whatspp group Career Opportunities admin USMAN BHATTi
Unit No. 10
SIMPLE HARMONIC MOTION & WAVES 1. 2. 3. 4. 5.
Answers Keys A 6. B 7. C 8. A 9. D
A B C B
QUICK QUIZ AND POINT TO PONDER “Quick Quiz” Q1. What is the displacement of an object in SHM, When the kinetic and potential energy are equal? Ans: When the displacing of an object in SHM is equal to 70% of the amplitude than the kinetic and optional energies equal. i.e. x = 0.7xo
Displacement CHECK YOUR UNDERSTANDING Tell whether or not these motions are example of simple harmonic motion? (a) Un and down motion of leaf in water pond Ans: It is an example of SHM. (b) Motion of a ceiling fan. Ans: It is not an example of SHM. (c) Motion of hands of clock. Ans: It is not an example of SHM. (d) Motion of Plucked string fixed at both its ends. Ans: It is an example of SHM. (e) Motion of honey bee. Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Ans: It is not an example of SHM. “Quick Quiz” Q2. Do mechanical waves pass through vacuum, that is, empty space? Ans: No, Mechanical waves cannot pass through vacuum because me mechanical waves are material waves and always require some medium for their propagation. “Quick Quiz”
Q3. What do the dark and bright fringes on the screen of ripple tank represent? Ans: The dark and bright fringes on the screen of ripple tank represent the crests and troughs of transverse waves. The crests appear as bright fringes and trough appear as dark fringes on the screen. Activity: 1. What happens to the angle of refraction when the water waves pass from deep to shallow part of the water? Ans: The angle of refraction decreases when water waves from deep to shallow part of the water? 2. Do the magnitudes of angle of incidence and angle of refraction equal? Ans: No, the magnitudes of angle of refraction are not equal. Conceptual Questions: Q1: If the length of a simple pendulum is doubled what will be the change in its time period? Ans: The formula of for the time period of simple pendulum is. T 2
1 g
If the length is doubled then l1 = 2l, so the new time period will be T 2
l g
T 2
2l g
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T 2 2
l g
T 2 2
l g
T 2T From the above equation it is clear that on doubling the length of simple pendulum time period of simple pendulum increases by 2 times. Q2: A ball is dropped from a certain height onto the floor and keeps bouncing is the motion of the ball simple harmonic? Explain. Ans: A ball is dropped from a certain height onto the floor which keeps bouncing does not
execute SHM because it does not fulfill the necessary condition of SHM.
AS WE KNOW THAT F
x
This relation is not true for the motion of a bouncing Ball. Q3:
A student performed two experiments with a simple pendulum. He / She used this bobs of different masses by keeping other parameters constant. To his / her astonishment the time period of the pendulum did not change, why?
Ans: We can find the time period of simple pendulum by formula give below: T 2
l g
From the above equation it is dear that time period of simple pendulum does not depend on the mass of the bob. So, on changing the mass of the bob time period remains unchanged. Q4:
What types of waves do not require any material medium for their propagation?
Ans: Electromagnetic waves do require any material medium for their propagation, like radio waves, light waves, x – rays etc. Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Q5:
Plane waves in the ripple tank undergo refraction when they move from deep to shallow water. What change occurs in the speed of the wave?
Ans: As we know that V f
It is clear from the above equation that speed of waves depends upon wavelength. When plane waves in the ripple tank undergo they move from deep to shallow water and their wavelength decreases. Hence, the speed of waves also decreases.
Numerical Problems Q.10.1: The time period of a simple pendulum is 2s. What will be its length on Earth? What will be its length on the moon if gm =ge/6? Where ge = l0ms-2. Data: Time period of simple pendulum = T = 2 sec g on earth = ge =10ms–2 g on moon = gm =1.67ms–2 Required: a) Length of pendulum on earth = le = ? b) Length of pendulum on moon = lm = ? Formula:
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T 2
l g
Solution: T 2
l g
Taking square on both sides T 2 4 2 l
l ge
T 2 ge . 4 2
........................................................(i)
a) On earth equ. (i) becomes: le
le
le
T 2 ge 4 2 4 10
43.14
2
40 4 9.86
le = 1.02m b) On moon equ. (i) becomes: T 2 gm 4 2 4 1.6 lm 4(3.14) 2
lm
lm
6.44 39.44
lm = 0.17m Result: a) Length of pendulum of earth = le = 1.02m b) Length of pendulum of moon= lm = 0.17m Q.10.2: A pendulum of length 0.99 m is taken to the moon by an astronaut. The period of the pendulum is 4.9 s. What is the value of g on the surface of the moon? Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Data: Length of pendulum on moon = lm = 0.99m Time period = T = 4.9sec Required: Value of g on moon = gm = ? Formula: T 2
l g
Solution: T 2
l gm
Squaring on both sides: T 2 4 2
l gm
g m 4 2
l T2
g m 43.14
2
0.99 4.92
g m 1.63ms 1
Result: Value of g of moon = gm = 1.63ms–2 Q.10.3: Find the time periods of a simple pendulum of 1 metre length, placed on Earth and on moon. The value of g on the surface of moon is l/6th of its value on Earth. Where ge is 10 ms–2 . Data: Length of pendulum = l = 1m Value of g on earth = ge = 10 ms–2 Value of g on moon = gm = ge / 6 = 10 /6 = 1.67ms–2 Required: a) Time period on earth = Te = ? b) Time period on moon = Tm = ?
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Formula: T 2
l g
Solution: We know that a) On earth Te 2
l ge
Te 23.14
1 10
Te 6.28 0.1
Te 6.28 0.316
Te 2 sec
b) On moon Tm 2
l gm
Tm 23.14
1 1.67
Tm 23.14 0.6172
Tm 4.9 sec
Result: a) Time period on earth = Te = 2 sec b) Time period on moon = Tm = 4.9sec Q.10.4: A simple pendulum completes one vibration in two seconds. Calculate its length when g = 10.0 ms2. Data: Time period = T = 2sec g = 10ms –2 Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Required: Length of pendulum = l = ? Formula: T 2
l g
Solution: We know that T 2
l g
Squaring on both sides T 2 4 2
l g
T 2g 4 2 4 10 l 2 43.14
l
l
40 4 9.86
l = 1.02m Result: Length of simple pendulum = l = 1.02m Q.10.5: If 100 waves pass through a point of a medium in 20 seconds, what is the frequency and the time period of the wave? If its wavelength is 6 cm, calculate the wave speed. Data: n = No of waves = 100 t = Time taken = 20 sec λ = Wavelength = 6cm = 6 / 100 = 0.06m Required: a) Frequency = f = ? Dedicated to My Sweet Father…… whatsapp group Career Opportunities
b) Time period = T = ? c) Wave speed = V =? Formula: n t a) 1 b) T f f
c) V f Solution: a) Frequency = ? f
No of Waves n Time taken t
100 20 f 5 Hz f
b) Time period = ? T
1 f
T
1 5
T=0.2sec
c) Wave speed= ? V = fλ = 5 × 0.06 = 0.3ms–1 Result: a) Frequency = f = 5Hz b) Time period = T = 0.25sec c) Waves speed of waves = V = 0.3ms–1
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Q.10.6: A wooden bar vibrating into the water surface in a ripple tank has a frequency 12Hz. The resulting wave has a wavelength of 3 cm. What is the speed of the wave? Data: Frequency = f = 12Hz λ = 3cm = 3/100 = 0.03 Required: Speed of waves = V = ? Formula: V = fλ Solution: V = fλ V = 0.03 × 12 V = 0.36ms–1 Result: Speed of waves = V = 0.36ms–1 Q.10.7: A transverse wave produced on a spring has a frequency of 190 Hz and travels along the length of the spring of 90 m, in 0.5s. a) What is the period of the wave? b) What is the speed of the wave? c) What is the wavelength of the wave? Data: Frequency = f = 190Hz Length = l = 90m Time taken = t = 0.5sec
Required: a) Time period = T= ? b) Speed of wave = v = ? c) Wave length = λ = ? Formula:
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1 f
a) T b) v
l t
c) v = fλ Solution: (a)
T
1 f
T
1 90
= 0.01sec T
1 5
(b)
v = l/t v = 90/0.5 v = 180ms–1 (c) v = fλ λ=v/f = 180 / 190 = 0.95m Result: a) Time period = T = 0.01sec b) Waves speed of waves = v = 180ms–1 c) Wave length = λ = 0.95m Q.10.8: Water waves in a shallow dish are 6.0 cm long. At one point, the water moves up a down at a rate of 4.8 oscillations per second. a) What is the speed of the water waves? b) What is the period of the water waves? Data: Length of wave = λ = 6cm = 6 /100 = 0.06m Frequency = f = 4.8Hz
Required: a) Speed of waves = v = ? Dedicated to My Sweet Father…… whatsapp group Career Opportunities
b) Time period of waves = T = ? Formula: a) V = fλ b) T = 1/f Solution: a) v = fλ v = 4.8 × 0.06 v = 0.29ms–1 b)
T=1/f T = 1 / 4.8 T = 0.21sec Result: a) Velocity = 0.29ms–1 b) Time period = 0.21 sec Q.10.9: At one end of a ripple tank 80 cm across, a 5 Hz vibrator produces waves whose wavelength is 40 mm. Find the time the waves need to cross the tank. Data: Length = l = 80cm = 80 /100 = 0.8m Frequency = f = 5Hz Wave length = λ = 40mm = 40 / 1000 = 0.04m Required: Time = t = ? Formula: l = vt
Solution: We know that v = fλ = 5 × 0.04 Dedicated to My Sweet Father…… whatsapp group Career Opportunities
= 0.2ms–1 Now; l = vt 0.2 × t = 0.8 t=
0.8 0.2
t = 4sec Result: Time taken = t = 4sec Q.10.10: What is the wavelength of the radio waves transmitted by an FM station at 90 MHz. where 1M = 106 , and speed of radio wave is 3 x 108 ms–1 . Data: Frequency = f = 90MHz = 90 × 106Hz Speed = v = 3 × 108ms–1 Required: Wave length = λ =? Solution: We know that v f
v f
3 108 90 106 λ = 3.33m
Result: Wave length = λ = 3.33m
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Unit No. 11
SOUND 1. 2. 3. 4. 5.
Answers Keys A 6. A 7. B C D
B C
QUICK QUIZ AND POINT TO PONDER “Quick Quiz” Q1. Indentify which part of these musical instruments vibrates to produce sound? a. Electric Bell Ans: Armature and hammer. b. Loud Speaker Ans: Diaphram c. Piano Ans: Piano strings d. Violin Ans: Violin Strings e. Flute Ans: Air particles in flute pipe. SELF ASSESSMENT 1. Explain how sound is produce by a school bell? Ans: When the hammer strikes the school bell it starts vibrating and hence produces sound. 2. Why sounds waves are called mechanical waves? Ans: Mechanical waves need medium for their propagation. Since sound waves also need a medium for their propagation, so they are also called mechanical waves. 3. Suppose you and your friends are on the moon. Will you be able to hear sound produced by your friend? Ans: We will not be able to hear any sound produced by my friend while standing on moon because there is no medium present at the surface of moon for the propagation of sound. “Quick Quiz” Q2. Why the voice of women is more shrill then that of men?
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Ans: Since the frequency of sound of women is higher than that of men so the voice of women is more shrill than that of men. . Q3. Which property of sound waves determines its (a) Loudness (b) Pitch? Ans: (a) Amplitude of vibrating body determines its loudness. (b) Frequency of sound determines its pitch. Q4. What would happen to the loudness of sound with increase in its frequency? Ans: Since loudness does not depends on frequency loudness will remain unchanged. Activity: 1. Develop an action plan to help you address any problem(s) with noise in you workplace considering the following points. a. Describe the problem (s). b. What are the sources of the problem(s)? c. Who are the people being affected? d. Your suggestions for the solutions. Ans: Problem: Industrial noise pollution. Sources of Problem: Metal fatigue in machinery. Affected People: Employees, workers and labours working in industry. Suggestions: Reduce nose to acceptable level by replacing old machinery with the new one. Reduce noise level by putting sound barriers. Make environment friendly for employees, workers and labours. Use hearing protection devices (ear plugs). Conceptual Questions: Q1: Why two can with a string stretched between them could be a better way to communicate than merely shouting through air? Ans: Two tin cane with a string stretched between them could be a better way to communicate than merely shouting through air because string is a solid and molecules in solids are closer than in liquids and gases, hence respond more quickly to a disturbance. Q2: We can recognize persons speaking with the same loudness from their voice. How is this possible?
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Ans: We can recognize person speaking with the same loudness from their voice due to difference in the quality of their sounds. Q3:
You can listen your friend round a corner, but you cannot watch him / her. Why?
Ans: We can listen to our friend round a corner without watching him/her due to the diffraction of sound around the corner. As sound waves travel through a medium in all directions and they spread or bend around the sharp edges or corners of obstacle (diffraction). Q4:
Why must the volume of a stereo in a room with wall-to-wall carpet be tuned higher than a room with a wooden floor?
Ans: The volume of a stereo in a room with wall-to-wall carpet be tune higher than in room with a wooden floor because reflection of sound is more prominent if the surface is rigid and smooth and less if the surface is soft and irregular. Since carpet is a soft porous material so it absorbs large amount of sound energy than wooden floor and thus quiet echoes. Q5:
A student says that the two terms speed and frequency of the wave refer to the same thing. What is your response?
Ans: No, speed and frequency of the wave do not refer to the same thing because speed is distance covered by wave per unit, time and frequency is number of waves passing through a point in unit time. Although the time factor in both quantities is common. Q6:
Two people are listening the same music at the same distance. They disagree on its loudness. Explain how this could happen?
Ans: Same music at the same distance. They disagree on its loudness because loudness of sound also depends upon the physical condition of ear of the listener. A sound appears louder to a person with sensitive or normal ear than to a person with defective ears. Q7:
Is there any difference between echo and reflection of sound?
Ans: 1.
Echo Reflection Incident sound wave must be bounced Incident sound wave can be bounced
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Q8:
back from a hand and rigid surface. back from a hand or soft surface. 2. Echo can be observed only when Reflection of sound many occur of any minimum distance between listener distance. and reflective surface is 17m. 3. Echo can be observed only when time No such requirement is necessary for difference between the original sound reflection of sound. and echo is 0.1 sec. Will two separate 50 dB sond together constitute a 100 dB sound? Explain.
Ans: Ye, two separate 50dB sound together can constitute a 100 dB sound when there is constructive interference between them and constructive interference occurs when two sounds having same frequency and moving simultaneously along the some direction interact with each other to enhance their effects. Q9:
Why ultrasound is useful in medical?
Ans: Due to the following characteristics of ultrasound they are usefully utilized in medical field: They carry more energy than audible sound waves. They have higher frequency than audible sound waves. They have very small wavelength than audible sound waves.
Numerical Problems Q.11.1: A normal conversation involves sound intensities of about 3.0 × 10-6 Wm-2. What is the decibel level for this intensity? What is the intensity of the sound for 100 d B? Data: Intensity of sound = I = 3.0 × 10–6 Wm–2 Intensity of faintest sound = Io = 10–12 Wm–2 Sound level = S.L. = L – Lo = 100dB Required a) Intensity level = L – Lo = ? b) Intensity = I = ? Formula: S.L. = L – Lo = 10 log
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I dB Io
Solution: We know that a) L Lo 10 log 10 log
I (dB) Io
3 10 6 10 12
= 10log (3.0 × 106)dB = 10 × 6.47dB = 64.7dB I
b) L Lo 10 log I
o
I 100 10 log 12 10
Taking antilog on both sides I
Antilog10 = 10 12 So, I = 0.01 w/m2 Result: a) Sound level = L – Lo = 64.8dB b) Intensity = I = 0.01 w/m2 Q.11.2: If at Anarkali bazar Lahore, the sound level is 80 dB, what will be the intensity level of sound there? Data: Sound level = S.L. = L – Lo = 80dB Intensity of faintest audible sound = Io = 10–12 w/m2 Required Intensity of sound = I=? Formula: I L – Lo = 10 log dB Io Solution: We know that Dedicated to My Sweet Father…… whatsapp group Career Opportunities
I L – Lo = 10 log dB Io I 80 10 log 12 10 80 log 1012 I ) 10
Taking antilog on both sides Antilog 8 = Antilog [log(1012 × I)] 108 = 1012 × I I
108 1012
I = 10–4 Wm–2 Result Intensity of sound = I = 10–4 Wm–2 Q.11.3: At a particular temperature, the speed of sound in air is 330 ms'1. If the wavelength of a note is 5 cm, calculate the frequency of the sound wave. Is this frequency lies in the audible range of the human ear? Data: Speed of sound = v = 330 ms–1 Wavelength = λ = 5cm = 5/100 = 0.05m Required Frequency = f = ? Formula: v = fλ
Solution: We know that V = fλ f f
v
330 0.05
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= 6600Hz = 6.6 × 103Hz Result Frequency = f = 6.6 × 103Hz Yes the frequency lies in the audible range of human ear Q.11.4: A doctor counts 72 heartbeats in 1 min. Calculate the frequency and period of the heartbeats. Data: No. of heartbeats = n = 72 Time = 1 mint = 60 seconds Required a) Frequency = f = ? b) Time period = T = ?
Formula: n t 1 b) T f a) f
Solution: We know that a) f
f
No. of heartbeat Time taken
72 60
f = 1.2Hz b) T
T
1 f
1 1 .2
T = 0.833sec Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Result: a) f = 1.2Hz b) T = 0.833sec Q.11.5: A marine survey ship sends a sound wave straight to the sea bed. It receives an echo 1.5 s later. The speed of sound in sea water is 1500 ms'1. Find the depth of the sea at this position. Data: Time taken = t = 1.5sec Speed = V = 1500 ms–1 Required Depth of seabed = h = ? Formula: S = vt Solution: We know that S = vt = 1500 × 1.5 = 2250m The echo distance must be half S 2 2250 h 2
h
h 1150m
Result: Depth of sea water = h =1150m Q.11.6: A student clapped his hands near a cliff and heard the echo after 5 s. What is the distance of the cliff from the student if the speed of the sound, v is taken as 346ms –1 ? Data: Time taken = t = 5Sec Speed = V = 346 ms–1 Required Distance = h= ? Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Formula: S = vt Solution: We know that S = vt = 346× 5 = 1730m
The echo distance must be half S 2 1730 h 2
h
h 865m
Result: Depth of sea water =h= 865m Q.11.7: A ship sends out ultrasound that returns from the seabed and is detected after 3.42sec. If the speed of ultrasound through seawater is 1531 ms'1, what is the distance of the seabed from the ship? Data: Time taken = t = 3.42sec Speed = V = 1531 ms–1 Required Depth of sea water = h = ? Formula: S = vt Solution: We know that S = vt = 153 × 3.42 = 5236.02m The echo distance must be half
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S 2 5236.02 h 2
h
h 2618m
Result: Depth of seabed =h= 2618 m Q.11.8: The highest frequency sound humans can hear is about 20,000 Hz. What is the wavelength of sound in air at this frequency at a temperature of 20° C? What is the wavelength of the lowest sounds we can hear of about 20 Hz? Assume the speed i sound in air at 20°C is 343 ms–1. Data: Highest frequency = f1 = 20000Hz Lowest frequency = f 2= 20Hz Speed of sound =v= 343ms–1 Required a) Wavelength at f1 =λ1= ? b) Wavelength at f2 = λ2= ? Formula: v =fλ Solution: We know that v =fλ a ) 1
1
V f1
343 20000
= 1.7 × 10–2m b) 2
2
v f2
343 20
= 1.7 × 10 –4m Result: Dedicated to My Sweet Father…… whatsapp group Career Opportunities
a) λ1 = 1.7 × 10–2 m b) λ2 = 1.7 × 10–4 m Q.11.9: A sound wave has a frequency of 2 kHz and wavelength 35 cm. How long will it take to travel 1.5km? Data: Frequency = f =2kHz=2 × 103Hz Wavelength = λ = 35cm=35/100 = 0.35m Distance = s=1500m Required Time = t = ? Formula: S = vt Solution: We know that v =fλ = 2 × 103×0.35 m v= 700ms–1 V f2r
As S = vt 1500 = 700 × t t
1500 700
t = 2.1sec Result: Time = t = 2.1 seconds
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Unit No. 12
GEOMETRICAL OPTICS 1. 2. 3. 4. 5.
Answers Keys C 6. C 7. B 8. C 9. B 10.
D B A B B
QUICK QUIZ AND POINT TO PONDER “Can You Tell” Q1. In this picture you can see clearly the image of a lion formed inside the sound water. Can you tell which phenomenon of physics is involved here? Ans: Reflection is the phenomenon of physics which is involved here. Q2. In large shopping centers, convex mirrors are used for security purpose. Do you know why? Ans: Convex mirrors are used for security purposes in large shopping centers because they can pick images and reflect them to other convex mirrors placed at strategic positions. Curved surfaces of the convex mirrors allow the viewer to see a much wider range of vision than a normal mirror. Q3. Why the position of fish inside the water seems to be at less depth than that of its actual position? Ans: The position of a fish inside the water seems to be at less depth than that of its actual positions due to refraction of eight. “Self-Assessment” Q: Will the bending of the light be more or less for a medium with high refractive index? Ans: The bending of the light will be more for medium with high refractive index because these two are in direct relation with each other. “Self-Assessment” Q: Where a pen is placed in front of o convex lens if the image is equal to the size to the pen? What will be the power of the lens diapers?
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Ans: When a pen is placed at double distance of foal length i.e. at 2F then the image of the pen is equal to the actual size of the pen. Power of lens
1 Focal Length
Power of lens
1 diopter f
Quick Quiz Q1. How the size of the Pupil of our eye will change: a. In dim light? b. In bright light? Ans: (a) In dim light pupil becomes enlarged and allows the maximum light to come in the eye. (b) In bright light pupil contracts to control the intensity of light. Conceptual Questions: Q1: A man raises his left hand in a Plane mirror; the image facing him is raising his right hand. Explain why? Ans: A man raises his left hand in a plane mirror the image facing him is raising his right hand because light rays are reflected in a mirror, causing us to see a inverted image. Q2: In your own words, explain why light waves are refracted at a boundary between two materials? Ans: Light waves are refracted at the boundy between two material because the density of the material is going to be changed at this point. When light waves enter from one transparent medium to another, due to change in density speed of light changes. The speed of light is different materials due to difference in their densities. The greater the optical density of the medium, the slower the speed of light and vice versa. . Q3:
Explain why a fish under water appears to be at a different depth below the surface than it actually is? Does it appear deeper or shallower?
Ans: A fish under water appears to be at a different depth below the surface than it actually is due to refraction of light. It will appear shallower because apparent depth is always less than that of real depth. Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Q4:
Why or why not concave mirrors are suitable for make up?
Ans: Concave mirrors are suitable for make up because when a person stands between the principal focus and the pole of a concave mirror, he/she sees on enlarged, erected and virtual image which reveals the mirror features of his/her face.
Q5:
Why is the driver’s side mirror in many cars convex rather than plane to concave?
Ans: The driver’s side mirror in many cars convex rather than plane or concave because the image formed by the convex mirror is always virtual and erect. Hence, the driver get a wider rear view of the automobiles behind and to see the vehicles following them. Q6:
when an optician’s testing room is small, he used a mirror to help him test the eye sight of his patients. Explain why?
Ans: if an optician’s testing room is small, he uses a mirror to help him test the eye sight of his patients to increase the test distance. When an optician performs an eye sight the ‘Shellen Charts’ (letter charts) are not only used to measure visual acuity (sharpness). Distance acuity is measured more often then near acuity because at a long distance accommodation is relaxed so that refraction can be more accurate at a longer test distance, the effect of small changes in the subjects positions is less important and can be ignored. So in smaller room, the use of mirror is recommended to increase the test distance.
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(Numerical Problems) Q.12.1: An object 10.0 cm in front of a convex mirror forms an image 5.0 cm behind the mirror. What is the focal length of the mirror? Data: Distance of object = p = 10cm Distance of image = q = –5 cm Required Focal length = f = ? Formula: 1 1 1 f p q
Solution: 1 1 1 f 10 5 1 1 f 10
f 10cm
Result: f= -10 cm Q.12.2: An object 30.0 cm tall is located 10.5 cm from a concave mirror with focal length 16.0cm. (a) Where is the image located? (b) How high is it? Data: Object height = ho = 30cm Distance of object =p= 10.5cm Focal length = f=16cm Required a) Distance of image = q = ? b) Image height = hi = ? Formula: a)
1 1 1 f p q
b)
h1 q ho p
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Solution: We know that 1 1 1 f p q
a)
1 1 1 q 16 10.5 1 10.5 16 q 161.5 1 5 .5 q 168
168 5.5
q
q = -30.54cm We know that b)
hi q ho p
q ho p 30.54 hi 30 10.5 hi
hi= 87.26cm Result: a) q = -30.54cm b) hi = 87.26cm Q.12.3:
An object and its image in a concave mirror are of the same height, yet inverted, when the object is 20.0 cm from the mirror. What is the focal length of the mirror? Data: Distance of object = p = 20cm Distance of image = q = 20cm Required:
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Focal length = f =? Formula: 1 1 1 f p q
Solution: We know that 1 1 1 f p q 1 1 1 f 20 20 1 2 f 20 1 1 f 10
f = 10cm Result: f = 10cm Q.12.4:
Find the focal length of a mirror that forms an image 5.66cm behind a mirror of an object placed at 34.4 cm in front of the mirror. Data: Distance of object = p = 34.4cm Distance of image = q = – 5.66 cm Required: Focal length = f = ? Formula: 1 1 1 f p q
Solution: We know that 1 1 1 f p q
1 1 1 f 34.4 5.66
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1 0.029 0.177 f 1 0.148 f
f = – 6.777cm Result: f = – 6.777cm Q.12.5: An image of a statue appears to be 11.5 cm behind a convex mirror with focal length 13.5 cm. Find the distance from the statue to the mirror. Data: Distance of image = q = – 11.5 cm Focal length = f =13.5cm Required: Distance of object = p = ? Formula: 1 1 1 f p q
Solution: We know that 1 1 1 f p q 1 1 1 p f q
1 1 1 p 13.5 11.5 1 1 1 p 13.5 11.5 1 25 p 155.25
P = 6 .21cm Result:
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Distance of object = p = 6.21cm Q.12.6:
An image is produced by a concave mirror of focal length 8.70 cm. The object is 13.2 cm tall and at a distance 19.3 cm from the mirror, (a) Find the location and height of the image, (b) Find the height of the image produced by the mirror if the object is twice as far from the mirror. Data: Object length = ho = 13.2cm Distance of object = p1 = 19.3cm Focal length = f = 8.70cm Required: a) i) Location of image = ? ii) Height of image = q = ? b) hi = ? if p2 = 2p1 Formula: a)
1 1 1 f p q
hi q ho p1 b)
hi q ho p2
Solution: We know that a) i )
1 1 1 f p q
1 1 1 q f P 1 1 1 q 8.7 19.3 1 19.3 8.7 q 167.9 1 10.6 q 167.9
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q
167.9 10.6
q = 15.83cm a) ii)
hi q ho p1
Putting the values 15.84 13.2 19.3 209.09 hi 19.3 hi 10.83cm hi
b) When the object become twice p2 = 19.3 ×2 p2 = 38.6cm hi q ho p2
hi
15.84 13.2 19.3
hi = 5.42cm Result: a) i) q = 15.83cm ii) Image height = 10.83cm b) Image height = 5.42cm Q.12.7: Nabeela uses a concave mirror when applying makeup. The mirror has a radius of curvature of 38.0 cm. (a) What is the focal length of the mirror? (b) Nabeela is located 50 cm from the mirror. Where will her image appear? (c) Will the image be upright or inverted? Data: Radius of the curvature = R = 38cm Distance of object = p = 50 cm Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Required: Focal length = f = ? Distance of image = q = ? Formula: R 2 1 1 1 b) f p q a) f
Solution: We know that a) f
f
R 2
38 2
f = 19cm using the formula b)
1 1 1 f p q
1 1 1 q 19 50 1 50 19 q 950
q
950 31
q = 30.64cm Result: a) f = 19cm b) q = 30.64cm c) Because magnification m = q/p Is positive then image will be upright.
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Q.12.8:
An object 4 cm high is placed at a distance of 12 cm from a convex lens of focal length 8 cm. Calculate the position and size of the image. Also state the nature of the image. Data: Height of object = ho = 4cm Distance of object = p = 12 cm Focal length = f = 8cm Required: a) Size of image = q = ? b) Position of image = hi = ? c) Nature of image = ? Formula: a)
1 1 1 f p q
b)
hi q ho p
Solution: Using the formula a)
1 1 1 f p q
1 1 1 q P q 1 1 1 q f P 1 1 1 q 8 12 1 12 8 q 196
1 4 q 96
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q
96 4
q = 24cm b)
hi q ho p
24 4 12 hi 8cm
hi
Result: a) q = 24cm b) hi = 8cm c) Image is real, inverted & magnified. Q.12.9:
An object 10 cm high is placed at a distance of 20 cm from a concave lens of focal length 15 cm. Calculate the position and size of the image. Also, state the nature of the image. Data: Size of object = ho = 10cm Distance of object = p = 20 cm Focal length = –15 cm Required: a) Position of image = q = ? b) Size of image = hi = ? Formula: a)
1 1 1 f p q
b)
h1 q ho p
Solution: Using the formula a)
1 1 1 f p q
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1 1 1 q 15 20 1 43 q 60
1 7 q 60
60 7
q
q = – 8.75cm
b)
hi q ho p
8.57 10 20 hi 4.28cm
hi
Result: a) q 8.75cm
b) hi 4.28cm
Q.12.10: A convex lens of focal length 6 cm is to be used to form a virtual image three times the size of the object. Where the lens must be placed? Data: Magnification = m = 3 Focal length = 6cm Required: Position of object = p = ? Formula: 1 1 1 f p q
Solution: Using the formula 1 1 1 f p q
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1 1 1 f p 3p 1 1 1 f p 3p 1 3 1 f 3p 1 2 f 3p
3p = 12 p = 4cm Result: Distance of object = p = 4cm Q.12.11: A ray of light from air is incident on a liquid surface at an angle of incidence 35o Calculate the angle of refraction if the refractive index of the liquid is 1.25. Also calculate the critical angle between the liquid air inter-faces. Data: Angle of incidence = i = 35o Reflective index = n = 1.25 Required: a) Angle of refraction = r = ? b) Critical angle = c = ? Formula: Sini Sinr 1 b) n SinC a) n
Solution: We know that a)
Sini n Sinr
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Sin35o 1.25 0.57 Sinr 1.25
Sinr
Sinr = 0.456 r Sin - 1 (0.456) r = 27.13
For critical angle SinC
1 n
1 C Sin 1 1.25
C Sin 1 0.8 C 52.13ans
Result: a) Angle of refraction = r = 27.13o b) Critical Angle = C = 52.13o
Q.12.12: The power of a convex lens is 5D. At what distance the object should be placed from the lens so that its real and 2 times larger image is formed. Data: Power of lens = P = 5D Magnification = m = 2 Required: Distance of object = p = ? Formula: a) f
b)
1 p
1 1 1 f p q
Solution:
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a) f
1 5
f = 0.2 f 0.2 100
f = 20cm Using this formula b)
1 1 1 f p q
q = 2p 1 1 1 20 p 2 p
1 2 1 20 2p 2 p 3 20
2 p 60
p = 30cm Result: p = 30cm
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Unit No. 13
ELECTROSTATICS 1. 2. 3. 4. 5. 6.
Answers Keys B 7. B 8. B 9. A 10. B 11. C
D A D B B
QUICK QUIZ AND POINT TO PONDER “Self-Assessment” Q1. Do you think amount of positive change on the glass rod after rubbing with silk cloth will be equal to the amount of negative charge on the silk? Explain. Ans: yes, the amount of positive charge on the glass rod after rubbing with silk cloth will be equal to the amount of negative charge on the silk because when we rub the glass rod with the silk cloth, the loosely bound electrons of glass rod are transferred to silk cloth. This makes the glass rod positively charged and silk as negatively charged. As electric charge is always conserved in a closed system so the number of electrons loosed by the glass rod are exactly equal to the number of electrons gained by the silk cloth. Q2. What would happen if a neutral glass rod is brought near a positively charge class rod? Ans: Nothing would happen if a natural glass rod is brought near a positively charged glass rod because there will be no electrification (Conduction / induction of charge). “Point to Ponder” Q: Whey leaves of charged electroscope diverge if you touch its disk with a metal rod but they do not diverge if you touch the disk with a rubber rod? Ans: The leaves of charged electroscope diverge of we touch its disk with a metal rod because it is a good conductor but they do not diverge if we touch the disk with a rubber rod because it is a good insulator so there will no leakage of charge through it and the leaf divergence will not alter. Q: In a dry day if you walk in carpeted room and the touch some conductor you will get a small electric shock! Can you tell why does it happen? Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Ans: In a dry day when someone walks in a carpeted room, friction between the feet and the synthetic fabric of the carpet causes charge to build up in his body. If he then touches some conductor, he gets a small electric shock. This occurs due to the movement of electric charges between the fingers and the conductor. Quick Quiz Q:
If we double the distance between two charges what will be the change in the force between the charges? Ans: According to Coulomb’s Law F
1 r2
It is clear from the above equation that if we double the distance between the two charges, becomes ¼th of the original force. “Point to Ponder” Q: A strong electric field exists in the vicinity of this “Faraday cage”. Yet the person inside is not affected. Can you tell why? Ans: A strong electric field exists in the vicinity of this “Faraday cage”. Yet the person inside is not affected because the interior of a hollow charged case is a field free region i.e. intensity is zero. Quick Quiz Q: Is the equivalent capacitance of parallel capacitors larger or small than the capacitance of any individual capacitor in the combination? Ans: As we know that Ce C1 C2 C3 ....Cn It is clear from the above equation that the equivalent capacitance of parallel capacitors is larger than the capacitance of any individual capacitor in the combination. “Quick Quiz” Q: Capacitor black DC current but allows AC current through a circuit. How does this happen? Ans: Capacitor blocks DC current because in case of DC capacitor behaves as an insulator and only store charge, but in case of AC capacitor behaves likes a conductor and allows AC current to pass through a circuit because of the continuously reversing polarity. “Conceptual Questions”: Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Q13.1: An electrified rod attracts pieces of paper. After a while these papers fly away! Why? Ans: A man raises his left hand in a plane mirror the image facing him is raising his right hand because light says are reflected in a mirror, causing us to see a inverted. Q13.2: How much negative charge has been removed from a positively charged electroscope if it has a charge of 7.5 × 10–11 C? Data: Q = 7.5 × 10–11 C e = 1.69 × 10–19C n=? Formula: Q = ne Solution: As we know that Q = ne Putting the values in the above equation 7.5 × 10–11 = n × 1.69 × 10–19 7.5 10 11 n 1.69 10 19 n = 4.68 × 108 electrons Result: Hence 4.68 × 108 number of negative charges has been removed from a positively charged electroscope if it has a charge 7.5 × 10–11 C. Q13.3: In what direction will a positive charge particle move in an electric field? Ans: The positively charged particle will move in the direction of electric field intensity. Because direction of electric field lines or from high potential to low potential. Q13.4: Does each capacitor carry equal charges in series combination? Explain? Ans: yes, each capacitor carries equal charge in series combination because the battery supplies + Q charge to the left plate of the first capacitor and due to induction Q charge is induced on the right plate of first capacitor. This charge induces + Q charge on the left plate of second capacitor and due to induction – Q charge is induced on the right plate of second capacitor. This process continues as a result of which each capacitor has same charge in series combination. Q13.5: Each capacitor in parallel combination has equal potential difference between its two plats. Justify the statement. Ans: Each capacitor in parallel combination has equal potential difference between its two plates because in parallel combination all the left plates of each capacitor are connected to one terminal of the battery and the right ones are connected to the other
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terminal. In this way each capacitor has a direct contact with the battery and draw the same potential. Q13.6: In the presence of charge necessary for the existence of electrostatic potential? Ans: Yes, the presence of charge is necessary for the existence of electrostatic potential because electrostatic potential because electric potential occurs when on electric field exists. Q13.7: Rubber tires get charged from friction with the rood. What is the polarity of the charge? Ans: When rubber tires move on the road, due to friction with the road surface, the loosely bound electrons of rubber are removed and rubber tires get positively charged. Since earth is at zero potential so electrons are shifted from rubber tires to the road. (earth). Q13.8: Perhaps you have see a gasoline truck trailing a metal chain beneath it? What purpose does the chain serve? Ans: A gasoline truck has a metal chain hanging from its rear side. This chain rolls on the road as the tanker moves. Due to friction with the air the air body of the truck charges any and tiny spark can cause havoc. This charge is continuously being transferred from the body of truck to the ground through this metallic chain. Thus the danger of spark is eliminated. Q13.9: If a high-voltage power line fell across your car while you were in the car whey should you not come out of the car? Ans: If a high-voltage power line fell across our car while we are in the car, we are advised not to come out of the car because there is a strong electric filed around the car but interior of the car is a filed free region (E = 0, ∆=0) as it serves as a “Faraday cage” but on coming out of the car, potential difference is establish with may cause a severe shock due to flow of current through our body. Q13.10: Explain why, a gloss rod can be charged by rubbing when held by hand but an iron rod cannot be charged by rubbing, if held by hand? Ans: A glass rod can be charged by rubbing when held by hand because glass rod is bad conductor and charge produced on it is trapped and cannot be conducted through our body whereas an iron rod cannot be charged by rubbing, if held by hand, because iron is a good conductor charge through our body. Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Q.13.1:
ELECTROSTATICS Numerical Problems The charge of how many negatively charged particles would be equal to 100μC. Assume charge on one negative particle is 1.6 × 10C–19 C? Data: Total charge = Q = 100μC =100× 10–6C Charge on an electron = e–1 = 1.6 × 10–19 C Required: No. of negative particles = ? Formula: Q = ne Solution: Using this formula Q ne
n
Q e
100 10 6 C n 1.6 10 19 C 10 2 10 6 1019 n 1.6 1 1015 1.6 1 1016 16
= 0.0625×1016 = 6.25 × 1019 particles Result: No. of negative particles = n = 6.25×1019 particles. Q.13.2:
Two point charges q,= 10 μC and q2= 5 μC are placed at a distance of 150 cm. What will be the Coulomb's force between them? Also find the direction of the force. Data:
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First point charge = q1 = 10μC Second point charge = q2 = 5μc Distance = r = 150cm
150 = 1.5m 100
Required: Magnitude of force = F = ? Direction of Force = ? Formula: F
Kq1 q 2 r2
Solution: As we know F
F
F
Kq1 q 2 r2 9 10 9 10 10 6 10 10 6
1.52
45 22.5
F 0.2 N
Direction of force is repulsive. Result: Magnitude of force = F = 0.2N Direction of force = Direction of force is repulsive. Q.13.3:
The force of repulsion between two identical positive charges is 0.8 N, when the charges are 0.1 m apart. Find the value of each charge. Data: q1 = q2 = q F=0.8N r = 0.1m Required: Value of each charge = q=? Formula:
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F
Kq1 q 2 r2
Solution: As we know q1 q 2 r2
F k
As; q1 = q2 = q F k
r2 Kq 2 F 2 r q2
Fr 2 k
q2
0.8 N 0.1 9 10 9
2
0.008 9 10 9
q
0.008 9 109
q 9.4 10 7 C q 9.4 10 7 C
Result: Value of each charge = q1 = q2 = q 9.4 10 7 C
Q.13.4:
Two charges repel each other with a force of 0.1 N when they are 5 cm apart. Find the forces between the same charges when they are 2 cm apart. Data: F= 0.1N r1 = 5cm = 5/100 r1 = 0.05m r2 = 2cm = 2/100
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r2 = 0.02m Required: Force = F2 = ? Formula: F1
Kq1q 2
F2
Kq1q 2
r1
2
r2
2
……………..1 …………….11
Dividing equation 1 by equation 2 Kq1q 2 2
F1 r 1 F2 Kq1q 2 2 r2
r12 F2 r22 F1 F2
F1 r12 r22
Solution: As we know F2
F1 r12 r22
0.1 0.05 F2 0.022
2
F2 = 0.62N Result: F2 = 0.62N Q.13.5:
The potential at a point in an electric field is 104 V. If a charge of +100 μC is brought from infinity to this point. What would be the amount of work done on it? Data: Electric potential =V = 104V Charge = q = 100μC
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= 100×10–6 = 10–4 C Required: Work done = W = ? Formula: V
W q
Solution: As we know V
Q.13.6:
W q
W = qV Putting the values W = 10 –4 ×104 W = 1J Result: Amount of work = W = 1J A point charge of +2C is transferred from a point at potential 100V to a point at potential 50V, what would be the energy supplied by the charge? Data: Charge = q = +2C Potential at point A = VA = 100V Potential at point B = VB = 50V Required: Energy supplied by charge = ? Formula: E = q(VA – VB) Solution: As we know E = q(VA – VB) Putting the values E = 2(100 – 50) = 2(50)
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E = 100J Result: Energy supplied = E =100J Q.13.7:
A capacitor holds 0.06 coulombs of charge when fully charged by a 9 volt battery. Calculate capacitance of the capacitor. Data: Charge = Q = 0.06C Voltage = V = 9V Required: Capacitance = C = ? Formula: Q = CV Solution: As we know Q = CV Q V 0.06 C 9 C
Q.13.8:
C = 6.67 × 10–3F Result: Capacitance = C = 6.67 × 10–3F A capacitor holds 0.03 coulombs of charge when fully charged by a 6 volt battery. How much voltage would be required for it to hold 2 coulombs of charge? Data: Charge = Q1= 0.03C Charge = Q2 = 2C Voltage = V1 = 6V Required: Voltage =V2= ? Formula:
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Equation Number 1 Q1 = CV1 Equation Number 2 Q2 = CV2 Dividing equation 1 by equation 2 Q1 CV1 Q2 CV2
So, Q1 V1 Q2 V 2
Solution: Since Q1 V1 Q2 V 2 0.03 6 2 V2
V2
Q.13.9:
6 2 0.03
V2 = 400 V Result: Voltage = V = 400V Two capacitors of capacitances 6 µF and 12 µF are connected in series with 12V battery. Find the equivalent capacitance of the combination. Find the charge and the potential difference across each capacitor. Data: Capacitance = C1 = 6µF = 6 × 10–6 F Capacitance = C2 = 12µF = 6 × 10–6 F Voltage = V= 12V Required:
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a) Equivalent Capacitance = Ceq= ? b) Charge on each capacitor = Q = ? c) Potential difference across each capacitor: i) V1 = ? ii) V2 = ? Formula a)
1 1 1 1 Ceq C1 C2 C3
b) Q = CV c) Q = CV Solution: a) Since capacitors are connected in series 1 1 1 Ceq C1 C2
Putting the values 1 1 1 Ceq 6F 12F 1 2 1 Ceq 12F 1 3 Ceq 12F Ceq
b)
c)
12 F 3
Ceq = 4µF Q = CV Q = 4 × 10–6 × 12 Q = 48μF i) Q = C1V1 48C V1 6 C V1 = 8μF ii) Q = C2V2 48C V2 12C
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V2 = 4V Result: a) Ceq = 4μF b) Q = 48μC c) i) V1 = 8V ii) V2 = 4V Q.13.10: Two capacitors of capacitances 6µF and 12µF are connected in parallel with a 12, battery. Find the equivalent capacitance of the combination. Find the charges and the potential difference across each capacitor. Data: Capacitance = C1 = 6μF = 6 × 10–6 F Capacitance = C2 = 12μF = 6 × 10–12 F Voltage = 12V Required: a) Equivalent Capacitance = ? b) Charge on each capacitor = ? i) Q1 = ? ii) Q2 = ? c) Potential difference = V = ? Formula: a) Ceq = C1 + C2 b) Q = CV c) Q = CV Solution: a) In parallel combination Ceq = C1 + C2 = 6μF + 12μF = 18μF b) i) Charge on capacitor C1 will be Q1 = C 1 V Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Q1 = 6×12 Q1 = 72μC ii) Charge on Capacitor C2 will be Q2 = 12×12 = 144μC c) The capacitors are connected in parallel so will remain same V = 12V Result: a) Ceq = 18μF b) i) Q1 = 72μC ii) Q2 = 144μC c) V = 12V
Unit No. 14
CURRENT ELECTRICITY 1. 2. 3. 4. 5.
Answers Keys D 6. C 7. B 8. C 9. B
D A C B
QUICK QUIZ AND POINT TO PONDER “Quick Quiz” Q1. How long does it take a current of 10mA to deliver 30C of charge? Ans: Data:I = 10mA = 10 × 10–3A Q = 30C Dedicated to My Sweet Father…… whatsapp group Career Opportunities
t =? Formula: I
Q t
Solution: As we know that I
Q t
Putting the values in the above equation: 10 10 3
t
30 t
30 10 10 3
t = 3000Sec Result: Hence 10mA current will take 3000 Sec to deliver 30C to charge. Q2. Which metal is used as the filament of an electric bald? Explain with reason. Ans: When current flows through the metal, the heat energy generated in the metal increases the temperature of the metal which goes on increasing as time passes (Joule’s Law). This rising temperature may melt the metal. So the metal used as the filament must have such a melting point that it does not melt at high temperature that is why ‘Tungsten’ metal is used as the filament of an electric bulb because it has the highest melting point among all the metals and does not melt even when it is glowing. “Point to Ponder” Q: The current versus voltage graph of resistor is a straight line with constant slope. The graph for light bulb is curved with a decreasing slope? What can you infer from this? Ans: The graph for light bulb is curved with the decreasing slope with infers that light bulb is a non-ohmic material and resistance of light bulb rises (current decreases) as it get hotter.
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Q: A bird can sit harmlessly on high tension wire. But is must not reach and grab neighbouring wire. Do you know why? Ans: A bird can sit harmlessly on high tention wire, But it must not reach and graph neighbouring wire, a potential difference would be established and a large amount of current would from through the bird causing its death. “Brain Teaser” Q: Connect a battery to a small 2.5v light bulb and observe the brightness of the bulb. Now add another light bulb in series with the first bulb. Observe the relative brightness of the bulbs compared to when only one bulb was lit. Repeat the process with two or three additional bulb in series. Using Ohm’s laws explain what happened to the brightness of each bulb? Ans: When we add another bulb in series with the first bulb the relative brightness of the bulbs reduces as compared to when only reduces as compared to when only one bulb was lit and when two or three additional bulbs are added in series, we observe that the relative brightness of the bulbs reduces but all the bulbs in the series circuit will have equal but reduced brightness. According to ohm’s law the total voltage in series circuit divides itself among the individual component. As voltage is the energy with which electrons flow and this energy goes on decreasing successively in the upcoming components so it decreases the brightness. Activity Q:
Connect as battery to a small 2.5V light bulb and observe the brightness of the bulb. Connect a second light in parallel with the first and observe the brightness of the bulbs. Now add a third bulb in parallel with the others and note the
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brightness of the bulbs. Does the brightness of the bulbs differ from the bulbs connect in the series with the battery? Explain. Ans: When we connect a second light parallel with the first bulb we observe the same brightness and on adding third bulb in parallel with the others even then brightness does not change. Yes, of course the brightness of the bulbs connected in parallel differs from the bulbs connect in the series with the battery. Brightness of each bulb connected in series will be equal but with decreased intensity and on the other hand in parallel combination brightness will be more as compared to bulbs in series and intensity remains the same. The reason in that in parallel combination potential drop across all the bulbs is same but in series it changes. “Self Assessment” Q: A light bulb is suitable on for 40 Sec. if the electrical energy consumed by the bulb during this time is 2400 J, find the power of the bulb. Ans: Data: t = 40Sec w = 2400J p=? Formula: P
w t
Solution: As we know that P
w t
Putting the values in the above equation: P
2400 40
P = 60w Result: Hence power of the bulb is 60w. . “Point to Ponder” Q: How many faults can you find after studying this picture?
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Ans: The fault after studying the given picture is that the green coloured earth wire and brown coloured live wire are not properly connect live wire are not properly connect with the pins of the three pin plug. The correct way of wiring a three pin main plug is show in the figure given below. “CONCEPTUAL QUESTIONS” Q14.1: Why in metal charge is transferred by free electrons rather than by positive charge? Ans: I metals (conductors) there exist a metallic bond between the atoms, due to which a pool of free electrons is generated. These free electrons are not bounded to nuclei and are free to move around randomly. They have weak force among them and nucleus. When such charges are exposed to external electric field, they physically move in specific direction and thus constitute current. On the other hand positive charges are actually the deficiency of the electrons and are not responsible for the transfer of charge. Q14.2: What is the difference between a cell and a battery? Ans: CELL BATTERY A cell is a single unit at the base A group of cells connect together is voltage. called battery. It is identified in the circuit by the It is identified in the circuit by the symbols. symbol as Q14.3: Can current flow in circuit without potential difference? Ans: According to Ohm’s law. VαI The above expression shows that the amount of current passing through the conductor is directly proportional to the potential difference. Hence, no current will flow in a circuit without potential difference because the flow of current continues as long as these are a potential difference.
Q14.4: Two points on an object are at different electric potentials. Does charge necessarily flow between them?
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Q14.5: Ans: Yes, if two points on an object are at different electric potential then charge will flow from higher potential to potential because flow of charge continues as long as there is a potential difference. . Q14.6: In order to measure the current in a circuit why ammeter is always connect is series? Ans: In order to measure the current in a circuit ammeter is always connected in series so that the total current flowing in the circuit should pass through the ammeter. Also the resistance of an ammeter is very flow so it does not charge the total value of current passing circuit when connected in series. Q14.7: In order to measure the voltage in a circuit why voltmeter is always connected in parallel? Ans: In order to measure the voltage in a circuit, voltmeter is always connected in parallel because the voltage a across the parallel components of a circuit is always same and the voltage drop across the component (across which voltage is to be measured) will be equal to the voltage of the voltmeter. Also the resistance of the voltmeter is very high so it draws a little current from the circuit but does not alter the required voltage. Q14.8: How many watt-hours are there in 1000J? Ans: As we know that: 1KWh 3.6 10 6 J 1 1J Kwh 3.6 10 6 1 1J 1000wh 3.6 10 6
1J 0.000278wh
So, 1000J 0.000278 1000wh 1000 J 0.278wh
Hence, there are 0.278 watt-house in 1000J. Q14.9: From your experience of watching cars on the roads at night, are automobile headlamps connected in series or parallel? Ans: The automobile headlamps are always connected in parallel because if one of the headlamps burns out due to some technical fault the other should continue to work properly.
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Q14.10: A certain flash-light can use a 10 Ohm bulb or a 5 Ohm bulb. Which bulb should be used to get the brighter light? Which bulb will discharge the battery first? Ans: As we know that: P I 2R From the above equation it is clear that the bulb having more resistance will have more power and brightness. According to Joule’s law: w I 2 Rt
From the above equation it is clear that the bulb having more resistance will dissipate more energy and will discharge the battery earlier. Hence 10 Ohm bulb used in flash-light will glow brighter and also discharge the bulb earlier. Q14.11: It is impracticable to connect an electric bulb and an electric heater in series. Why? Ans: It is impracticable to connect an electric bulb and an electric heater in series due to the following reasons: i. If one of the appliances fuses then the other will not operate as the circuit breaks and the only path for the flow of current becomes incomplete. ii. In series combination supplied voltage divides itself among the appliances and becomes insufficient to run the appliances. iii. In series combination the equivalent resistance of the circuit becomes very high and the current reduces to a low value such that unable to run the appliances. Q14.12: Does a fuse in a circuit control, the potential difference or the current? Ans: If fuse is a type of sacrificial over-current protection device so it controls the currents in a circuit up to safety limit not the potential difference.
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Numerical Problems Q.14.1: A current of 3mA is flowing through a wire for 1 minute. What is the charge flowing through the wire? Data: Current = I = 3mA = 3 × 10–3A Time = t = 1 min = 60Seconds Required: Charge = Q = ? Formula: I
Q T
Solution: We know that I
Q T
Q=I×t Q = 3 ×10–3 ×60 Q = 180 × 10–3 C Q = 180 mC Result: Q = 180 mC Q.14.2: At 100,000 Ω, how much current flows through your body if you touch the terminals of a 12 V battery? If your skin is wet, so that your resistance is only 1000Ω, how much current would you receive from the same battery? Data: R1 = 100000Ω V = 12V R2 = 1000Ω Required: a) I1 = ? b) I2 = ? Formula: Dedicated to My Sweet Father…… whatsapp group Career Opportunities
V=IR
Solution: We know that a) V=I1R1 I1
V R1
I1
12 100000
I1 = 1.2×10–4 b) V = I2R2 I2
V R2
I2
12 1000
I2 = 1.2 × 10–2 A Result: a) I1 = 1.2×10–4 b) I2 = 1.2 × 10–2 A Q.14.3: The resistance of a conductor wire is 10 MΩ. If a potential difference of 100 volt is applied across its ends, then find the value of current passing through it in mA. Data: Resistance = 10MΩ = 10 × 106Ω Potential difference = 100 Volts Required: Current = I = ? Formula: V=IR Solution: We know that V=IR Dedicated to My Sweet Father…… whatsapp group Career Opportunities
V R 100 I 10 10 6 I
I = 10–5 I = 10–2 × 10–3 I = 10–2 mA I = 0.01mA Result: Current = I = 0.01mA
Q.14.4: By applying a potential difference of 10V across a conductor a current of 1.5A passes through it. How much energy would be obtained from the current in 2 minutes? Data: Potential difference = V = 10Volts Current = I = 1.5Amp Time = t = 2min = 2× 60 = 120Sec Required: Energy = W = ? Formula: W = I2Rt Solution: We know that W = I2Rt = I(IR)t V = IR
= I(V)t =IVt = 1.5 × 10 × 120 W = 1800J Result: Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Energy = W = 1800J Q.14.5: Two resistances of 2kQ and 8kΩ are joined in series, if a 10 V battery is connected across the ends of this combination, find the following quantities: a. The equivalent resistance of the series combination. b. Current passing through each of the resistances. c. The potential difference across each resistance. Data: Value of first resistance : R1 = 2KΩ = 2 × 103Ω Value of second resistance = R2 = 8KΩ = 8 × 103Ω Voltage = V = 10V Required: a) Equivalent resistance = Req = ? b) Current = I = ? c) Potential difference : i. V1 = ? ii. V2 = ? Solution: a) Req = R1 + R2 b) V = IReq c) V = IR Solution: a) Req = R1 + R2 = 2KΩ + 8 KΩ = 10KΩ b) Circuit in series so the current remain same I = I1 = I 2 V = IReq I
V Req
I
10 10 ×103
= 1 × 10-3A Dedicated to My Sweet Father…… whatsapp group Career Opportunities
I = 1mA c) i) V1 = IR1 = 1 × 10–3 × 2 × 103 V1 = 2V I = 1mA ii) V2 = IR2 = 1 × 10–3 × 2 × 103 = 8V Result: a) Req = 10KΩ b) I = 1mA c) i) V1 = 2V ii)V2 = 8V Q.14.6: Two resistances of 6KΩ and 12kO are connected in parallel. A 6V battery is connected across its ends, find the values of the following quantities: a) Equivalent resistance of the parallel combination. b) Current passing through each of the resistances. c) Potential difference across each of the resistance. Data: R1 = 6KΩ = 6 × 103Ω R2 = 2KΩ = 12 × 103Ω V = 6V Required: a) Equivalent Resistance = Req = ? b) Current through each resistance i) I1 = ? ii) I2 = ? c) Potential difference = V = ? Formula: a)
1 1 1 Re q R1 R 2
b) V = IR Solution:
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a)
1 1 1 Req R1 R2 1 1 1 3 Req 6 10 12 103 2 1 12 10 3 3 12 10 3 1 1 Req 4 10 3
Req = 4 × 103Ω Req = 4KΩ b)
i) Quantity of current in first resistance = I l Il
V R1
6 6 10 3
I1 = 1 × 10–3 I1 = 1mA ii) Quantity of current in Second R I 2 I2
V R2
6 12 10 3
I2 = 0.5 × 10–3 I2 = 0.5mA c)
Circuit is parallel, so the voltage remain same V = 6V Result: a) Req = 4KΩ b) i) I1 = 1mA ii) I2 = 0.5mA Dedicated to My Sweet Father…… whatsapp group Career Opportunities
c) V = 6V Q.14.7: An electric bulb is marked with 220V, 100W. Find the resistance of the filament of the bulb. If the bulb is used 5 hours daily, find the energy in kilowatt-hour consumed by the bulb in one month (30 days). Data: Voltage of bulb = V = 220V Power of bulb = P = 100W Daily use of bulb = t = 5h No of days = 30 days Required: a) Resistance of bulb = R = ? b) Energy consumer by bulb = E = ? Formula: a) P = V2/R b) E
Pt Kw 1000
Solution: a)
V2 R V2 R P P
2202
100 48400 100
= 484Ω b) Energy in hour =
100 5 30 1000
=15 Kwh Result: a) R = 484Ω
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Power Time 1000
b) E = 15Kwh Q.14.8: An incandescent light bulb with an operating resistance of 95 Q is labelled "150 W." Is this bulb designed for use in a 120V circuit or a 220V circuit? Data: Resistance = R = 95Ω Power = P = 150W Required: For which V bulb designed = ? (220W – 120V) Formula: P = I2 R Solution:
&
V = IR
We know that P R 150 I2 95 2 I 1.5784
I2
I 2 1.5784
I = 1.2568 A As we know V = IR V = (1.2565)(95) V = 119.37V V = 120V Result: So bulb is designed for 120V Q.14.9: A house is installed with a) 10 bulbs of 60 W each of which are used 5 hours daily. b) 4 fans of 75 W each of which run 10 hours daily. c) One T.V. of 250 W which is used for 2 hours daily. d) One electric iron of 1000 W which is used for 2 hours daily.
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If the cost of one unit of electricity is Rs.4. Find the monthly expenditure of electricity (one month =30 days). Data: a) Power of 10bulb = 60 × 10 = 600W b) Power of fans 4 = 75 × 4 = 300W c) Power of 1iron = 1 × 1000 = 1000W d) Power of 1 T.V. = 1 × 250 = 250W Required: Monthly cost of electricity house = ? Formula: Energy
Power hour 1000
Solution: a) Energy consumed by bulb Power hour 1000 600 5 30 1000
= 90 units b) Energy consumed by fans Power hour 1000 300 10 30 1000
= 90 units c) Energy consumed by T.V. Power hour 1000 250 2 30 1000
= 15 units
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d) Energy consumed by Iron Power hour 1000 1000 2 30 1000
= 60 units So, Total consumed energy = 90 + 90 + 60 + 15 = 255 units Price of electricity = 255 × 4 = 1020Rs. Result: Total Price = 1020 Rs. Q.14.10: A 100 W lamp bulb and a 4 kW water heater are connected to a 250 V supply. Calculate (a) the current which flows in each appliance (b) the resistance of each appliance when in use. Data: P1 = 100w P2 = 4Kwatt = 4 × 103W V = 250V Required: a) i) I1 = ? ii) I2 = ? b) i) R1 = ? ii) R2 = ? Formula: a) P = VI b) V = IR Solution: a) i) P1 VI1 P1 V 100 I1 250
I1
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I1
2 5
I1 = 0.4A ii) P2 = VI2 I2
P2 V
4 103 250 4000 I2 250
I2
b)
I2 = 16A i) V = I1R1 R1
V I1
R1
250 0.4
R1 = 625Ω ii) V = I2R2 R2
V I2
R2
250 16
R2 = 15.125 Result: a) i) I1 = 0.4A ii) P2 = 16A b) i) R1 = 625Ω ii) R2 = 15.125Ω Q.14.11: A resistor of resistance 5.6 Q is connected across a battery of 3.0 V by means of wire of negligible resistance. A current of 0.5 A passes through the resistor. Calculate (a) Power dissipated in the resistor (b) Total power produced by the battery. Dedicated to My Sweet Father…… whatsapp group Career Opportunities
(c) Give the reason of difference between these two quantities. Data: R = 5.6Ω V = 3.0V I = 0.5A Required: a) P = ? b) Pbattery = ? c) Reason of difference Formula: a) P = I2R b) P = VI Solution: a) P = I2R = (0.5)2(5.6) P = 1.4W b) Pbattery = VI = (3)(0.5) Pbattery = 1.5W Result: a) P = 1.4W b) Pbattery = 1.5W c) Some power is lost by the internal resistance of the battery. Unit No. 15
ELECTROMAGNETISM 1. 2. 3. 4. 5.
Answers Keys D 6. B 7. D 8. A 9. C
B D B C
QUICK QUIZ AND POINT TO PONDER “Activity”
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Q1. Suppose direction of current passing through two straight wires is same. Draw the pattern of magnetic field of current due to each wire would the wire attract or repel each other? Ans: The wires will attract each other because magnetic fields in between the wires cancel each other creating a weak magnetic field region. On the outer sides field region. On the outer sides of the both wires, there is a strong magnetic field and force acts from stronger magnetic field region to weaker magnetic field region. Hence both the wires are attracted towards each other. “CONCEPTUAL QUESTIONS” Q15.1: Suppose someone handed you three similar iron bars and told you one was not magnet but the other two were. How would you find the iron bar that was not magnet? Ans: Bring one end of 1st iron bar close to the end of the 2nd iron bar. If these end of bars atract each other then change the end of 1st iron bar and again bring it close to the 2nd iron bar. If again these two bars show attraction then one of them is not a magnet. Now, bring one end of the 3rd bar close to one end of the 1st bar. If there is attraction then change the end of the 1st iron bar and again bring it close to the 3rd iron ban. If they show repulsion then these 1st and 3rd gars are magnets and 2nd is simple iron rod but if they show again attraction then the 1st bar is simple iron bar and 2nd and 3rd are magnets.
Q15.2: Suppose you have a coil of wire and a bar magnet. Describe how you could use them to generate an electric current? Ans: As we know that an e.m.f is induced is a relative motion between the coil and the magnet and this value of induced e.m.f is directly proportional to the rate of change of number of magnetic lines through it (Faraday’s Law). If we place a coil in the magnetic field of a bar magnetic, some of the magnetic lines of force will pass through it. If the coil is far away from the magnet only few lines of force will pass through the coil. However if the coil is close to the magnet, a large number of lines of force will pass through it. Dedicated to My Sweet Father…… whatsapp group Career Opportunities
This moves we can change the number of magnetic lines of force through a coil by moving it in the magnetic field. This change in the number of magnetic field lines will induce an em.f. in the coil. This is the basic principle of production of electricity and working of a transformer. Q15.3: Which device is used for converting electrical energy into mechanical energy? Ans: AD.C motor is used for converting electrical energy into mechanical energy. Q15.4: Suppose we hang a loop of wire so that it can swing easily. If we now put a magnet into the coil, the coil will start swinging, which way will it swing relative to the magnet and why? Ans: If we put a magnet into the hanging coil, then the coil will start swinging and it will swing in the same direction as the direction of motion of magnet as show in the figure. Reason: According to the lenz’s law the direction of an induced current in a circuit is always such that passes the cause that produces it. So when the magnet is moved forward to coil then, due to change of flux, an induced current is produced in anti-clockwise direction (according to the right hand rule) which creates North Pole on the same side as the magnet. The two north poles then repel each other which cause the coil to the swing. Q15.5: A conductor wire generates a voltage while moving through a magnetic field. In what direction should the wire be moved, relative to the field to generate the maximum voltage? Ans: If want to generate the maximum voltage in the conductor wire then it must be moved perpendicular to the magnetic field.
Q15.6: What is the difference between a generator and motor? Ans: Generator Motor A generator converts mechanical A motor converts electrical energy energy into electrical energy. into mechanical energy. In generator, firstly coil is rotated In motor, firstly electricity is supplied
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then electricity is generated. then coil rotates. Slip rings are used in generators. Split rings are used in motors. At least one of the coil and magnetic Both of the coil and magnet should should be energized be energized. Its principle is changing flux includes Its principal is torque acting on a e.m.f or current in the coil. current carrying coil placed in a magnetic field. Q15.7: What reserves the direction of electric current in the armature coil of D.C. motor? Ans: Split rings reverse the direction of current in the armature of D.C. motor. . Q15.8: A wire lying perpendicular to an external magnetic field carries a current in the direction shown in the diagram below. In What direction will the wire move due to resulting magnetic force? Ans: According to the Fleming’s left hand rule stretch the thumb, fore figure and middle finger of left hand mutually perpendicular to each other. If fore figure points in the direction of the magnetic field the middle figure in the direction of current then the thumb would indicate the direction of force acting on the conductor. Thus on the basis of this rule the wire will move in downward direction. Q15.9: Can a transformer operate on direct current? Ans: No, a transformer cannot operate on direct current because it working principle is mutual induction which changing current is required in primary coil to generate the current in the secondary coil. But in D.C., current does not change so it cannot produce induced current in the secondary coil.
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Numerical Problems Q.15.1: A transformer is needed to convert a mains 240 V supply into a 12 V supply, if there are 2000 turns on the primary coil, then find the number of turns on the secondary coil. Data: VP = 240V VS = 12V NP = 2000 Required: NS = ? Formula: N S VS N P VP
Solution: As we know that NS
VS N P VP
NS
12 2000 240
NS = 100 Result: NS = 100 Q.15.2: A step-up transformer has a turn ratios of 1:100. An alternating supply of 20 V is connected across the primary coil. What is the secondary voltage? Data: VP = 20V Ns / Np = 1/100 Required: VS = ? Formula: N S VS N P VP
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Solution: As we know that NS VS
VS NP VP
100 20 1
VS = 2000V Result: VS = 2000V Q.15.3: A step-down transformer has a turn’s ratio of 1: 100. An AC voltage of amplitude 170 V is applied to the primary. If the current in the primary is 1.0 mA, what is the current in the secondary? Data: VP = 170V Ns / Np = 1/100 IP = 1 mA = 1 × 10 – 3 A Required: IS = ? Formula: N S VS N P VP
&
Solution: As we know that VS
NS VP NP
VS
1 170 100
VS = 1.7V For ideal transformer PP = PS Dedicated to My Sweet Father…… whatsapp group Career Opportunities
I S VP I P VS
I S VP I P VS IS
I PV P VS
110 3 170 IS 1.7
IS = 0.1A Result: IS = 0.1A Q.15.4: A transformer' designed to convert the voltage from 240 V A.C. mains to 12 V, has 4000 turns on the primary coil. How many turns should be on the secondary coil? if the transformer were 100% efficient, what current would flow through the primary coil when the current in the secondary coil was 0.4A Data: VP = 240V NP = 4000 VS = 12V IS =0.4A Required: a) NS = ? b) IP=? Formula: a)
N S VS N P VP
b)
I S VP I P VS
Solution: a) As we know that N S VS N P VP NS
VS NP VP
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12 4000 240
NS
NS = 200 b) for Ideal transformer PP = PS I S VP I P VS IP
I S VS VP
IP
0.4 12 240
I = 0.02A Result: NS = 200 IS = 0.02A Q.15.5: A power station generates 500 MW of electrical power which is fed to a transmission Sine. What current would flow in the transmission line if the input voltage is 250 kV? Data: V = 250kV = 250 × 103V Power = P = 500 × 106W Required: Current = I = ? Formula: P=VI Solution: As we know that I
P V
I
500 10 6 250 10 3
I = 2kA I = 2000A Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Result: I = 2000A
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Unit No. 16
BASIC ELECTRONICS 1. 2. 3. 4. 5.
Answers Keys D 6. D 7. C D C
A B
QUICK QUIZ AND POINT TO PONDER “Point to ponder” Q1. When a magnet is brought near to the screen of a television tube, picture of the screen is distorted. Do you know why? Ans: The picture on the TV screen is formed by the help of beam of electrons and when a magnet is brought near to the screen of a television tube, picture on the screen is distorted because when a moving charge enters in the magnetic field. Thus, the target of the electron beam will be disturbed. As a result, the picture on the TV screen becomes distorted. “Quick quiz” Q1. Assume you have an OR gate with two inputs, A and B. Determine the output, C for the following. (a) A = 1, B= 0 (b) A = 0, B = 1 If either input is one, what is the output? Ans: For an OR gate with two inputs A and B, the outputs are as given in the table: Case a b c
Input (A) 1 0 1
Input (B) 0 1 1
Out Put (C = A + B) 1 1 1
“CONCEPTUAL QUESTIONS” Q16.1: Name two factors which can enhance thermionic emission? Ans: Following are some factors which can enhance thermionic emission. Dedicated to My Sweet Father…… whatsapp group Career Opportunities
The nature of the metal. The temperature of the surface grater the temperature greater is the rate of thermionic emission. The surface area of the body. If the surface area of the body is larger than larger will be the thermionic emission. Q16.2: Give three reasons to support the evidence that cathode rays are negatively charged. Ans: Following are the reasons to support the evidence that cathode rays ware negatively charge electrons. When a beam of electrons pass through an electric field it bends towards the positive plate. When a beam of electrons pass through a magnetic field it bends towards the North Pole. Cathode rays produce chemical change, because they have reducing effect as electrons. Q16.3: When electrons pass through two parallel plates having opposite charges they are deflected towards the positively charged plate. What important characteristic of electrons can be inferred from this? Ans: When electrons pass through two parallel plates having opposite charges they are deflected towards the positively charged plate. So we can infer the important characteristic of electrons that they are negatively charged particles.
Q16.4: When a moving electron enters the magnetic field, it is deflected from its straight path. Name two factors which can enhance the electron deflection? Ans: When a moving electron enters the magnetic field, it is deflected from its straight path. Following are the two factors which can enhance the deflection of electron. By increasing the strength of magnetic field as FB . F evBSin By increasing the speed of electron as FV . F evBSin Q16.5: In what ways is an oscilloscope a voltmeter? Ans: If we want to use the cathode ray oscilloscope as a voltmeter then firstly we have to calibrate it. To do so, a battery of know e.m.f. is connected to the y-input of the oscilloscope and deflection of the bright spot of light on the screen is measured. This gives us the sensitivity of the Y-plates. Then the battery is disconnected and a Dedicated to My Sweet Father…… whatsapp group Career Opportunities
voltage which is to be measured is connected to the y-input. As the x-axis is calibrated in volts and x-axis in time as shown in figure, we can easily find the instantaneous and peak value of voltage. In this way the oscilloscope act as a voltmeter. Q16.6: How can you compare the logic operation x-A.B with usual operation of multiplication? Ans: If we compare logic operation (X = A.B) with usual operation of multiplication, we obtain the some results with same inputs. The outputs of both logic operation and multiplication are 0 if one of the inputs is 0. If inputs both logic operation and multiplication are 1 then output will be 1. This can be verified from the truth table given below. Inputs A
Logic operation Multiplication (x = A.B) (X = A × B) B 0 0 0 0 1 0 0 0 0 1 0 0 1 1 1 1 Note: if the inputs are other than 0 or 1 then logic operation will fail to give output because logic operation only works on Boolean inputs. Q16.7: NAND gate is the reciprocal of AND gate. Discuss. Ans: NAND gate is simply an AND gate followed by a Not gate. The output of NAND gate is 0 when both of its inputs are 1. It can also be verified from truth table given below. Input
And gate X = A.B (A) (B) 0 0 0 0 1 0 1 0 0 1 1 1 This shows that NAND gate is the reciprocal of AND gate. Q16.8: Show that the circuit given as below acts OR gate. Dedicated to My Sweet Father…… whatsapp group Career Opportunities
NAND X A.B
1 1 1 0
Ans:
When a NOR gate is coupled with NOT gate then it acts as an OR gate. This can be verified from the truth table given below. Input (A)
x A B
(B)
0 0 1 1 0 0 0 1 0 1 1 0 Hence the above circuit act as an OR gate.
Outputs
y x
0 1 1 1
Q16.9: Show that the circuit given as below acts AND gate. Ans:
In this circuit two NOT gates are working as input terminal of NOR gate. So the collective circuit will act as an AND gate. This can be verified from the truth table given below:
x
Input A
B
0 1 0 1
0 0 1 1
1
1 0 1 0
A
x 1 1 0 0
Hence the above circuit acts as an AND gate.
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2
B
x x1 x2
Output
1 1 1 0
0 0 0 1
y x
Unit No. 17
INFORMATION AND COMMUNICATION 1. 2. 3. 4. 5.
Answers Keys C 6. B 7. D C C
D B
QUICK QUIZ AND POINT TO PONDER “Point to ponder” Q1. What is the impact of ICT in education? Ans: ICT has so many favorable impacts in education as it help in: Extending the research of high quality education to all. Nurturing powerful communities of learning. Enabling relevant, personalized and engaged learning. Giving teachers greater insight and more time. Supporting agile, efficient and connected education system.
“CONCEPTUAL QUESTIONS” Q17.1: Why optical fiber is more useful tool the communication process? Ans: Optical fiber is more useful tool for the communication process due to the following reasons. They are highly flexible, light weight and much cheaper as compared to copper cables. The transmitting capacity of optical fiber is thousands times greater than that of radio waves. Signals travel in the optional fiber and come out without any loss of intensity. A single strand of optical fiber can transmit thousands of telephone calls at the same time without interfering each other. Q17.2: Which is more reliable? Floppy or a hard disk? Ans: Hard Dis is more reliable than a floppy disk due to following reasons: Dedicated to My Sweet Father…… whatsapp group Career Opportunities
A typical floppy has strong capacity of between 1 and 3MB, whereas a hard disk might hold hundreds or thousands of MB’s of information. Information can be transferred quickly to and from a hard disk much faster than with a floppy. Floppies are reliable only for short-term data storage. Whereas Hard Disk are used for long term data storage. A little external magnetic field can easily destroy data stored on floppy but the data stored on hard dis is even not disturbed a little even not disturbed a little in the presence of a strong external magnetic field. Q17.3: What is the difference between RAN and ROM memories? Ans: RAM Rom It is elaborated as “Random Access It is elaborated as “read only Memory”. memory” In reference with the processor the The processor cannot access directly information stored in RAM is easily that is stored in ROM in order to accessed. access the ROM information first the information will be transferred into RAM it gets executed by the processor. Both read and write operation can be The ROM memory only allow the performed over the information. user to read information user cannot make any changes to it. RAM memory is used to store the ROM memory is used to stored temporary memory. permanent information and cannot be deleted. The accessing speed of RAM is Speed of ROM is slower in faster. It assists the processor to boost comparison with RAM. ROM cannot up the speed. boost up the speed of processor.
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Unit No. 18
ATOMIC AND NUCLEAR PHYSICS Answers Keys A 6. B 7. B 8. D 9. B
1. 2. 3. 4. 5.
None B A D
CONCEPTUAL QUESTIONS: Q1: It is possible for an element to have different types of atoms explain? Ans: Yes, it is possible for an element to have different types of atoms because naturally there exist some atoms of a particular element having same atomic number and different atomic mass number. They are called isotopes of that element. 1 1
For example element of hydrogen has three different types of atoms i.e. protium H 2 1
3 1
, deuterium H and tritium H . Q2: What nuclear reaction will release more energy, the fission or the fusion reaction? Ans: Nuclear fusion reaction will release more energy as compared to nuclear fission reaction. The energy released by nuclear fusion is three to four time greater than the energy released by nuclear fission. As the binding energy per nucleon of a heavy nucleus is much more than the binding energy. Per nucleon for the two combining light nuclei. Therefore a lot of energy per nucleon is released in fusion than fission. Q3:
Which has more penetration power, alpha particle or gamma ray photon?
Ans: Gamma ray photon has more penetration power than alpha particle due to large spend and neutral nature and it can penetrate at least 2 Km in air. But on the other hand alpha particle has less penetrating power due to its strong interacting or ionizing power and it can penetrate only few centimeters in air.
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Q4:
What is the different between natural and artificial radioactivity?
Ans: Natural Radio Activity Artificial Radioactivity The spontaneous emission of radiation The process of coverting stable nuclei by unstable nuclei is called natural into unstable unclei by bombarding radio activity. them with protons, neutrons or alpha particles. All natural radioactivie nuclear All artificial radioactive nuclear reaction are spontaneous. reaction are non-spontaneous. Natural radioactivity comes from Artificial radioactivity comes from radioactive elements present in nature. elements created in nuclear reactors and accelerators. Elements having atomic mass number Elements having atomic mass number more than 82 show natural less than 82 show artificial radioactivity. radioactivity when bombarded with neutrons. Q5:
How loon would you likely have to wait to watch any sample of radioactive atoms completely decay?
Ans: We have to wait on infinite time to watch any sample of radioactive atoms to decay completely. According to law of integration, no radioactive element can completely decay because in this way an infinite time is required for all the atoms to decay. Q6:
Which type of natural radioactivity leaves the number of protons and the number of neutrons in the nucleus unchanged?
Ans: Gamma decay is type of natural radioactivity which leaves the number of protons and the number of neutrons in the nucleus unchanged. A Z
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X Az X OOY
(Class 10th)
Physics
Numerical Q.18.1: The half-life of
16 7
N is 7.3 s. A sample of this nuclide of nitrogen is observed for 29.2s.
Calculate the fraction of the original radioactive isotope remaining after this t i m e .
Data: Half-life of 167 N = 7.3 Time Observed = 29.2 sec Required: Remaining fraction of radioactive isotope =? Formula: N=Remaining fraction = original(N0)
1 2t
Solution: N=Remaining fraction = original(N0)
1 2t
As we know t = half-life period = time / half life
29.2 =4 7.3
So, 1 2t N 1 4 No 2
N No
N 1 N o 16
Result: Thus fraction for original radioactive isotope remaining after 4 half-lives will be 1/16th Q.18.2: Cobalt-60 is a radioactive element with half-life of 5.25 years. What fraction of ttti original sample will be left after 26 years?
Data: Half-life of C – 14 = 5.25 years Time = t = 26 years Required: Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Remaining fraction = N = ? Formula: N=Remaining fraction = original(N0)
1 2t
Solution: t = half-life period = time / half life t
26 4.95 5 5.25
So, During 26 years, five half-lives are escaped 1 2t 1 N No 5 2 N 1 N o 25
N No
N 1 N o 32
Result: Thus fraction for original radioactive isotope remaining after 5 half-lives will be 1/32th Q.18.3: Carbon-14 has a half-life of 5730 years. How long will it take for the quantir, carbon-14 in a sample to drop to one-eighth of the initial quantity?
Data: Half-life of C – 14 = 5730 years. N 1 No 8
Required: Time =? Formula: Time = No. of half-lives × half life Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Solution: As the quantity of C – 14 drop to 1 / 8th of the original quantity So, 1 1 8 23
So, No. of half-life = t = 3 Time = No. of Half-lives × half-lives Time = 3 × time ½ Time = 3 × 5730years Time = 1.72 × 104 years Result: Time = 1.72 × 104 years Q.18.4: Technetium-99m is a radioactive element and is used to diagnose brain, thyroid liver and kidney diseases. This element has half-life of 36 hours. If there is 200 mg 1 this technetium present, how much will be left in six hours.
Data: Half-life = T ½ = 6 hours Time = 36 hours Original quantity = 200 mg Required: Remaining fraction = N = ? Formula: N=Remaining fraction = original(N0) Solution: During 36 hours, t = half-life period = time / half life
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1 2t
36 =6 6
So, N=Remaining fraction = original(N0) 200
1 2t
1 26
N = 3.125mg Result: Remaining amount = N = 3.125mg Q.18.5: Half-life of a radioactive element is 10 minutes. If the initial count rate is 368 coui per minute, find the time for which count rates reaches 23 counts per minute.
Data: Half-life = T ½ = 10 min Initial count rate = 368 per min Final count rate = 23per min Required: Time taken = ? Formula: Time Taken
Initial count Half Life Final Count 2
Solution: Time Taken
Initial count Half Life Final Count 2
Time Taken
368 10 23 2
Time taken = 40 minutes Result: Time taken = 10 minutes
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Q.18.9: Ashes from a campfire deep in a cave show carbon-14 activity of only one-eighth the activity of fresh wood. How long ago was that campfire made?
Data: Half-life of Carbon (C – 14) = T1/2 = 5730 years No. of half-life = N/ No =
1 8
Required: Time = T = ?
Formula: T = T1/2 (half-life) × t (No. of half lif) N 1 t No 2
Solution: As we know N 1 t No 2
Given is N 1 No 8 N 1 3 No 2
So t=3 Now T = T½ × t = 5730 × 3 T = 17190 years Result: Dedicated to My Sweet Father…… whatsapp group Career Opportunities
Time = T = 1790 years
Dedicated to My Sweet Father…… whatsapp group Career Opportunities
whatsapp group Career Opportunities