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# 202110208-APEX-STUDENT-WORKBOOK-PHYSICAL_SCIENCE-G09-PART2

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Physical Science Workbook_9_P_2.pdf 1 18-10-2019 14:10:21 Name: ___________________________________ Section: ________________ Roll No.: _________ School: __________________________________

Table of Contents 1 31 7 REFLECTION OF LIGHT AT CURVED SURFACES 60 8 GRAVITATION 88 9 FLOATING BODIES 117 10 WORK AND ENERGY 146 11 SOUND 160 12 UNITS AND GRAPHS 161 PROJECT BASED QUESTIONS ADDITIONAL AS-BASED PRACTICE QUESTIONS TABLE OF CONTENTS

7. REFLECTION OF LIGHT AT CURVED SURFACES SESSION 1 REFLECTION OF LIGHT AT SPHERICAL SURFACES i. Curved Surfaces: A rounded surface which is not flat ii. Converging light rays: The rays are meeting at a point. iii. Diverging light rays: The light rays do not meet at a point or appear to meet. iv. Mirror: A surface of glass coated with a metal amalgam, which reflects light rays and shows the clear image of an object. v. Intensity of light: Amount of light produced by a specific lamp source vi. Incident ray: A ray of light that strikes a surface in a particular angle vii. Reflected ray: A ray of light corresponding to a given incident ray that represents the light reflected by the surface viii. Concave Mirror: A converging mirror has a reflecting surface that is bent inward. SESSION 1. REFLECTION OF LIGHT AT SPHERICAL SURFACES 1

ix. Convex mirror: A diverging mirror in which the reflective surface bulges towards the light source x. Angle of incidence: The angle between the incident ray and the perpendicular or normal to the surface xi. Angle of reflection: The angle made by a reflected ray with the perpendicular to the reflecting surface xii. Normal: A line perpendicular or at the right angle to the mirror surface at the point of incidence xiii. Ray diagram: A diagram that traces the path that light takes in order for a person to view a point on the image of an object xiv. Pole: The midpoint (Geometrical centre) of the spherical mirror is called pole. xv. Center of curvature: It is the centre point of the sphere of which the spherical mirror is a part. xvi. Focus: A point on principal axis where all the parallel rays of light, after reflection through the spherical mirror, meet or appear to meet xvii. Radius of curvature: The distance between pole (P) and center of curvature (C) of the spherical mirror xviii. Focal length: The distance between the pole (P) and Focus (F) of the spherical mirror xix. Principal axis: An imaginary line passing through the pole and the centre of curvature of a spherical mirror i. Concave surfaces are bent inward and convex surfaces bent outward. ii. Concave surfaces converge light rays and convex surfaces diverge. iii. The light rays after reflection through concave mirror converge to a focus point. iv. The light rays after reflection through a convex mirror diverge. v. A spherical mirror follows the laws of reflection of light. vi. The parallel light rays are converged at the focus of the concave mirror. vii. The radius of curvature is twice that of the focal length. SESSION 1. REFLECTION OF LIGHT AT SPHERICAL SURFACES 2

Q1. “Laws of reflection of light are not applicable to curved surfaces” Is it correct? [Refer to TB page 124 Q1] A. No, the statement is not correct. The laws of reflection are applicable to all surfaces, whether they are curved or plane. Q2. How do you find the focal length of a concave mirror? [Refer to TB page 124 Q2] A. Hold a concave mirror such that sunlight falls on it. Take a small paper and slowly move it in front of the mirror and find out the point where you get the smallest and brightest spot. The rays coming from the Sun parallel to the principal axis of the concave mirror converge to a point. This point is called the focus or focal point F of the concave mirror. The distance between the focal point F and the pole of the concave mirror is known as the focal length of the concave mirror and it is denoted by ‘f ’. The radius of the curvature R is twice the focal length f. R = 2f Q3. State the differences between a convex and concave mirror. [Refer to TB page 124 Q5] A. Convex mirror Concave mirror Convex mirror has an outer reflecting surface. Concave mirror has an inner reflecting surface Convex mirror is a diverging mirror. Concave mirror is a converging Convex mirror diverges to the mirror. parallel light beam. Concave mirror converges to the Convex mirror always forms a parallel light beam at the focus. virtual (not formed on the screen) and erect image but the size of The image formed by a concave mirror is enlarged when the object is held close to the mirror. And the nature of the image is SESSION 1. REFLECTION OF LIGHT AT SPHERICAL SURFACES 3

the image is always smaller than erect and virtual. As the distance the size of the object. of the object increases, the size of the image decreases and the nature of the image is inverted and real (formed on the screen). Q1. By observing steel vessels and different images in them; Surya, a third class student, asked his elder sister Vidya some questions. What may be those questions? [Refer to TB page 124 Q12] A. i. Why is the image appearing small on the bulging-out surface of the steel vessel? ii. Why is the image appearing larger on the bent-in surface of the steel vessel? iii. Can we use the vessels’ surface as a mirror? iv. How can we see the images on the surface of steel vessels? Q1. How do you find the focal length of a concave mirror in the lab? [Refer to TB page 125 Q14] A. i. A concave mirror obeys the laws of reflection. ii. Hold a concave mirror such that sunlight falls on it. iii. The rays coming from the Sun parallel to the principal axis of the concave mirror converge to a point. This point is the focal point of the concave mirror. iv. Take a small-sized paper and slowly move it in front of the mirror and find out the point where you get the smallest and brightest spot. v. This spot is the image of the Sun. vi. Now measure the distance between the pole and the focus. This is the focal length of the mirror. SESSION 1. REFLECTION OF LIGHT AT SPHERICAL SURFACES 4

Q2. How can you show the diverging and the converging of light by using a laser light? [Refer to TB page 125 Q1, Try These] A. i. Take a rectangular (3’’x 6”) acrylic sheet. It works as a mirror to reflect light rays. ii. Hold the acrylic sheet with your palm. iii. Now, bend your palm slowly as the sheet bends inwards in the shape of a concave mirror. iv. Ask your friend to focus the laser light. v. Observe the reflected light on the wall. vi. The reflected light rays are converged at one place because the acrylic sheet is bent inwards and acts as a concave mirror. vii. Now, slowly bend your palm as the sheet bends outwards in the shape of a convex mirror. viii. Again ask your friend to focus the light. ix. Observe the reflected light on the wall. x. The reflected light rays are not converged. That means that they are diverged with low intensity. So it acts as a convex mirror. Q1. Collect information about the history of spherical mirrors in human civilization. Display it in your class room. [Refer to TB page 125 Q2, Try These] A. Student’s response (Hint: i. The first mirrors used by people were most likely pools of water or water. ii. The earliest manufactured mirrors were pieces of polished stones such as obsidian, a naturally-occurring volcanic glass. iii. Mirrors of copper were crafted in Mesopotamia from around 6000 B.C. iv. Bronze mirrors were manufactured from around 2000 B.C. v. Parabolic mirrors were described and studied in classical antiquity by the mathematician Diodes in his work on burning mirrors. SESSION 1. REFLECTION OF LIGHT AT SPHERICAL SURFACES 5

vi. Parabolic mirrors were also described by the physicist Ibn Sahl in the 10th century and Ibn al Haytham discussed concave and convex mirrors in both cylindrical and spherical geometries. vii. Ptolemy conducted a number of experiments with curved, polished iron mirrors. He also discussed plane, convex, concave and spherical mirrors in his optics. viii. In China, people began making mirrors with the use of silver mercury amalgams as early as 500 A.D. ix. In the 16th century, Venice, a big city famed for its glass making expertise, became a centre of mirror production using this new technique. x. The invention of the silvered-glass mirror was credited to German chemist Justus Von Liebig in 1835.) Q2. Think about the objects which act as concave or convex mirrors in your surroundings. Make a table and display it in your classroom. [Refer to TB page 125 Q3, Try These] A. Student’s response (Hint: Objects Image Concave Mirror/ Convex Mirror Bent inwards surface of spoon Convex Bent outwards surface Concave of spoon Surface of steel flask Convex mirror SESSION 1. REFLECTION OF LIGHT AT SPHERICAL SURFACES 6

Rear view mirror of Convex mirror vehicle Curved surface of Concave mirror torch light Headlight of vehicle Concave mirror ) Q3. How will our image be in concave and convex mirrors? Collect photographs and display in your classroom. [Refer to TB page 125 Q4, Try These] A. Student’s activity SESSION 1. REFLECTION OF LIGHT AT SPHERICAL SURFACES 7

SESSION 2 IMAGES FORMED BY SPHERICAL MIRRORS i. Object distance: The distance between the pole of the spherical mirror and the position of the object on principal axis ii. Image distance: The distance between the pole of the spherical mirror and the position of the image formed on principal axis iii. Erect image: An erect image is one that appears right-side up. iv. Inverted image: An inverted image means that the image is upside down. v. Virtual Image: An image that is formed when the outgoing rays from a point on an object always diverge vi. Real image: An image that is formed when the rays of light after reflection actually meet at some point vii. Magnification: The ratio of the size of the image to the size of the object viii. Solar cooker: It is a device which uses the energy of direct sunlight to heat and cook food materials. SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 8

2.3 Solved Examples Q1. An object 4 cm in size is placed 25 cm in front of a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and size of the image. A. According to sign convention: focal length (f) = -15cm object distance (u) = -25cm object height (ho) = 4 cm image distance (v) = ? image height (hi) = ? Substituting the above value in the equation, 1/f = 1/u + 1/v 1/-15 = 1/-25+ 1/v 1/v = 1/25 – 1/15 1/v = -2/75 v = -37.5cm So, the screen should be placed at 37.5 cm from the pole of the mirror. The image is real. Magnification m =hi/ho = -v/u By substituting the above values, hi/4 = -(-37.5)/ (-25) hi = - (37.5 X4)/25 hi = - 6cm So, the image is inverted and enlarged. SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 9

2.4 Key Concepts i. A ray parallel to the principal axis, after reflection, always passes through the focus of the mirror. ii. A ray that passes through the focal point of the mirror will travel parallel to the axis after reflection. iii. The light ray that passes through the centre of curvature always gets reflected along the same path. iv. For different positions of the object, we can get different positions and different sizes of the image for spherical mirrors. v. A virtual image cannot be formed on screen. vi. A real image forms on the screen. vii. A TV antenna dish has a concave surface. viii. Rear view mirrors are convex mirrors. ix. Mirror formula: 1/f = 1/v + 1/u; where f = focal length, v = image distance and u = object distance. x. We should use sign convention rules while using formulae of the mirrors. xi. Magnification, m = size of the image (hi) / size of the object (ho) or m = - distance of the image ( v) / distance of the object (u) Q1. Where will the image form when we place an object, on the principal axis of a concave mirror at a point between the focus and centre of curvature? [Refer to TB page 124 Q3] A. The image will form beyond the centre of curvature (C) when the object is placed on the principal axis of a concave mirror at a point between the focus and centre of curvature. SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 10

Q2. Find the distance of the image when an object is placed on the principal axis at a distance of 10 cm in front of a concave mirror whose radius of curvature is 8 cm. [Refer to TB page 124 Q4] A. Let the distance of the image = v Distance of the object, u = 10 cm Radius of curvature, R = 8 cm As we know that, R = 2f; where f = focal length So, f = R/2 Or, f = 8/2 cm or f = 4 cm According to mirror formula, 1/f = 1/v + 1/u Substituting all the values, we get 1/4 = 1/v + 1/10 or 1/v = 1/4 – 1/10 Or, 1/v = 3/20 or v = 20/3 = 6.67 cm Hence the image will be formed at a distance of 6.67 cm. Q3. Distinguish between real and virtual images. [Refer to TB page 124 Q6] A. Differences between real and inverted images are as follows. Real Image Virtual Image A real image is formed when the A virtual image is formed when the rays of light after reflection actually rays of light after reflection appear meet at some point. to meet at a point. A real image can be formed in a A virtual image cannot be formed in screen. a screen. A real image is always inverted and A virtual image is always erect and formed in front of the mirror. formed behind the mirror. SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 11

A real image is formed by only A virtual image can be formed by concave mirrors. concave, convex and plane mirrors. Q4. How do you get a virtual image using a concave mirror? [Refer to TB page 124 Q7] A. Take a concave mirror and place it on the V-stand. i. Find the focal point. ii. Now place the object (candle) between the mirror and the focal point. iii. Now take a screen and move that away from the mirror, behind the mirror. iv. The image formed is virtual, erect and enlarged, and is formed behind the mirror. Q5. What do you know about the terms given below related to spherical mirrors? [Refer to TB page 124 Q8] a) Pole b) Center of curvature c) Focus d) Radius of curvature e) Focal length f) Principal axis g) Object distance h) Image distance i) Magnification A. a) Pole: The midpoint (geometrical centre) of the spherical mirror (either concave or convex) is called pole (P). b) Center of curvature: It is the centre point of the sphere of which the spherical mirror is a part. It is represented by C. c) Focus: A point on the principal axis where all the parallel rays of light, after reflection through the spherical mirror, meet or appear to meet, is known as the focus or focal point (F) of the spherical mirror. SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 12

d) Radius of curvature: The distance between the pole (P) and center of curvature (C) of the spherical mirror is known as the radius of curvature (R). It is twice the focal length (f), i.e. R = 2f. e) Focal length: The distance between the pole (P) and Focus (F) of the spherical mirror is known as the focal length (f) of the mirror. f) Principal axis: An imaginary line passing through the pole and the centre of curvature of a spherical mirror is known as the principal axis of the mirror. g) Object distance: The distance between the pole of the spherical mirror and the position of the object on the principal axis is known as object distance (u). h) Image distance: The distance between the pole of the spherical mirror and the position of the image formed on the principal axis is known as image distance. i) Magnification: It is the ratio of the size of the image to the size of the object. It is represented by the m. m = size of the image (hi) / size of the object (ho) or m = - distance of the image ( v) / distance of the object (u) Q6. Write the rules for sign convention. [Refer to TB page 124 Q9] A. Sign convention Rule: i. All distances should be measured from the pole. ii. The distances measured in the direction of incident light, to be taken positive and those measured in the direction opposite to incident light to be taken negative. iii. Height of the object (Ho) and height of image (Hi) are positive if measured upwards from the axis and negative if measured downwards. Q7. The magnification produced by a convex mirror is -1. Do you agree? [Refer to TB page 124 Q10] A. i) No, this statement is incorrect because a convex mirror always forms a virtual image behind the mirror and according to sign convention of the mirror, the distance measured in the direction of the incident light from the pole of the mirror, to be taken positive. So its magnification should be positive always. ii) Also, the convex mirror forms diminished image of the object, so its magnification should be less than 1 always. SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 13

Q1. Imagine that spherical mirrors were not known to human beings. Guess the consequences. [Refer to TB page 124 Q11] A. i. Concave mirrors are used in head lights of automobiles so that the light spreads on a wider surface and creates wider vision. Convex mirrors are used in vehicles’ rear mirror. If this was not known to human beings, then we would be unable to see behind us and this would increase accidents on the road. ii. Solar cookers use concave mirrors as reflectors to converge Sun rays to increase the temperature of the box. If concave mirrors were not known to us then we would not be able to use the solar energy as a fuel. iii. We would not be able to observe small things as big and the dentists who use concave mirrors to see our teeth would face a huge problem. Q1. How do you form a diminished image by a concave mirror on a screen? [Refer to TB page 125 Q13] A. i. For forming a diminished image by a concave mirror on a screen, keep a candle in front of a concave mirror beyond the centre of curvature. ii. A ray of light R1 passing through the focus of the concave mirror becomes parallel to the principal axis after reflection. iii. A ray of light R2 which is parallel to principal axis passes through the focus F after reflection. iv. The reflected rays meet at A. v. Since any ray coming from any point on the axis and travelling along the axis will get reflected on the axis itself. So, the base of the image is going to be on the axis. vi. If the object is placed vertically on the axis, the image is going to be vertically on the axis. vii. Hence, the image formed in the diagram will be inverted and diminished as shown in the image. SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 14

Q2. What do you infer from the experiment which you did with the concave mirror and where you measured the distance of the object and the distance of the image? [Refer to TB page 125 Q15] A. i. We can infer that there is a relation between object distance and object size and image distance and image size of an object placed in front of a concave mirror. ii. As we change the position of the object in front of a concave mirror, the position and the size of the image also changes. Hence we can say that for different positions of the object, we can get different positions and different sizes of the image. iii. Images formed by a concave mirror are as follows. Position of the Position of the Enlarged/ Erect/ Real/ Object Image Diminished Inverted Virtual Between mirror and Behind the Enlarged Erect Virtual focal point F mirror On focal point F At infinity - -- Enlarged Inverted Real Between Focal point Beyond C Same size Inverted Real F and centre of On C curvature C On centre of curvature C SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 15

Beyond centre of Between F and Diminished Inverted Real curvature C C iv. As the distance of the object from the mirror decreases, object distance increases and the size of the image also increases. At the centre of curvature, we can see from the above table, image distance and object distance are the same and at focus, the image forms at infinity or we can say that no image is formed. Q1. Draw a ray diagram to show the process of converging of the parallel beam by a concave mirror by taking four parallel incident rays. [Refer to TB page 125 Q16] A. i. To draw a ray diagram to converge four parallel beams by a concave mirror, first draw a concave mirror MM1 and four parallel incident lines AG, BH, DI and EJ on mirror. ii. Now, draw the normal at incident point to the parallel light ray. iii. Measure incident angle and draw the reflected ray with the reflected angle which is the same as the incident angle (according to the law of reflection). iv. Follow the same procedure for all the incident rays. v. You will observe that in the ray diagram, the parallel light rays are converged at the focus of the mirror. SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 16

Q2. Draw suitable rays by which we can guess the position of the image formed by a concave mirror. [Refer to TB page 125 Q17] A. i. All the rays that are parallel to the axis get reflected such that they pass through the focal point of the mirror. ii. All the rays that come from the tip of the object and go through the focal point and fall on the mirror, after reflection travel parallel to the principal axis. iii. All the rays coming from the tip of the object and going through the centre of curvature to meet the mirror will get reflected along the same line. iv. Along with these rays, the ray which comes from the object and reaches the pole of the mirror is also useful in drawing the ray diagram. SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 17

Q3. Show the formation of image with a ray diagram when an object is placed on the principal axis of a concave mirror away from the centre of curvature. [Refer to TB page 125 Q18] A. If the object is placed beyond the center of curvature, the image formed will be inverted and diminished. Q4. Make a solar heater/cooker and explain the process of making. [Refer to TB page 125 Q19] A. Process of making solar heater/cooker: Material required: Dish antenna, Fevicol, glue, Aluminum foil, black colour coated bowl, scissors Procedure: i. Take the aluminum foil sheet. ii. Cut the sheet into isosceles triangle-shaped pieces with a base of 5 cm and other two sides with the length equal to the radius of the dish antenna. iii. Paste these triangular pieces to the inner side of the antenna. iv. Now, the solar cooker is ready. v. Then identify where the sun rays are converged. vi. Take a bowl with water and place it at the focus of the solar cooker. SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 18

Q5. To form the image on the object itself, how should we place the object in front of a concave mirror? Explain with a ray diagram. [Refer to TB page 125 Q20] A. i. To form the image on the object itself, the object should be kept at the centre of curvature of a concave mirror. ii. An object such as candle has been placed at the centre of curvature C on the concave mirror. iii. A ray of light R1 which is parallel to principal axis passes through the focus F after reflection. iv. A ray of light R2 passing through the focus of the concave mirror becomes parallel to the principal axis after reflection. v. The reflected rays meet at A. vi. So the real image is formed at point A of the object. vii. The image formed is real, inverted and the same size as the object. Q1. How do you appreciate the role of spherical mirrors in daily life? [Refer to TB page 125 Q21] SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 19

A. i. A concave mirror focuses parallel sun rays at the focal point of the mirror. So with a small concave mirror we can heat up and burn paper. ii. By using a concave mirror, we can make a solar cooker and cook food by keeping solar cooker in the open place as its concave shape should face the sunlight. Place the bowl at the focus of the sun rays. In this way we can cook food. iii. Concave mirrors are kept in the cars headlight to get parallel light beam. iv. A concave mirror enlarges the image when the object is held close to it such as less than the focal length. And the image formed is erect. This property of mirror is used in shaving mirrors and mirrors used by dentists. v. Concave mirrors are used in TV dish antenna as reflectors. vi. Convex mirrors are used in rear view mirrors of cars because a convex mirror provides a larger field of view. Q2. How do you appreciate the use of reflection of light by a concave mirror in the making of TV antenna dishes? [Refer to TB page 125 Q5, Try These] A. i. If you see TV antenna dishes, you will observe many concave surfaces. ii. They contain concave mirrors to receive signals from distinct communication satellites. iii. The concave shape of a dish antenna helps to reflect the signal to the focus of the dish. iv. A device known as feed horn is mounted at the focal point which gathers the signals and sends them to a processing unit. v. Hence, I appreciate the use of reflection of light by a concave mirror in the making of TV antenna dishes. Q1. How do doctors use concave mirrors? [Refer to TB page 125 Q22] SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 20

A. i. Concave mirrors enlarge the image of an object when the object is placed near the mirror. ii. In this case the image formed is erect. iii. This property is used in the mirrors used by dentists. iv. They place the mirror near the teeth and can easily see the problem in the mirror. Q2. Why do we prefer a convex mirror as a rear view mirror for the vehicles? [Refer to TB page 125 Q23] A. i. Convex mirrors are used in rear-view mirrors of the vehicles because a convex mirror provides a larger field of view. ii. They give an erect, virtual, full-size diminished image of distant objects with a wider field of view. iii. Thus, convex mirrors enable the driver to view a much larger area. Q3. Complete the table - 2 which is related to experiment done by a concave mirror. [Refer to TB page 125 Q24] A. Position of Position Real the Object of the Enlarged/diminished Erect/inverted or Image Between Virtual mirror and F Behind the mirror Enlarged Erect Virtual At focal point At infinity - -- Between F Beyond C Enlarged Inverted Real and C Same size Inverted Real On C Diminished Inverted Real At centre of Between F curvature and C Beyond C Q4. A convex mirror with a radius of curvature of 3 m is used as the rear-view mirror of an automobile. If a bus is located at 5 m from the mirror, find the SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 21

position, nature and size of the image. [Refer to TB page 125 Q6, Try These] A. i. Radius of curvature of the mirror, R = 3 m ii. So, focal length of the mirror, f = R/2 = 3/2 m = 1.5 m (Focus lies behind the mirror, i.e., in the direction of light. So its sign is taken as positive.) iii. Location of the bus from the mirror, u = -5 m (Object is always placed in front of the mirror. Hence, the sign of object is taken as negative.) iv. Let the position of the mirror = v v. Using mirror formula, 1/f = 1/v + 1/u Or, 1/v = 1/f – 1/u Or, 1/v = 1/1.5 – 1/-5 or 1/v = 13/15 Or, v = 15/13 = 1.15 m vi. Magnification, m = -v/u = - 1.15/-5 = 0.23 vii. Hence, the image is smaller than the size of object and it is virtual and erect as it is formed behind the mirror. Q5. An object is placed at a distance of 10 cm from a convex mirror of focal length 15cm. Find the position and nature of the image. [Refer to TB page 125 Q7, Try These] A. i. Object distance, u = -10 cm (Object is always placed in front of the mirror. Hence the sign of object is taken as negative.) ii. Focal length of the convex mirror, f = 15 cm (Focus lies behind the mirror, i.e., in the direction of light. So its sign is taken as positive.) iii. Let the position of the object be v. iv. Using mirror formula, 1/f = 1/v + 1/u Or, 1/v = 1/f -1/u Or, 1/v = 1/15 – 1/-10 or 1/v = 5/30 or 1/v = 1/6 Or, v = 6 cm v. As v is positive, the image formed behind the mirror. Thus it is virtual and erect. Q6. Write answers to the following questions based on the data given in the table -3. [Refer to TB page 125 Q8, Try These] SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 22

1) What changes will come gradually in the size of an image, when we move an object away from the concave mirror? 2) In which situation are inverted images formed by a concave mirror? 3) If the centre of curvature of a concave mirror is 10 cm, where should the object be placed to get an image at the centre of curvature? A. 1) When we move an object away from the concave mirror, the size of the mirror gradually decreases. 2) When object is placed between i. Focal point and centre of curvature, ii. On the centre of curvature iii. Beyond centre of curvature 3) The object should be placed at 10 cm, i.e., on the centre of curvature to get image on the centre of curvature. SESSION 2. IMAGES FORMED BY SPHERICAL MIRRORS 23

CCE Based Practice Questions AS1 – Conceptual understanding Very Short Answer Type Questions 1. Fill in the blanks. [Refer to Session 7.2] (i) In a concave mirror, if the object is placed at focus, the image formed is at _________________________. (ii) In a concave mirror, if the object is placed between the centre of curvature and the focus, the image formed is at _____________________. (iii) In a concave mirror, if the image is formed at the centre of curvature, the size of the image is __________________________ than the object. (iv) In a concave mirror, if the object is placed between the mirror and the focus, the image is ________________________ in nature. 2. State true or false. [Refer to Session 7.2] (i) Concave mirrors diverge light rays. [] (ii) The mid-point of the mirror is called its focus. [] (iii) The distance between the focus and the centre of curvature is called the radius of curvature. [] (iv) Focal length is twice the radius of curvature. [] Short Answer Type Question 3. Answer the following question in 3-4 sentences. (i) [(Session 7.2)] Name the type of mirror used in the following devices. (a) Headlights of a car (b) Side/rear-view mirror of a vehicle (c) Solar furnace Support your answer with reasons. (ii) [(Session 7.2)] List the sign convention followed in drawing a ray diagram. CHAPTER 7. REFLECTION OF LIGHT AT CURVED SURFACES 24

Long Answer Type Questions 4. Answer the following questions in 6-8 sentences. (i) [(Session 7.2)] State two positions in which a concave mirror produces an enlarged image of a given object. List two differences between the two images. AS2-Asking questions and making hypothesis Short Answer Type Questions 5. Answer the following questions in 3-4 sentences. (i) [(Session 7.2)] On a magic mirror, Ruhima, a 2nd class student finds that her head is very big, her torso is very small and her legs are of same size. On observing her body, she asked her sister some questions. What may those questions be? (ii) [(Session 7.2)] Imagine that the concave mirror is not known to human beings. Guess the consequences. AS3-Experimentation and ﬁeld investigation Long Answer Type Questions 6. Answer the following question in 6-8 sentences. (i) [(Session 7.1)] Design an activity using concave mirror to prove that it’s converging in nature. Also state a method to find its rough focal length. (ii) [(Session 7.1)] What happens when the rays of sun are focused at a point on the paper by using concave mirror? What is the position and nature of the image? Discuss. AS4-Information skills and projects Very Short Answer Type question 7. Answer the following question. (i) [(Session 7.2)] Complete the table by filling in whether each object acts as a concave or a convex mirror. CHAPTER 7. REFLECTION OF LIGHT AT CURVED SURFACES 25

Image Mirror Short Answer Type Questions 8. Answer the following questions. (i) [(Session 7.2)] Think about the objects which act as concave mirrors and convex mirrors in your surroundings. Make a table with pictures. (ii) [(Session 7.2)] Collect information about the uses of convex mirrors in daily life. Long Answer Type Question 9. Answer the following question. (i) [(Session 7.2)] Complete the table based on where the image is formed in a concave mirror. Incident Ray Reflected Ray Parallel to principal axis Parallel to principal axis Through the centre of curvature CHAPTER 7. REFLECTION OF LIGHT AT CURVED SURFACES 26

Incident at the pole AS5-Communication through drawing and model making Short Answer Type Question 10. Answer the following question. (i) [(Session 7.2)] A set of parallel rays fall on a convex mirror. At what direction will these rays reflect? Through a ray diagram represent the reflected rays. Long Answer Type Question 11. Answer the following question. (i) [(Session 7.2)] If the image formed by mirror for all positions of the object placed in front of it is always virtual and diminished, state the type of the mirror. Draw a ray diagram in support of your answer. AS6-Appreciation and aesthetic sense, values Short Answer Type Question 12. Answer the following question in 3-4 sentences. (i) [(Session 7.2)] Explain the use of reflection of light by a concave mirror in the making of a torch. Long Answer Type Question 13. Answer the following question in 6-8 sentences. (i) [(Session 7.2)] Explain the use of reflection of light by a concave mirror in the making of a solar cooker. How do you cook food using a solar cooker? AS7-Application to daily life, concern to bio diversity Very Short Answer Type question CHAPTER 7. REFLECTION OF LIGHT AT CURVED SURFACES 27

14. Answer the following question in 1-2 sentences. (i) [(Session 7.1)] What is the radius of curvature of a concave mirror if its focal length is 8 cm? Short Answer Type Question 15. Answer the following question in 3-4 sentences. (i) [(Session 7.2)] At what distance from a concave mirror of focal length 20 cm should an object 4 cm long be placed in order to get an erect image that is 12 cm tall? Long Answer Type Questions 16. Answer the following question in 6-8 sentences. (i) [(Session 7.2)] An object is placed at a distance of 25 cm from a converging mirror of focal length 20 cm. Discus the effect on the nature and position of the image if the position of the object changes from 25 cm to 15 cm. Justify your answer without using mirror formula. (ii) [(Session 7.2)] A 5 cm needle is placed 15 cm away from a convex mirror of focal length 15 cm. Give the location of the image formed and its magnification. Describe what happens to the image as the needle is moved farther from the mirror. Objective AS1-Conceptual understanding 17. Choose the correct answer. (i) Which of the following statements is correct? (A) The image formed in a concave mirror is real when the object placed beyond the centre of curvature. (B) The image in a convex mirror is always of the same size as the object. (C) The image formed is inverted if the object is placed between the mirror and the focal point in a concave mirror. (D) The image formed in a concave mirror is laterally inverted. (ii) What kind of mirror is used in vehicles to see the traffic on the rear side? (A) convex mirror (B) plane mirror CHAPTER 7. REFLECTION OF LIGHT AT CURVED SURFACES 28

(C) concave mirror (D) either concave mirror or convex mirror (iii) Which of the following is correct for a concave mirror? (A) When the object is at focal point, the image is formed at focal point. (B) It forms both highly enlarged and highly diminished images. (C) When the object is placed at the centre of curvature, the image is real, inverted and positioned at C only. (D) The image formed is always virtual and erect. (iv) In torches, search lights and headlights of vehicles, the bulb is placed (A) between the pole and focus of the reflector. (B) very near to the focus of the reflector. (C) between the focus and the centre of curvature of the reflector. (D) at the centre of curvature of the reflector. (v) Which type of reflective surface was used by Archimedes? (A) concave surface (B) convex surface (C) plane surface (D) none of these AS3-Experimentation and ﬁeld investigation 18. Choose the correct answer. (i) Ravi is standing in front of a special mirror. He finds that the image of his head is smaller, that of his torso is the same size and that of his legs is bigger. The following is the order of combinations for the special mirror from the top. (A) plane, convex, concave (B) convex, concave, plane (C) concave, plane, convex (D) convex, plane and concave (ii) Which of the following best describes the image formed by a concave mirror when the object distance from the mirror is less than the focal length (f)? (A) virtual, upright and enlarged (B) real, inverted and reduced (C) virtual, upright and reduced (D) real, upright and enlarged AS4-Information skills and projects CHAPTER 7. REFLECTION OF LIGHT AT CURVED SURFACES 29

19. Choose the correct answer. (i) Which of the following produces a virtual, magnified and upright image? (A) torch (B) flashlight (C) shaving mirror (D) magnifying glass AS6-Appreciation and aesthetic sense, values 20. Choose the correct answer. (i) In which of the following, can we appreciate the role of convex mirrors in daily life? (A) in the rear-view mirror of a car (B) in a TV dish antenna as a reflector (C) in using it to heat up and burn paper (D) in a solar cooker (ii) A concave mirror is kept in such a way that a parallel light beam is obtained. In which of the following is this phenomenon used? (A) in a solar cooker (B) in the head light of a car (C) in a TV dish antenna (D) all of these CHAPTER 7. REFLECTION OF LIGHT AT CURVED SURFACES 30

8. GRAVITATION SESSION 1 INTRODUCTION AND UNIFORM CIRCULAR MOTION 1.1 Mind Map 1.2 Terminology i. Uniform circular motion – The motion of a body along a circular path with constant speed. ii. Centripetal acceleration – The acceleration that changes only the direction of velocity of a body is called “centripetal acceleration”. iii. Centripetal force – The net force which change only the direction of the velocity of a body is called “centripetal force”. 1.3 Key Concepts i. The force of attraction between two objects is called gravitational force. ii. Motion of a body with constant speed in a circular path is called uniform circular motion. iii. The acceleration which causes changes in direction of velocity of a body is called centripetal acceleration and it is directed always towards the center of the circle during the uniform circular motion of a body. iv. The net force required to maintain a body in uniform circular motion is known as ‘centripetal force’ and its magnitude is given by F = mv2 /R. SESSION 1. INTRODUCTION AND UNIFORM CIRCULAR MOTION 31

1.4 Conceptual Understanding Q1. A car moves with constant speed of 10 m/s in a circular path of radius 10 m. The mass of the car is 1000 kg. Who or what is providing required centripetal force for the car? How much is it? [Refer to TB page 139 Q1] A. The centripetal force for the car is provided by the mass of the car. Centripetal force, F = mv2 R Mass of car (m) = 1000 kg Speed of car (v) = 10 m/s R, radius of circular path = 10 m F=? F = 1000×102 = 104N 10 Centripetal Force (F) = 104 N Q2. Two cars having masses m1 and m2 move in circles of radii r1 and r2 respectively. If they complete the circles in equal time, what are the ratios of their speeds and centripetal accelerations? [Refer to TB page 139 Q4] A. Masses of cars: m1 and m2 ; Radius of circles : r1 and r2 Given that their time period is equal. Therefore T1 = T2 T1 = 2πr1 and T2 = 2πr2 v1 v2 T1 = T2 ⇒ 2πr1 = 2πr2 v1 v2 SESSION 1. INTRODUCTION AND UNIFORM CIRCULAR MOTION 32

Ratio of speeds v1 = 2πr1 = r1 v2 2πr2 r2 Centripetal acceleration v2 r Ratio of centripetal acceleration Q3. A scooter weighing 150 kg together with its rider moving at 36 km/hr is to take a turn of radius 30 m. What force on the scooter towards the center is needed to make the turn possible? Who or what provides this? [Refer to TB page 140 Q4, Try These] A. F = mv2 = 150×(36× 5 2 = 5 × (10)2 = 500N R 18 ) 30 This force is provided by the friction between the tyres of the car and road. 1.5 Application to Daily Life, Concern to Bio Diversity Q1. A small metal washer is placed on the top of a hemisphere of radius R. What minimum horizontal velocity should be imparted to the washer to detach it from the hemisphere at the initial point of motion? [Refer to TB page 140 Q4, Try These] A. Radius of hemisphere = R Mass of hemisphere = M SESSION 1. INTRODUCTION AND UNIFORM CIRCULAR MOTION 33

Let the radius and mass of washer = r, m Distance between hemisphere and washer = R + r The centripetal force required to rotate the washer = mv2 R The gravitational force of washer due to hemisphere is GMm R2 But the necessary centripetal force must be equal to the gravitational force mv 2 = GMm r R2 ⇒ v2 = GMm × r r2 m Here r is the distance, r= R+r v2 = GM (R+r) When R and r are compared, r<<R, which is negligible. v2 = GM (R) We know the relation between G and g as g = GM (R2) ⇒G = gR2 M ∴ v2 = gR2 × M M R ⇒ v2 = gR √ or, v = gR SESSION 1. INTRODUCTION AND UNIFORM CIRCULAR MOTION 34

SESSION 2 UNIVERSAL LAW OF GRAVITATION AND FREE FALL 2.1 Mind Map 2.2 Terminology i. 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 between them. ii. Weight –Weight of a body is the force of attraction on the body due to earth. It is given by the product of mass of the body and the acceleration due to gravity; W = mg. iii. Weightlessness – The state of the freely falling body. iv. Free–fall – A body is said to be in free fall when only gravity acts on it. (its acceleration is ‘g’). SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 35

2.3 Solved Examples Q1. Example 1: What is the time period taken for one revolution of a satellite near the earth’s surface? Neglect height of the orbit of satellite from the surface of ground. (Refer to TB page 131) 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 =GRM2m according to Newton’s law of gravitation. 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 133 ) SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 36

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 page133) 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 2 20 m/s. What time does it take to reach the ground? (g = 10 m/s )( Refer to TB page 133) SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 37

+ _ 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 38

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 134) 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 39

2.5 Conceptual Understanding Q1. 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 140 Q1, Try These] A. v=u + at 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 + l0 x 1.5 = 15 Speed after 1.5 sec = 15 m/s Distance covered by apple s = ut + ½ at2 u = 0; t=1.5 sec; a=10 m/s2 s = (0 x 1.5) + ½ x 10 x (1.5)2 = 0 + ½ x l0 x 2.25 = 11.25 Distance covered by the apple = 11.25 m Q2. A body is projected with a speed of 40 m/s vertically up from the ground. What is the maximum height reached by the body? What is the entire time of motion? What is the velocity at 5 seconds after the projection ? Take g= 10m/s2 [Refer to TB page 139 Q2] A. u = 40m/s ; a = –10 m/s2 (negative sign because going up) SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 40

v= 0 m/s (at top) 41 s=h= ? ; t = ? Let us take the formula v2 – u2 = 2as. We have, 02 –(40)2 = 2 x (–10) x h –1600 = –20 h H = 1600 / 20 = 80 Maximum height reached = 80 m (ii) From the formula, v = u + at we have, 0 = 40 – g x t (When thrown upwards g is negative.) –40 = –10 x t Total time = 4 seconds. (Time of ascent = Time of descent) Total time = time of ascent + time of descent = 4 + 4 = 8 seconds Entire time of motion = 8 seconds Or Entire time of motion (T)=2u/g=2x40/10=8s Entire time of motion is 8 seconds. At 5 seconds the body is in downward direction. u= 0 m/s, a= g= 10m/s2 , t= 5–4=1 sec v= u +at = 0 + 10×1 = 10 m/s Therefore the velocity at 5 seconds is 10 m/s downward. SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL

Q3. A ball is dropped from a height. If it takes 0.2 s to cross the last 6m before hiting the ground, ﬁnd the height from which it is dropped. Take g = 10 m/s2 . [Refer to TB page 140 Q2, Try These] 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 + 1/2 at2 , where a = g (acceleration due to gravity) = 10 m/s2 , t = 0.2 s Thus 6 = 0.2 u + 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. v2 – u2 = 2gs or, v2 – u2 = 2gx or, 292 = 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 Q4. 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 m/s2 ) [Refer to TB page 139 Q3] A. u = 50 m/s v2 – u2 = 2gs 0 – 502 = – 2 (10) s SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 42

s = 125 m i.e., Maximum height = 125 m Time to reach maximum height, S = 1/2 (gt2 ) 125 = 1/2 (10) t2 t = 5s At maximum height speed = 0 Q5. 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 139 Q5] A. m1 =m2 =10 kg; D= 10 cm = 10-2 m F = G M1 M2 d2 = G 10×10 0.1×0.1 = G 102 10−2 = G × 104 N Q6. 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 140 Q3, Try These] A. Let, m= 7.4×1022 kg r= 1740km = 1740×103 m SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 43

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 = 1.630 m/s2 Now, ge = 9.8 m/s2 (free–fall acceleration of earth) Therefore, gm = 1.630 = 1 ge 9.8 6 This shows that free–fall acceleration of moon is 1/6th of earth’s free–fall acceleration. Q7. The bob of a simple pendulum 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 140 Q5, Try These] A. Mass of the bob = 100g= 0.1kg Length of the string = 1m Speed of the bob, v= 1.4m/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 SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 44

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 2.6 Asking Questions and Making Hypothesis Q1. What path will the moon take when the gravitational interaction between the moon and earth disappears? [Refer to TB page 139 Q7] 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. Q2. Can you think of two particles which do not exert gravitational force on each other? [Refer to TB page 139 Q8] A. No, everything in the universe exerts a force on everything else even if inﬁnitesimally small. SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 45

2.7 Application to Daily Life, Concern to Bio Diversity Q1. A boy is throwing balls into the air one by one in such a way that when the ﬁrst ball thrown reaches maximum height he starts to throw the second ball. He repeats this activity. To what height do the ball rise if he throws twice in a second? [Refer to TB page 140 Q7, Try These] A. The boy throws the second ball when the ﬁrst ball reaches its maximum height. He throws twice in a second. Time of ascent of ﬁrst ball is 1/2sec. After 1/2 sec, the ﬁrst ball starts to fall down and the second ball starts from ground. Let the distance travelled = s meters, Initial velocity = u m/s Time of ascent (t1 ) = 1/2 sec t1 = u ⇒ 1 = u ⇒ u= 5 m/s g 2 10 Distance travelled in 1/2 is S = ut + 1 gt2 2 S = 5 × (1/2) + 10×(1/2)2 2 = 5 + 5 2 4 = 15 m 4 They reach a height of 15/4 meters SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 46

Q2. 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 140 Q9, Try These] A. Both the apple and the earth exert equal forces but in opposite directions. Force of the apple on the earth is accelerating the Earth. Q3. A ball is dropped from a balloon going upto a speed of 5 m/s. If the balloon was at a height 60 m at the time of dropping the ball, how long will the ball take to reach the ground? [Refer to TB page 140 Q8, Try These] A. At t=0, the stone was going up with a velocity of 5 m/s. After that, it moves as free falling body, with downward acceleration ‘g’. If it reaches the ground at time t, s=–60 m, u=5 m/s, a=g= 10 m/s2 s=ut + 1/2 gt2 −60 = 5 × t + 1 × −10 × t2 2 −60 = 5t − 5t2 t2 –t–12 =0 t2 –4t+3t–12=0 t(t–4)+3(t–4) = 0 (t–4)(t+3)=0 t–4=0, t+3=0 t=4 t= –3 Time will not be negative. Hence t = 4s Therefore, the ball will reach the ground after 4s. SESSION 2. UNIVERSAL LAW OF GRAVITATION AND FREE FALL 47

SESSION 3 CENTRE OF GRAVITY 3.1 Mind Map SESSION 3. CENTRE OF GRAVITY 48

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