept of evaporation explain why dogs pant during hot summer days? [Refer to TB page 208 Q1] A. Dogs pant during hot summer days to reduce their internal temperature. When dogs pant, the water molecules acquire energy and get evaporated as a result of which the interior of the dogs’ body gets cooled. Q2. [AS1] What do you observe in the surroundings in terms of cooling or heating when water vapour is getting condensed? [Refer to TB page 208 Q3] A. The surroundings get heated primarily because condensation is a warming process. When vapour touches a cool object, the heat from the molecules in vapour flow out to attain thermal equilibrium with the object it is in contact with. 3.6 Suggested Experiments Q1. [AS3] Perform an experiment to prove that the rate of evaporation of a liquid depends on its surface area and vapour already present in surrounding air. Write a report. [Refer to TB page 209 Q2] A. Take a few drops of spirit say 1 ml) in two petri dishes (a shallow glass or plastic cylindrical lidded dish used in the laboratory) separately. Keep one of the dishes containing spirit under a ceiling fan and switch on the fan. Keep another dish with its lid closed. Observe the quantity of spirit in both dishes after 5 minutes. We will notice that spirit in the dish that is kept under the ceiling fan disappears, whereas you will find some spirit left in the dish that is kept in the lidded dish. The reason behind this change is evaporation. The molecules of spirit that is kept in petri dish, continuously move with random speeds in various directions. As a result, these molecules collide with other molecules. During the collision, they transfer energy to other molecules. When the molecules inside the liquid collide with molecules at the surface, the molecules at the surface acquire energy and may fly off from the surface. SESSION 3. EVAPORATION, CONDENSATION, HUMIDITY AND DEW AND... 48