simple harmonic motion & waves...q.10.2: a pendulum of length 0.99 m is taken to the moon by an...

Author By: Adeel Zafar Physics (Class 10th)

Dedicated to My Sweet Father……

Unit No. 10

SIMPLE HARMONIC MOTION & WAVES

Answers Keys

1. A 6. A

2. B 7. B

3. C 8. C

4. A 9. B

5. D

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.


(d) Motion of Plucked string fixed at both its ends.

Ans: It is an example of SHM.

(e) Motion of honey bee.




“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.

g

T1

2

If the length is doubled then l1 = 2l, so the new time period will be

g

lT 2

g

lT

22



g

lT 22

g

lT 22

TT 2

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

xF

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:

g

lT 2

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.



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

fV

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:



g

lT 2

Solution:

g

lT 2

Taking square on both sides

eg

lT 22 4

)......(...................................................4 2

2

igeT

l

a) On earth equ. (i) becomes:

2

2

4e

e

gTl

214.34

104el

86.94

40

el

le = 1.02m

b) On moon equ. (i) becomes:

2

2

4m

m

gTl

2)14.3(4

6.14ml

44.39

44.6ml

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?



Data:

Length of pendulum on moon = lm = 0.99m

Time period = T = 4.9sec

Required:

Value of g on moon = gm = ?

Formula:

g

lT 2

Solution:

mg

lT 2

Squaring on both sides:

mg

lT 22 4

2

24T

lgm

2

2

9.4

99.014.34mg

163.1 msgm

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 = ?



Formula:

g

lT 2

Solution:

We know that

a) On earth

e

eg

lT 2

10

114.32eT

1.028.6eT

316.028.6 eT

sec2eT

b) On moon

m

mg

lT 2

67.1

114.32mT

6172.014.32mT

sec9.4mT

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



Required:

Length of pendulum = l = ?

Formula:

g

lT 2

Solution:

We know that

g

lT 2

Squaring on both sides

g

lT 22 4

2

2

4

gTl

214.34

104l

86.94

40

l

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 = ?



b) Time period = T = ?

c) Wave speed = V =?

Formula:

a) t

nf

b) f

T1

c) fV

Solution:

a) Frequency = ?

t

n

takenTime

WavesofNof

20

100f

Hzf 5

b) Time period = ?

fT

1

5

1T

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



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:



a) f

T1

b) t

lv

c) v = fλ

Solution:

(a) f

T1

90

1T

= 0.01sec

5

1T

(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 = ?



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



= 0.2ms–1

Now;

l = vt

0.2 × t = 0.8

t = 2.0

8.0

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 10

8 ms

–1 .

Data:

Frequency = f = 90MHz

= 90 × 106Hz

Speed = v = 3 × 108ms–1

Required:

Wave length = λ =?

Solution: We know that

fv

f

v

6

8

1090

103

λ = 3.33m

Result:

Wave length = λ = 3.33m



Unit No. 11

SOUND

Answers Keys

1. A 6. B

2. A 7. C

3. B

4. C

5. D


“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?



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).


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?



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:

Echo Reflection

1. Incident sound wave must be bounced Incident sound wave can be bounced



back from a hand and rigid surface. back from a hand or soft surface.

2. Echo can be observed only when

minimum distance between listener

and reflective surface is 17m.

Reflection of sound many occur of any

distance.

3. Echo can be observed only when time

difference between the original sound

and echo is 0.1 sec.

No such requirement is necessary for

reflection of sound.

Q8: 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 = dBI

I

o

log10




a) )(log10 dBI

ILoL

o

12

6

10

103log10

= 10log (3.0 × 106)dB

= 10 × 6.47dB

= 64.7dB

b)

o

oI

ILL log10

1210log10100

I

Taking antilog on both sides

Antilog10 = 1210

I

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:

L – Lo = dBI

I

o

log10




L – Lo = dBI

I

o

log10

1210log1080

I

)10log10

80 12 I

Taking antilog on both sides

Antilog 8 = Antilog [log(1012

× I)]

108 = 1012

× I

12

8

10

10I

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

vf

05.0

330f



= 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:

t

nfa )

fTb

1)


takenTime

heartbeatofNofa

.)

60

72f

f = 1.2Hz

fTb

1)

2.1

1T

T = 0.833sec



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:


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

2

Sh

2

2250h

mh 1150

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= ?



Formula:

S = vt


= 346× 5

= 1730m


2

Sh

2

1730h

mh 865

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:


Speed = V = 1531 ms–1

Required

Depth of sea water = h = ?

Formula:

S = vt


= 153 × 3.42

= 5236.02m




2

Sh

2

02.5236h

mh 2618

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

1

1)f

Va

20000

3431

= 1.7 × 10–2

m

2

2)f

vb

20

3432

= 1.7 × 10 –4

m

Result:



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

rfV 2

As

S = vt

1500 = 700 × t

700

1500t

t = 2.1sec

Result: Time = t = 2.1 seconds



Unit No. 12

GEOMETRICAL OPTICS

Answers Keys

1. C 6. D

2. C 7. B

3. B 8. A

4. C 9. B

5. B 10. B


“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.


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?



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.

LengthFocal

lensofPower1

diopterf

lensofPower1

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.


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.



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.



(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:

qpf

111

Solution:

5

1

10

11

f

10

11

f

cmf 10

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:

qpf

a111

)

p

q

h

hb

o

1

)



Solution:

We know that

qpfa

111)

5.10

1

16

11

q

5.116

165.101

q

168

5.51

q

5.5

168

q

q = -30.54cm

We know that

p

q

h

hb

o

i )

oi hp

qh

305.10

54.30ih

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 image = q = 20cm

Required:



Focal length = f =?

Formula:

qpf

111

Solution:

We know that

qpf

111

20

1

20

11

f

20

21

f

10

11

f

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:


Formula:

qpf

111

Solution:

We know that

qpf

111

66.5

1

4.34

11

f



177.0029.01

f

148.01

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:

qpf

111

Solution:

We know that

qpf

111

qfp

111

5.11

1

5.13

11

p

5.11

1

5.13

11

p

25.155

251

p

P = 6 .21cm

Result:



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:

qpf

a111

)

1p

q

h

h

o

i

2

)p

q

h

hb

o

i

Solution:

We know that

qpfia

111))

Pfq

111

3.19

1

7.8

11

q

9.167

7.83.191

q

9.167

6.101

q



6.10

9.167q

q = 15.83cm

1

))p

q

h

hiia

o

i

Putting the values

2.133.19

84.15ih

3.19

09.209ih

cmhi 83.10

b) When the object become twice

p2 = 19.3 ×2

p2 = 38.6cm

2p

q

h

h

o

i

2.133.19

84.15ih

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



Required:


Distance of image = q = ?

Formula:

2

)R

fa

qpfb

111)

Solution:

We know that

2)

Rfa

2

38f

f = 19cm

using the formula

qpfb

111)

50

1

19

11

q

950

19501

q

31

950q

q = 30.64cm

Result:

a) f = 19cm

b) q = 30.64cm

c) Because magnification m = q/p Is positive then image will be upright.



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


Focal length = f = 8cm

Required:

a) Size of image = q = ?

b) Position of image = hi = ?

c) Nature of image = ?

Formula:

qpf

a111

)

p

q

h

hb

o

i )

Solution:

Using the formula

qpfa

111)

qPq

111

Pfq

111

12

1

8

11

q

196

8121

q

96

41

q



4

96q

q = 24cm

p

q

h

hb

o

i )

412

24ih

cmhi 8

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


Focal length = –15 cm

Required:

a) Position of image = q = ?

b) Size of image = hi = ?

Formula:

qpf

a111

)

p

q

h

hb

o

1)

Solution:

Using the formula

qpfa

111)



20

1

15

11

q

60

341

q

60

71

q

7

60q

q = – 8.75cm

p

q

h

hb

o

i )

1020

57.8

hi

cmhi 28.4

Result:

cmqa 75.8) cmhb i 28.4)

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:

qpf

111

Solution:

Using the formula

qpf

111



ppf 3

111

ppf 3

111

pf 3

131

pf 3

21

3p = 12

p = 4cm

Result:


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:

Sinr

Sinina )

CSinnb

1)

Solution:

We know that

nSinr

Sinia )



25.1

35oSinSinr

25.1

57.0Sinr

Sinr = 0.456

(0.456) 1-Sin r

27.13 = r

For critical angle

nSinC

1

25.1

11SinC

8.01 SinC

ansC 13.52

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:

p

fa1

)

qpfb

111)

Solution:



5

1) fa

f = 0.2

1002.0 f

f = 20cm

Using this formula

qpfb

111)

q = 2p

pp 2

11

20

1

p2

12

20

1

2032 p

602 p

p = 30cm

Result:

p = 30cm



Unit No. 13

ELECTROSTATICS

Answers Keys

1. B 7. D

2. B 8. A

3. B 9. D

4. A 10. B

5. B 11. B

6. C



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?



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

2

1

rF

It is clear from the above equation that if we double the distance between the two

charges, becomes ¼th of the original force.


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

nCCCCCe ....321

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”:



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

19

11

1069.1

105.7

n

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



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.



ELECTROSTATICS

Numerical Problems

Q.13.1: 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× 10–6

C

Charge on an electron = e–1

= 1.6 × 10–19

C

Required:

No. of negative particles = ?

Formula:

Q = ne Solution:

Using this formula

neQ

e

Qn

C

Cn

19

6

106.1

10100

6.1

101010 1962

n

1510

6.1

1

1610

16

1

= 0.0625×10

16

= 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:



First point charge = q1 = 10μC

Second point charge = q2 = 5μc

Distance = r = 150cm 100

150 = 1.5m

Required:

Magnitude of force = F = ?

Direction of Force = ?

Formula:

2

21

r

qKqF

Solution:

As we know

2

21

r

qKqF

2

669

5.1

10101010109 F

5.22

45F

NF 2.0

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:



2

21

r

qKqF

Solution:

As we know

2

21

r

qqkF

As; q1 = q2 = q

2r

qqkF

2

2

r

KqF

k

Frq

22

9

2

2

109

1.08.0

Nq

9109

008.0

9109

008.0

q

Cq 7104.9

Cq 7104.9 Result:

Value of each charge = q1 = q2 = Cq 7104.9

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



r2 = 0.02m

Required:

Force = F2 = ?

Formula:

2

1

211

r

qKqF ……………..1

2

2

212

r

qKqF …………….11

Dividing equation 1 by equation 2

2

2

21

2

1

21

2

1

r

qKq

r

qKq

F

F

1

2

2

2

2

1

F

F

r

r

2

2

2

112

r

rFF

Solution:

As we know

2

2

2

112

r

rFF

2

2

202.0

05.01.0 F

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



= 100×10–6

= 10–4

C

Required:

Work done = W = ?

Formula:

q

WV

Solution:

As we know

q

WV

W = qV

Putting the values

W = 10 –4

×104

W = 1J

Result:

Amount of work = W = 1J

Q.13.6: 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)



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

V

QC

9

06.0C

C = 6.67 × 10

–3F

Result:

Capacitance = C = 6.67 × 10–3

F

Q.13.8: 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:



Equation Number 1

Q1 = CV1

Equation Number 2

Q2 = CV2

Dividing equation 1 by equation 2

2

1

2

1

CV

CV

Q

Q

So,

2

1

2

1

V

V

Q

Q

Solution:

Since

2

1

2

1

V

V

Q

Q

2

6

2

03.0

V

203.0

62 V

V2 = 400 V Result:

Voltage = V = 400V

Q.13.9: 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:



a) Equivalent Capacitance = Ceq= ?

b) Charge on each capacitor = Q = ?

c) Potential difference across each capacitor:

i) V1 = ?

ii) V2 = ?

Formula

a) 321

1111

CCCCeq

b) Q = CV

c) Q = CV

Solution:

a) Since capacitors are connected in series

21

111

CCCeq

Putting the values

FFCeq 12

1

6

11

FCeq 12

121

FCeq 12

31

FCeq 3

12

Ceq = 4µF

b) Q = CV

Q = 4 × 10–6

× 12

Q = 48μF

c) i) Q = C1V1

C

CV

6

481

V1 = 8μF

ii) Q = C2V2

C

CV

12

482



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= C1V



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

Answers Keys

1. D 6. D

2. C 7. A

3. B 8. C

4. C 9. B

5. B


“Quick Quiz”

Q1. How long does it take a current of 10mA to deliver 30C of charge?

Ans: Data:-

I = 10mA

= 10 × 10–3

A

Q = 30C



t = ?

Formula: -

t

QI

Solution: -

As we know that

t

QI

Putting the values in the above equation:

t

301010 3

31010

30

t

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.


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.



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



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:

t

wP

Solution:

As we know that

t

wP

Putting the values in the above equation:

40

2400P

P = 60w

Result:

Hence power of the bulb is 60w.

. “Point to Ponder”

Q: How many faults can you find after studying this picture?



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

voltage.

A group of cells connect together is

called battery.

It is identified in the circuit by the

symbols.

It is identified in the circuit by the

symbol as

Q14.3: Can current flow in circuit without potential difference?

Ans: According to Ohm’s law.

V α I

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?



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:

JKWh 6106.31

KwhJ6106.3

11

whJ 1000106.3

11

6

whJ 000278.01

So,

whJ 1000000278.01000

whJ 278.01000

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.



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:

RIP 2

From the above equation it is clear that the bulb having more resistance will have

more power and brightness.

According to Joule’s law:

RtIw 2

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.



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 × 10–3

A

Time = t = 1 min

= 60Seconds

Required:

Charge = Q = ?

Formula:

T

QI

Solution:

We know that

T

QI

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:



V=IR

Solution:

We know that

a) V=I1R1

1

1R

VI

100000

121 I

I1 = 1.2×10

–4

b) V = I2R2

2

2R

VI

1000

122 I

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



R

VI

61010

100

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

IR = V

= I(V)t

=IVt

= 1.5 × 10 × 120

W = 1800J

Result:



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 = I2

V = IReq

eqR

VI

103×10

10I

= 1 × 10-3

A



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) 21

11

Re

1

RRq

b) V = IR

Solution:



a) 21

111

RRReq

33 1012

1

106

11

eqR

31012

12

31012

3

3104

11

eqR

Req = 4 × 103Ω

Req = 4KΩ

b) i) Quantity of current in first resistance = 1

lR

VI

3l106

6I

I1 = 1 × 10–3

I1 = 1mA

ii) Quantity of current in Second 2

2R

VIR

I2 31012

6

I2 = 0.5 × 10–3

I2 = 0.5mA

c) Circuit is parallel, so the voltage remain same

6V = V

Result:

a) Req = 4KΩ

b) i) I1 = 1mA

ii) I2 = 0.5mA



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) KwtP

E1000

Solution:

a) R

VP

2

P

VR

2

100

2202

100

48400

= 484Ω

b) Energy in hour =1000

TimePower

1000

305100

=15 Kwh Result:

a) R = 484Ω



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 = I2R & V = IR

Solution:

We know that

R

PI 2

95

1502 I

5784.12 I

5784.12 I

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.



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 1000

hourPower

Solution:

a) Energy consumed by bulb

1000

hourPower

1000

305600

= 90 units

b) Energy consumed by fans

1000

hourPower

1000

3010300

= 90 units

c) Energy consumed by T.V.

1000

hourPower

1000

302250

= 15 units



d) Energy consumed by Iron

1000

hourPower

1000

3021000

= 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) 11 VIP

V

PI 1

1

250

1001 I



5

21 I

I1 = 0.4A

ii) P2 = VI2

V

PI 2

2

I2250

104 3

250

40002 I

I2 = 16A

b) i) V = I1R1

1

1I

VR

4.0

2501 R

R1 = 625Ω

ii) V = I2R2

2

2I

VR

16

2502 R

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.



(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

Answers Keys

1. D 6. B

2. B 7. D

3. D 8. B

4. A 9. C

5. C


“Activity”



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.


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 2

nd 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 2

nd

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 3

rd iron ban. If

they show repulsion then these 1st and 3

rd gars are magnets and 2

nd is simple iron rod

but if they show again attraction then the 1st bar is simple iron bar and 2

nd and 3

rd 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.



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

energy into electrical energy.

A motor converts electrical energy

into mechanical energy.

In generator, firstly coil is rotated In motor, firstly electricity is supplied



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

should be energized

Both of the coil and magnet should

be energized.

Its principle is changing flux includes

e.m.f or current in the coil.

Its principal is torque acting on a

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.



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:

P

S

P

S

V

V

N

N

Solution:

As we know that

P

PS

SV

NVN

240

200012SN

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:

P

S

P

S

V

V

N

N



Solution:

As we know that

P

P

S

S NV

VN

201

100SV

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:

P

S

P

S

V

V

N

N &

S

P

P

S

V

V

I

I

Solution:

As we know that

P

P

SS V

N

NV

170100

1SV

VS = 1.7V

For ideal transformer

PP = PS



S

P

P

S

V

V

I

I

S

PP

SV

VII

7.1

170101 3

SI

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) P

S

P

S

V

V

N

N

b) S

P

P

S

V

V

I

I

Solution:

a) As we know that

P

S

P

S

V

V

N

N

P

P

SS N

V

VN



240

400012SN

NS = 200

b) for Ideal transformer

PP = PS

S

P

P

S

V

V

I

I

P

SSP

V

VII

240

124.0 PI

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

V

PI

3

6

10250

10500

I

I = 2kA

I = 2000A



Result:

I = 2000A



Unit No. 16

BASIC ELECTRONICS

Answers Keys

1. D 6. A

2. D 7. B

3. C

4. D

5. C


“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 Input (A) Input (B) Out Put (C = A + B)

a 1 0 1

b 0 1 1

c 1 1 1


Q16.1: Name two factors which can enhance thermionic emission?

Ans: Following are some factors which can enhance thermionic emission.



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 BF . evBSinF

By increasing the speed of electron as VF . evBSinF

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



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 Logic operation

(x = A.B)

Multiplication

(X = A × B) A 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

NAND

BAX . (A) (B)

0 0 0 1

0 1 0 1

1 0 0 1

1 1 1 0

This shows that NAND gate is the reciprocal of AND gate.

Q16.8: Show that the circuit given as below acts OR gate.



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 Outputs

(A) (B) BAx xy

0 0 1 0

1 0 0 1

0 1 0 1

1 1 0 1

Hence the above circuit act as an OR gate.

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:

Input Ax 1 Bx 2 21 xxx Output

xy A B

0 0 1 1 1 0

1 0 0 1 1 0

0 1 1 0 1 0

1 1 0 0 0 1

Hence the above circuit acts as an AND gate.



Unit No. 17

INFORMATION AND COMMUNICATION

Answers Keys

1. C 6. D

2. B 7. B

3. D

4. C

5. C


“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.


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:



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

Memory”.

It is elaborated as “read only

memory”

In reference with the processor the

information stored in RAM is easily

accessed.

The processor cannot access directly

that is stored in ROM in order to

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

performed over the information.

The ROM memory only allow the

user to read information user cannot

make any changes to it.

RAM memory is used to store the

temporary memory.

ROM memory is used to stored

permanent information and cannot be

deleted.

The accessing speed of RAM is

faster. It assists the processor to boost

up the speed.

Speed of ROM is slower in

comparison with RAM. ROM cannot

boost up the speed of processor.



Unit No. 18

ATOMIC AND NUCLEAR PHYSICS

Answers Keys

1. A 6. None

2. B 7. B

3. B 8. A

4. D 9. D

5. B

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.

For example element of hydrogen has three different types of atoms i.e. protium

H

1

1

, deuterium

H

1

2and tritium

H

1

3.

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.



Q4: What is the different between natural and artificial radioactivity?

Ans:

Natural Radio Activity Artificial Radioactivity

The spontaneous emission of radiation

by unstable nuclei is called natural

radio activity.

The process of coverting stable nuclei

into unstable unclei by bombarding

them with protons, neutrons or alpha

particles.

All natural radioactivie nuclear

reaction are spontaneous.

All artificial radioactive nuclear

reaction are non-spontaneous.

Natural radioactivity comes from

radioactive elements present in nature.

Artificial radioactivity comes from

elements created in nuclear reactors

and accelerators.

Elements having atomic mass number

more than 82 show natural

radioactivity.

Elements having atomic mass number

less than 82 show artificial

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.

YXX O

O

A

z

A

Z



Numerical Q.18.1: The half-life of N16

7 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 N16

7 = 7.3

Time Observed = 29.2 sec

Required:

Remaining fraction of radioactive isotope =?

Formula:

N=Remaining fraction = original(N0) t2

1

Solution:


1

As we know

t = half-life period = time / half life

3.7

2.29 = 4

So,

tNoN

2

1

42

1

oN

N

16

1

oN

N

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:



Remaining fraction = N = ?

Formula:


1

Solution:


595.425.5

26t

So,

During 26 years, five half-lives are escaped

tNoN

2

1

52

1 oNN

25

1

oN

N

32

1

oN

N

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.

8

1

oN

N

Required:

Time =?

Formula:

Time = No. of half-lives × half life



Solution:

As the quantity of C – 14 drop to 1 / 8th

of the original quantity

So,

32

1

8

1

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:


1

Solution:

During 36 hours,




6

36 = 6

So,


1

62

1200

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:

2

LifeHalf

CountFinal

countInitialTakenTime

Solution:

2

LifeHalf

CountFinal

countInitialTakenTime

2

10

23

368TakenTime

Time taken = 40 minutes

Result:

Time taken = 10 minutes



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 = 8

1

Required:

Time = T = ?

Formula:

T = T1/2 (half-life) × t (No. of half lif)

t

oN

N

2

1

Solution:

As we know

t

oN

N

2

1

Given is

8

1

oN

N

32

1

oN

N

So

t = 3

Now

T = T½ × t

= 5730 × 3

T = 17190 years

Result:



Time = T = 1790 years

simple harmonic motion & waves...q.10.2: a pendulum of length 0.99 m is taken to the moon by an...

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