202110243-TRIUMPH-STUDENT-WORKBOOK-PHYSICAL_SCIENCE-G09-PART1

2.3 Solved Examples Q1. Example 1: What is the time period taken for one revolution of the satellite near the earth’s surface? Neglect height of the orbit of satellite from the surface of ground. [Refer to TB page 73] A. The force on the satellite due to earth is given by F = GmM , R2 where M: Mass of earth, m: Mass of satellite, R: Radius of earth. Let ‘v’ be the speed of the satellite v = 2πR ⇒ T = 2πR T v Required centripetal force is provided to satellite by the gravitational force hence, Fc = mv2 R But Fc = GMm according to Newton’s law of gravitation. R2 i.e., GMm = m(2πR)2 R2 T 2R ⇒ T2 = 4π2R3 , as mass of the earth (M) and G are constants the value of T depends GM only on the radius of the earth. ⇒ T 2 ∝ R3 Substituting the values of M, R and G in above equation we get, T = 84.75 minutes. Thus the satellite revolving around the earth in a circular path near to the earth’s surface takes 1 hour and 24.7 minutes approximately to complete one revolution around earth. Q2. Example 2: A body is projected vertically up. What is the distance covered in its last second of upward motion? (g = 10 m/s2 ) [Refer to TB page 75] SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 149

A. The distance covered by the object in the last second of its upward motion is equal to the distance covered in the first second of its downward motion. Hence s = 1 gt2 = 1 × 10 × 1 =5m 2 2 Q3. Example 3: Two bodies fall freely from different heights and reach the ground simultane- ously. The time of descent for the first body is t1 = 2s and for the second t2 = 1s. At what height was the first body situated when the other began to fall? (g = 10 m/s2 ) [ Refer to TB page 75] A. The second body takes 1 second to reach ground. So, we need to find the distance traveled by the first body in its first second and in two seconds. The distance covered by first body in 2s. h1 = 1 gt2 = 1 × 10 × 22 = 20 m 2 2 The distance covered in 1 s, h2 = 5 m. The height of the first body when the other begin to fall h = 20 – 5 = 15 m. Q4. Example 4: A stone is thrown vertically up from a tower of height 25 m with a speed of 20 m/s. What time does it take to reach the ground? (g = 10 m/s2 ) [Refer to TB page 75] SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 150

+ _ A. Sign convention must be used to solve this problem. It is shown in figure. We consider the upward direction as positive and downward direction as negative with respect to a point of reference. In the above example the point of projection is considered as the point of reference. Then, u = 20 m/s a = g = –10 m/s2 s = h = –25 m From equation of motion s = ut + 1 at2 2 –25 = 20t – 1 × 10 × t2 2 –25 = 20t – 5 t2 –5 = 4t – t2 i.e., t2 – 4t – 5 = 0 Solving this equation, we get (t – 5 ) (t + 1) = 0 SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 151

t = 5 or –1 Therefore, t = 5 s Q5. Example 5: Find the time taken by the body projected vertically up with a speed ‘u’ to return to the ground. [Refer to TB page 76] A. Let us take the equation s = ut + 1 at2 2 For entire motion, s = 0 a = –g u=u 0 = ut – 1 gt2 2 1 gt2 = ut 2 t = 2u g 2.4 Key Concepts i. Newton’s universal law of gravitation: Every object in this universe attracts every other body. The force of attraction between the two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance be- tween them. ii. All the bodies falling due to gravity have the same acceleration equal to 9.8 m/s2 near the surface of the earth. iii. A body is said to be in free fall when only gravity acts on it. iv. The force with which a body is attracted by the earth is called its weight, w = mg. v. When a body is in free fall condition it will be in a state of weightlessness. SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 152

2.5 Reflection on Concepts Q1. [AS1] Explain the universal law of gravitation. [Refer to TB page 81 Q3] A. Newton generalized the force of gravitation and said, \"It acts between any two bodies in the universe”. The universal law of gravitation states that every body in the universe attracts other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The direction of the force of attraction is along the line joining the centers of the two bodies. Let two bodies of masses M1 and M2 be separated by a distance of ‘d’. Then the force of gravitation between them. Fgrav ∝ M1 M2 d2 Fgrav = G M1 M2 d2 G is a proportionality constant, called universal gravitational constant and found by Henry Cavendish to be G = 6.67 × 10−11 Nm2Kg−2 The value of G is equal to the magnitude of force between a pair of 1 kg masses that are 1 metre apart. 2.6 Application of Concepts Q1. [AS1] What is the speed of an apple dropped from a tree after 1.5 s? What distance will it cover during this time? Take g=10 m/s2. [Refer to TB page 81 Q2] A. v=u + at SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 153

u, Initial velocity of apple = 0 a, acceleration due to gravity = 10 m/s2 t, time = 1.5 s v, speed of apple = ? v = 0 + 10 × 1.5 = 15 Speed after 1.5 sec = 15 m/s Distance covered by apple s = ut + 1 at2 2 u = 0; t=1.5 sec; a= 10 m/s2 S = (0 × 1.5) + 1 × 10 × (1.5)2 2 = 0 + 1 × 10 × 2.25 = 11.25 2 Distance covered by the apple = 11.25 m Q2. [AS1] A ball is dropped from a height. If it takes 0.2 s to cross the last 6m before hitting the ground, find the height from which it is dropped. Take g = 10 m/s.2 [Refer to TB page 81 Q6] A. Let height from which ball is dropped = H If we consider initial velocity of ball during the last 6m distance travelled by ball = u Then using, s = ut + 1 at2, where a = g (acceleration due to gravity) = 10 m/s2, t = 0.2 s 2 Thus 6 = 0.2u + 0.5 × 10 × 0.04 Or, u = 29 m/s Now to acquire the velocity of 29 m/s the ball should be dropped from height ‘x’ under the influence of ‘g’ and thus we can make use the following equation. SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 154

v2 – u2 = 2gs or, v2 – u2 = 2gx 2 or, 29 = 2 × 10 × x or, x = 42.05 m Thus total height from which the ball was dropped is given by H = x + 6 or H = 42.05 + 6 = 48.05 m Q3. [AS1] A ball is projected vertically up with a speed of 50 m/s. Find the maximum height, the time to reach maximum height and the speed at the maximum height. (Take g = 10 2 m/s ) [Refer to TB page 81 Q3] A. u = 50 m/s v2 – u2 = 2gs 0 – 502 = – 2 (10) s s = 125 m i.e., Maximum height = 125 m Time to reach maximum height, S = 1 (gt2) 2 125 = 1 (10)t2 2 t = 5s At maximum height speed = 0 Q4. [AS1] Two spherical balls of mass 10 kg each are placed with their centres 10 cm apart. Find the gravitational force of attraction between them. [Refer to TB page 81 Q4] A. m1 = m2 = 10 kg; D = 10 cm = 10-2 m SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 155

F = G M1 M2 d2 = G 10×10 0.1×0.1 = G 102 10−2 = G × 104N Q5. [AS1] Find the free–fall acceleration of an object on the surface of the moon, if the radius of the moon and its mass are 1740 km and 7.4 x 1022 kg respectively. Compare this value with free–fall acceleration of a body on the surface of the earth. [Refer to TB page 81 Q5] A. Let, m = 7.4×1022 kg r = 1740km= 1740 × 103 m G = 6.673 × 10-11 Nm2 kg-2 (Constant value) We know that, gm = Gm r2 gm = (6.673×10−11)(7.4×1022) (1740×103)2 gm =1.630m/s2 Now, ge =9.8m/s2 (free–fall acceleration of earth) Therefore, gm = 1.630 = 1 ge 9.8 6 SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 156

This shows that free–fall acceleration of moon is 1/6th of earth’s free–fall acceleration. Q6. [AS1] The bob of a simple pendulam of length 1 m has mass 100 g and a speed of 1.4 m/s at the lowest point in its path. Find the tension in the string at this moment. [Refer to TB page 81 Q7] A. Mass of the bob = 100g= 0.1kg Length of the string = 1m Speed of the bob, v= 1.4 m/s Let the tension in the string be T. The forces acting on the bob are a) Weight of the bob mg, downwards b) Tension in the string ‘T’ upwards Weight of the bob F = mv2 l Tension in the string T= g cos T Therefore according to Newton’s third law T = mv2 + mgcosθ l T = m( v2 + g cosθ) l = 0.1( 1.4×1.4 + 9.8 × cos0o) 1 = 0.1 × (1.96 +9.8) = 0.1 × 11.76 = 1.176 N SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 157

Q7. [AS2] What path will the moon take when the gravitational interaction between the moon and earth disappears? [Refer to TB page 81 Q8] A. If the gravitational interaction between earth and moon disappears, then the moon will travel in a direction tangential to its orbit, since there shall be no centripetal force acting on it. Q8. [AS2] Can you think of two particles which do not exert gravitational force on each other? [Refer to TB page 81 Q9] A. No, everything in the universe exerts a force on everything else even if infinitesimally small. 2.7 Higher Order Thinking Skills Q1. [AS7] An apple falls from a tree. An insect in the apple finds that the earth is falling towards it with an acceleration g. Who exerts the force needed to accelerate the earth with this acceleration? [Refer to TB page 82 Q2] A. Force on the apple due to the earth. Both the apple and the earth exert equal forces but in opposite directions. Objective Questions (1) The weight of an object whose mass 1 kg is (Pg 82;Q 4) (A) 1 Kg/m2 (B) 9.8 m/Sec2 (C)9.8 N (D)9.8 N/m2 Correct Answer: C (2) The state of a freely falling body is (pg 82;Q 5) (A) Heavy weight (B) Less weight (C)Weight less (D)Constant weight Correct Answer: C SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 158

SESSION 3 CENTRE OF GRAVITY 3.1 Mind Map SESSION 3. CENTRE OF GRAVITY 159

3.2 Terminology i. Centre of gravity – The center of gravity is the point where total weight of the body acts. ii. Stability – The state of the body when both center of mass and center of gravity coincide. 3.3 Key Concepts i. The point through which the total weight of the body acts is called the centre of gravity. ii. If the weight vector passes through the base of the body then the body will be in equilibrium. 3.4 Reflection on Concepts Q1. [AS1] Explain some situations where center of gravity of man lies outside the body. [Refer to TB page 81 Q4] A. Centre of gravity of a human being is located interior to the second sacral vertebra. Centre of gravity of man lies outside the body in the following situations: 1. While walking with one leg. 2. While carrying a load like bucket full of water with one hand. 3. While doing sit - ups. Q2. [AS1] Calculate the acceleration of the moon towards earth’s center. [Refer to TB page 81 Q2] A. We can calculate the speed of the moon using the equation, V = 2πR T Thus the acceleration of the moon towards the centre of the earth am = v2 = 4π2R R T2 SESSION 3. CENTRE OF GRAVITY 160

Substituting the values of R and T in above equation we can get am = 0.27 cm/s2 Q3. [AS2] Where does the center of gravity of the atmosphere of the earth lie? [Refer to TB page 81 Q5] A. The earth’s atmosphere is about 11,000 km thick, but most of its bulk is contained in the first 11 km above the earth’s surface. Since the earth and its atmosphere are roughly spherical, the centre of the earth is also the centre of gravity of the earth’s atmosphere. Q4. [AS7] Explain why a long pole is more beneficial to the tight rope walker if the pole has slight bending. [Refer to TB page 81 Q6] A. Two principles of physics contribute to a funambulist’s stability: i. The pole increases the funambulist’s moment of inertia. ii. The downward bend of the pole lowers the funambulist’s centre of gravity. iii. When an artist finds that he is falling towards left, he shifts the pole towards right, so that his centre of gravity stay undisturbed. 3.5 Application of Concepts Q9. [AS7] Why is it easier to carry the same amount of water in two buckets, one in each hand than in a single bucket? [Refer to TB page 81 Q10] A. It is because in the later case centre of gravity of our body shifts towards the bucket and there is a tendency that the line joining the centre of gravity and centre of equilibrium may fall outside our feet. However in the former case the centre of gravity not only gets lowered, but also it is at such a point that the line joining the centre of gravity and centre of equilibrium falls within our feet. Hence one bucket in each hand gives a stable equilibrium. 3.6 Higher Order Thinking Skills Q1. [AS7] A man is standing against a wall such that his right shoulder and right leg are in contact with the surface of the wall along his height. Can he raise his left leg at this position without moving his body away from the wall? Why? Explain. [Refer to TB page 82 Q1] A. i. He can’t raise his left leg in this position without moving his body away from the wall. ii. As we try to raise the left leg the centre of gravity doesn’t fall inside the base. So we have to move our body away from the wall to make the weight vector of the body to fall within base. SESSION 3. CENTRE OF GRAVITY 161

3.7 Suggested Experiments Q1. [AS3] Conduct an experiment to find the Centre of Gravity of an object and write a report. [Refer to TB page 82 Q1] A. i. Take a metre scale. ii. Suspend it from various points. iii. We observe that it bends to one side. iv. Suspend it from the midpoint. v. We observe that the scale will be in horizontal position without bending to any side. vi. At this point, the scale behaves as if its entire weight is concentrated at this point. vii. The support given at this single point gives support to the entire scale. viii. This point is its centre of gravity. Q2. Conduct an experiment to find 2s/t2value for a freely falling body and also find the value of ‘g’. [Refer to TB page 82 Q2] A. Students’ activity. 3.8 Suggested Projects Q1. [AS3] Collect the information about the base area and stability of some objects with different shapes and write a report. [Refer to TB page 82 Q1] A. Students’ activity. Q2. [AS3] Collect the information about the path of revolution of moon around the earth and write a report. [Refer to TB page 82 Q2] A. Students’ activity. Objective Questions (1) The value of Universal Gravitaitonal Constant is (Pg 82;Q 3) (A) 6.67 x 10-11 N.m2 Kg-2 (B) 9.8 m/ Sec 2 (C)6.67 x 10-12 N.m2 Kg-2 (D)981 m/ Sec2 Correct Answer: A SESSION 3. CENTRE OF GRAVITY 162

—— CCE Based Practice Questions —— AS1-Conceptual Understanding is a motion of the body with Very Short Answer Type Questions 1. Fill in the blanks. [Refer to Session 5.1 ] (i) a constant speed in circular path. (ii) is always directed towards the center of the circle during the uniform circular motion of a body. (iii) The force of attraction between two objects is called force. (iv) = gR . (v) The net force required to maintain a body in uniform circular motion is known as . (vi) The magnitude of centripetal force is given by . (vii) The net force which can change only the direction of the velocity of a body is called . (viii) The acceleration which causes changes in the direction of velocity of a body is called . (ix) Banking of roads provides the necessary . 2. State true or false. [Refer to Session 5.2 ] (i) Gravitational constant was found by Henry Cavendish. [] CHAPTER 5. GRAVITATION 163

(ii) The value of ‘g’ decreases as we go upwards from the earth’s surface. ] [ ] (iii) We can calculate the speed of the moon using the equation V = 44R . 7T [ (iv) S.I. unit of weight is Newton. [] (v) Laws of gravitation are not applicable to point masses. [] (vi) Weight of a body and mass of a body is same in all cases. [ ] 3. Fill in the blanks. [Refer to Session 5.2 ] (i) All the bodies have the same acceleration equal to near the surface of the earth. (ii) If the distance between two bodies is doubled, the force of attraction, F, between them will be . (iii) The value of the universal gravitational constant is . (iv) According to the the sun is at centre and planets move around it. (v) The force with which a body is attracted by the earth is called its . (vi) The mass of the earth is . (vii) A cricket ball is thrown vertically upwards with a initial velocity of 40 m/s. If g = 10m/s2 the maximum height reached by the ball is . (viii) SI unit of weight is . CHAPTER 5. GRAVITATION 164

(ix) In the formula F = GM 1M2 , G stands for the d2 . 4. State true or false. [Refer to Session 5.3 ] (i) A boy sitting on a chair can directly get up from the chair without any support. ] [ (ii) Centre of gravity always lies at the centre of an object. [] (iii) We cannot get up from a chair without bending our body. [] (iv) The centre of gravity of any freely suspended object lies directly beneath the point of suspension. [] (v) If the line through the centre of gravity falls outside the base then the object will not be stable. [] (vi) A body is in equilibrium when the weight vector goes through the base of the body. [] (vii) If the weight vector passes through the base of the body then the body will be in equilibrium. [] (viii) Centre of gravity is the point where total weight of the body acts. [ ] (ix) Location of the centre of gravity is not important for stability. [ ] (x) Centre of gravity is simply the average position of weight distribution. ] [ CHAPTER 5. GRAVITATION 165

5. Fill in the blanks. [Refer to Session 5.3 ] (i) The centre of gravity of a triangular lamina lies at its . (ii) Centre of gravity of a circular ring lies at its . (iii) The simple method of locating the centre of gravity of an object is . (iv) The downward bend of the pole lowers the centre of gravity. (v) The center of gravity of a uniform object, is at its . Short Answer Type Questions 6. Answer the following questions in 3-4 sentences. (i) [(Session 5.1)] What is a force? (ii) [(Session 5.1)] What is centripetal acceleration? What is its direction? CHAPTER 5. GRAVITATION 166

7. Answer the following questions in 3-4 sentences. (i) [(Session 5.2)] What is free fall acceleration? (ii) [(Session 5.2)] What is the difference between mass and weight? CHAPTER 5. GRAVITATION 167

(iii) [(Session 5.2)] A body is projected vertically up on a planet whose acceleration due to gravity is 8 m/s2. What is the distance covered in its last second? 8. Answer the following questions in 3-4 sentences. (i) [(Session 5.3)] Define centre of gravity. (ii) [(Session 5.3)] What do you mean by stability? 168 CHAPTER 5. GRAVITATION

(iii) [(Session 5.3)] Where do the centres of gravity of a sphere and a triangular lamina lie? (iv) [(Session 5.3)] Why must you bend forward when carrying a heavy load on your back? CHAPTER 5. GRAVITATION 169

Long Answer Type Questions 9. Answer the following questions in 6-8 sentences. (i) [(Session 5.1)] A ball thrown up vertically returns to the thrower after 6 sec. Find: a) The velocity with which it was thrown up. b) The maximum height it reaches. c) Its position after 4 s. 10. Answer the following questions in 6-8 sentences. 170 (i) [(Session 5.2)] Explain the universal law of gravitation. CHAPTER 5. GRAVITATION

(ii) [(Session 5.2)] Discuss the various factors on which the value of ‘g’ depends. CHAPTER 5. GRAVITATION 171

AS2-Asking questions and making hypothesis Short Answer Type Questions 11. Answer the following questions in 3-4 sentences. (i) [(Session 5.2)] If there is an attractive force between all objects, why do we not feel ourselves gravitating towards massive buildings in our vicinity? CHAPTER 5. GRAVITATION 172

12. Answer the following questions in 3-4 sentences. (i) [(Session 5.3)] Can an object have more than one centre of gravity? Long Answer Type Questions 13. Answer the following questions in 6-8 sentences. (i) [(Session 5.2)] You buy a bag of sugar of weight ‘x’ at a place on the equator and take it to Antarctica. Would its weight be the same or does it vary? Why? AS3-Experimentation and field investigation 173 Short Answer Type Questions CHAPTER 5. GRAVITATION

14. Answer the following questions in 3-4 sentences. (i) [(Session 5.3)] Write the standard equations when an object is dropped from a height. Long Answer Type Questions 15. Answer the following questions in 6-8 sentences. (i) [(Session 5.2)] Describe an experiment to measure the weight of a freely falling body. What are the observations? CHAPTER 5. GRAVITATION 174

AS4-Information skills and projects Short Answer Type Questions 16. Answer the following questions in 3-4 sentences. (i) [(Session 5.2)] Complete the table. S.No. M mR R2 F=GMm/R2 1 8 32 64 2G 29 6 AS5-Communication through drawing and model making Long Answer Type Questions 17. Answer the following question. (i) [(Session 5.3)] Draw diagram to show the position of center of gravity in each of the following objects: a. Rectangular lamina b. Triangular lamina c. Ring CHAPTER 5. GRAVITATION 175

AS6-Appreciation and aesthetic sense, Values Long Answer Type Questions 18. Answer the following questions in 6-8 sentences. (i) [(Session 5.3)] Explain why a small piece of stone is not attracted towards another big piece of stone on the earth’s surface. Also, the earth attracts an apple. Does the apple also attract the earth? If it does, why does the earth not move towards the apple? CHAPTER 5. GRAVITATION 176

AS7-Application to daily life, concern to bio diversity Short Answer Type Questions 19. Answer the following questions in 3-4 sentences. (i) [(Session 5.2)] Find out the weight of a freely falling ball which has a mass of 2 kg and is attracted by earth at 9.8 m/s2. (ii) [(Session 5.2)] The mass of two freely moving asteroids are 2000 kg and 5000 kg respectively. The distance between these two asteroids is 5000 m. Find out the force of attraction between these two asteroids. (G = 7 × 10−11Nm2/kg2 ) CHAPTER 5. GRAVITATION 177

20. Answer the following questions in 3-4 sentences. (i) [(Session 5.3)] How can you get up from a chair without bending your body or legs? (ii) [(Session 5.3)] Write a short note on weightlessness. CHAPTER 5. GRAVITATION 178

Objective Questions AS1-Conceptual Understanding 21. Choose the correct answer. (i) Weight of 10 kg mass is equal to (A) 9.8 N (B) 98 N (C)980 N (D)1/9.8 N (ii) As we go from the equator to the poles, the value of ‘g’ (A) remains the same (B) decreases (C) increases (D)none of these (iii) All bodies whether large or small fall with the (A) same force (B) same acceleration due to gravity (C) same velocity (D)same momentum (iv) At which of the following locations is the value of ‘g’ the largest? (A) On top of Mount Everest (B) On top of Qutub Minar (C)At a place on the equator (D)A camp site in Antarctica (v) The force of gravitation between two bodies does not depend on (A) their separation (B) the product of their masses (C)the sum of their masses (D)gravitational constant AS2-Asking questions and making hypothesis 22. Choose the correct answer. (i) If the earth suddenly shrinks to half its present size, the value of acceleration due to gravity will CHAPTER 5. GRAVITATION 179

(A) become twice (B) remain unchanged (C)become half (D)become four times (ii) The weight of the body would not be zero (A) at the centre of the earth (B) during a free fall (C)in interplanetary space (D)on a frictionless surface AS3-Experimentation and field investigation 23. Choose the correct answer. (i) A stone dropped from the roof of a building takes 4 s to reach the ground. The height of the building is (A) 19.6 m (B) 39.2 m (C)156.8 m (D)78.4 m AS4-Information skills and projects 24. Choose the correct answer. (i) The earth attracts a body with a force of 10 N; with what force does that body attract the earth ? (A) 10 N (B) 1 N (C)2 N (D)1/10 N (ii) A body is projected upwards with a velocity of 100 m/s. It will strike the ground in approximately (A) 10 s (B) 20 s (C)15 s (D)5 s CHAPTER 5. GRAVITATION 180

6. IS MATTER PURE? SESSION 1 TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLLOIDS AND SUSPENSIONS 1.1 Mind Map 1.2 Terminology i. Pure substances – Pure substances are defined as substances that are made of only one type of atom or only one type of molecule (a group of atoms bonded together). ii. Mixture – A mixture is a combination of two or more pure substances in which each pure substance retains its individual chemical properties. iii. Heterogeneous mixture – A mixture in which the components are not uniformly dis- tributed is called a heterogeneous mixture. iv. Homogeneous mixture – A mixture in which the components are uniformly distributed is called a homogeneous mixture. v. Suspension – A suspension is a heterogeneous mixture in which solute–like particles settle out of a solvent–like phase some time after their introduction. SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 181

vi. Solution – A solution is a homogeneous mixture of two or more substances. The major component of a solution is called the solvent, and the minor, the solute. vii. Emulsions – Emulsions are mixtures which consist of two liquids that do not mix and they settle into layers when they are left undisturbed. viii. Colloidal dispersions – Colloids are heterogeneous mixtures in which the particle size is too small to be seen with the naked eye, but is big enough to scatter light. The colloid has the dispersed phase and the medium in which they are distributed is called the dispersion medium. ix. 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. x. Concentration of solution – Concentration of solution is defined as the amount of solute dissolved in a specific (fixed) amount of solvent. Units of concentration are Molarity, Molality, Normality. xi. Tyndall effect – The Tyndall effect, also known as Tyndall scattering, is light scattering by particles in a colloid or else particles in a very fine suspension. 1.3 Solved Examples Q1. A solution contains 50 g of common salt in 200g of water. Calculate the concentration in terms of mass by mass percentage of the solution. [ Refer to TB page 87] A. Mass of solute (salt) = 50 g Mass of solvent (water) = 200 g Mass of solution = Mass of solute + Mass of solvent = 50 g + 200 g = 250 g Mass percentage of a solution = Mass o f solute × 100 = 50 × 100 = 20% Mass o f solution 250 1.4 Key Concepts i. All matter around us can be classified into two groups – pure substances and mix- tures. ii. A mixture is composed of different substances mixed in any proportion and not chem- ically combined. SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 182

iii. A mixture in which the components are uniformly distributed is called a homogeneous mixture. iv. A mixture in which the components are not uniformly distributed is called a heteroge- neous mixture. v. A homogeneous mixture of two or more substances is called a solution. Solution = solute + solvent vi. The substance which is present in large quantity is called solvent and the substance which is in less quantity is called solute. vii. The amount of solute present in unit volume or per unit mass of a solution or solvent is called the concentration of the solution. viii. Materials that are insoluble in a solvent and have particles that are visible to naked eyes form a suspension. A suspension is a heterogeneous mixture. ix. Emulsions are mixtures which consist of two liquids that do not mix and they settle into layers when they are left undisturbed. x. A heterogeneous mixture in which the size of constituent particles is too small to be individually seen by naked eyes is called a colloid. But these particles are big enough to scatter light. xi. The scattering of a beam of light by the particles of a colloidal solution is called Tyndall effect. 1.5 Reflection on Concepts Q1. [AS1] Explain the following giving examples: (a) Saturated solution (b) Pure substance (c) Colloid (d) Suspensions. [Refer to TB page 99 Q2] A. (a) Saturated solution: If we take 50 ml of water and add a spoonful of salt to it and stir well, the salt dissolves in water and no trace of it is left behind. If we continue adding more and more of salt keeping the solution at room temperature, at a certain point no more of the salt goes into the solution but settles at the bottom. It is called a saturated solution. SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 183

Definition: 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 (b) Pure substance: The general term “Pure substance” means a substance free from adulteration. But in chemistry it means that the substance is homogeneous and the composition doesn’t change no matter which part of the substance has been taken as a sample for examination. E.g. Gold bar (c) Colloid: A mixture which contains very small particles of a substance but appears to be homogeneous and the particles of which scatter a beam of visible light is called a colloid. E.g. Milk (d) Suspension: If we add a little chalk powder to 20 ml of water, we find that the particles of chalk do not dissolve, but remain suspended throughout the volume of wa- ter. It is a heterogeneous mixture and the particles are visible to the naked eye. Such heterogeneous mixtures are called suspensions. E.g. Syrups, Chalk powder added to water 1.6 Application of Concepts Q1. [AS1] Classify the following substances in the below given table: ink, soda water, brass, fog, blood, aerosol sprays, fruit salad, black coffee, oil and water, boot polish, air, nail polish, starch solution, milk. [Refer to TB page 99 Q3] SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 184

A. Solution Suspension Emulsion Colloidal dispersion Black Fruit salad Oil and Fog coffee water Soda Nail polish Ink water Brass Milk Liquid Starch solution Boot polish Air Aerosol Sprays Blood Q2. [AS1] Determine the mass by mass percentage concentration of a 100 g salt solution which contains 20 g salt. [Refer to TB page 99 Q1] A. Percentage of solution = (Mass of solute × 100) ÷ Mass of solution Mass of solute = 20 g Mass of solution = 100 g So, percentage of solution = (20 × 100) ÷ 100 = 20 % 20 % NaCl solution. Q3. [AS1] Calculate the concentration in terms of mass by volume percentage of the solution containing 2.5 g of potassium chloride in 50 ml of potassium chloride solution? [Refer to TB page 99 Q2] A. Mass of solute = 2.5 g Volume of KCl solution = 50 ml Mass/volume % = (Mass of solute × 100) ÷ Volume of solution = 2.5 × 100 ÷ 50 = 5% SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 185

1.7 Higher Order Thinking Skills Q1. [AS1] How would you confirm that a colourless liquid given to you is pure water? [Refer to TB page 99 Q1] A. i. A pure substance has a characteristic melting point or boiling point at a given pressure. For example, pure water is always colourless, odourless and tasteless and boils at 373 K at normal atmospheric pressure. Fractional Distillation can be used to find whether a given solution is pure or not. On fractional distillation, the pure liquid will completely evaporate as it will have its characteristic boiling point and condense thus separate. Nothing will be left in the distillation flask. While if it is impure. the liquid will distill off leaving behind the impure part. ii. Observe the smell. We should not find any smell. iii. Observe with naked eyes, we should not find any suspended particles or fumes or air bubbles. iv. Pass a beam of light. It should not scatter. If the liquid satisfy all the above conditions then the given colourless liquid is pure water. 1.8 Suggested Experiments Q1. [AS3] Take a solution, a colloid, and a suspension in three different beakers. Pass a beam of light from the side of the beaker and test whether they show the Tyndall effect or not. [Refer to TB page 100 Q2] A. The scattering of light by particles is called Tyndall effect. It is shown by suspensions and colloidal solutions. A true solution does not show Tyndall effect as the particle size is too small to scatter light. SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 186

Q2. [AS3] Which of the following will show Tyndall effect? How can you demonstrate Tyndall effect in them? [Refer to TB page 100 Q1] 1. Salt solution 2. Milk 3. Copper sulphate solution 4. Starch solution A. Milk and starch solution will show Tyndall effect as they are colloidal solutions. By passing a light through the solution, we can demonstrate Tyndall effect in them. 1.9 Suggested Projects Q1. [AS4] Make a list of solids, liquids and gases from your surroundings. (These substance maybe organic or chemical). Separate mixtures from them and classify them into solutions, colloids and suspensions. A. Students’ activity. Objective Questions (1) Which is not formed by the physical mixture of two substances (Pg 100; TB Q 2) (A) Mixture (B) Compound (C) Colloid (D) Suspension Correct Answer: B SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 187

(2) The substance which is relatively less in quantity in a solution is (Pg 100; TB Q 3) (A) Solute (B) Solvent (C) Solubility (D) Concentration Correct Answer: A (3) The amount of solute present in a saturated solution at constant temperature is known as its (Pg 100;TB Q 4) (A) Solubility (B) Concentration (C)Volume percentage (D)Weight percentage Correct Answer: A (4) If the quantity of solute is more in a solution then the solution is said to be (Pg 100; TB Q 5) (A) Saturated solution (B) Dilute solution (C)Concentrated solution (D)Unsaturated solution Correct Answer: C (5) The phenomenon of scattering of a visible light by the particles of colloid is known as (Pg 100; TB Q 6) (A) Tyndall effect (B) Chromatography (C) Sublimation (D) Reflection Correct Answer: A SESSION 1. TYPES OF SUBSTANCES - MIXTURES, SOLUTIONS, COLL... 188

SESSION 2 SEPARATING THE COMPONENTS OF A MIXTURE 2.1 Mind Map 2.2 Terminology i. Evaporation –This phenomena of change of a liquid into vapours at any temperature below its boiling point is called evaporation. ii. Centrifuge –Centrifuge is a machine that uses a force pulling dense objects away from the center to separate particles or to draw off moisture. An example of a centrifuge is a machine that separates cream and milk. iii. Immiscible liquids – An immiscible liquid is one which doesn’t dissolve but forms a layer over another liquid and can be separated easily. iv. Miscible liquids – A liquid is said to be miscible if it dissolves completely in another liquid and is difficult to separate like alcohol is miscible in water. v. Chromatography –A technique for the separation of a mixture by passing it in solution or suspension through a medium in which the components move at different rates. vi. Distillation – Process in which the components of a substance or liquid mixture are separated by heating it to a certain temperature and condensing the resulting vapors. vii. Fractional distillation – The process of separating the constituents of a liquid mixture by heating it and condensing separately the components according to their different boiling points. SESSION 2. SEPARATING THE COMPONENTS OF A MIXTURE 189

2.3 Key Concepts i. Heterogeneous mixtures can be separated into their constituents by using simple physical methods like filtration, handpicking, sieving etc. ii. Homogeneous mixtures and solutions can also be separated into their constituents, but needs advance methods and more complex machinery. 2.4 Reflection on Concepts Q1. [AS1] Which separation techniques will you apply for the separation of the following? [Refer to TB page 99 Q1] a. Sodium chloride from its solution in water. b. Ammonium chloride from a mixture containing sodium chloride and ammonium chlo- ride. c. Different pigments from an extract of flower petals. d. Oil from water. e. Fine mud particles suspended in water. A. a. Sodium chloride from its solution can be separated by evaporation. b. Ammonium chloride is a substance which undergoes sublimation. So, on heating the mixture and cooling the vapours of sublimated ammonium chloride we can separate the constituents of this mixture. c. Different pigments from an extract of flower petals can be separated by using chro- matography. d. Oil from water can be separated by using a separating funnel. As oil and water are immiscible, the denser liquid(water) settles at the bottom while the lighter liquid (oil) floats on it. We can make the heavier water run down into a beaker from the separating funnel and later we can collect oil in another beaker. SESSION 2. SEPARATING THE COMPONENTS OF A MIXTURE 190

e. Fine mud particles suspended in water can be separated by decantation. If we keep the water containing mud particles in a vessel and allow it to settle for some time, the mud particles collect at the bottom of container and the pure water settles above these mud particles. 2.5 Higher Order Thinking Skills Q1. [AS7] Write the steps you would use for making tea. Use the words given below and write the steps for making tea: Solution, Solvent, Solute, Dissolve, Soluble, Insoluble, Filtrate, and Residue. [Refer to TB page 99 Q2] A. (1) About 25 ml of water is taken in a vessel and heated on the flame of a gas stove. (2) Just when the water begins to boil, two teaspoonful of tea powder is added to the water and is continued heating for one to two minutes. The water is the solvent and the tea powder is the solute here. (3) The tea powder is insoluble in water. It is brewed in water and the product is tea decoction. (4) Then a strainer is used to separate/ filter out the insoluble tea powder from the solution and the liquid is collected in a separate glass. This is called filtrate, which is the black tea. (6) The residue left in the strainer is undissolved tea powder and is thrown away. (7) Appropriate amount of sugar is added to the black tea. Sugar is soluble in black tea. (8) Add appropriate amount of boiled milk. The tea is ready. Objective Questions (1) The machine used to separate the massive particles and light particles from a mixture is (Pg 100;TB Q1) (A) Atwood machine (B) Centrifuge (C)Filter paper (D)Separating funnel Correct Answer: B SESSION 2. SEPARATING THE COMPONENTS OF A MIXTURE 191

(2) Immiscible liquids can be separated by (Pg 100;TB Q 7) (A) Distillation process (B) Fractional distillation (C) Chromatography (D)Separating funnel Correct Answer: D (3) Miscible liquids can be separated by (Pg 100; TB Q 8) (A) Distillation process (B) Fractional distillation (C) Chromatography (D)Separating funnel Correct Answer: A SESSION 2. SEPARATING THE COMPONENTS OF A MIXTURE 192

SESSION 3 TYPES OF PURE SUBSTANCES 3.1 Mind Map 3.2 Terminology i. Elements – An element is a form of matter that cannot be broken down by chemical reactions into simpler substances. ii. Compounds – A compound is a substance composed of two or more different types of elements, chemically combined in a fixed proportion. 3.3 Key Concepts i. Pure substances are the substances which cannot be separated by any further methods of separation. ii. Pure substances can be either elements or compounds. An element is a form of matter that cannot be broken down by chemical reactions into simpler substances. SESSION 3. TYPES OF PURE SUBSTANCES 193

A compound is a substance composed of two or more different types of elements, chemically combined in a fixed proportion. iii. The properties of compounds differ from their constituent elements. But in a mixture the properties of its constituent elements or compounds are exhibited. 3.4 Reflection on Concepts Q1. [AS1] Classify the following into elements, compounds and mixtures. [Refer to TB page 99 Q3] (a) Sodium (b) Soil (c) Sugar solution (d) Silver (e) Calcium carbonate (f) Tin (g) Silicon (h) Coal (i) Air (j) Soap (k) Methane (l) Carbon dioxide (m) Blood (n) Sea Water. A. Elements Compounds Mixtures Sodium Calcium Carbonate Soil Silver Methane Sugar Solution Tin Carbon Dioxide Coal Silicon Soap Air, Blood Sea Water Objective Questions (1) The scientist who invented the elements like Sodium, Magnesium, Boron, Chloride etc is (Pg 100;TB Q9) (A) Isaac Newton (B) Henning brand (C)Sir Humphry Davy (D)Robert Boyle Correct Answer: C SESSION 3. TYPES OF PURE SUBSTANCES 194

—— CCE Based Practice Questions —— AS1-Conceptual Understanding Very Short Answer Type Questions 1. State true or false. [ ] [Refer to Session 6.3 ] [ ] [ ] (i) Pure substance is classified into elements and compounds. [ ] [ ] (ii) Soil is a compound. (iii) Tin is an element. (iv) Calcium carbonate is a mixture. (v) Blood is a compound. CHAPTER 6. IS MATTER PURE? 195

2. Match the following. [(Session 6.2)] [] Column B Column [] a. Distillation i. Separation of salt from water [] b. Separating funnel ii. Separation of salt by ammonium chloride [] c. sublimation iii. Separation of dyes from ink d. Evaporation iv. Separation of two miscible liquids [] e. Chromatography v. Separation of immiscible liquids 3. Answer the following questions in one sentence. [Refer to Session 6.2 ] (i) How do we use the following methods in our daily life? a) Centrifugation – b) Distillation – c) Chromatography – d) Separating funnel – e) Evaporation – CHAPTER 6. IS MATTER PURE? 196

(ii) Which of the following methods would you use to separate: a) Sodium chloride from solution of water– b) Butter from curd– c) Oil from water– d) Iron pins from sand– e) Fine mud particles suspended in water– 4. State true or false. [Refer to Session 6.1 ] (i) If the amount of solute in a solution is less than saturation level, it is called saturation level. [] (ii) If the amount of solute present is more, it is called concentrated solution. ] [ (iii) When a solid is completely dissolved in a solution, it is said to be a suspension. ] [ (iv) Mixtures are of two types. [] (v) A solution is a homogeneous mixture of two or more substance. [ ] CHAPTER 6. IS MATTER PURE? 197

5. Match the following. Column B [ ] a. Solvent [(Session 6.1)] Column A i. Components of mixture are uniformly distributed ii. Component of solution that dissolves [ ] b. Diluted other component in it iii. Amount of solute present in saturated [ ] c. Saturated solution at certain temperature iv. No more solute can be dissolved in so- [ ] d. Homogenous lution v. In a solution, amount of solute present [ ] e. Solubility is little Short Answer Type Questions 6. Answer the following questions in 3-4 sentences. (i) [(Session 6.3)] Define pure substance. (ii) [(Session 6.3)] Define element and compound. CHAPTER 6. IS MATTER PURE? 198