SSLC : SCERT Physics
Chapter 1: Sound Waves
Let's Assess
1. Which of the following statements is correct? a) Sound and light are transverse waves. b) Sound and light are longitudinal waves. c) Sound is a longitudinal wave and light is a transverse wave. d) Sound is a transverse wave and light is a longitudinal wave.
◦ Explanation: Sound travels through a medium forming alternating compressions and rarefactions, meaning particles vibrate parallel to the wave's direction, making it a longitudinal wave. Light waves, being electromagnetic waves, are transverse waves, where particles (fields) vibrate perpendicular to the direction of propagation.
2. The upper limit of frequency of sound that a bat can hear is 120 kHz. If so, what is the maximum wavelength of sound it can hear? Consider the speed of sound as 350 m/s.
◦ Given:
▪ Frequency (f) = 120 kHz = 120,000 Hz
▪ Speed of sound (v) = 350 m/s
◦ Formula: The relation between speed, frequency, and wavelength is v = fλ.
◦ Calculation:
▪ λ = v / f
▪ λ = 350 m/s / 120,000 Hz
▪ λ ≈ 0.0029 m
3. A graphic illustration of two waves travelling at a speed of 3.2 m/s is given. a) Find out the frequency, period, and wavelength of each wave.
◦ Wave (a) [Figure 1.31(a)]:
▪ Wavelength (λ): From the figure, one complete wave cycle (distance between two consecutive crests or troughs) is 4 m.
▪ Speed (v): 3.2 m/s
▪ Frequency (f): f = v / λ = 3.2 m/s / 4 m = 0.8 Hz
▪ Period (T): T = 1 / f = 1 / 0.8 Hz = 1.25 s
◦ Wave (b) [Figure 1.31(b)]:
▪ Wavelength (λ): From the figure, one complete wave cycle is 2 m.
▪ Speed (v): 3.2 m/s
▪ Frequency (f): f = v / λ = 3.2 m/s / 2 m = 1.6 Hz
▪ Period (T): T = 1 / f = 1 / 1.6 Hz = 0.625 s
4. Which of the following frequency can be heard by humans? a) 5 Hz b) 2000 Hz c) 200 kHz d) 50 kHz
◦ Explanation: For a person with normal hearing, the audible range of sound frequency is about 20 Hz to 20,000 Hz (20 kHz).
▪ 5 Hz is infrasonic (below 20 Hz).
▪ 2000 Hz falls within the audible range.
▪ 200 kHz and 50 kHz are ultrasonic (above 20,000 Hz).
5. A wave has a frequency of 2 kHz and a wavelength of 35 cm. How far does this wave travel in 0.5 s?
◦ Given:
▪ Frequency (f) = 2 kHz = 2000 Hz
▪ Wavelength (λ) = 35 cm = 0.35 m
▪ Time (t) = 0.5 s
◦ Step 1: Calculate the speed of the wave (v).
▪ v = fλ
▪ v = 2000 Hz × 0.35 m = 700 m/s
◦ Step 2: Calculate the distance travelled (d).
▪ d = v × t
▪ d = 700 m/s × 0.5 s = 350 m
6. What is the frequency of a wave that produces 50 crests and 50 troughs in 0.5 s?
◦ Explanation: A complete oscillation (or cycle) consists of one crest and one trough.
◦ Given:
▪ Number of oscillations = 50 (since 50 crests and 50 troughs)
▪ Time = 0.5 s
◦ Formula: Frequency (f) = Number of oscillations / Total time
◦ Calculation:
▪ f = 50 / 0.5 s = 100 Hz
7. Which of the following is different regarding the waves given in the figures 1.31 (a) and 1.31 (b)? (frequency, amplitude, wavelength)
◦ The source states that "An illustration of two waves of the same amplitude passing through a medium at the same time interval is given".
◦ From the analysis in question 3, Wave (a) has a wavelength of 4 m and a frequency of 0.8 Hz. Wave (b) has a wavelength of 2 m and a frequency of 1.6 Hz.
◦ Therefore, frequency and wavelength are different, while amplitude is the same.
8. The distance between two adjacent troughs of a transverse wave is 2 m. Find the frequency if its speed is 20 m/s.
◦ Explanation: The distance between two consecutive troughs (or crests) in a transverse wave is defined as its wavelength (λ).
◦ Given:
▪ Wavelength (λ) = 2 m
▪ Speed (v) = 20 m/s
◦ Formula: Frequency (f) = v / λ
◦ Calculation:
▪ f = 20 m/s / 2 m = 10 Hz
9. When sound passes through a medium, …………….. travels. (the particles in the medium / the wave / the source of sound / the medium)
◦ Explanation: Wave motion is defined as the continuous propagation of energy from one part to the other parts through oscillations. The disturbance (wave) spreads, but the particles in the medium only oscillate about their equilibrium position without significant displacement along with the disturbance.
10. Two pith balls are suspended near the two prongs of a tuning fork fixed on a table so as to touch the prongs. A person plays a piano sitting near this system. a) In this case the pith balls move slightly. What is the reason? (forced vibration / echo) * Reason: Forced vibration. This occurs when an object vibrates due to an external vibrating object. Here, the sound from the piano causes the table and tuning fork to vibrate. b) While playing certain notes on the piano, the pith balls are thrown to a maximum distance. Which phenomenon is responsible for this? (reverberation / resonance) * Phenomenon: Resonance. If the natural frequency of the forcing object (piano note) matches that of the forced object (tuning fork), they vibrate with maximum amplitude.
Extended Activities
1. Plan an activity that illustrates the resonance of sound.
◦ Activity: Hacksaw Blade Resonance Experiment
1. Setup: Use a device with two sets of three identical hacksaw blades (e.g., lengths 13 cm and 17 cm) fixed between two wooden blocks. Label one blade 'A'.
2. Procedure: Excite blade A by tapping it gently with your finger.
3. Observation: Observe that other blades will vibrate. Specifically, any other blades (e.g., C and E in Figure 1.5(a) if they match blade A) whose natural frequency is equal to that of blade A will vibrate with the maximum amplitude.
4. Conclusion: This demonstrates resonance, the phenomenon where objects vibrate with maximum amplitude when the natural frequency of the forcing object (blade A) is equal to the natural frequency of the forced object (blades C and E).
2. Prepare and present a seminar paper on the topic: 'Ultrasonic Waves and their Applications.'
◦ Definition: Ultrasonic waves are sound waves with frequencies greater than 20,000 Hz (20 kHz), which are beyond the human hearing range.
◦ Properties: They travel through media and can reflect off surfaces.
◦ Applications:
▪ Navigation and Hunting (e.g., Bats): Bats use ultrasonic waves to navigate and locate prey in the dark.
▪ Medical Applications (Diagnosis and Treatment):
• Crushing Kidney Stones: Used to break small stones in the kidneys.
• Physiotherapy: Utilized for therapeutic purposes.
• Ultrasonography: Used to create images of internal organs like kidneys, liver, gallbladder, and uterus. The waves travel through tissues, reflect off areas of varying density, and are converted into electrical signals to form an image.
▪ Cleaning: Effective for cleaning irregularly shaped machine parts, spiral tubes, and electronic components.
▪ SONAR (Sound Navigation and Ranging): A device that uses ultrasonic waves to determine the distance to underwater objects, such as the seabed or submarines. For example, by emitting an ultrasonic wave and measuring the time it takes to return after reflection, the distance to an underwater object can be calculated (e.g., distance to a rock = (speed × time) / 2).
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Chapter 2: Lenses
Let's Assess
1. The focal length of a convex lens is 20 cm. An object of height 3 cm is located at a distance of 60 cm from its optic centre on the optic axis. a) Calculate the height of the image. * Given: * Focal length (f) = +20 cm (for convex lens) * Object height (h_o) = +3 cm * Object distance (u) = -60 cm (object placed on the left, opposite to incident ray direction) * Lens Equation (1/f = 1/v - 1/u): * 1/20 = 1/v - 1/(-60) * 1/20 = 1/v + 1/60 * 1/v = 1/20 - 1/60 = (3 - 1)/60 = 2/60 = 1/30 * v = +30 cm (Image is formed on the other side, real) * Magnification Equation (m = h_i/h_o = v/u): * h_i / 3 cm = (+30 cm) / (-60 cm) * h_i / 3 = -0.5 * h_i = 3 × (-0.5) = -1.5 cm b) What are the characteristics of the image obtained? * Real: Since v is positive, the image can be projected on a screen. * Inverted: Since h_i is negative. * Diminished: Since |h_i| (1.5 cm) is less than |h_o| (3 cm). * Position: Between F and 2F on the other side of the lens (object at 3F, image at 1.5F).
2. The focal length of a lens is 20 cm. a) An object is placed 30 cm away from the lens. Calculate how far the screen should be placed to get a clear image. * Assumption: For a clear image to be formed on a screen, it must be a real image. Real images are formed by convex lenses. * Given: * Focal length (f) = +20 cm (for convex lens) * Object distance (u) = -30 cm * Lens Equation (1/f = 1/v - 1/u): * 1/20 = 1/v - 1/(-30) * 1/20 = 1/v + 1/30 * 1/v = 1/20 - 1/30 = (3 - 2)/60 = 1/60 * v = +60 cm * Result: The screen should be placed 60 cm from the lens on the other side. b) If the height of the object is 1.2 cm, what will be the height of the image appearing on the screen? * Given: * Object height (h_o) = +1.2 cm * v = +60 cm, u = -30 cm (from part a) * Magnification Equation (m = h_i/h_o = v/u): * h_i / 1.2 cm = (+60 cm) / (-30 cm) * h_i / 1.2 = -2 * h_i = 1.2 × (-2) = -2.4 cm * Result: The height of the image will be 2.4 cm (inverted).
3. The focal length of a convex lens is 100 mm. An object of height 15 mm is located 60 mm from the optic centre on its optic axis. a) Draw its ray diagram on a graph paper and find the position and height of the image. * Given: * Focal length (f) = +100 mm (convex lens) * Object height (h_o) = +15 mm * Object distance (u) = -60 mm * Ray Diagram Description: Since the object distance (60mm) is less than the focal length (100mm), the object is placed between the principal focus (F) and the optic centre (O) of the convex lens. 1. A ray from the top of the object parallel to the principal axis passes through the principal focus (F) on the other side after refraction. 2. A ray from the top of the object passing through the optic centre (O) goes undeviated. 3. These refracted rays diverge. When traced backward, they appear to intersect at a point on the same side of the lens as the object. * Calculations: * Image Position (v) using 1/f = 1/v - 1/u: * 1/100 = 1/v - 1/(-60) * 1/100 = 1/v + 1/60 * 1/v = 1/100 - 1/60 = (3 - 5)/300 = -2/300 = -1/150 * v = -150 mm (Image is virtual, on the same side as the object) * Image Height (h_i) using m = h_i/h_o = v/u: * h_i / 15 mm = (-150 mm) / (-60 mm) * h_i / 15 = +2.5 * h_i = 15 × 2.5 = +37.5 mm * Characteristics: The image is Virtual, Erect, and Magnified (as h_i is positive and greater than h_o). b) Calculate the magnification if the distance to the object is 20 mm. * Given: * Focal length (f) = +100 mm (convex lens) * Object distance (u) = -20 mm * Calculate Image Position (v): * 1/100 = 1/v - 1/(-20) * 1/100 = 1/v + 1/20 * 1/v = 1/100 - 1/20 = (1 - 5)/100 = -4/100 = -1/25 * v = -25 mm * Calculate Magnification (m = v/u): * m = (-25 mm) / (-20 mm) = +1.25
4. Four statements are given regarding the image formed by a concave lens. Find and choose the correct answer. i. It will be diminished and inverted ii. It will be diminished and virtual iii. It will be magnified and virtual iv. It will be diminished and erect a) Only the second statement is true b) Only the first statement is true c) Second statement and fourth statements are true d) Only the third statement is true
◦ Explanation: A concave lens always forms an image that is virtual, erect, and diminished. Therefore, statements ii and iv are correct.
5. A concave lens has a focal length of 50 cm. What will be its power?
◦ Given:
▪ Focal length (f) = -50 cm = -0.5 m (for concave lens, focal length is negative)
◦ Formula: Power (P) = 1 / f (where f is in metres)
◦ Calculation:
▪ P = 1 / (-0.5 m) = -2 Dioptre (D)
◦ Answer: c) -2 D
6. Find the most appropriate statement related to a telescope. a) The objective lens has a shorter focal length and the eyepiece lens has a longer focal length. b) The objective lens has a longer focal length and the eyepiece has a shorter focal length. c) Objective lens and eyepiece lens are concave lenses. d) Objective lens will be concave lens and eyepiece lens will be convex lens.
◦ Explanation: In a refracting telescope, the objective lens is designed to have a longer focal length and greater aperture, while the eyepiece has a shorter focal length and lesser aperture. Both are typically convex lenses.
7. When an object is placed in front of a lens, the image formed is inverted. a) Is it real or virtual? * If an image is inverted, it is always real. (Virtual images are always erect). b) What will you do if you want another image of this obtained image to be real, erect and of the same size? * The provided sources do not explicitly describe a single-lens setup that can convert an inverted, real image into a real, erect, and same-size image. Generally, a real, erect image of the same size cannot be formed by a single lens from a real object.
8. When an object is placed at the principal focus of a lens, an image that is erect and diminished is obtained. a) What kind of lens is this? * A concave lens always forms virtual, erect, and diminished images, regardless of the object's position (except at infinity). When an object is placed at its principal focus, the image formed is virtual, erect, and diminished, located between F and the lens. b) Draw the ray diagram of the image formation. * Ray Diagram Description (for a concave lens with object at F): 1. Draw a concave lens and its principal axis, marking the principal focus (F) on both sides. Place the object at F on one side. 2. Draw a ray from the top of the object parallel to the principal axis. After refraction by the concave lens, this ray will appear to diverge from the principal focus (F) on the same side as the object. 3. Draw a second ray from the top of the object passing through the optic centre (O). This ray passes undeviated. 4. When the two refracted rays are extended backward, they intersect between F and O on the same side as the object. This intersection point forms the virtual, erect, and diminished image.
9. The image (IM) obtained when an object is placed in front of a lens is depicted. (Figure 2.28 shows an erect, magnified, virtual image.) a) If PQ is a lens in the figure, what type of lens does PQ represent? * The image formed is magnified and erect [Figure 2.28]. This type of image (virtual, erect, magnified) is characteristic of a convex lens when the object is placed between its principal focus (F) and the optic centre (O). b) Complete the ray diagram and find the position of the object. * Ray Diagram Completion: To obtain a virtual, erect, magnified image with a convex lens, the object must be placed between the principal focus (F) and the optic centre (O). 1. Trace a ray from the top of the image (M) backward through the optic centre (O) to the principal axis. This line should pass through the top of the object. 2. Trace a second ray from the top of the image (M) backward to the lens, such that after passing through the lens, it became parallel to the principal axis. This original ray (before refraction) must have passed through the principal focus (F) on the same side as the object. 3. The intersection of these two rays will reveal the position of the object, which will be between F and O. c) The height of the object is ........ than the height of the image (greater / lesser). * Since the image (IM) is magnified [Figure 2.28], the height of the object is lesser than the height of the image.
10. Match the items in the columns A, B and C appropriately.
◦ A | B | C
◦ Magnification | h_i / h_o | hi positive (for erect image)
◦ Power of lens | 1 / f | dioptre
◦ Inverted image | (related to v/u) | negative (magnification sign)
◦ Real image | (not directly in B) | hi negative (as real images are inverted)
◦ Erect image | (not directly in B) | hi positive
Extended Activities
1. You may know people who use spectacles for various purposes. Collect, tabulate and analyse information regarding the type of lens used in different types of spectacles, the power of lens, age of the users and the problems faced by them.
◦ This is a research and data collection activity that requires information from outside the provided sources. The sources indicate:
▪ Convex lenses (positive power) are used for correcting long sightedness (hypermetropia), where individuals struggle to see nearby objects clearly (near point beyond 25 cm). They are also used for presbyopia, an age-related loss of accommodation due to decreased efficiency of ciliary muscles.
▪ Concave lenses (negative power) are used for correcting short sightedness (myopia), where individuals struggle to see distant objects clearly (far point not at infinity).
2. Collect a transparent polythene bag. Fill it with water and tie to get it almost in the shape of a sphere. Use it as a convex lens to form various sized images of a burning candle.
◦ This is a practical demonstration activity, not a question to be solved. The activity aims to demonstrate that a spherical water-filled object can act as a convex lens due to the refraction of light through its curved surfaces, enabling it to converge light rays and form images, similar to a glass convex lens.
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Chapter 3: The World of Colours and Vision
Let's Assess
1. Find the most appropriate answer. Name the optical phenomena taking place when light rays pass through water droplets to form a rainbow. a) internal reflection b) refraction c) refraction and internal reflection d) none of these
◦ Explanation: A rainbow is formed due to the combined effect of refraction, dispersion, and internal reflection of sunlight within water droplets. Option (c) best represents the primary physical processes involved.
2. Which of the following pairs of colours can produce white light? a) magenta, blue b) yellow, green c) red, green d) magenta, green
◦ Explanation: When a secondary colour of light is combined with its complementary primary colour, they produce white light.
▪ Magenta is formed by combining Red + Blue primary colours of light.
▪ The primary colour missing from Magenta is Green.
▪ Therefore, Magenta + Green = White light.
3. When a ray of white light enters obliquely and passes through a prism i) does not undergo refraction ii) undergoes dispersion iii) undergoes dispersion and deviation iv) not subjected to any of these Choose the correct option. a) ii & iii b) iv c) i & iv d) none of these
◦ Explanation: When white light passes through a prism, it undergoes refraction, causing it to deviate towards the base of the prism. Simultaneously, it splits into its component colours (VIBGYOR) due to different wavelengths deviating by different amounts, a phenomenon called dispersion. Thus, both dispersion and deviation occur.
4. Fill in the blanks appropriately: a) Cyan colour + Red → white light (Cyan = Blue + Green. Blue + Green + Red = White) b) Blue colour + Yellow → white light (Yellow = Red + Green. Blue + Red + Green = White) c) Magenta colour + green colour → white light (Magenta = Red + Blue. Red + Blue + Green = White) d) Magenta colour + cyan colour + yellow colour → Dark (These are primary dyes, and their combination results in dark/black).
5. Give scientific explanations based on scattering for the following: a) Red light is used for emergency lamps. * Explanation: Red light has the longest wavelength among the colours in the visible spectrum. Light with longer wavelengths undergoes less scattering when it encounters atmospheric particles. This allows red light to travel a greater distance through the atmosphere without significant reduction in intensity, making it highly visible and suitable for warning signals. b) The sky of the moon appears dark even during the day. * Explanation: The moon has no atmosphere. Scattering of light, which makes the sky appear blue on Earth, requires particles in a medium (like air) to scatter sunlight. Since there are no atmospheric particles on the moon, no scattering occurs, and the sky remains dark even when the sun is shining. c) The deep sea appears blue. * Explanation: Water molecules and suspended particles in the sea scatter shorter wavelengths of light (like blue) more effectively than longer wavelengths (like red and yellow). As sunlight penetrates deeper into the water, the red and yellow components are absorbed, while the blue light is scattered and reflected towards the observer, making the deep sea appear blue.
6. Complete the path of a light ray falling on the glass prism (Fig. 3.31).
◦ Description:
1. When a ray of light (e.g., white light) enters the prism from air, it undergoes refraction and deviates towards the base of the prism.
2. If it's white light, it will also undergo dispersion, splitting into its component colours (VIBGYOR). Violet light deviates the most (shortest wavelength), and red light deviates the least (longest wavelength).
3. When the light rays exit the prism back into the air, they undergo a second refraction, again deviating further towards the base, but the emergent rays will be separated into a spectrum of colours.
7. What are the radiations that are seen on either side of visible light in the electromagnetic spectrum? Write one use of the radiation with a shorter wavelength than visible light.
◦ Radiations on either side of visible light in the electromagnetic spectrum:
▪ Longer wavelength side: Infrared rays, Microwaves, Radio waves.
▪ Shorter wavelength side: Ultraviolet rays, X-rays, Gamma rays.
◦ Use of radiation with a shorter wavelength than visible light:
▪ Ultraviolet radiation (UV): Helps to produce vitamin D in the human body.
8. The near point of a person with hypermetropia is 40 cm. a) Can this person read the letters in a book held 25 cm away? * No. For a healthy eye, the least distance of distinct vision (near point) is 25 cm. For a person with hypermetropia (long sightedness), their near point is greater than 25 cm (in this case, 40 cm). They cannot see objects clearly if they are closer than their near point. b) Can this person see an object at infinity? * Yes. Long sightedness (hypermetropia) means a person can see distant objects clearly but has difficulty with nearby objects. The far point for a healthy eye and for hypermetropic eyes is considered to be infinity. c) How can this defect of the eye be rectified? * Long sightedness (hypermetropia) can be rectified using a convex lens with suitable power.
9. Water is colourless, but it appears white in waterfalls. Why?
◦ Explanation: In a waterfall, water is broken into countless tiny droplets and creates numerous air bubbles. These small water surfaces and air bubbles cause the incident white light to undergo multiple reflections and scattering in all directions. When all component colours of white light are scattered equally (which happens when the scattering particles are larger than the wavelength of light), the combined effect is perceived as white.
10. Based on the colour of illumination in the room, how can the coloured objects given in the table be seen? Complete the table.
◦ Principle: An opaque object reflects light of its own colour and colours associated with adjacent wavelengths, absorbing all others. A filter transmits its own colour and its component colours.
◦ Table Completion:
▪ Colour of the object | Light in the room | Colour of the object seen
▪ Green | Red | Dark (Green objects reflect green, absorb red)
▪ Blue | Green | Dark (Blue objects reflect blue, absorb green)
▪ Red | Yellow | Red (Yellow light = Red + Green. Red objects reflect red, absorb green)
▪ Magenta | Green | Dark (Magenta objects = Red + Blue. They reflect red and blue, absorb green)
▪ Cyan | Yellow | Green (Cyan objects = Blue + Green. They reflect blue and green, absorb red. Yellow light = Red + Green. So, only green is reflected)
▪ Yellow | Magenta | Red (Yellow objects = Red + Green. They reflect red and green, absorb blue. Magenta light = Red + Blue. So, only red is reflected)
Extended Activities
1. Prepare a note on the differences between colours of light and dyes.
◦ Colours of Light (Additive Primary Colours - RGB):
▪ Primary Colours: Red, Green, Blue (RGB).
▪ Mixing Principle: When these primary colours of light are combined, they add up to produce other colours.
▪ Result of Full Combination: Mixing all three primary colours of light (Red + Green + Blue) in equal intensity produces White light.
▪ Secondary Colours: Formed by mixing two primary colours (e.g., Red + Green = Yellow; Red + Blue = Magenta; Blue + Green = Cyan).
▪ Complementary Pairs: A primary colour and the secondary colour formed by the other two primary colours combine to form white light (e.g., Yellow + Blue = White).
◦ Colours of Dyes/Pigments (Subtractive Primary Colours - CMY):
▪ Primary Colours: Cyan, Magenta, Yellow (CMY).
▪ Mixing Principle: Dyes work by absorbing certain wavelengths of light and reflecting/transmitting others. When dyes are mixed, they subtract colours from white light.
▪ Result of Full Combination: Mixing all three primary dyes (Cyan + Magenta + Yellow) produces Dark (or black). This is because collectively they absorb almost all wavelengths of light.
▪ Secondary Colours: Formed by mixing two primary dyes (e.g., Cyan + Yellow = Green; Cyan + Magenta = Blue; Yellow + Magenta = Red).
▪ Application: Primary dyes are used in painting and printing.
2. Construct Newton's colour disc and operate.
◦ This is a practical activity, not a question to be solved.
◦ Construction: Paint a circular disc with the seven colours of the visible spectrum (VIBGYOR) in their correct proportions and sequence.
◦ Operation & Observation: When the disc is rotated very rapidly, it appears almost white.
◦ Reason: This phenomenon is explained by the persistence of vision. The human eye retains the visual impression of an object for about 1/16th of a second. When the disc spins quickly, the different colours pass across the retina faster than the persistence time, causing the brain to combine all the colour impressions, resulting in the perception of white light.
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Chapter 4: Magnetic Effect of Electric Current
Let's Assess
1. A conducting wire AB is bent into a loop as shown in the figure. A battery is connected to the ends of the conductor. a) When the switch is turned on, find the direction of the magnetic field around the conductor at points A and B. * Requires Figure 4.29 (not provided, but implied setup). Assuming AB is a straight segment of the loop and the current flows, say, upwards through A and downwards through B. * The direction of the magnetic field around a current-carrying conductor can be found using the Right Hand Thumb Rule. * Right Hand Thumb Rule: If you imagine holding the conductor with your right hand such that your thumb points in the direction of the electric current, then your curled fingers will indicate the direction of the magnetic field lines around the conductor. * Therefore, the magnetic field lines will form concentric circles around the wire segments A and B. The specific direction (clockwise or anticlockwise) at points A and B would depend on the direction of current in each segment as determined by the battery connection and the orientation of the loop. b) State the law used for this. * The law used is the Right Hand Thumb Rule. c) Explain how to find the direction of the magnetic field in a conducting loop. * To find the direction of the magnetic field in a conducting loop (or coil/solenoid), you can apply the Right Hand Thumb Rule to each segment, or more generally for the entire loop: * Imagine curling the fingers of your right hand in the direction of the current flow around the loop. * Your thumb will then point in the direction of the North pole of the magnetic field produced by the loop. This also indicates the direction of the magnetic flux lines inside the coil (out of the "North" end and into the "South" end). * Alternatively, if the current viewed from one end is clockwise, that end will be the South pole (field lines go in). If the current is anticlockwise, that end will be the North pole (field lines come out).
2. The direction of the magnetic field around a current carrying conductor AB is marked. Find the direction of the electric current through the conductor and state the law that supports this.
◦ Requires Figure 4.30 (not provided, but usually shows magnetic field lines). Assuming Figure 4.30 shows the magnetic field lines around AB as anticlockwise when viewed from top.
◦ Direction of Electric Current: By applying the Right Hand Thumb Rule in reverse: if the magnetic field lines are anticlockwise (as indicated by your curled fingers), your thumb must be pointing upwards. Therefore, the electric current through conductor AB is upwards.
◦ Law that supports this: Right Hand Thumb Rule.
3. In figure 4.29, AB is a conducting rod that is free to move. (Figure 4.29 not provided, but implying a setup for motor principle.) a) When the bell switch is turned on, in which direction will the metal rod AB move? * This requires the application of Fleming's Left Hand Rule. Without Figure 4.29, the specific direction of the magnetic field and current cannot be determined, thus the direction of motion cannot be given definitively. * Fleming's Left Hand Rule: Hold your thumb, first finger, and second finger of your left hand mutually perpendicular. If the first finger points in the direction of the magnetic field (N to S), and the second finger points in the direction of the electric current, then your thumb will indicate the direction of the force (motion) experienced by the conductor. b) What should be done to keep the direction of motion of the rod unchanged while changing the direction of the current? * If the direction of the current is reversed, the direction of the force (and thus motion) on the conductor will also be reversed. To keep the direction of motion unchanged, the direction of the magnetic field must also be reversed simultaneously. This means reversing the poles of the magnet causing the field.
4. What is the energy conversion that takes place in a moving coil loudspeaker?
◦ In a moving coil loudspeaker, electric energy is converted into sound energy. Electrical audio signals cause the voice coil to vibrate (mechanical energy), which then causes the diaphragm to vibrate and reproduce sound waves.
5. Name two devices that work on the principle of a motor.
◦ Two devices that work on the motor principle are:
1. Fans
2. Mixies
▪ Other examples: electric motor itself, cranes using electromagnets, Maglev trains, MRI scanners (though MRI scanners mainly use strong fields, not necessarily motors for their primary function).
6. Choose the correct statement regarding the magnetic polarity of a current carrying solenoid and write it down. a) If the current in one end of the solenoid is clockwise, then that end is north pole. b) If the current in one end of the solenoid is clockwise, then that end is south pole. c) If the current in one end of the solenoid is anticlockwise, then that end is south pole. d) None of the above.
◦ Explanation: When current flows through a solenoid, if you look at one end and the current appears to be flowing in a clockwise direction, that end acts as a South pole. If the current appears to be flowing in an anticlockwise direction, that end acts as a North pole.
7. Observe the diagram. a) Identify the device shown in the diagram. b) To rotate the armature in a clockwise direction, which terminal of the battery should be connected to the point X? c) What is the necessity of using a split ring commutator in this device?
◦ Requires Figure 4.32 (schematic diagram of an electric motor). a) Identify the device shown in the diagram.
▪ The device shown is an Electric Motor. b) To rotate the armature in a clockwise direction, which terminal of the battery should be connected to the point X?
▪ Assuming the magnetic field is from N (left) to S (right). For clockwise rotation, the force on side AB (top segment) should be downwards, and the force on side CD (bottom segment) should be upwards.
▪ Applying Fleming's Left Hand Rule: If the force on AB is downwards (Thumb), and the magnetic field is right (First Finger), then the current in AB (Second Finger) must be into the page (A to B).
▪ For the current to flow from A to B (into the page), point X must be connected to the positive terminal of the battery, as current flows from positive to negative. c) What is the necessity of using a split ring commutator in this device?
▪ The split ring commutator is crucial for an electric motor because it reverses the direction of the current through the armature coils after every half rotation (180 degrees). This ensures that the forces on the sides of the armature always act in the same rotational direction (e.g., always pushing one side down and the other side up), allowing the armature to rotate continuously in a single direction.
8. What is the function of the diaphragm in a moving coil loudspeaker? a) To amplify sound signals. b) To convert mechanical energy into sound waves. c) To separate high frequency sound signals. d) To increase the strength of the magnetic field.
◦ Explanation: In a moving coil loudspeaker, the voice coil vibrates due to the interaction with the magnetic field. This vibration is mechanically transmitted to the diaphragm, which then vibrates the surrounding air, thereby reproducing the sound waves.
9. A conductor is held above and parallel to a magnetic needle. a) What causes the magnetic needle to deflect when the switch is turned on? * When an electric current passes through a conductor, it creates a magnetic field around the conductor. This magnetic field interacts with the existing magnetic field of the magnetic needle, causing the needle to deflect due to the magnetic force exerted on it. This is known as the magnetic effect of electricity. b) Suggest two ways to reverse the direction of this deflection. * The direction of the magnetic field around the conductor depends on the direction of the current. Therefore, to reverse the deflection: 1. Reverse the direction of the electric current flowing through the conductor. 2. Change the relative position of the conductor (e.g., move the conductor from being above the magnetic needle to below it, or vice versa).
10. Observe the diagrams [Fig. 4.34 (a), (b)]. a) In both cases, does the north pole of the magnetic needle deflect clockwise or anticlockwise, when the switch is turned on? * Requires Figure 4.34 (a) and (b). Assuming current flows from left to right (A to B) in the conductor in both cases. * Figure 4.34 (a) (Conductor above needle): If the current is from A to B (left to right) and the conductor is above the magnetic needle, the north pole of the magnetic needle will deflect anticlockwise. * Figure 4.34 (b) (Conductor below needle): If the current is from A to B (left to right) and the conductor is below the magnetic needle, the north pole of the magnetic needle will deflect clockwise. * Answer: The deflection direction is different in each case. b) Justify your answer. * Justification: The direction of the magnetic field lines around a current-carrying conductor follows the Right Hand Thumb Rule. When the conductor is placed above the magnetic needle, the magnetic field produced by the current at the needle's position has a certain direction. When the conductor is placed below the needle, the magnetic field direction at the needle's position effectively reverses, causing the north pole of the needle to deflect in the opposite direction, even if the current direction remains the same.
11. AB is a copper wire. An acrylic sheet is kept above the south pole of a magnet. Two copper wires are placed above the sheet in such a way that they are parallel. A battery and a switch are connected to the wires. AB is placed above them.
◦ Requires Figure 4.35 (setup showing current in AB in a magnetic field). In Figure 4.35, the magnetic field lines from the South pole of the magnet will be directed upwards (into the South pole). Assuming the battery connection causes current in AB to flow from right to left. a) In which direction will the copper wire roll when the switch is turned on? (towards Q / towards P)
▪ Applying Fleming's Left Hand Rule:
• First Finger (Magnetic Field): Upwards (coming out of the South pole).
• Second Finger (Current): From right to left (across AB).
• Thumb (Force/Motion): With the first finger pointing up and the second finger pointing left, the thumb points towards P (away from the viewer).
▪ The copper wire will roll towards P. b) What happens if the direction of the current is reversed?
▪ If the direction of the current is reversed (from left to right), the direction of the force experienced by the conductor will also be reversed. Thus, the copper wire would roll towards Q.
12. Observe figure 4.36.
◦ Requires Figure 4.36 (schematic diagram of a moving coil loudspeaker). a) Identify the device shown in the schematic diagram.
▪ The device is a Moving Coil Loudspeaker. b) What is its working principle?
▪ Its working principle is the Motor Principle (or the magnetic effect of electric current). A current-carrying conductor (the voice coil) placed in a magnetic field experiences a force and vibrates. c) What is the energy conversion taking place in this device?
▪ Electric energy is converted into Sound energy. (Electrical signals are converted to mechanical vibrations, which then produce sound). d) Name the labelled parts.
▪ Based on Figure 4.28 (b) and text:
• Voice coil
• Paper diaphragm
• Field magnet e) Name another device that works on the same principle.
▪ An Electric Motor (or a fan, mixie).
13. A wooden block contains mercury between the north and south poles. A freely rotating toothed wheel is in contact with the mercury. When an electric current is passed through the wheel,
◦ Requires Figure 4.37 (Barlow's Wheel). a) in which direction is the wheel rotating? (clockwise direction / anti clockwise direction)
▪ Applying Fleming's Left Hand Rule: Assume the current flows from the center of the wheel radially outwards through the spokes into the mercury, and the magnetic field is horizontal (e.g., from N to S, left to right, across the wheel).
▪ If current is radially outwards (Second Finger) and magnetic field is horizontal (First Finger), the force on the spokes (Thumb) will be tangential, causing rotation. For a typical Barlow's wheel setup with current downwards into mercury and field right-to-left, the wheel rotates clockwise. b) Justify your answer.
▪ Justification: This device operates on the motor principle. When an electric current passes through the conducting toothed wheel, the portion of the wheel in contact with the mercury acts as a current-carrying conductor placed within the magnetic field of the poles. According to Fleming's Left Hand Rule, a force is exerted on this conductor. This force is tangential to the wheel, causing it to rotate continuously. The direction of rotation is determined by the relative directions of the current and the magnetic field.
Extended Activities
1. Construct and operate a device to prove the principle of a motor using two permanent magnets, a piece of copper wire, conducting wires, and a cell.
◦ This is a practical demonstration activity, not a question to be solved. The source provides guidance for this in Figure 4.24, involving a coil (armature) placed within a magnetic field (from a ring magnet) connected to a battery and switch. When the switch is turned on, the coil rotates due to the forces experienced by the current-carrying conductors in the magnetic field, demonstrating the motor principle.
2. Dismantle a scrap loudspeaker. Identify its parts and arrange them on a paper with labels. Explain why the voice coil in it is very thin.
◦ This is a practical activity, not a question to be solved.
◦ Parts to Identify (from source): Voice coil, paper diaphragm, field magnet.
◦ Explanation for why the voice coil is very thin: While the source doesn't explicitly state why it's thin, we can infer from its function:
▪ The voice coil needs to vibrate rapidly and precisely in response to varying electrical audio signals to reproduce sound accurately.
▪ A thin (and thus light) voice coil has lower inertia, allowing it to respond quickly to high-frequency changes in the electrical signal. This ensures that the diaphragm, to which it's attached, can produce faithful sound reproductions across a wide frequency range.
▪ Additionally, a thin wire allows for a greater number of turns within a compact space, which can increase the magnetic field strength and the force experienced by the coil for a given current, enhancing its sensitivity
