232

232
FORM THREE PHYS
HANDBOOK
[
Wit
hwel
ldr
awndi
agr
ams
,sol
vedexampl
esandques
tionsf
orexer
cis
e]
(2015 Edition)

Mwalimu Lucy Anyanga

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page1

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 232

Tabl
eofCont
ent
s
ACKNOWLEDGEMENT Page2

BRI
EFPERSONALPROFI
LE Page2

GUI
DELI
NESI
NMYLI
FE Page2

Chapt
er1 LI
NEARMOTI
ON Page3

Chapt
er2 NEWTON’
SLAWSOFMOTI
ON Page12

Chapt
er3 WORK,
ENERGY,
POWERAND Page19
MACHI
NES

Chapt
er4 REFRACTI
ONOFLI
GHT Page28

Chapt
er5 GASLAWS Page37

Chapt
er6 WAVESI
I Page43

Chapt
er7 CURRENTELECTRI
CITYI
I Page51

Chapt
er8 HEATI
NGEFFECTOFAN Page61
ELECTRI
CCURRENT

Chapt
er9 ELECTROSTATI
CSI
I Page64

Chapt
er QUANTI
TYOFHEAT Page71
10

Acknowl
edgement
Fi
rstandf or
emostIt
hankt
heAl
might
yGodf
ort
hegi
ftofl
i
fe,ener
gy,knowl
edgeandski
l
lst
o
pursuet
hiswork.
Iam ver
ygrat
efultotheenti
reNyabururuGir
ls’Hi
ghschoolfr
aterni
tyf
orgener
ousl
ysuppor
ti
ngme
al
lroundasIwor kedont hi
smateri
al.Imustspecifi
call
yappreci
atet
heH.O.DPhysi
csNyabur
uru
Gir
ls’
Mr.Al
bertO.Onditif
orthesupportandencouragement
.
Thesuppor tbyMatongoSecondarySchoolSci
encedepar
tmentmembers,Mr.OnyanchaandMr.
MisatiofPhysi
cs,Mr.Ondi
ekiofChem andMadam Abi
gael
,Pr
isci
l
laandJaelofChem/Biomustbe
appreciat
ed.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page2
 232

Thecareandbestwi shesIr
ecei
vedfrom mymot herJoyceMokei
raandmysi
bli
ngsdeser
ve
speci
alat
tent
ion.Theywereagr
eatsour
ceofencour
agement.

Li
nest
hati
nfl
uenceact
ivi
ti
esi
nmyl
i
fe
1.Godi
sal
wayst
her
etoassi
stpr
ovi
dedyouaskf
orHi
m.
2.Ati
tsbest
,Physi
csel
i
minat
escompl
exi
tybyr
eveal
i
ngunder
lyi
ngsi
mpl
i
cit
y.
3.Ther
eisnomet
hodofchangi
ngyourf
ateexceptt
hroughhar
dwor
k.
4.Cohesion wi
thi
mmedi
ate nei
ghbour
s and det
ermi
nat
ion al
ways bet
ter
s youri
mmedi
ate
envi
ronment.
Br
iefPer
sonalPr
ofi
l
e(Cel
lPhone–0726593003)
Chweya,N.E.i
saPhysics/
Chemist
ryteacher
.Hei
saFirstCl
assHonorsB.
Edgraduat
efrom Moi
Uni
versi
ty(
Chepkoi
l
el)
.Healsohasprof
oundknowl
edgei
ncomputerappl
i
cat
ionsandgr
aphics.

Chapt
erOne LI
NEARMOTI
ON

Speci
fi
cobj
ect
ives Cont
ent

Byt
heendoft
hist
opi
cthel
eanershoul
dbeabl
eto: 1. Dist
ance,di
splacement
,speed,vel
oci
ty,
accel
erat
ion(experi
mental
treat
mentrequi
red)
a) Defi
nedistance,
displ
acement
,speed,
vel
oci
tyand
accel
erat
ion 2. Accel
erat
ionduet
ogr
avi
ty

b) Descr
ibeexper
iment
stodet
ermi
nev
eloci
tyand  f
reef
all
,
accel
erat
ion
 Si
mpl
ependul
um met
hod
c) Det
ermi
neaccel
erat
ionduet
ogr
avi
ty
3. Mot
ion-t
imegr
aphs
d) Pl
otandexpl
ainmot
ion-t
imegr
aphs
 Di
spl
acement
-t
imegr
aphs
e) Appl
ytheequat
ionofuni
for
mlyaccel
erat
edmot
ion
 Vel
oci
ty-t
imegr
aphs
f
) Sol
venumer
ical
quest
ions.
4. Equat
ionsofuni
for
mlyaccel
erat
edmot
ion

5. Pr
obl
emsonuni
for
mlyaccel
erat
edmot
ion

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

I
ntr
oduct
ion i
i
i.Speed

 Thi
stopi
cdeal
swi
thst
udymot
ionofbodi
esi
na Thi
sistherat
eofchangeofdi
stancecover
edbya
st
rai
ghtl
i
ne. movingbodywit
hti
me.Speedisascalarquant
it
y.
di
sta
ncec
overe
d
Ter
msAssoci
atedwi
thLi
nearMot
ion s
pee
d=
ti
metake
n

i
.Di
stance Forabodymov
ingwi
thanon-
uni
for
m speed,

 Di
stancei
stheact
uall
engt
hcov
eredbyamov
ing tot
aldi
sta
ncec
overe
d
a
ver
ages
pee
d=
body.I
thasnospecif
icdi
rect
ionanditi
sther
eforea tota
lti
metake
n
scal
arquant
it
y.TheSIuni
tofdist
anceismeter(m)
 Instant
aneous speed r
efer
st othe rat
e of
changeofdi stanceofamov i
ng bodyata
point(aninst
ant).TheSIuni
tofspeedisthe
i
i
.Di
spl
acement
,s
met
r ms-1)
epersecond(
 Thisisthedistancecover
edbyamov i
ngbodyina
specif
ied di
recti
on.Displacement isther
ef ea
or
vectorquant
it
y.TheSIunitofdispl s met
acementi r
e i
v.Vel
oci
ty
(m)
 Thisist her at
eofchangeofdi spl
acement
I
ll
ust
rat
ingdi
stanceanddi
spl
acement withtime.Itcanalsobedefi
nedasthespeed
i
naspeci fi
eddi r
ecti
on.Vel
oci
tyi
sther
ef ea
or
Consi
derthe di
agr
am bel
ow showing moti
on ofa
bodyst
art
ingf
rom poi
ntAandmovinginthedi
rect
ion
vectorquanti
ty.
shown.
cha
ngei
ndi
spl
aceme
nt
ve
loc
ity=
ti
met
aken

Forabodymov
ingwi
thav
ary
ing
v
eloci
ty,

tota
ldispl
ace
me nt
a
ver
ageve
loc
ity= .
a) AtpointB,dist
ance cov
ered i
s AB whil
et he total
timetake
n
di
spl
acementoft
hebodyisABinthedi
rect
ionAB
 TheSIuni
tofv
eloci
t st
yi hemet
repersecond
b) Atpoi
ntC,di
stancecover
edisAB+BCwhi
l
ethe ms-1)
( .
di
spl
acementi
sACi nthedir
ect
ionAC

c) Whenbackatst
arti
ngpoi
ntA,di
stancecover
edi
s
AB+BC+CAwhi lethedi
spl
acementiszer
o. v
)Accel
erat
ion

 Thi
sisthechangeofv
eloci
typeruni
tti
me.I
t
sav
i ect
orquant
it
y.

cha
ngeinvel
oci
ty
Ac
cel
era
tion=
t
imeta
ken

fi
nal
vel
oci
ty-i
nit
ialve
loc
ity
Ac
cel
era
tion=
ti
metaken

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page4
 232

Sol
uti
on

v-u -1
55556×104ms ,v= 0ms
0×105kmh =5.
u=2.
-1 -1

a= e,vi
,wher sthefi
nalvel
oci
ty,u i
sthei
nit
ial
t -1
vel
ociyandti
t sthet
imetaken. v-u ( 55556×104)ms
0-5. -2
a= ;
a= 777778×106ms
=-2.
t 0×10-2s
2.
I
nstant
aneousaccel
erat
ion i
stheacceler
ati
onofa
The
refor
eret
arda
t i
onis-(-2.
777778×106ms )
-2
bodyatapoi . Decel
nt erat
ionorret
ardati
on i
sthe =2. 6
777778×10 ms
-2

negati
veaccel
erat
ioni
nwhichabodymoveswi t
ha
decreasi
ng vel
oci
ty wit
h ti
me. The SI uni
t of Exer
cise
-
2
accel
erat
i smet
oni repersquar
esecond(
ms )
.
1) Avanonast raightroadmoveswi t
haspeedof180
-1
Exampl
es kmh f 0r45minutes,andthencl
i
mbsanescarpment
wihaspeedof72kmh-1f
t or30minut
es.Cal
cul
ate:

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

1.Abodymoves30m dueeastin4seconds,t
hen40 I
. Theaver
agespeedoft
hevan
m duenor
thi
n8seconds.Det
ermi
ne:
I
I. Theaver
ageaccel
erat
ionpr
oduced
a) Thet
otaldi
stancemovedbyt
hebody.
2) Agi r
lruns40m duesouthin40secondsandt
hen20
b) Thedi
spl
acementoft
hebody. m duenorthi
n10seconds.Cal
cul
ate:

c) Theaver
agespeedoft
hebody. I
. heraver
agespeed

d) Theaver
agevel
oci
tyoft
hebody. I
I. heraver
agevel
oci
ty

Sol
uti
on I
II
. herchange i
n vel
oci
tyf
ort
he whol
e
j
ourney

I
V. Theaccel
erat
ionpr
oducedbyt
hegi
rl
.

Measuri
ng Speed,Vel
oci
tyand Accel
erat
ion Usi
ng
Ti
ckerTi
mer

 Atickertimerhasanarm whichvibr
atesregularl
y
a
) Tot
aldi
sta
nce=t
ota
lle
ngt
hcove
red
due t o changi
ng curr
entint he mains supply
30m +40m =70m (al
ternati
ng current
).As the arm v i
brates,it
makesdot sont hepapertapewhichi smov ing
undert hear m.Successi
vedot saremar kedat
t
hesamei
nter
valoft
ime
b) Tot
aldi
spl
ace
me s(302+402)m
nti
 Mostt i
ckerti
meroperatesatafrequencyof50
2500m =50monbe
ari 87o
ng36.
hert
z( 50Hz)i.e.50cycl
espersecondi .
e.they
t
ota
ldi
sta
ncec
ove
red make 50 dot s persecond.The ti
me interval
c) Aver
ages
pee
d= 1
t
ota
lti
met
ake
n betweent woconsecut
ivedot
sis: =0. 02sfora
50
70
833ms-1
= =5.
50Hzt ickert
imer.Thi
st i
meint
ervaliscaleda
l
12 ti
ck.
tota
ldispl
ace
me nt
d) ve
rag
eve
loc
ity=
total
timetake
n

50
Sampl
esecti
onsoft apesar
easshownbelow.The
167ms-1onbe
= =4. arng3
i 87o
6. arr
owshowsthedi
recti
oninwhi
cht
hetapei
spull
ed.
12

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page6
 232

2. Thespeedofaofbodyr ol
li
ngonanincl
inedpl
aneis10
ms-1 whent i
mei s0satti
met=10st hespeedoft he
-
1
bodyi sfoundtobe25ms .I ft
hebodyi
smovingi nthe
same di r
ecti
on thr
oughout,cal
cul
atet he average
accelerat
ionofthebody

Sol
uti
on
v-u
a=
t

( ms-1
25-10)
a=
10-0)s
(

-1
15ms -2
a= =1.
5ms
10s

3. Aparti
clemovingwi
thavel
ocit
yof2. 0X105kmh-1i
s
broughttor
esti 0X10-2s.cal
n2. culatet
heaccel
erat
ion
ofthebody,hencet
heret
ardat
ion.

Exampl
es I
. Cal
cul
atet
heaver
agevel
oci
tyf
ort
hemot
ion

1. Atapeispull
edthroughatickert
imerwhichmakesone I
I. What does t
he ar
ea under t
he st
rai
ght l
i
ne
dotevery second.Ifitmakes t hr
ee dots and t
he r
epresent
?
di
stancebetweent hefi
rstandt hethi
rddotis16cm,
fi
ndthevel
ocityofthet
ape. I
II
. What is the diff
erence bet
ween successi
ve
sect
ionsoftape?
Sol
uti
on
I
V. Cal
cul
atet
heaccel
erat
ionoft
het
rol
l nms-1
eyi

Mot
ionGr
aphs
f=1Hz  Gr aphscan beused torepr
esentv ari
ati
on of
Ti
mebe
twe
enc
ons
ecut
ivedot
s=1s
,tot
alt
ime=1sX2 di
s t
ance,speed,v
e l
oci
tyora ccel
erat
ion ofa
=2s mov i
ngbodywitht
ime.Whenusedt hi
swayt hey
ar
ecal l
edmoti
ongraphs
Di
sta
ncebe
twe
en1s
tand3r
ddot
s=16c
m =0.
16m

t
otal
displ
ace
me nt 0.
16m
a
ver
ageve
loc
ity= = =0.
08ms-1
t
otalt
imetake
n 2s Di
spl
acement–Ti
meGr
aphs

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

3. Thetapeinthefigur
ebelow wasproducedbyat icker
ti
merwithafrequencyof100Hz.Findtheaccel
eration
ofthet
rol
leyt
hatwaspull
ingthetape. Sol
uti
on 1. St
ati
onar
ybody

 Displ
acementdoesnotchangewithti
me,since
displ
acementi
sav ect
orquant
it
ytheposit
ionof
thebodymaybenegat iv
eorposit
iver
elat
iveto
beobserver
.
Sol
uti
on

1
t
imebe
twe
enc
ons
ecut
ivedot
s= =0.
01s
100

0.005m
i
nit
ial
vel
oci
ty,
v= 5ms-1
=0.
0.01s

0.025m
f
ina
lve
loc
ity= 5ms-1
=2.
0.01s

t
imet
ake
n=4×0.
01=0.
04s

v-u (2. 5)
5-0. ms-1 200
a
cce
ler
ati
on= = = =50ms-2
t 0.
04s 0.
04

Exer
cise 2. Abodymov
ingwi
thuni
for
mvel
oci
ty

1. Thef i
gurebel
owshowsapi eceoftapepull
edt ough  For a body mov
hr i
ng with uniform v elocit
y,
atickerti
merbyat rol
l
eydownani ncl
inedplane.The displacementchangesunif
orml
yov erequaltime
fr
equencyofthetickert
imeri
s50Hz i
nt er
v al
s.The graph ofdispl
acementagai nst
ti
mei sast r
aightli
newhosesl opeorgr adient
I
. Whatt ypeofelectr
iccur
renti
susedtooper at
ethe
representsthe vel
ocit
y ofthe body whi ch is
ti
ckert
imer?
const ant
.
I
I. Cal
cul
atethe aver
age vel
oci
tyf
or t
he t
rol
l
ey
bet
weenAandB

2. Thefigurebel
ow showsat apechar tfr
om t
hepaper
tapeobtai
ned(
frequencyoft
ickert
imer50Hz)

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page8
 232

3. Abodymov
ingwi
thv
ari
abl
evel
oci
ty

 a.Forabodymov i
ngwi hav
t eloci
tyi ncr
easing
unif
ormlywi t
ht ime,thedisplacement -t
imegr aph
i
sacur veofincr easi
ngslopesi ncet hedistance
thebodycov er
si ncreasesforequal t
imei nt
ervals.
Theslopeoft hegr aphatanygi venpoi ntgives
i
nstantaneousv eloci
tyofthebodyi .
e.

∆s
i
nst
ant
ane
ousve
loc
ity=
∆t

2. A body mov
ing wi
th i
ts v
eloci
ty changi
ng
uni
for
mly

 The gradi
entoft hi
s graph i
s a strai
ghtli
ne;
meaning t
hatv el
ocit
ychanges uni f
ormlyover
equaltimeint
erval
s.Thisgradientgraphgiv
es
∆V
const
antaccel
erati
oni.
e.acc
ele
ra t
ion=
∆t

 b.Forabodymov ingwi hav

t eloci
tydecreasing
unif
ormlywi
thti
me,t hedispl
acement -t
imegr aph
i
scur veofdecr
easingslopesi ncet hedistance
thebodycov
ersdecreasesforequalti
mei nterval
s
asshownbelow.
3. Abodymovi
ngwi
thi
tsv
eloci
tychangi
ngnon-
uni
for
mly

 a) For a body mov ing with an i ncreasi

ng
acceler
ati
on, meani ng that its v elocit
y i s
i
ncreasingatani ncreasi
ngrate,t
hev elocit
y-t
ime
graph is curve ofincreasi
ng slope as shown
al
ongside.Thesl opeoft hegraphatanypoi nt
gi
vest heinstant
aneousaccel er
ationoft hebody
atthatpoint.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Vel
oci
ty–Ti
meGr
aphs

1. Bodymov
ingwi
thuni
for
mvel
oci
ty

 The slope/
gradientoft he gr
aph i
s zero and
ther
eforeaccelerat
ionofthebodymov ingwith
unif
ormv el
oci
tyiszero.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page10
 232

 b. For a body mov i

ng wi h a decr
t easi
ng
accelerat
ion, meaning that its v eloci
ty is
i
ncr easi
ngatadecr easi
ngrate,thevelocit
y-t
ime Sol
uti
on
graph is curv
e ofdecreasi
ng slope as shown
 ab-t
hevelocit
yoft hecarincreasesuni
for
mly
below. fr
om r
est(i
.e.i
taccel
erat
esuni
formly)

 bc-theveloci
tyofthecardecr
easesuni
for
mlyt
o
rest(i
.e.i
tdecel
erat
esunif
ormly)

 cd-vel
oci
tyofthecarincreasesuni
for
mlybuti
n
opposi
te dir
ecti
on (acceler
ates uni
for
mly i
n
opposi
tedi
rect
ion)

 de-vel
oci
tyofthecardecreasesuniforml
ybutsti
l
l
i
nsameopposi tedir
ecti
on( decel
eratesuni
for
mly
i
nopposit
edirecti
on)

2,A carst art

ing fr
om r estacceleratesunif
ormlyfor5
minutestoreach30ms-1.I tcont
inuesatt hi
sspeedforthe
next20mi nut
esandt hendeceleratesuniforml
ytocomet o
stopin10mi nutes.Ontheaxespr ovided,sket
chthegraph
Ar
eaunderv
eloci
ty–t
imegr
aph ofvelocit
yagai nstti
mef orthemot ionoft hecarand
hence,f
indthet ot
aldi
stancecoveredbyt hecar.Sol
uti
on
 Considerabodyst ar
ti
ngfrom r
estmov ingwi t
h
constantaccel
erat
ionfort
ime,t
.Thevelocit
y-t
ime
graphforthebodyisasshownalongsi
de.

Tot
aldi
sta
nce=t
ota
lar
eaunde
rtheg
raph

1
Tot
aldi
sta
nce= ( 60×30(
35+20) i.
e.a
reaofat
rape
zium)
2

Tot
aldi
sta
nce=49500m

Exer
cise

1. Thefi
gurebel
ow showsthedi
spl
acementt
imegr
aph
ofmoti
onofaparti
cle.

di
sta
ncet
rave
lle
d=a
ver
ageve
loc
ity×t
ime

= (O+V
2 ) 1
×t= vt
2

 Theref
ore,areaAundertheveloci
ty–ti
megr aph
repr
esentsthedi
stancecover
edbyt hebodyaft
er
ti
metseconds.
St
atet
henat
ureoft
hemot
ionbet
ween:(
i)AandB(
ii
)B

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Exampl
es andC(
ii
i)CandD

1. I
nter
prett
hegraphbel
owr
epr
esent
ingmot
ionofacar
f
rom poi
ntat
oe.
2. Thefi
gurebelow showsavel oci
tyti
megraphfort
he
moti
onofacer t
ainbody.Descri
bethemotionoft
he
bodyi
ntheregion:(i
)OA(ii
)AB( i
i
i)BC.

3. Thef i
gurebel
ow showst heaccel
erat
iontimegraph
foracertai
nmotion.Ontheaxesprovi
ded,sket
chthe
displ
acement-
ti
megr aphf
orthesamemot i
on.

Therefore, for a body moving with uni

for
m
accelerati
on,anyoftheset
hreeequat
ionsmaybe
used.

1.
v=u+a
t

12
2.s=ut+ at
2

3.v2=u2+2a
s

Exampl
es

1. A carstar
tsfrom r
estwi
thuni
for
m accel
erat
ionof
5ms-2.How l
ongdoesitt
aketocoveradi
stanceof
400m?

Sol
uti
on

a=5ms-2

S=400m

Equat
ionsofUni
for
mlyAccel
erat
edMot
ion t=?

 Consi
derabodymov inginastrai
ghtli
newit
h U=0ms-1
uni
for
m accel
erat
ion a ,so t
hat it
sv el
oci
ty

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page12
 232

i
ncr
easesf
rom anini
ti
alv ueut
al oaf
inalv
aluev 12
Be
ste
qua
tionf
orus
eis
;s=ut+ at
2
i
ntmetandi
i ti
sdispl
acedbys;
1 2
400=0×t+ ×5×t
2

52
Der
ivat
i he1stequat
onoft ion 400= t
2

cha
ngeinvel
oci
ty ∆v
a
cce
rat
ion,
a= = 400×2
t
imeta
ken ∆t t= =12.
65seconds.
5
v-u
a=
t

a
t=v-u 2. A bodyi sunif
ormlyaccel
erat
edf r
om r
esttoafi
nal
velociy of 100ms-1 i
t n 10seconds.Cal
cul
atet
he
v=u+a
t distancecover
ed.

Sol
uti
on

u=0
Der
ivat
ionof2ndequat
ion
v=100ms-1
di
spl
ace
ment
,s=a
ver
ang
eve
loc
ity×t
ime
t=10s
u+v
s=( )×t
2 s=?

butf
rome
qua
tion1,
v=u+a
t r
ighte
qua
tioni
sv2=u2+2as

(u+u+a
t
)
1002=02+2a
s
s= t
2
v-u
Buta=
t
1 2
(
2ut+a
t )
2 100-0
=10ms-2
10
12
s=ut+ at
2 10000m2s-2
s= =500m
2×10

3. Abodywhosei ni
ti
alvel
oci
tyi s30ms-1 moveswit
ha
-
2
Der
ivat
ionof3rdequat
ion const
antr
etar
dati
onof3ms . Cal
cul
atet heti
metaken
fort
hebodytocometorest
.
di
spl
ace
ment=a
ver
ageve
loc
ity×t
Sol
uti
on
u+v
s=( )×t
2 u=30ms-1

butf
rome
qua
tion1,
v=u+a
t a= -3ms-1
v-u
t= t=?
a

v+u v-u v=0ms-1

s=( )×( )
2 a
Ri
ghte
qua
tioni
sv=u+a
t
v2-uv+uv-u2
s= 0=30-3t
2a

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

s=v2-u2
2a -30=-3t

v2=u2+2a
s t=10s

4. Abodymovi ngwi t
hauni f
orm acceler
ationof10ms-2
cover
sadi stanceof320m.I fit
si nit
ialvel
oci
tywas
-1
60ms , cal
culatei
tsfi
nalvel
oci
ty.

a=10ms-2

s=320m

u=60ms-1

v=?

2
Ri
ghte
qua
tioni
sv =u2+2a
s b)Ri
( ghte
qua
tioni
s

2
v2=602+2×10×320 v =u2+2g
s

2
v= 3600+6400 20 =02+2×10×s

= 10000 400
s= =20m
20
=100ms-1

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page14
 232

Mot
ionunderGr
avi
ty 2. Abulletshotver
ti
call
yupwardsr
isest
oamaxi
mum
hei
ghtof1000m.Deter
mine:
Acceler
ati
on (In Free Fal
l
) and Decel
erat
ion (
In
Vert
icalPr
oject
ion)DueToGr av
ity a) Thei
nit
ialvel
oci
tyoft
hebul
l
et

 Whenabodyi spr oject

edv erti
call
yupwar ds,it b) Thet i
meoff
li
ghtofthebul
let
decel
eratesuni
formly(negati
veaccelerat
ion)due Soluti
on
gravi
tat
ionalpull.When t hi
s body f al
lsf rom
maximum hei ght,i t accelerat
es uni for
mly (a)Atmaximumveloc
ity,
vel
ocit
yisz
e r
o
downwar dsandthisi
scalledfreefal
l. -1
v=0ms
 Consi
derabal
lthr
ownv
ert
ical
l
yupwar
dsf
rom
u=?
ms-1
t
he ground.The gr
aph below shows how the
v
eloci
tyofthebal
lchangeswi t
htimefrom when g=-10ms-1
i
tleav
est hegr
oundunti
lithit
sthegroundagain.
be
caus
ethebul
letde
cel
era
tesupwa
rds
Ai
rresi
stance i
s assumed t
o be negl
i
gibl
e.
Downwardv el
ocityistakentobeposi t
ive.The s=1000m
gr
adientofthi
sgr aphisaconstantwhosev al
ue
sgr
i avit
ati
onalacceler
ati
on,g. Us
ingt
hee
qua
tionv2=u2+2g
s

0=u2+2×(-10)×1000

2
0=u -20000

42ms-1
u= 20000=141.

(
b)Tot
alt
imet
hebul
leti
sina
ir

12
us
ingt
hee
xpr
ess
ions=ut+ gt,
2

di
spl
ace
menti
s0m

1 2
 Equat
ionsofuni
for
mlyacceler
atedbodi
esal
so 0=141. t
42×t+ (-10)
2
appl
yinmoti
onundergr
avi
ty.
141.
42=5t
1.v=u+g
t
141.
42
t= =28.
28s
12 5
2.s=ut+ gt
2
Exer
cise

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

3.v2=u2+2g
s 1. Ast
onei
srel
easedf
rom acl
i
ffof180m hi
ghcal
cul
ate

a) Thet
imei
ttakest
ohi
tthegr
ound

b) The vel
oci
ty with whi
ch i
t hi
ts t
he wat
er
(t
akeg=10ms-1)

2. A bodyisproject
edver
tical
l
yupwar
dwi t
hani ni
tial
vel
ocit
yu.itretur
nstothesamepointofpr
oject
ion
aft
er8s.Pl
ot:
Exampl
es
a) Thespeedt
imegr
aph
1. Astoneisreleasedver
ti
cal
l
ydownwar
dsf
rom ahi
gh
cl
i
ff.Det
ermine b) Thevel
oci
tyt
imegr
aphf
ort
hebody

a) i
tsvel
oci
tyaf
tert
woseconds 3. A bodyisthrown ver
ti
cal
l
yupwar
dswi
th an i
nit
ial
vel
oci
tyofu.showt
hat:
b) Howf
ari
thast
ravel
l
edaf
tert
woseconds.
u
I
. Ti
met
akent
oreachmaxi
mum hei
ghti
st=
Sol
uti
on g

a
()u=0 I
I. Ti
mef
li
ght(
ti
met
akenf
ort
hebody(
proj
ect
il
e)
2u
t
ofal
lbackt
opoi
ntofpr
oject
ion)i
st=
t=2s
econds g

u2
g=10ms-1 I
II
. Maxi
mum hei
ghtr
eachedi
sHmax=
2g
Ri
ghte
xpr
ess
ioni
s
I
V. Vel
oci
tyofr etur
ni sequali
n magni
tude t
o
v=u+g
t vel
oci
tyofpr
ojecti
on

v=0+10×2 4. Astoneisproj
ectedverti
cal
lyupwardwi
thavel
oci
ty
-
1
of30ms fr
om theground.Cal
cul
ate:
v=20ms-1
I
. Thet
imei
ttakest
oreachmaxi
mum hei
ght
.

I
I. Thet
imeoff
li
ght
.

I
II
. Themaxi
mum hei
ghtr
eached.

I
V. Thevel
oci
tywi
thr
eachi
tlandst
hegr
ound.

Hor
izont
alpr
oject
ion Exampl
e

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page16
 232

Someexampl
esofhor
izont
alpr
oject
ioni
ncl
ude: 1. Anar row isshothor izont
all
yfrom thet opoft he
bui
ldingandi tlands200m f r
om thef ootoft he
1. Aj
etf
rom awat
erpi
pehel
dhor
izont
all
y bui
ldingafter10s.Assumingthattheai
rresist
anceis
negli
gibl
e,calcul
ate:
2. Abul
l
etf
rom agunhel
dhor
izont
all
y
a. Thei
nit
ialvel
oci
tyoft
hear
row.
3. A t
ennisbal
lwheni
trol
l
sfr
om t
het
enni
s
t
able. b. t
hehei
ghtoft
hebui
l
ding

Sol
uti
on
4. Ast
onet
hrownhor
izont
all
y.
a
.ini
ti
alhor
izont
alve
loc
ity,
u=?
5. Anar
rowr
eleasedhor
izont
all
yfr
om bow.
R=200m

t=10s
 Considerabodypr oj
ectedhorizontall
ywithan
i
niti
alhori
zontalvel
ocit
yu.Thebodymai ntai
ns Fr
omt
hee
xpr
ess
ionR=ut
;
thatini
ti
alhorizont
alv el
ocit
ybutsi ncei
talso
200m =u×10
experi
encesfr
eef al
lduet ograv
ity,i
tdescr
ibesa
curvedpathasshownbel ow. 200
u= =20ms-1
10

12
b. f
romt
hee
xpr
ess
ion;
h= gt
2

1
h= ×10×102
2

=5×100=500m

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Exer
cise

1. Abal
listhrownfr
om thet
opofaclif
f20m hi
ghwi
th
ahor
izont
alvel
oci
tyof10ms-1,
cal
cul
ate:

a. Thet
imet
akenbyt
hebal
ltost
ri
ket
hegr
ound

b. Thedist
ancefr
om t
hef
ootoft
hecl
i
fft
owher
e
thebal
ll
ands.

c. Theverti
calvel
oci
tyatt
het
imei
tst
ri
kest
he
ground.

2. Ast oneisthr
ownhor izontal
lyfr
om t hebui
ldingthat
i
s45m,hi ghaboveahor izont
algr ound.I
thi t
st he
 Thepat hfol
lowedt hebodyproject
edhor i
zontall
y groundatapoi ntwhichis60m f rom thefootoft he
(projecti
le)is cal led t
raj
ect
ory.The maxi mum buil
ding.Cal
cul
atetheiniti
alvel
ocityofthestone.
horizontal di stance cov er
ed by t he body
3. Abal
listhrownfr
om thet
opofaclif
f20m hi
ghwi
th
projected horizont al
lyiscall
ed the range.The
ahor
izont
alvel
oci
tyof10ms-1,
cal
cul
ate:
verti
calaccel erationisduet ogr avi
tywhi l
et he
hori
zontalaccel
erati
on iszero si
ncethebody I
. Thet
imet
akenbyt
hebal
ltost
ri
ket
hegr
ound
maintai
nsit
sinit
ialhori
zont
alvel
ocit
ythr
oughout
I
I. Thedist
ancefr
om t
hef
ootoft
hecl
i
fft
owher
e
themotion. thebal
ll
ands.

 Thedi
spl
acementoft
heproj
ect
il
eatanygi
ven I
II
. Theverti
calvel
oci
tyatt
het
imei
tst
ri
kest
he
12 ground.
ti
meti
sgivenbys=ut+ at
2
4. Ast oneisthr
ownhor i
zontal
lyforthebuil
dingthatis
45m,hi gh above a hor
izontalground.Ithitst he
groundatapoi ntwhichis60m f r
om thefootoft he
buil
ding.Cal
culat
etheini
ti
alveloci
tyofthebuil
ding.
Hor
izont
aldi
spl
acement
,R

 Si
ncehor
izont
alaccel
erat
ion,
a,i o,R=ut
szer

NB:
REMEMBERTHESI
MPLEPENDULUM
Thev
ert
ical
displ
acement
,h
EXPERI
MENT.
 I
nit
ialv
eloci
tyforvert
icaldi
spl
acementi
szer
o.
Thi
smeanst hatv
ert
icaldi
spl
acementi
s;

12
s=0× + gt Apendulum i
sasmallheavybody
2
suspendedbyal
ighti
nextensi
blest
ri
ng.
12
s= gt
2

12
h= gt
2

Note:Theti
mef orhorizont
aldispl
acementi
sequal
to t
imeforvert
icaldispl
acementatanygiven
point
.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page18
 232

Revi
sionExer
cise

1. A body movi ng at an i niti

al vel
oci
tyu ( ms-1) i
. St
atet
he di
rect
ion i
n which t
he par
celmoved
acceler
atesata( -
2
ms )f ortsecondsandattai
nsafinal i
mmedi
atel
yitwasrel
eased.
vel
ocityv(ms-1)
.Representthismotiononthevel
ocity
i
i
. Cal
cul
atetheti
metakenbytheparcelt
oreacht
he
againstt
imeaxesshownbel ow.
gr
oundfrom t
hemaximum hei
ght.

i
i
i. Calcul
atethevel
oci
tyoft
hepar
celwheni
tst
ri
kes
2. Thedat
ainthetabl
ebel
owr
epr
esent
sthemot
ionover theground.
aper
iodof7seconds
i
v. Calcul
atethemaximum hei
ghtabovet
hegr
ound
Ti
mei
n 0 1 2 3 4 5 6 7 theparcelr
eached.
s
v. Whatwas t
he heightofthe hel
i
copt
eratt
he
Di
nm 0 2 40 60 80 9 105 110 i
nst
antt
heparcelwasdr
opped.
0 5

a) Plotongr aphpaperagraphofdisplacement(y-
axis)agai
nstti
me. 5. Astonei sthrownhorizontal
lyfrom abui l
dingthatis
50m hi ghaboveahor i
zontalground.Thest onehits
b) Describethemotionoft
hevehicl
ef orthefi
rst4 thegroundatapoi nt,whichi s65m from thef ootof
seconds. thebuil
ding.Cal
cul
atetheiniti
alofthestone.

c) Det
erminetheveloci
ti
esat4.5sand6.5s.Hence
orother
wisedeterminetheaverageaccel
erat
ion
oft
hevehicl
eoverthisti
meinter
val. 6. Thefi
gur
erepresent
sdotsmadebyat i
cker
-t
imer
.The
dot
sweremadeataf r
equencyof50dot
spersecond.
(Di
agr
am notdrawntoscal
e)

3. Aball-
beari
ngXi sdroppedverti
cal
lydownwar
ds,f
rom
theedgeofat abl
eandi ttakes0.5stohitt
hefloor
bel
ow.Anot herbeari
ngYl eavestheedgeoft
hetable
hori
zontal
l
ywithavelocit
yof5m/ s.fi
nd:
i
. Whatist
ime i
nter
valbet
ween t
wo consecut
ive
i
. Thet
imet
akenf
orbear
ingYt
oreacht
hef
loor
. dot
s?

i
i
. The hori
zontaldi
stance t
ravel
ed by Y bef
ore i
i
. Thefi
rstdotf
rom thel
eftwasmadeattimet=0.
hi
tt
ingthefl
oor. Copy the di
agr
am and i ndi
cat
e usi
ng ar
rows
poi
nti
ng downwardsthe dotsmade att= 0.
1s,
i
i
i. Thehei
ghtoft
het
abl
e-t
opabovet
hef
loorl
evel
. 0.
2s,0.
3s.

i
i
i. Determinetheaver
agevel
oci
ti
esofthetapeover
ti
mei nt
erval
s-0.
02sto0.
02s,
0.08st
o0.12s,
0.18s
4. A heli
copter
,whi ch was ascending verti
call
y ata to0.22sand0.28st
o0.32s
steadyvel
ocit
yof20m/ s,releasedapar celthattook
20secondtoreachtheground. i
v. Draw asuitabl
egraphandf
rom i
tdet
ermi
net
he
accel
erat
ionofthetape.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Chapt
erTwo NEWTON’
SLAWSOFMOTI
ON

Speci
fi
cObj
ect
ives Cont
ent

Byt
heendoft
hist
opi
c,t
hel
eanershoul
dbeabl
eto: 1. Newton’
s laws of mot i
on (
exper
iment
al
t
reat
mentofi
ner
ti
arequi
red)
a) St
ateNewt
on’
slawsofmot
ion
2. Conservati
onal of l
inear momentum: el
ast
ic
b) Descr
ibesi
mpl
eexper
iment
stoi
l
lust
rat
einer
ti
a col
li
sion, inel
ast
ic colli
si
ons, r
ecoilvel
oci
ty,
i
mpulse( obli
quecol
l
isionsnotr
equir
ed)
c) St
atet
hel
awofconser
vat
ionofl
i
nearmoment
um
3. F=ma
d) Def
ine el
ast
ic col
l
isi
on, i
nel
ast
ic col
l
isi
on and
i
mpulse 4. Fri
cti
onal
for
ce

e) Der
ivet
heequat
ionF=ma  Adv
ant
agesanddi
sadv
ant
ages
f
) Descr
ibet
heappl
i
cat
ionoff
ract
ional
for
ce
 Vi
scosi
ty
g) Def
inev
iscosi
ty
 Ter
minal
vel
oci
ty(
qual
i
tat
ivet
reat
ment
)
h) Expl
aint
ermi
nal
vel
oci
ty
5. Problem on Newton’slaw ofconser
vat
ion of
l
inearmoment um (excl
udeprobl
emsonel ast
ic
i
) Sol
venumeri
calprobl
emsinvolv
ingNewton’
slaws
coll
i
sions)
andthel
awofconserv
ati
onofli
nearmomentum.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page20
 232

I
ntr
oduct
ion 2. Thebot t om woodenbl ockcaneasi l
ybepul l
ed
outoft hestackwithoutdistur
bingothersplaced
 Ef
fectsofforceonmoti
onofabodyar ebasedon on i t
. The ot her blocks remai n i
n a pi l
e
Newton’
sthreelawsmot
ion.I
nthi
stopi
c,t
hesel
aws undisturbedexceptthattheirposit
ionislowered
ar
elookedinto. becauseofdownwar dpul loft hegravit
ati
onal
for
ce.

Newt
on’
sFi
r TheLawofI
stLaw( ner
ti
a)

 I
tstatesthat“abodyr emainsi
nit
sstat
eofrestor
ofunifor
m mot ioni nast rai
ghtl
ineunl
essacted
uponbyanext ernalforce”
.Thisl
aw i
salsocal
led
thel
awofi ner
ti
a

3. Whenamovi ngt rai

noracarst opssuddenl y,
Def
init
ionofIner
ti
a passengersaret hr
ow nf or
w ar
d.T hepassenge r
i
namovi ngvehicleisalsoinast ateofmotion.
 I sthetendencyofabodyt oremaini
nitsstateof Hence,when t he vehicl
e stops suddenl
y,t he
resti
fitwasatrestori
nit
sstateofmoti
onifitwas upperpartofhi
sbodycont i
nuestomove.Thisi s
i
nmot ion. whyiti
snecessar ytowearseatbelt.

4. Acycl
i
stonalevelgr
oundcont
inuest
omovef
or
someti
mewit
houtpedali
ng.
Exampl
esofIner
ti
a(i
.e.Exampl
esofNewt
on’
sFi
rstLaw
i
nPract
ice)

1. Whenacar dsuppor t
ingacoinonagl asst
umbl eris Def
ini
ti
on ofForce asperNewton’
sFi rstLaw of
suddenlyfl
icked,i
tisobservedthatt
hecardf l
i
esof f Moti
on
butthecoinf al
lsi
nt hetumbler
.Thisisbecauset he
cointendst omai ntainit
sstateofr est(
itfall
si n  Forceisdefi
nedasthatwhichproducesmotion
glassbecauseanot herfor
ceactsonit.Whichisthat i
nbodyatrestorwhi
chal
ter
sitsexist
ingst
ateof
otherf
orce?) moti
on.

Moment
um Rel
ati
onshi
pbet
weenFor
ce,
MassandAccel
erat
ion

 Momentum ofabodyisdef
inedast
hepr
oductofi
ts Consi
deraforceFactingonabodyofmassm f
ora
massandveloci
ty.Forabodyofmassm i
nkgand t i
met.ifi
tsv
eloci
tychangesf
rom ut
ov,t
hen;
v
eloci
t nms-1,
yvi
c
hang
einmome
ntum =f
ina
lmome
ntum-i
nit
ial
mome
ntum

mome
ntum =ma
ss(
kg)×ve
loc
it ms-1)
y( ∆p=mv-mu

P=mv

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

 SIuni
tofmoment st
um i he ki
l
ogr
am met
re per Ra
teofc
hang
eofmome
ntum =
cha
ngei
nmoment
um
;
time
kgms-1)
second ( .Moment s av
um i ect
orquant
it
y
since ithas both magni
tude and di
recti
on.The ∆p mv-mu
=
direct
ion ofmoment um i
s same as t hatofthe t t
veloci
tyofthebody. mv-mu
Fr
omNe
wton’
sse
condl
aw,F∝
t

v-u
F∝m ( t); butv-tu=a
Exampl
es
t
her
efor
e,F∝ma
1. Whatisthemomentum ofar
aci
ngcarofmass500kg
s
o,F=kma
.whe
reki
sac
ons
tant
ofpr
oti
ona
lit
y.
dr
ivenat270km/
h?
Whe N,a=1ms-2,m =1kg,
nF=I the
n.k=1
Sol
uti
on
he
nce
,F=ma,
v=270kmh-1=75ms-1;m =500kg
Thi
sisa
nex
pre
ssi
onf
orNe
wton'ss
econdl
aw
mome
ntum =ma
sskg
( )×ve
loc
ity(
ms-1)

mome
ntum =500kg× 75ms-
1

Def
ini
ti
onofanewt
onasperNewt s2nd l
on’ aw of
=37500kgms-1
moti
on
2. Fi
ndt
hemoment
um of:
 A newt on i
st he for
ce whi
ch pr
oduces an
-
1 -2
a. Anobj
ectofamass100gmovi
ngat20ms accel
erati
onof1ms wheni tact
sonabodyof
mass1kg.
Sol
uti
on

mome
ntum =ma
sskg
( )×ve
loc
ity(
ms-1)
Exampl
es
100kg×20ms
= 0. -
1

1. Whatforceisneededt
ost
opa500kgcarmovi
ngat
=2kg
ms-1
180kmh-1i
n12.
5m?
b. Anobjectofmass2.
0kgwhi
chf
all
sfr
om r
estf
or10s
(
Moment
um aft
er10s).
Sol
uti
on
Sol
uti
on
,v=0ms-1,
m =500kg
u=0ms-1,t=10s
,g=10ms-2,v=?
u=180kmh-1=50ms-1,s=12.
5m,F=?
us
ingt
hee
xpr
ess
ionv=u+g
t,
v2-u2
v=0+10×10 Fr
omt
hee
qua
tion,
v2=u2+2a
s,a=
2s
=100ms-1
02-502
1x100=100kg
mom=o. ms-
1 a= ,
2×12. 5
Exer
cise
a=-100ms-2

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page22
 232

1. Whatisthemomentum ofar
aci
ngcarofmass500kg Fr
omNe
wton's2ndl
aw,F=ma
,F=500×-100
dr
ivenat270km/
h?
F=-50000N.
2. Anappl
eofmass100gf al
lsadi
stanceof2.
5m t
othe
gr
oundf
rom abr
anchofatr
ee. wha
tdoe
sthene
gat
iveme
an?

a) Calculat
ethespeedatwhi chithi
tsthegroundand 2.An ext ernalfor
ce appl
ied to a ballofmass 160g
thet i
metakenforitt
ofal
l
.(I
gnor
eairr
esi
stance)
. i
ncreasesi t
svelocityf
rom 2.5cms -1 t
o275cms-1 i
n10
seconds.Calcul
atethef
orceappli
ed.
b) Calculat
et hemoment um oftheapplejustbefor
e
hit
tingtheground

Sol
uti
on

m =160g=0.
160kg
,u=2.
5cms-1=0.
025ms-1,

Newt
on’
sSecondLawofMot
ion ms-1=2.
v=275c 75ms-1,F=?
N

v-u
 I
tstat
esthat“therateofchangeofmoment um ofa F=ma,F=m t ( )
bodyi
sdirectl
yproporti
onalt
otheresul
tantext
ernal
for
di
cepr
r
ect
i
oduci
onoft
ngt
hef
hechangeandt
orce”.
akespl
acei nt
he F=0.
160
2.
75-0.
10
025
=0.
0436N ( )
Re
sul
tantf
orc
eac
ting∝r
ateofc
hang
eofmome
ntum

Exer
cise

1. Whati
sthemassofanobj
ectwhi
chi
saccel
erat
edat5
-
1
ms byafor
ceof200N?

2. Agunfir
esabulletofmass10.0ghori alat50ms-1at
zont
afi
xedtargetofsoftwood.Thebull
etpenet
rat
es50cm
i
ntoatar
get.Cal
culat
e

a) Timetakenbyt
hebul
l
ett
ocomet
oresti
n
thewood

b) Theaver
ageret
ardi
ngf
orceexer
tedbyt
he
woodonthebul
l
et.

3. At rol
leyofmass1.5kgispull
edal
ongbyanelasti
ccord
givenanacceler
ati -
2
onof2ms .Findthefri
cti
onalf ce  Not
or e:Impul
seoccurswhenbodiescoll
ideand
actingonthet
rol
leyift
hetensi
oni
nthecor
dis5N. t
he i
mpulsi
ve f
orce i
sthe one whi
ch causes
destr
uct
ionduringcolli
sion.Theti
mef orwhi ch
4. St
ateNewt
on’
ssecondl
awofmot
ion.Hence,showt
hatF thi
simpulsi
vefor ceactsdetermi
nest heext ent
=ma.
ofdamagecaused.I ft i
meofi mpacti sl ong,
5. Def
inet
henewt
on(
uni
toff
orce) damage islesst han when time ofimpacti s
short
.The f ol
l
owi ng are some exampl es of
6. Acarofmass1500kgi
sbroughtt
orestfr
om avel
oci
tyof desi
gnsmadet o pr
olong ti
meofi mpactand
-
1
25ms bya constantfor
ce of3000N.Determi
ne the ther
efor
ereducedamagebyi mpul
siveforce.
changeinmoment
um pr
oducedbythefor
ceandtheti
me

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

t
hati
ttakest
ocomet
orest
. 1. Eggsar
epackedi
nspongycr
ates

7. Ahammerofmass800gpr oducesaforceof400Nwhen 2. Smar

tphonesar
eputi
nsof
thol
der
s
i
tst
rikest
heheadofanail
.Descri
behowitispossi
blef
or
t
hehammert odri
vesthenai
lint
oapi eceofwood,yeta 3. vehi
clesar
efi
tt
edwi
thsaf
etyai
rbags
wei
ghtof400Nresti
ngontheheadofthenai
lwouldnot
4. some vehi
cles have col
l
apsi
ble bumper
s
8. Aresult
antfor
ceFactsonabodyofmassm causi ngan andst
eeri
ng
accel
erat
iona1ont
hebody.Whenthesamef orceactson
abodyofmass2m,i tcausesanaccel
erat
iona2.Expr
ess 5. Hi
ghjumper
susual
l
ylandonsof
tgr
ound
a2int
ermsofa1. et
c.

I
mpul
se

 I
mpul
sei
sdef
inedast
hepr
oductoff
orceact
ingon Exampl
es
a body and the ti
me in which t
he f or
ce act
s. 1. Determi
nethechangei
nmomentum pr
oducedwhena
I
mpulsivefor
cereferstotheforcewhichactsona forceof4000N actsonabodywhichisatrestfor
bodyforaver
yshorttimeduri
ngacoll
ision. 0.003minut
es

 I
faf ceF act
or sonabodyofmassm fort
ime,
t,
t
hent
heimpul
seoft
hef
orcei
sgi
venby
:
Sol
uti
on

F=4000N, t
ime
,t=0.
003mi
nut
es=0.
18s
I
mpul
se=f
orc
e×t
ime
I
mpul
se=c
hang
einmome
ntum =Ft
I
mpul
se=Ft
c
hangei
nmome ntum =4000×0.
18
(
SIuni
tofi
mpul
sei
sthene
wtons
econd(
Ns)
) =720kgms-1or720Ns

mv-mu
Butf
romNe
wton’
sse
condl
awofmot
ion,F= 2. Acarofmass400kgst ar
tsfr
om r
estonahor
izont
al
t
t
rack.Fi
ndthespeed4saf t
erst
art
ingi
fthet
racti
ve
Ft=mv-mu,
butFt=i
mpul
sea
ndmv-mu f
orcebytheengi
neis500N.

i
sthec
hang
einmome
ntumoft
hebody

Sol
uti
on

 Thi
simpl
i hatt
est hei
mpul
seoff
orceact
ingona I
mpul
se=c
hang
einmome
ntum,Ft=m(
v-u)

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page24
 232

body dur
ing some t
ime i
nter
vali
s equalt
ot he 500×4=400(
v-0)
changeinmomentum pr
oducedint
hatbodyi
nthat
500×4
ti
me. v= =5ms-1
400

 Theareaundert
heplotoffor
ceFagai
nstt
ime( t
)
repr si
esent mpul
seorchangeinmoment
um dur
ing
acol
l
isi
on.
Exer
cise

1.Anappleofmass100gfal
lsadist
anceof2.
5m t
othe
groundfr
om abr
anchofat
ree.

I
.Cal
cul
atethespeedatwhichithit
st hegroundandt
he
ti
metakenfori
ttof
all
.(I
gnoreairresi
stance)
.

I
I. Assumi ngt heappl
et akes100mi l
li
secondstocomet o C. Li
ftmovi ngdownwar dswi t
haccelerati
on
restCal cul
atet heaveragef orceexperi
enced bythe
apple.  Thedownwar daccelerat
ionisnegati
v eandt his
i
s why t he one f eels li
ghter when l i
fti s
2. Thet ablebelowshowst hevaluesoftheresult
antf
orce,F, accel
erat
ing downwar ds.Ther ef
orer eadi
ng on
andt het i
metf orabullettr
aveli
nginsi
det hegunbarrel
themachine( appar
entwei ghtofthebodyi nlif
t)
aft
ert hetri
ggerispull
ed.
i
s:
For
ce,
F(N) 36 34 300 240 17 11
0 0 0 0 P’=mg-ma
Not
es:
Time,(t
) 3 4 8 1 17 22
(mil
l
isecond 2 I
. Ifthe li
ftmoves wit
h const
antveloci
ty,the
s)
machine wil
lread wei
ghtoft he bodysince
I
. Pl
otagr
aphofFor
ce,
F,agai
nstt
imet
. accel
erati
onwil
lbezer
o.

I
I. Det
ermi
nef
rom t
hegr
aph: I
I. I
fa = g( f
ree fal
l)the body wil
lexperi
ence
wei
ght
lessnesssincethereact
ionfrom t
helif
t
a) The t
ime requir
ed forthe bul
lettotravelt
he ont
hebodywi l
lbezero.
l
ength ofthe barrelassuming t
hatthe for
ce
becomeszerojustattheendofthebar
rel. Exer
cise

b) Thei
mpul
seoft
hef
orce. 1. Aladyofmass80kgst andsonweighi
ngmachi
neina
l
if
t.Deter
minethereadingont heweighi
ngmachi
ne
c) Gi
venthatt
hebull
etemergesf
rom t
hemuzzleof
whentheli
ftmoves:
t
hegunwi t
haveloci
tyof200m/s,cal
cul
atethe
massofthebul
l
et. a) downwardsataconstantvel
oci
t 0ms-1
yof2.
3. Abodyofmass5kgi sej
ect
edver t
ical
lyfrom t
hegr
ound
b) downwar
dswi
thanaccel
erat
ionof3ms-2
whenafor
ceof600Nactsonitfor0.1s.Calcul
ate the
vel
oci
tywi
thwhi
chthebodyl
eavesthegr ound.
c) upwar
dswi
thanaccel
erat
ionof3ms-2
4. A hi
ghjumperusual
l
yl andsonat hi
cksof
tmatt
ress. 2. A man ofmass 80 kg st ands on a li
ftwhich i
s
Expl
ainhow t
hematt
resshelpsi
nreduci
ngt
hef
orceof -
2
accel
erat
ing upwards at 0.
5 ms . i f g=10N/kg
i
mpact.
deter
minethereact
ion ont
hemanbyt heflooroft
he

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

l
i
ft.

Newt
on’
sThi
rdLawofMot
ion

Itst
atesthat“foreveryact i
on,ther
eisanequalbut
opposi
tereacti
onforce”. Thi
smeanst hati
fabodyP TheLawofConser v
ati
onofLinearMomentum
exer
tsaforceonanot herbodyQ,Qexertsanequaland
opposi
teforceonP.I tisclearthatiti
sduetoact on  Thi
i slaw st
atest ,“
hat f
orasyst em ofcol
l
idi
ng
(for
ceexert
edbyf ootonground)andreact
ion(
for
ce bodies, t
heirt ot
al l
i
near moment um is a
exert
ed byear
th on foot
)thatwe are abl
eto wal
k constant
,provi
dednoexter
nalf
orcesareact
ing”
.
forwar
d.
 I
fabodyAofmassmA i nit
ial
l
ymov i
ngwitha
vel
oci
tyofuA coll
i
desli
neall
ywi t
habodyBof
massmB init
ial
l
ymov i
ngwi t
hav el
oci
tyuB and
t
heirvel
ociti
es aft
ercoll
i
sion arev A and vB

r
espect
ivel
y,t
hen:

Tot
all
ine
armome
ntumbe
for
eimpa
ct

Wei
ghti
nal
i
ft =Tot
all
ine
armome
ntuma
fte
rcol
li
si
on

a. Li
ftatr
est mAuA +mBuB =mAvA +mBvB

 Li
ftmachi
ner
eadst
hewei
ghtoft
hebodyi
nli
ft Not
e:ve
loc
ityi
save
ctorqua
nti
tya
ndmus
tbe
si
nceact
ionandr
eact
ionar
eequalandopposi
tei
.e.
t
rea
teda
ppr
opr
iat
elyi
nca
lcul
ati
ons
R=mg.

b. Li
ftmovi
ngupwar
dswi
thaccel
erat
iona
Ty
pesofCol
l
isi
ons
Ther esultantupwardfor
ceFproducesthe
acceleration(F=ma) I
. El
ast
icCol
l
isi
on
resultantupwa r
dforc
eF
=total upwardfor
ceP-downwadforce(wei
g )W
ht  Thi
sisacolli
sioninwhi chbothkinet
icenergy
andmomentum ar econserved.I
fthebodiesA
F=P-W ⇒ P=F+w a
nd∴ andBabov
ecollideel
asti
cal
ly;
P=mg+ma mAuA +mBuB =mAvA +mBvB

 Thi
siswhatthel
if
tmachi
newi
l
lread(
ther
eact
ion 1 1 1 1
oft
heli
ftont
hebody)
. m u2+ m uB2= m vA2+ mBvB2
2AA 2B 2A 2

 Not
e:I
nthi
scoll
isi
on,bodi
esseparat
eandmove
i
nsameordif
fer
entdir
ecti
onsaf
tercol
li
si
on.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page26
 232

I
I. I
nel
ast
icCol
l
isi
on I
. mb+mw)
mbub+mwuw =( v

 Thi
sisacoll
isioninwhichmomentum i
sconserved 0.
020×50+1.
980×0=(
0. v
980)
020+1.
butki
neti
cener gyisnot.I
nt hi
scol
l
isi
on,col
li
ding
1
esf
bodi useandmovet ogetheri
nonedi
rect
ionwith 5ms-1
v= =0.
2
acommonv elocit
y.
I
I. a
tma
ximumhe
ight
,al
lK.
Eisc
onve
rte
dtoP.
E
 I
fthe col
li
sion ofbodi
es A and B abov
e col
l
ide
i
nel
ast
ical
l
y,then; 1 2
mv =mg
h,mc
anc
elsout
2
mAuA +mBuB =mAvA +mBvB
1
52=10×h
×0.
1 1 1 1 2
m u2+ m uB2 ≠ m vA2+ mBvB2
2AA 2B 2A 2
0.
125
h= =0.
0125m
 Note:t
het otalkinet
icenergyaftert
heimpactis 10
al
wayslesst hanthetotalki
net
icener
gybef
orethe
Exer
cise
i
mpactandt helossisdueto:
1. A lorr
yofmass 3000 kg t r
avell
ing ata constant
I
. Ener
gyusedi
ndef
ormat
ionofbodi
es vel
ociyof72kmh-1 col
t li
deswi t
hast ati
onarycarof
mass600kg.Thei mpacttakes1.5secondsbeforethe
I
I. Ener
gyt
ransf
ormedt
oheat
,soundandeven two move togetherata const antveloci
tyf or15
l
ight seconds.Cal
cul
ate:
Exampl
es i
. Thecommonvel
oci
ty
1. Two tr
oll
eysofmasses2kgand1. 5kgaretravel
ing i
i
. Thedi
stancemovedaf
tert
hei
mpact
towar
dseachotherat0.
25m/sand0.
40m/
srespecti
vely.
Thetr
oll
eyscombineoncol
li
sion. i
i
i. Thei
mpul
sivef
orce
I
. Cal
cul
atet
hevel
oci
tyoft
hecombi
nedt
rol
l
eys. i
v. Thechangei
nki
net
icener
gy
I
I. In what di
rect
ion do t
he t
rol
l
eys move af
ter 2. Abull
etofmass15gi sshortfrom agun15kgwi
tha
col
l
isi
on?
muzzl
eveloci
tyof200ms-1.I
fthebull
eti
s20cm l
ong,
Sol
uti
on cal
cul
ate:

I
. t
heaccel
erat
ionoft
hebul
l
et

I
I. t
her
ecoi
lvel
oci
tyoft
hegun

3. Expl
ainhow:

Tot
all
ine
a rmoment
umbef
oreimpact i
. Rocketpr
opul
siont
akespl
ace
=Totall
inearmoment
umaf
tercol
li
si
on
i
i
. Gar
denspr
inkl
erwor
ks
mA +mB)
mAuA +mBuB =( v.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

⇒2.
5×0.
25+1.
5×(-0.
40)=(
2. 5)
5+1. v

0.25
v= 00625ms-1
=0. Fr
ict
ional
For
ce
4.0

 Fri
cti
onalf
orcereferst otheforcethatopposes
ortendstoopposer elati
vemotionbetweentwo
I
II
. Theymovei
nthedi
rect
iontowhichtr
oll
eyofmass surf
acesincontact.Fr i
cti
onalfor
ceact sinthe
2.
5kgwasmovingt
obeforecol
l
isi
on dir
ecti
onopposi
tetot hatofthepull
i
ngf or
ce.

Ty
pesofFr
ict
ional
For
ce

I
. St
ati
c/Li
mit
ingFr
ict
ionalFor
ce

2. Abullet0fmass20gt r
avel
l
inghorizont
alyataspeedof  Thi
l sistheforcerequi
redtojustst
artthebody
50ms-1 embedsi
tsel
finablockofwoodofmass1980g sli
ding.Theforceisdi
rect
lypropor
tionalt
othe
suspendedfr
om ali
ghti
next
ensibl
estri
ng.Find: reacti
onforceont hebodybythesur f
ace.
FS=µsR,
I
. Thevelocit
yoft
hebul
l
etandbl
ocki
mmedi
atel
y
aft
ercoll
isi
on whe
reµsi
sthec
oef
fi
ci
entofs
tat
icorl
imi
ti
ng

I
I. Thehei
ghtt
hroughwhi
cht
hebl
ockr
ises f
ri
cti
ona
lfor
ce

Sol
uti
on

I
I. Ki
net
ic/Sl
i
dingFr
ict
ionalFor
ce

 Thisisthef
orcerequi
redt okeept hebodyslidi
ng
ormov i
ng ata const antspeed.I topposes
mot i
onbetweentwosur facest hatar
ei nrel
ativ
e
mot i
on.Sli
ding f
ri
cti
on is directl
ypr oport
ional
react
ionfor
ce. FK =µkR,

whe
reµki
sthec
oef
fi
ci
entofki
net
icor

s
li
dingf
ri
cti
ona
lfor
ce

Not
e:Consi
derabl
ockofwoodbei
ngpul
l
edusi
nga Exer
cise

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page28
 232

r
ubberbandonahor
izont
alsur
faceasshownbel
ow. 1. A bull
etofmass 10g t ravel
i
ng hori
zont
all
ywitha
vel
ocit
yof300m/ sstri
kesabl ockofwoodofmass
290gwhich rests on rough hori
zont
alfloor.Af
ter
impacttheymovet ogetherandcomet orestaf
ter
tr
avel
ingadistanceof15m.

I
. Calcul
atethecommonvel
oci
tyoft
hebul
l
etand
theblock.

I
I. Cal
cul
atet
heaccel
erat
ionoft
hebul
l
etandt
he
bl
ock.

I
II
. Cal
cul
ate the coef
fi
cient of sl
i
ding f
ri
cti
on
bet
weenthebl
ockandthefloor
.

2. Underadr
ivi
ngf
orceof4000N,acarofmass1250kg
hasanacceler
ati
onof2. s2. Fi
5m/ ndt
hef
ri
cti
onal
for
ceact
ingonthecar
.

3. A bull
etofmass 22g t ravell
ing with a hor i
zontal
vel
ociyof300ms-1 st
t ri
kesabl ockofwoodofmass
378gwhichr est
sonar oughhor i
zontalsurface.Aft
er
theimpact,t
hebull
etandt heblockmovet ogetherand
comet orestwhent heblockhast ravell
edadi st
ance
of5m.Cal cul
atethe coeffici
entofsl iding fri
cti
on
betweentheblockandthef l
oor.
 Ther ubberbandst r
etchesf orsomet imebeforethe
block starts moving ( st
retch is due tolimit
ing 4. Ablockofamet alAhavingamassof40kgrequi
resa
fri
cti
onal).Thestretchingincreasestoapointwhen hor
izont
alfor
ceof100Ndr agi
twit
hunif
orm vel
ocit
ya
thebl ockst ar
tssl i
ding steadil
y( sl
idi
ng f
ri
cti
onal l
onghorizont
alsurf
ace.
forceli
mitsmot i
onatt hispoint).
µs>µk
I
. Cal
cul
atet
hecoef
fi
cientoff
ri
cti
on
 Fr
ict
ionalfor
cei susef
ul/advantageousinwal
ki
ng, I
I. Det
erminethef or
cer equi
redt
odr agasi
milar
movingv ehi
cles,br
aki
ng,writ
ing,li
ghti
ngamat ch bl
ockhavingamassof30kgal ongthesame
st
icketc. hor
izont
alsur
face,cal
culat
e.

 Fr
ict
ionalf
orcecanbedi
sadvant
ageousasi
tcauses I
II
. I
fthetwobl
ocksAandBar econnect
edwi
tha
wearandteari
nmov i
ngpartsofmachi
neandl
eads t
wobarandafor
ceof200Nisappl
iedt
opul
lthe
t
ogenerat
ionofunnecessar
ynoi
se. t
wolongt
hesamesur
face,
cal
cul
ate.

a) Thet
ensi
oni
nthet
owbar

Exampl
e b) Theaccel
erat
ion

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Asmoot hwoodenblocki
spl
acedonasmoot hwoodentabl
e. I
V. I
ft hetowbarisremovedandthe40kgbl
ocksof
Af or
ceof14Nshouldbeexer
tedont he4kgwoodenbl ockt
o met almovesaround asmoot hpathofradius
keepthebl
ockmovingwit
haconstantveloci
ty. 10m ataconst antspeedof24ms-
1cal
culate
thecentr
ipet
alf
orce.
a) Whati
sthecoef
fi
cientofki
net
icf
ri
cti
on,
µk
V. Att heendoft hecir
cularpath,the40kgmass
dropsver t
ical
l
yi nat r
ench10m highandf all
s
fr
eelydeterminethet imeittakestolandatthe
Sol
uti
on bottom ofthetr
ench.̀

FK =µkR, FK =µkmg Fact

ors Aff
ecti
ng Fri
cti
onalFor
ce bet
ween Two
Sur
facesinCont
act
14=µk×4×10
I
. Nor
malr
eact
ion– f
ri
cti
onalf
orcei
sdi
rect
ly
14
µk= =0.
35 pr
opor
ti
onal
tonor
mal
react
ion.
40
I
I. Nat
ureoft
hesur
faces-t
hemor
erougha
sur
facei
s,t
hel
argert
hef
ri
cti
onal
for
ce.
b) I
fa20Nbr i
ckispl
acedont hebl
ock,whatforcewi
ll
ber equir
edtokeeptheblockandbrickmovingwih Met
t hodsofmi ni
mizi
ngfri
cti
on
requiredconst
antspeed?
a) Usingr ol
l
ers-Theroll
ersarelai
ddownont
he
surfaceandtheobj
ectpushedovert
hem

Sol
uti
on b) Lubr
icat
ion-Appl
i
cat
ionofoi
lorgr
easet
othe
coeff
ici
entofkine
t i
cfr
icti µkr
on, ema
insc
ons
tant movingpart
s
FK =µkR⇒ FK =0.35×( 40+20)

=20N

c) Useofbal
lbeari
ng-Thi
sisappl
iedonr
otat
ingaxles. Fact
orsAf
fect
ingVi
scosit
yInFl
uids
Thebeari
ngal l
owsthemov ementofthesurface
overt
heother
. I. Densit
y-Thehi ghert
hedensi
tyoft
hef
lui
d,
t
hegreat
ertheviscousdragandt
her
efor
e
d) Ai
rcushi
oni
ng-Thi
sisdonebybl
owi
ngai
rint
othe t
hel
owertheter
minalvel
oci
ty.
space bet
ween sur
faces.Thi
s pr
event
s surf
aces
fr
om comingi nt
ocontactsi
nceairismatterand
occupi
esspace.

Vi
scosi
ty

 Thi
sisthe for
ce thatopposes r
elat
ive mot
ion
bet
weenl
ayer
softhefl
uid.

 Consi
derasmallbal
lbeari
ngintr
oducedgent
lyi
nto
gl
yceri
nei
nal ongcyl
indr
icalj
ar.Thefor
cesact
ing
onthebal
lar
easshownbel ow.
I
I. Temper at
ure -I n liquids, vi
scous drug
decreases(ter
mi nalvelocit
yincreases)wit
h
temperatur
e,whi lein gases vi scous dr
ug
i
ncreases( t
erminalvelocitydecreases)wit
h
temperatur
e.

Exer
cise

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page30
 232

1. The di agram shows a t al

l measur i
ng cyli
nder
containi
ngavi scousli
quid.Aver ysmal lst
eelballis
rel
eased f r
om r estatthe surf
ace oft he l
iqui
d as
shown.Sket chtheveloci
ty-ti
megr aphforthemot ion
ofthebal lfr
om theti
mei tisrel
easedtot heti
mej ust
beforeitr
eachest hebott
om ofthecyli
nder.

 The r esult
ant downward force (
mg-( )
u+Fr)
accel
eratesthebal
ldownwar
ds.

 Thev i
scousdr agincreaseswi t
hv el
oci
tyunt i
lthe 2. Two smal lspheri
calidenti
calst ones A and B are
sum ofupwardf or
cesequal t
hedownwardf orces, rel
easedf r
om thesamehei ghtabovethegr ound.B
fal
lsthroughairwhileAf al
lsthroughwat er
.Sket ch
u+Fr=mg thegraphsofveloci
tyagainsttime( t
)foreachst one.
Labelthegraphappropr
iat
ely.
 Atthispoi
ntther esult
antforceiszer
oandt heball
att
ainsaconstantvelocit
ycaledt
l ermi
nalvelocit
y.

 Ter
minalvel
oci
tyi
sdef
inedast
heconst
antvel
oci
ty
att
ainedbyabodyfall
ingi
naf l
uidwhenthesum of
upwardf orcesi
sequaltotheweightofthebody.
Graphical
ly:

Chapt
erThr
ee ENERGY,
WORK,
POWERANDMACHI
NES

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Speci
fi
cobj
ect
ives Cont
ent

Byt
heendoft
hist
opi
c,t
hel
eanershoul
dbeabl
eto: 1. For
msofener
gyandener
gyt
ransf
ormat
ions

a) Descr
ibeener
gyt
ransf
ormat
ions 2. Sour
cesofener
gy

b) St
atet
hel
awofconser
vat
ionofener
gy  Renewabl
e

c) Def
inewor
k,ener
gy,powerandst
atet
hei
rSI  Non-r
enewabl
e
uni
ts
3. Lawconser vat
ionofener
gy
d) Definemechanicaladvantagev
eloci
tyrat
ioand
eff
iciencyofmachines 4. Wor k,ener
gyandpower( wor
kdonebyr
esol
ved
forcenotrequi
red)
e) Solve numer i
cal problems invol
vi
ng wor k,
energy,powerandmachines. 5. Kineti
candpot ent
ial
ener
gy

6. Si
mpl
emachi
nes

7. Pr
obl
emsonwor
k,ener
gy,
powerandmachi
nes

Ener
gy  Not e:pot
enti
alenergyist heenergypossessed
byabodyduet oitsr elati
veposit
ionorstate
 Ener gyisthecapaci
tyt
odowor k.TheSIunitof whieki
l net
icenergyist heonepossessedbya
energyisthe j
oul
e(J)afert
t hephysi
cistJames bodyduetoit
smot i
on.
Prescot
tJoul
ewhowasalsoabrewer
.
Conser
vati
onandTransfor
mat ionofEnergy
Sour
cesofEnergy
TheLawofConser
vat
ionofEner
gy
 They are cl assi
fi
ed i
nto r
enewabl
e and
nonr
enewabl
esources.  Thislaw st
ates that“Ener
gy can neit
herbe
creat
ed nor dest r
oyed but can onl y be
i
. Renewabl
esour
ces tr
ansfor
medfrom onefor
mt oanother
.”
 Thesear esour ceswhosesupplycanber enewed Ener
gyTransformation
againandagai nforuse.Exampl
esare;wat
er,sol
ar,
wind,geother
malet c.  Anydevicet hatfacili
tatest
het r
ansf
ormationof
ener
gyf rom one f orm to anot
heris cal
led a
i
i
. Non- renewabl
esour ces tr
ansducer.Thef ol
lowi ngar
esomeexampl es:

 Thesear esour
cesofener gywhosesuppl ycannot I
nit
ial f
orm Fi
nal f or
m Tr
ansducer
ber enewedagai
nandagai nf oruse.Exampl
esare; ofener
gy ofener
gy
fossil
s,f
ir
ewood,nucl
earsourceetc.
Sol
ar Heat Sol
arpanel
For
msofEner
gy

 Thev
ari
ousf
ormsofener
gyi
ncl
ude: El
ect
ri
cal Ki
net
ic Mot
or

o Mechani
cal(
pot
ent
ialandki
net
ic) Ki
net
ic El
ect
ri
cal Dy
namo

o Chemi
cal–st
oredi
nbat
ter
iesandf
oods Sol
ar El
ect
ri
cal Sol
arcel
l

o El
ect
ri
cal Heat El
ect
ri
cal Ther
mocoupl
e
o Li
ght el
ect
ri
cal Sound Loudspeaker

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page32
 232

o Nucl
ear chemi
cal El
ect
ri
cal Bat
ter
y

o Wave  Not
e:Ener
gyt
ransf
ormat
ionsar
erepr
esent
edby
char
ts.

Exampl
es Exampl
es

Descri
betheenergytransf
ormat
ionthatt
akespl
acei
neach 1. Cal
cul
atetheamountofwor kdoneby:
ofthefol
l
owing:
a) Amachi nelif
tingaloadofmass50kgt
hrougha
a) Acarbatter
yisusedtol
ightabul
b vert
icaldi
stanceof2.4m

Sol
uti
on

b) Coali
susedt
ogener
ateel
ect
ri
cit
y wor
kdone
,W =f
orc
e,F×di
sta
nce

wor
kdone=mg×di
sta
nce

=(
50×10)
N ×2.
4m = 1200J

c) Apendul
um bobswi
ngt
oandf
ro
b) Al
abor
erwhocar
ri
esal
oadofmass42kgt
oa

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

hei
ghtof4.
0m

d) Wat
eratt he t
op ofa waterf
all fal
l
s and i
ts
t
emper
atur
eri
sesonreachi
ngt
hebott
om Sol
uti
on

wor
kdone
,W =f
orc
e,F×di
sta
nce

wor
kdone=mg×di
sta
nce

=(
42×10)
N ×4.
0m = 1680J

2. Amanofmass70kgwal ksupatrackincl
i
nedatan
Wor
kandEner
gy eof300 t
angl othehor
izont
al.I
fhewalks20m,how
muchworkdoeshedo?
 Wor
kisdef
inedasthepr
oductofforceanddist
ance
mov
edinthedirect
ionofappl
icati
onoft heforce.
Sol
uti
on
wor
kdone
,W =f
orc
e,F×di
sta
ncemove
din

t
hedi
rec
tionof

t
hea
ppl
iedf
orc
e,d

W =F×d

 Wor kisther
eforesaidtobedonewhenanappl i
ed wor
kdone
,W =f
orc
e,F×di
sta
nce
forcemakest hepoi ntofappl
icat
ionofthefor
ce
mov ei nt
hedirect
ionofthefor
ce.Nowor ki
sdone wor
kdone=mg×di
sta
nce
,h
whenaper sonpushesawal lunti
lhesweatsor =(
70×10)
N ×(
20s
in30)
m = 7000J
carry
ingabagofcementonhi sheadforhour
swhil
e
standing. Exer
cise

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page34
 232

 TheSIuni
tofwor
kist
hej
oul
e(J)
. 1. Agir
lofmass40kgwalksupafli
ght10steps.I
feach
1j
oul
e(J
)=1ne
wtonme
tre(
Nm) st
epis40cm hi
gh,
cal
culat
etheworkdonebythegir
l.

N/B:Joulei
stheworkdonewhenthepoi
ntof 2. A bodyisacteduponbyavar yingfor
ceFovera
appl
icati
onofaforceof1newt
onmovesthrough1 di
stance of35 m as shown i
nt he f
igur
e bel
ow.
metreinthedi
rect
ionoft
hefor
ce.

Not
es:

I
. Wor
k done i
s equi
val
entt gy conv
o ener ert
ed
whi
l
edoingwor k.
Cal
cul
atet
het
otalwor
kdonebyf
orce
I
I. Thear
eaunderf
orce-
dist
ancegr
aphr
epr
esent
s
wor
kdonebyt
hef
orceorener
gyconver
ted. 3. Sometimesworkisnotdoneevenift
hereisanappl
i
ed
for
ce.Descr
ibesomesituat
ionswhenthi
scanhappen.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Var
iat
ion ofK.
E and P.
EforA Body Pr
oject
ed
Upwards
Gr
avi
tat
ionalPot
ent
ialEner
gy
 Considerabodyofmassm pr ojectedv erticall
y
 Thisistheenergypossessedbyabodyduet oit
s upwards.Gr avi
t ati
onalf or
ce i st he onlyf orce
heightabovesomesur face.Considerabl
ockof acti
ngoni t,assumi ngnegl i
gibl
eai rr
esi stance.
massm r ai
sedthroughtheheighththegr
ound.At Asi traiseskinet i
cenergydecr easessi ncet he
thatheighttheblockhasgravi
tati
onalpot
enti
al vel
ocity decr eases ( t
he body decel erates
energy. upwards) .Atthesamet i
me,t hepot entialener gy
ofthebodyi ncr easesandbecomesmaxi mum at
thehighestpoi nt ,whereK.Ei szer o.Ast hebody
fal
lsfrom thehi ghestpoi nt,P.Edecr easeswhi l
e
K.Eincreases.Thecur vesforv ari
at i
onofK. Eand
P.Eoft hebodywi t
htimeareshownbel ow.

Pot
ent
ial
ene
rgy,
P.Eg
aine
d=wor
kdonei
nra
isi
ngt
hebl
ockP.
E

=we
ightoft
hebl
ockXhe
ight
P.
E=mg h

El
ast
icPot
ent
ialEner
gy

 Thi
si st he energy st
ored in a str
etched or
compr
essedspring. Theenergyisequaltowor k
 Ther
efor
e, at any gi
ven poi
nts;
donei
nstr
etchi
ngorcompr essi
ngthespr
ing.
t
otal
ene
rgy,
E=P.
E+K.
E=Cons t
ant
.
wor
kdone=For
ce×di
sta
ncemove
dindi
rec
tionoff
orc
e

wor
kdone=Ave
rag
eFor
ce×c
hang
einl
eng
thofs
pri
ng
Exampl
es
o+F
(
compr
ess
ionore
xte
nson)wor
i kdone= ×e
;
2 Astoneofmass2.5kgi
srel
easedf
rom ahei
ghtof5.
0m
abovethegr
ound:
F
wor
kdone= ×e,
2
a) Cal
cul
atet
heveloci
tyoft
hest
onej
ustbef
orei
tst
ri
kes
butF=ke
,whe
reki
sthes
pri
ngc
ons
tant∴ t
heground.

b) Atwhatveloci
tywillthestonehi
tthegr
oundifa
wor
1
2( )
kdone= kee
,
const
antai
rresi
stancef or
ceof1.0Nactsonitasit
fall
s?
1 2
wor
kdone= ke
2
Sol
uti
on
Ki
net
icEner
gy,
K.E
1
a h= mv2
)mg
2
 Consi derabodyofmassm bei ngact
eduponbya
steadyf orceF.thebodyaccel
erat
esunif
ormlyfr
om 1
2.
5×10×5. 5×v2
0= ×2.
resttof i
nalvel
ocit
yvinti
metseconds.Ifi
tcover
sa 2
distances;
100
v= =10ms-1
wor
kdonei
nac
cel
era
tingt
hebody
1

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
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 232

=t
heK.
Ega
ine
dbyt
hebody 1
b)Re
sul
tantf
ore×h= mv2 ;
c
2
=For
ce,
F×di
sta
nce
,s
1
mg-r
( esi
sta
nce
)h= mv2
Met
hod1 2

1
K.
E=F×s; (
2.5×10-1. 5×v2
0)×5= ×2.
2
E=ma×(
K. ave
rag
eve
loc
ity×t
i )
me
240
v= 798ms-1
=9.
v-u u+v 2.5
K.
E=m
t ( )
×(
2
×t
)

K.
E=m (v-t0)×(0+v
2 )=mv
×t ×
vt
t 2

1
E= mv2
K.
2

Met
hod2 Exer
cise

1. A stoneofmass5kgmovest hroughahori
zontal
K.
E=F×s;K.
12
E=ma× ut+ a
2
t ( ) di
stance10m from rest
.Ift
heforceact
ingonthe
st
onei s8N,
cal
culat
e:
v-u
K.
E=m ( t)×(0×t+12(v-tu))
t 2

a) t
hewor
kdonebyt
hef
orce

K.
E=m (v-t0)×((0×t+12(v-t0))
t )= ×
mv vt
2
t 2
b) t
heki
net
icener
gygai
nedbyt
hest
one

1 c) t
hevel
oci
tyoft
hest
one
E= mv2
K.
2

2. Calcul
atetheamountofener gyneededbyacat apul
tto Power
-
1
thr
owast oneofmass500gwi thaveloci
tyof10ms
 Poweri sdef nedast
i her ateofdoi ngwork(i
.e.
3. Atennisballisdroppedfr
om aheightof1.8m.itrebounds workdoneperuni tt i
me).Si ncewor kdonei s
toaheightof1.25m. equiv
alenttoener gyused,andener gycannotbe
a) Descr ibetheenergychangeswhichtakeplace destr
oy ed orc reat
ed butc onver
tedf r
om one
for
mt oanot her,
b) Wi t
hwhatvel ocit
ydoestheballhi
ttheground?
 Powercanal sobedef nedast
i herateofener
gy
c) Wi t
hwhatvel ocit
ydoestheballl
eavetheground? conversi
onORt herateoftransferofenergy
.
4. Aballrol
l
sonat abl
einast rai
ghtl
ine.Apartfrom the workdone ene
rgyconvert
ed
Powe
r= orPowe
r=
tr
ansi
ti
onalki
neti
cenergy,st
atetheotherf
orm ofkineti
c ti
metaken timetake
n
ener
gypossessedbytheball
.
 TheSIuni st
tofpoweri hewat
t;namedaf
tert
he
5. Abodyhas16Joulesofkinet
icener
gy.Whatwoul
dbei
ts physi
cistJamesWat
t.
ki
net
icener
gyi
fit
sveloci
tywasdouble? 1wa
ttW)=1j
( oul
epe
rse
cond(
Js-1)

6. Aforceof8N st
ret
chesaspri
ngby10cm. How much Rel
ati
onshi
pbet
weenpowerandvel
oci
ty
wor
ki sdonei
nst
ret
chi
ngt
hisspr
ingby13cm?

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

7. Asimplependulum isreleasedfrom restanditswings workdone

Power= ;
towar
dsitslowestposition.Ift
hespeedatt helowest ti
metaken
posi
ti
onis1.
0m/ s,cal
culatethevert
icalhei
ghtoft
hebob
Forc
e×displ
ace
me ntofpoi
ntofa
ppl
ic
ati
onoff
orc
e
wheniti
srel
eased. Powe
r=
t
imet
ake
n
8. Amet albal
lsuspendedver
ti
cal
l
ywithawir
eisdi
spl
aced
di
spl
ace
me nt
thr
oughanangl eθ asshowninthedi
agr
am bel
ow.The Powe
r=For
ce×
t
imetake
n
bal
lisrel
easedfr
om Aandswingbackt
oB.
Powe
r=For
ce×ve
loc
ity

Gi
vent hatt
hemaxi mum vel
oci
tyatthelowestpointB i
s
-1
2.
5ms .Findthehei
ghthfr
om whi
chthebal
lisr
eleased

9. A 30gbul l
etstri
kesat r
eet r
unkofdi ameter40cm at
Exampl
es
200ms-1 andleavesitf
rom theopposit
esideat100ms-1.
Fi
nd: 1. Anelectr
icmot orrai
sesa50kgmassataconst ant
vel
ocit
y.Cal
culatethepoweroft
hemotori
fitt
akes30
I
. Theki net
icener
gyoft hebul
letjustbef
orei
tstr
ikes
secondstor
aiset hemassthr
oughahei
ghtof15m
thet ree.
Sol
uti
on
I
I. Theki
net
icener
gyoft
hebul
l
etj
ustbef
orei
tleavees
fr
om t
hetr
ee. di
spl
ace
me nt
Powe
r=For
ce×ve
loc
ity;
Powe
r=mg×
t
imetake
n
I
II
. Theaver
agefor
ceact
ingont
hebul
l
etasi
tpasses
thr
ought
hetr
ee. 15m
powe
r=50N× =25W
30s
10.Theini
tialvel
oci
tyofabodyofmass20kgi s4ms-1.How
l
ongwoul daconstantfor
ceof5. 0N actonthebodyin 2.
A soldi
ercli
mbst othetopoft hewat chtowerin15
ordert
odoubleitski
neti
cenergy? minut
es.Ift
heworkdonebyt hesoldi
eragainstgr
avi
tyi
s
60kJ,whati
shisaver
agepowerincli
mbing?
11.Acompr essedspri
ngwithaloadattachedtooneendand
fi
xedattheotherandisrel
easedasshownbelow. Solut
ion

workdone
Powe
r=
ti
metaken

Sketchonthesameaxisthevar
iat
ionofpot
ent
i gy, power=60×1000=66.
alener 67W
ki
net
icener
gyandtot
alener
gywit
hti
me 15×60

12.Thef
igur
ebel
owshowshowt
hepot
ent
ialener
gy(
P.E)of Exer
cise

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

abal
lthr
ownver
ti
cal
l
yupwar
dsvar
ieswi
thhei
ght 1. A crane l
i
ftsaload of200 kg t
hrough a ver
ti
cal
di
stanceof3.
0m i
n6seconds.Det
erminet
he;

I
. Wor
kdone

I
I. Powerdevel
opedbyt
hecr
ane

2. Al
oadof100Ni
srai
sed20m i
n50s.Cal
cul
ate;

I
. Thegai
ninpot
ent
ialener
gy
Ont
hesameaxespl
otagr
aphoft
heki
net
icener
gyoft
heba
l
l I
I. Thepowerdevel
oped

3. Wat erf
all
sthroughaheightof60m atarateoffl
owof
10×105 lit
resperminute.Assumi
ngt hatther
eareno
energy losses, cal
cul
ate the amount of power
generat
edatt hebaseofthewaterfall
.(t
hemassof1
l
iterofwateris1kg)

4. I
f50li
tr
esofwateri
spumpedthroughahei
ghtof15m
i
n30seconds,whati
sit
spowerr at
ingoft
hepumpis
80%ef
fici
ent
?(t
hemassof1li
terofwateri
s1kg)

5. Asmallwi
ndpumpdevelopsanaver
agepowerof50N.
I
trai
seswaterfr
om aborehol
etoapoint12Nabove
t
hewaterl
evel
.Deter
minethemassofwaterdel
i
vered
i
nonehour.

Si
mpl
eMachi
nes  N/B Eff
ici
ency, j ustlike M.
A ofa machi ne
dependsonf r
ict
ionbetweenmovi ngpartsofa
 Amachineisadevicet
hatmakeswor
keasierORIs machineandweightoft hepart
sthathavetobe
anydevi
cebymeansofwhichafor
ceappl
iedatone l
ift
ed.Thesereasonsexpl ai
nwhyt heeff
ici
ency
poi
ntofitcanbeusedtoover
comeafor
ceatsome ofamachineisalwayslessthan100%.
ot
herpoi
ntofi
t.
Exampl
es
Ter
msAssoci
atedwi
thMachi
nes
1. Acertai
nmachineusesanef f
ortof400N t
orai
sea
1.Ef
for
t,E l
oadof600N.I ft
heeff
ici
encyofthemachi
neis75%,
det
erminei
tsvel
oci
tyr
ati
on.
 Thi
sisthef
orceappl
iedt
othemachi
ne.TheSIuni
t
ofef
for
tist
henewton(N)
. Sol
uti
on

l
oad
2.Load,
L M.
A=
eff
ort
 Thi
sist
hefor
ceexer
tedbyt
hemachi
ne.TheSIuni
t 600
M.
A= =1.
5
ofl st
oadi henewt
on(N). 400

M.
A 1.
5
3.Mechani
caladvant
age,
M.A η= ×100%⇒75= ×100%
V.
R V.
R

 Thi
sistherat
iooftheloadt
ot heeff
ort
.Ithasno 1.
5
V.
R= ×100% =2
uni
tssi
ncei
tisarat
iooftwof
orces. 75

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

l
oad 2. A crane l
i
ftsaload of200 kg t
hrough a ver
ti
cal
M.
A=
eff
ort di
stanceof3.
0m i
n6seconds.Det
erminet
he;

 Formostmachines,M.Aisgreaterthanonesince I
. Wor
kdone
l
oadisgreat
ert
haneffort
.InafewmachinesM.A.is
l
essthanone(i
.eefforti
sgreaterthanl
oad)e.
g.a I
I. Powerdevel
opedbyt
hecr
ane
bi
cycl
e. I
II
. Eff
iciencyoft
hecr
anegiventhati
tis
operatedbyanelect
ri
cmot orrat
ed
Fact
orsAf
fect
ingM.
AofaMachi
ne
1.
25kW.
I
. Fr
ict
ionbet
weenmovi
ngpar
tsoft
hemachi
ne-
The Sol
uti
on
great
er the f
ri
cti
on,t
he l
ess t
he mechani
cal
advantage I
.wor
kdone=mg
h

I
I. Par
tsoft
hemachi
net
hathavet
obel
i
fted– The wor
kdone=200×10×3.
0=6000J
heavi
er the wei
ght
,the l
ess t
he mechani
cal
wor
kdone
advant
age. I
I.Powe
r=
t
imet
ake
n
4.Vel
oci
tyRat
io,
V.R 6000
Powe
r= =1000W or
1kW
6
 I
tisther
atiooft
hedist
ancemovedbyt
heeff
ortDE
t
ot hedi
stancemov
edbyt oadDL i
hel nthesame poweroutput
I
II
.η= ×100%
t
ime. powe
rinput

di
sta
ncemove
dbyt
hee
ffor
t,
DE 1000
Ve
loc
ityr
ati
o,V.
R= η= ×100% =80%
di
sta
ncemove
dbyt
hel
oad,
DL 1250

DE Exer
cise
V.
R=
DL
1. Whenanel ect
ri
cpumpwhoseeffi
ciencyis70%rai
ses
 Note:I
ft womachi nesAandBwi t
hv el
oci
tyrat
iosV.RA watert
oahei ghtof15m,wat
eri
sdeliver
edatther
ate
andV.RB respect
ivelyar
ecombi
ned,ther
esult
antvel
oci
ty of350li
tr
espermi nut
e.
rat
ioV.Rwi l
lbegivenby:
I
. Whati
sthepowerr
ati
ngoft
hepump?
V.
R=VRA ×VRB
I
I. Whati
stheener
gyl
ostbyt
hepumppersecond?
5.Ef
fi
ciency,
η
2. Anel ectri
cpumpcanr aisewat erfr
om alower -l
evel
 I
tist
her atiooft heusefulworkdonebyt hemachi
ne reservoirtothehighlevel r
eser
v oi
rtotheh i
ghl evel
5
(
workout put)tot hetot
alwor kputintothemachi
ne reservoirattherateof3. 0x10 kgperhour .The
(
workinput )expressedasaper centage. verti
cal hei
ghtofthew at
e ri
sraised360m. I
fther ate
ofener gylossi
nform ofheatis200kW,determinet he
wor kdone
bythemac
hine(
workout
put) effi
ciencyofthepump.
η= ×100%
tot
alwor
kint
othema
chi
ne(
wor
kinput
)
3. Def
inetheeffi
ciencyofamachi
neandgi
vear
eason
Rel
ati
onship bet
weenMechani
calAdvant
age,Vel
oci
tyRat
io whyitcanneverbe100%
andEff
ici
ency
4. Apumpuses1gofami xt
ureofpet
rolandal
coholi
n
wor
kdoneonl
oad(workoutput
) t
herati
o4:1bymasst
orai
se1000kgofwaterfr
om a
η= ×100%
wor
kdonebye
ffor
t(workinput
) wel
l200m deep.

L×DL I
. Howmuchener
gyi
sgi
venby1gofmi
xtur
e?
η= ×100%
E×DE

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page40
 232

L DL I
I. I
ft hepumpi s40% ef
fi
cient,whatmassoft
his
η= × ×100%
E DE mixtur
eisneededtorai
set hewater?
DL 1
(
1gofal cohol=7000J,
ofpet r
ol=48000J)
L
but =M.
Aand =
E DE V.
R
5. I
namachi ne,thi
sl oadmoves2m whent heeffor
t
1 moves8m,ifanef
fortof20Nisusedt
or ai
sealoadof
η=M.
A× ×100%
V.
R 60N,whati
stheef
fici
encyoft
hemachine?

M.
A
η= ×100%
V.
R

Ty
pesofSi
mpl
eMachi
nes Exer
cise

1.Lever
s Thef
igur
ebel
owshowsal
ever

 Alev eri
sasi mplemachi newhoseoper at
ionr el
ies I
. Det
ermi
net
hef
orceFA i
neachcase
ont heprincipleofmoment s.I
tconsi
stsoftheef f
ort
I
I. Det
ermi
net
heM.
AandV.
Rineachcase
arm,loadar m and pi votfurcul
um.Theef f
ortarmEA
i
st heper pendiculardi st
anceoft hel
ineofactionof I
II
. Cal
cul
ateef
fi
ciencyi
neachcase
theef f
ortf rom t hepi vot.Theloadcan,LA i st he
perpendiculardi stanceoft heli
neofact i
onoft he
l
oadf rom piv ot.

 Consi
dert
hef
igur
eofsi
mpl
elev
ersshownbel
ow

Cl
assesofLever
s

A. Leverswi
thpi
votbet
weenl
oadandef
for
te.
g.
pl
iers,
hammeret
c.

B. Lever
swiththeloadbetweenpivotandef
for
t
e.
g.wheelbar
row,bot
tl
eopener
setc.
di
sta
ncemovedbytheeff
ort
,DE C. Lever
swiththeeff
ortbet
weent
heloadandt
he
V.
Rofl
ave
r=
di
sta
ncemovedbyloa
d,DL pi
vote.g.sweepi
ngbrooms,afi
shi
ngrodand
DE EA hammerarm.
Us
ingc
onc
eptofs
imi
lart
ria
ngl
es; =
DL LA

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

eff
orta
rm,EA 2.Wheelandaxl
e
The
ref
ore
,fort
hel
eve
r,V.
R=
l
oadarm,LA
 Consistsoflar
gewheelofr adiusRat tachedto
axl
eofr adi
usr,theeffor
tisappliedonthewheel
whil
et hel
oadisattachedtotheaxle.Anexample
ofthewheelandaxl eisthewi nchusedt odraw
waterfrom wel
l

Exampl
e

Thef
igur
ebel
owshowsal
ever
 I
nonecompl et
erevol
uti
onthewheelmovesthr
ougha
di
stance2πRwhil
etheloadmovesthr
ough2πr.

2πR R Radiusofwheel
V.
Rofwhe
ela
nda
xle= = ,V.
R=
2πr r Radiusofa
xle

a) Det
ermi
net
hef
orceFA

b) Det
ermi
net
heM.
AandV.
R

c) Cal
cul
ateef
fi
ciency
Exampl
e
Sol
uti
on
Awheelandaxleisusedt orai
seal
oadof300Nbyaf orce
a
)cl
ockwi
semome
nt=a
nti
cl
ockwi
semome
nt of50Nappliedtother i
m ofthewheel.Ift
her adi
iofthe
wheelandaxlear
e85cm and10cm r espect
ivel
y,cal
cul
ate
×5=75×2 theM. Randef
A,V. fi
ciency.
FA
Sol
uti
on
150
⇒FA = =30N l
oad
5 M.
A= ;
eff
ort

300
M.
A= =6
l
oad 50
b)M.
A= ;
eff
ort
Radiusofwheel 0.
85m
V.
R= ; V.
R= =8.
5
75 Radiusofa
xle 0.
10m
M.
A= =2.
5
30
M.
A 6
η= ×100%; η= ×100% =70.
59%
eff
orta
rm,EA V.
R 8.
5
V.
R= ;
l
oadarm,LA

5
V.
R= =2.
5
2

M.
A
c
)η= ×100%;
V.
R

2.
5
η= ×100% =100%
2.
5

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page42
 232

3.TheI
ncl
i
nedPl
ane V.
Rofs
cre
w=
ci
rcumf
ere
nceofthes
cre
whe
ad 2πR
=
pit
ch pi
tch
 Considertheplanebel
ow i
ncl
i eθt
nedatanangl o
thehorizont
al.
WhereRi stheradi
usoft hescr
ew head.A scr
ew
combinedwi t
hal everisusedasaj ackforli
ft
ing
heavyloadssuchascars.

Exampl
e

Thefigur
ebel owshowsacar-jackwit
hal everar
m of40
eff
ortdi
sta
nce L
V.
Roft
hei
ncl
ine
dpl
ane= = cm andapi tchof0.5cm.Iftheeffi
ciencyi
s60%,what
l
oaddis
tance h eff
ortwouldberequi
redt
oli
ftaloadof300kg.
h
buts
inθ= ⇒h=Ls
inθ
L

L 1
V.
Roft
hei
ncl
ine
dpl
ane= ; V.
R=
Ls
inθ si
nθ

Exampl
e

A manusest heincli
nedpl
anet oli
ft100kgloadthrougha
Sol
uti
on
ver
ticalheightof8.0m.Theincli
nedplanemakesanangl eof
400 withtheverti
cal.Ift
heeff
ici
encyoft hei
ncl
inedplaneis M.
A M.A
η= ×100% ⇒η= ×100%
85%, cal
culate V.
R 2πR
pi
tch
I
. Theeff
ortneededto movethel
oadupt
hei
ncl
i
ned
M.
A
pl
aneataconst
antvel
oci
ty. 60= ×100%
2π×0.
4
0.
005
I
I. The wor
k done againstfr
ict
ion i
nrai
sing t
he l
oad
thr
ought
heheightof8.0m 60× 2π×0.(
4 )
0.
005
⇒M.
A= =301.
63
100

Load(mg) 300×10
Sol
uti
on M.
A= ⇒301.
63=
Ef
fort Eff
ort
M.
A M.A 300×10
I
.η= ×100%; 85= ×100%; Ef
for
t= =9.
46N
V.
R 1 301.
63
si
n50
Exer
cise
85×1
s
in50
M.
A= =1.
110
100 Acarwei
ghi
ng200kgisli
ftedwit
haj ackof15mm pit
ch.I
f
t
hehandl
eis32cm f
rom t
hescrew,f
indtheappl
iedfor
ce.
loa
d( mg) 1000
M.
A= ;1.
110= ;
ef
fort ef
for
t

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
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 232

1000 5.Gear
s
e
ffor
t= =900.
9N
1.
110
 A gearisawheelwhi chcanr otat
eabouti t
s
centr
e and has equal
lyspaced t
eeth orcogs
I
I.15% ofwor
kinputi
sus
edt
oove
rcomef
ri
cti
on aroundit
.Twoormor egearsarear r
angedto
makeamachi newhichcanbeusedt ot r
ansmit
i
.
e.15% of(
eff
ort×di
sta
ncemove
dbye
ffor
t)
mot i
onfr
om onewheel t
oanother
.
15
100 (
× 900.
9×
s
i )
8
n50
=1411.
26j
oul
es

4.TheScr
ew

 I
fthedri
verwheel hasPteethandthedriven
wheelQteeth,
then,whent hedr
iverwheel makes
P
onerev
oluti
on,thedriv
enwheel makes
Q
revol
uti
on
 Thedi
stancebetweentwosuccessi
vet
hreadsofa
scr
ewiscall
edthepit
choft
hescrew. re
voluti
onsmadebythedr
ive
rwheel 1 Q
V.
R= ; =
revol
uti
onmadebythedri
venwhe
el P P
Q
 I
n one compl
eter
evol
uti
on,t
he scr
ew mov
es
for
war
dorbackwar
dthr
oughadi
stanceequalt
o numberoft
eethindri
venwheel
V.
Rofag
ears
yst
em=
onepi
tch. numberofthe
ethinthedri
verwheel

Exer
cise

Acertai
ngearhas30t eethanddr
ivesanotherwi
th75t
eeth.
How manyr evol
uti
onswi llt
hedr
ivergearwhenthedr
ivi
ng
gearmakes100revolut
ions?

6.Pul
l
eys

 A pull
ey i s a wheel wi t
h a gr oove for
accommodati
ngast r
ingorrope.V.
Rf orapulley
syst
em i
st he numberofropessupporti
ng l
oad.
Ther
eist
het
hreecommonpul
l
eysy
stems.

A)Si
ngl
eFi
xedPul
l
ey
NBForablockandtackl
ewithanoddnumberof
pul
l
eysi
tisconveni
enttohavemor
epull
eysfi
xed

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page44
 232

t
hanmovabl
e.

Thev
eloci
tyr
ati
oofsi
ngl
efi
xedpul
l s1.
eyi
Exampl
e

Ablockandtackl
esyst
em i
susedtol
if
tamassof400kg.
B)Si
ngl
eMovabl
ePul
l
ey I
fthi
smachinehasaveloci
tyr
ati
oof5andanef
fi
ciencyof
75%

a) Sket
ch a possi
ble ar
rangementoft
he pul
l
eys
showinghowtheropei
swound

 Sol
uti
on

Theupperpul
l
eyi
n(b)makesi
tpossi
blef
oref
for
ttobe
appli
ed downward.Thevel
oci
tyr
atio ofeach ofthe
pull
eys(a)and(b)is2si
ncetworopesaresupport
ing
theload. b) Cal
cul
atet
heef
for
tappl
i
ed.

 Sol
uti
on

M.
A M.
A
C)Bl
ockandt
ackl
e η=
V.
R
×100% ⇒75=
5
×100%

Thi
ssy st
em consi
stsoftwo ormor
eset
sofpul
l
ey 75×5
M.
A= =3.
75
bl
ocks.Bel
owareexampl
es. 100

Load 400×10
M.
A= ;3.
75=
Effor
t Eff
ort

4000
⇒Ef
for
t= =1066.
7N
3.
75

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
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 232

Exer
cise

1. Aneff
ortof125Nisusedt
olif
taloadof500Nthrough
ahei
ghtof2.5m usi
ngapull
eysystem.I
fthedi
stance
movedbytheeff
orti
s15m,calcul
ate

a) Thewor
kdoneont
hel
oad

b) Thewor
kdonebyt
heef
for
t

c) Theef
fi
ciencyoft
hepul
l
eysyst
em.

2. Drawalockandt ackl
epul
l
eysyst
em ofV.
R6t oshow
how the pul
l
eycan be used torai
se a l
oad L by
appl
yi
nganef f
ortE

7.Pul
l
eyBel
ts Exampl
e

Theradiusoftheeffor
tpist
onofahydrauli
cli
ftis2.1cm
whi
lethatofthel
oadpistoni
s8.
4cm.Thi
smachi neisused
torai
seal oadof180kgataconstantveloci
tythrougha
hei
ghtof5m.Gi ven t
hatthemachi
nei s75% ef fi
cient
,
cal
cul
ate:

a) Theef
for
tneeded

b)Theener
gywast
edi
nusi
ngt
hismachi
ne.

Sol
uti
on

M.
A
 Theyaref ound i
n posh mil
l
s,sewi
ng machines, a
()η= ×100% but
,
V.
R
motorengi
nesetc.I
fther
adiusoft
hedri
verpul
l
eyis
Randt hatoft
hedriv
enpulleyisr,t
hebelttur
nsa cross-s
ectiona
re a
ofl
oa dpis
ton πRL2
V.
R= =
distanceof2πR whenthedr i
vi
ngwheelmakest he c ross-sec
tionareaofe
ffortpi
ston πRE
2

revoluti
on.The l
oad wheel(dri
ven wheel
)att he
0842
π×0.
samet i
memakes V.
R= 2
=16
π×0.
021
2πR R M.A 75×16
= re
vol
uti
ons 75= ×100% ⇒M.
A= =12
2πr r 16 100

numbe
rofrevol
uti
onsma
debyeff
ort 1 r 180×10
V.
R= = = L
numberofr
evol
uti
onsmadebyl
oad R R E= ⇒E= =150N
M.
A 12
r

ra
diusofdr
ive
nwhe el
V.
R=
radi
usofdr
iverwhe
el (
b)25%ofworkinputisequi
val
entt
oener
gywast
ed,
i.
e.15%of
(
(workout
put
)/Ef
fi
ciency×100)

8.Hydr
aul
i
cmachi
nes 100
× (
25 1800×5
75 )
×100 =3000N

Exer
cise

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page46
 232

1. Thedi
agr
am bel
owshowshydr
aul
i
cbr
akesyst
em.

a) St
atet
hreepr
oper
ti
esoft
hehydr
aul
i
cbr
akeoi
l
.

b) A forceof20N isappli
edonafootpedaltoa
2
pistonofareas50cm andthi
scausesst
oppi
ng
forceof5000N.

I
. Pr
essur
eint
hemast
ercyl
i
nder
Whentheef f
ortpi
stonmov
esdownwar
dst
he
l
oadpi
stoni
spushedupwar
ds. I
I. Ar
eaoft
hesl
avepi
ston
vol
umeofl
iquidt
hatl
eavesef
fortcyl
inder I
II
. Vel
oci
tyr
ati
ooft
hesyst
em
=vol
umeofli
qui
dthatent
ersloadcyli
nder

di
sta
ncemovedbyeff
ort×cr
oss-sect
ionofef
for
tpi
ston
2. Studythefi
gurebel
ow ofahydr
aul
i
cli
ftandanswer
=dista
ncemovedbyload×cr
oss-s
ect
ionare
aofpis
ton thequest
ionbel
ow.

di
sta
ncemove
dbyeff
ort cross-s
ectiona
re a
ofl
oa dpis
ton
=
dis
tanc
emovedbyl
oad c r
oss-sec
tionareaofe
ffortpi
ston

Ve
loc
ityr
ati
o=

Theareasofcross-sect
ionsofthepist
onsandthelengt
h
ofthearm ar
easi ndi
cat
ed.FindFo,M.Aandef
fi
ciencyof
themachine.

Revi
sionExer
cise

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

1. The f i
gure below shows a syst em of gear s for
transmi
tti
ngpower .GearAhas200t eet
handactast he
drivi
nggear.GearsBandCwi th40t eet
hand100t eeth
respect
ivel
yar e mounted on t
he same axl
e and they
transmi
tmot i
ontot hel
astgearDwhichhas50teet
h.

I
. I
nwhatdirect
ion(
s)wouldgearCandDr ot
ate?I
f
gearAi
srotat
edincl
ockwisedi
rect
ion.

I
I. Fi
ndt
hevel
oci
tyr
ati
ooft
hegearsyst
em

2. Thef i
gurebelow showstherearwheelofabicycl
eand
thecrankwheelA,connect
edtothespr
ocketBbyachain.
I
fwheelAhas40t eethwhi
leBhas25teethandther
adi
us
oftherearwheeli
s42cm, cal
culat
e:

a) t
hevel
oci
tyr
ati
ooft
hemachi
ne,

b) thedist
ancetr
avel
ledbyt
hebi
cycl
einoner
evol
uti
on
ofthecrankwheel
.

3. Thepull
eysystem abovehasaMAOF3cal culat
e: i)t
he
tot
alworkdonewhenal oadof60N i
sraisedthrougha
hei
ghtof9m.i
i)theeffi
ciencyoft
hemachi
ne.

Chapt
erFour REFRACTI
ONOFLI
GHT

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page48
 232

Speci
fi
cObj
ect
ives Cont
ent

Bytheendoft
hist
opi
cthel
eanershoul
dbeabl
e 1. Refr
acti
onofli
ght–lawsofr
efr
act
ion(
exper
iment
al
t
o: tr
eatmentr
equi
red)

a) Descr i
be si mple exper i
ments t o ill
ust
rat
e 2. Determinat
ionofrefr
acti
veindex
refr
actionofl i
ght
 Snell’
slaw
b) Statethel awsofrefract
ionofl
ight
 Real/apparentdepth
c) VerifySnell’
slaw
 Criti
calangl
e
d) Definerefracti
veindex
3. Dispersi
onofwhi t
elight(Exper
iment
alt
reat
menti
s
e) Determi neexperi
ment al
lyther
efract
ivei
ndex requi
red)

f) Descri
beexper i
mentt oi ll
ustr
atedispersionof 4. Totali
nter
nalreflecti
onandi t
seffects:cr
it
icalangl
e
whit
elight
5. Appli
cati
onsoft otalint
ernalr
efl
ection
g) Expl
aintot
alinter
nal r
eflect
ionanditseffects
 Pr ism per i
scope
h) Stat
etheapplicat
ionsoft otal
inter
nalrefl
ecti
on
 Opt i
calfibre
) Sol
i v
e numericalpr oblems involv
ing refr
activ
e
i
ndexandcr i
ti
calangle. 6. Probl
emsonr ef
ract i
veindexandcriti
calangle

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Def
ini
ti
onofRef
ract
ionofLi
ght  Ref
ract
ionofl
i
ghti
sther
easonast
owhy
;

 Refracti
on ofl i
ghtr eferst ot he change i n a) ast
ickappear
sbentwhenpar
tofi
tisi
nwat
er
dir
ectionoflightatt hei nterf
ace asi ttrav
els
fr
om onemedi um toanot heratanangl e,for
exampl e,arayofl i
ghtf rom airt owat er
.The
cause ofr ef
ract
ion ofl ightist he change in
vel
ocityofli
ghtasittravelsfrom onemedi um to
another.The change i n veloci
tyi s due t o
vari
ationofopti
caldensityofmedi a.

b) acoininabeakerofwaterappear
sneart
he
sur
facet
hani
tact
ual
l
yis,

c) apoolofwaterappearsmoreshall
owwhen
vi
ewedmor eobli
quelyetc.(
Student
stoper
for
m
t
hispract
ical
l
y)

 Ar ay thattr
avel
s perpendicul
artoi nt
erf
ace
proceedsacr
ossthei
nterfacenotdevi
atedsi
nce
theangleofi
nci
dencetothenor mali
szero. Opti
caldensi
ty(
transmi
ssi
ondensi
ty)andr
efr
act
ionof
l
ight

 Arayoflighttr
av el
li
ngf r
om anopt icall
ylessdense
medi
um t o an opt i
call
y denser medi um bends
t
owardsthenormalaf terref
racti
one.g.ar ayfr
om ai
r
t
oglassblockasi n(a)below.Theangl eofinci
dence
i
nthi
scasei sgreaterthanangleofrefracti
on.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page50
 232

 Ar ayofli
ghttr
avell
ingfrom anopt i
call
ydense si
ni
=cons t
ant.
medium to an optical
l
yl ess dense medium si
nr
bendsawayfrom thenormalafterrefr
acti
one.
g.
arayfrom gl
asstoairasin(b)below.Theangle Thec onst
antiscal
ledrefra
cti
veindex.
ofinci
denceinthiscasei slesst hanangleof
 Agraphofs i
niagainstsi
nri sast r
aightl
i
nepassi
ng
ref
ract
ion.
t
hrought heor i
ginandsl opeoft hegraphgiv
esthe
r
efract
iveindexofamat er i
al.

 NB:some medi
a ar
e phy
sical
l
y denserbut
opticall
yl essdenset hanotherse.g.kerosenei s Ref
racti
v eIndex,n
phy si
callyl essdensebutopt ical
lydensert han
wat er.  Ref ractivei
ndexi sdefi
nedas:t herat
iooft
hesineof
theangl eofincidence(i)tothesineoftheangleof
Exerci
se ref
ract i
on(r)forar aypassingf r
om onemedium to
another .
1. Definethet ermref
racti
on
Consi derarayofl ighttrav
ell
ingfrom medium 1to
2. Draw adi agram toshow refr
acti
onforar ayoflight
medi um 2.
across the f ol
l
owing boundar
ies i
nt he or
derthey
appear
 Fort het womedi a,
a) ai
r-wat
er
s
ini
=r
efr
act
ivei
nde
x,1n2
b) wat
er-gl
ass s
inr

c) gl
ass-ai
r (
read as t
he r
efr
acti
ve i
ndex ofmedi
um 2 wi
th
d) gl
ass-ai
r-wat
er r
espectt
omedium 1)

3. Thef
igur
ebel
owshowshowr
efr
act
ionoccur
s.  Bythepr
inci
pleofr
ever
sibi
l
ityofli
ght,
arayt
rav
ell
ing
fr
om medium 2tomedium 1alongthesamepathis

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

r
efr
act
edmaki
ngt
hesameangl
es.

Whi choft hetwomediaisopti

call
ydenser
?
Expl
ain. s
inr
=2n1.rist
heangl
eofinci
dencei
nthi
scase.
s
ini
4. Expl
ainwitht hehelpofadi
agr
am whypenci
lpl
aced
part
lyinwat erappear
sbent
. s
ini 1
but, =1n2=
s
inr s
inr
s
ini

1
1n2=
.n
2 1
LawsofRef
ract
ion
s
inr
Law1 =2n1.ri
stheangl
eofi
nci
dencei
nthi
scase.
s
ini
 The i
nci
dentray
,the r
efr
acted r
ay and t
he
s
ini 1
nor
malalll
iei
nthesameplaneatthepoi
ntof but
, =1n2=
i
nci
dence. sinr s
inr
s
ini
Law2(
Snel
l
’sl
aw)
1
1n2=
 I
tstatesthat;“
ther
ati
oofthesineoftheangle .n
2 1
ofincidence (
i)tothe si
ne ofthe angl
e of
ref
racti
on( r
)isaconstantforagivenpairof
media”.

Not
e: 3.Usetheinf
ormati
ongiveninthef i
gures(
a)and(
b)bel
owt
o
cal
cul
atetheref
ract
ivei
ndexanw andtheangl
eθ
Absol
uter
efr
act
ivei
ndexi
sther
efr
act
ivei
ndexofa
mater
ialwit
hr especttovacuum.Av
acuum hasa
r
efr
activ
eindexof1.000.

Exampl
es

1.A r ay ofl i
ghtpasses t hrough airint
o a certai
n
transparentmat er
ial.I
ft he angl es ofinci
dence and
refract
ion are 600 and 350 respecti
vel
y,calcul
atethe
refract
iveindexofthemateri
al

Sol
uti
on
4.St
atet
hepr
inci
pleofr
ever
sibi
l
ityofl
i
ght
.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page52
 232

Ref
ract
iveI
ndexi
nTer
msofVel
oci
ty

 Lighttravel
sfasterinanoptical
l
ylessdensemedi um
thaninanopt icall
ydensermedium.Considerarayof
l
ightcr ossing the boundaryfrom medium1 wi t
h
speedv1 t o median2 wi t
hspeed v2,wher e v1 i
s
si
ni greaterthanv2asshownbel ow.
Re
fra
cti
vei
nde
x=
si
nr

si
n60
Re
fra
cti
vei
nde
x= =1.
510
si
n35

 Ref
ract
iveindex,1n2 ofmedi
um 2wi
thr
espectt
o
medium 1i
sgivenas:
2.Gi
venthatr
efract
iveindexofgl
assis1.5,cal
cul
atethe
angl
eofinci
denceforar ayofli
ghttravel
l
ingfr
om airto
s
ini v1
gl
assift
heangleofrefr
acti
onis100. 1n2= =
s
inr v2
Sol
uti
on
ve
loc
ityofl
igt
hinme
dium1
1n2=
ve
loc
ityofl
igt
hinme
dium2

 Absolut
erefr
acti
veindexnistheref
ract
ivei
ndexof
themedium whenli
ghtistr
avell
i
ngfr
om thevacuum
si
ni
tothemedium.
Re
fra
cti
vei
nde
x=
si
nr Abs
olut
ere
fra
cti
vei
nde
x,

si
ni ve
loci
tyofl
igthi
nva
c uum
1.
5= n=
sn10o
i veloc
ityofl
igt
hinmedium

Ang
leofi
nci
denc
e,i= S
in-1 1.
5×s
i ( 09o
n10o =15. )  Consi
dert
hedi
agr
am bel
ow;

3.Cal
cul
atether
efr
acti
veindexforl
i
ghtt
ravel
l
ingf
rom
gl
asst
oairgi
vent
hatang=1.572. fv1i
 I sthev el
ocit
yofl i
ghti
nmedium 1ofr
efr
acti
ve
Sol
uti
on ndexn1 andv2 t
i hev el
ocit
yofl
i
ghtinmedium 2of
r
efr
acti
veindexn2then;

s
inθ1 n2 v1
1n2= = = ;
1 s
inθ n1 v2
2
= =0.
6361
1.
572
⇒n1s
inθ1=n2s
inθ2;
Exer
cise

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

1.Arayofl
ightst
ri
ki
ngat
ranspar
entmat
eri
ali
sref
ract
ed ⇒ns
inθ=c
ons
tant
asshownbelow.
 Ther
efor
e,t heproductoftheref
ract
ivei
ndexofa
medium andtheanglear aymakeswitht
henormal
i
nthemedi um i
saconstant.

Cal
cul
atet
her
efr
act
ivei
ndi
ces:

a) 1n2

b) 2n1

2.Cal culat
et he angl
e ofr ef
ract
ion f
ora r
ayoflight
str
ikinganai rglassinter
facemaki eof500wi
nganangl th
theinterf
ace( ang=1.
526)

Exampl
es 5. Thefigurebelow showsaglasspri
sm ofrefr
act
iveindex
1.
5wi thequi
later
altr
iangl
ecr
osssect
ion.Fi
ndtheangleof
1.Giventhattherefr
act
ivei
ndexofdi amondi s2.
51and devi
ati
onD.
theveloci
tyoflightinairi
s3.0x108 ms-1,cal
cul
atet
he
vel
ocityofl
ighti
ndiamond.

Sol
uti
on

vel
ocit
yofli
gthi
nai
r
r
efr
act
ivei
nde
xofdi
amond =
ve
loc
ityofl
igt
hindi
amond

3.0×108ms-1
2.
51= ; 6. Thespeedofl
i
ghti
nmedi
um m1 i 0x108ms-1 andt
s2. he
ve
loc
ityofl
igt
hindi
amond
medi 5x108ms-1.Cal
um m2 1. cul
atet
her
efr
act
ivei
ndexof
0×108ms-1
3. medi
um m2wit
hrespectt
om1
⇒ve
loc
ityofl
igt
hindi
amond= 195×108ms-1
=1.
2.
51
7. Cal
cul
ateangl
eθ bel
ow,gi
vent
hatr
efr
act
ivei
ndi
cesof
3 4
gl
assandwat
erar
e and r
espect
ivel
y.Rayi
sfr
om wat
er
2 3
2.Gi
venthatthevelocit
yofl i
ghtinwateri
s2.27x108ms-1 t
ogl
ass
8 -
1
andi
nglassis2.1x10 ms , calcul
ateangl
eθbelow.

Sol
uti
on 8. Ar ayofl
i
ghtisi
nci
dentonapar af
fi
ngl
assi
nter
faceas
shownint
hefi
gur
ebelow.Cal
cul
ater
.
s
inθwater vwater
=
si
nθ vglass
g
las
s

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page54
 232

si
nθ 2.27×108
= ;
s
in22 2.1×108

i
⇒θ=sn-1 (27×108
2.
1×108
2.
×s
in22 )
o
=23.
89

3.Arayofli
ghti
sinci
dentonawat er
-gl
assint
erf
aceas
shownbel
ow.Cal
cul
ater.(
taker
efr
acti
veindi
cesofgl
ass
3 4
andwat
er and r
espect
ivel
y)
2 3

Sol
uti
on

nws
inθw =ngs
inθg

4 3
s
in40= s
2
i
nr Ref
ract
iont
hroughsuccessi
vemedi
a
3

( )
4  Consi
derar ayofl i
ghttr
avel
l
ing t
hrough mult
ipl
e
s
in40×2
i
r=sn-1 3
l
ayersoftr
ansparentmediawhoseboundariesare
3
par
all
elt
oeachotherasshownbel
ow.
o
=34.
85

Exer
cise
s
ini s
inr
1
3 1n2= 2n
; 3=
9. I
fther
efr
act
ivei
ndexofgl
assi
s ,cal
cul
atet
he s
inr
1 s
inr
2
2
r
efr
act
ivei
ndexoft
hemedi
um i
nthef
igur
ebel
ow
sni s
i nr
i 1 s
ini
1n2x2n3= × = …….
.
(1)
s
inr1 si
nr2 s i
nr2

si
nr2 s
ini
3n1 = ;⇒ 1n3= …………(
2)
s
ini si
nr2

10.Expl
ainwhyt
hel
i
ghtbendswheni
ttr
avel
sfr
om one

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

medi
um t
oanot
her
.  Ther
efor
e,f
rom equat
ions(
1)and(
2)1n
; 3=1n2x2n3

Forksuccessi
vemedi
aar
rangedwi
thboundar
ies
par
all
el1n
; k=1n2x2n3…………….
k–1nk

Exampl
es. Exer
cise

4 3
1. Ther
efr
act
ivei
ndexofwat
eri
s andt ass . 1.Ar
hatofgl ayofl
i
ghtt
ravel
sfr
om ai
rint
omedi
um 1and2asshown.
3 2
Cal
cul
atet
her
efr
act
ivei
ndexofgl
asswi
thr
espectt
o
wat
er.

Sol
uti
on

wn n xang
g=w a

1 3
wng= × ;
4 2
3 Cal
cul
ate;

3 3 I
. Ther
efr
act
ivei
ndexofmedi
um 1
wng = × =1.
125
4 2
I
I. Cr
it
icalangl
eofmedi
um 1

I
II
. Ther ef
ract
ivei
ndexofmedi
um 2r
elat
ivet
omedi
um
2. Ar
ayofl
i
ghtt
ravel
sfr
om ai
rthr
oughmul
ti
plel
ayer
s (1n2)
oft
ranspar
entmedi
a1,2and3whoseboundar
iesar
e
2.Arayofl
i
ghtfrom ai
rtravelssuccessi
vel
ythroughparal
l
el
par
all
elasshowni
nthefi
gur
ebel
ow.
l
ayer
sofwat
er,oi
l
,glassandt heni
ntoairagai
n.Theref
racti
ve
46 3
i
ndi
cesofwat
er,oi
l
,andgl
assar
e , and r
espect
ivel
ythe
35 2
angl
eofi
nci
dencei
nai
ris60%

a) Draw adiagr
am toshow how t
her
aypassest
hrough
themult
iplel
ayer
s

b) Cal
cul
ate:

I
. Theangl
eofr
efr
act
ioni
nwat
er

I
I. The angl
e of i
nci
dence att
he oi
lgl
ass
Cal
cul
ate : i
nter
face

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page56
 232

a) Angl
eθ Ref
ract
iveI
ndexi
nTer
msofReal
andappar
entdept
h

b) Ther
efr
act
ivei
ndexofm2  Anobj ectunderwat erorglassblockappear st obe
nearerthesur facethanitactuall
yi swhenv iewed
c) Speed ofl i
ghtin m1( speed ofl i
ghtin ai
r
8 normallyoralmo stnormal
ly.Thi
s i
s duet oref
racti
on
=3.0×10ms-1)
ofli
ght.
d) Theref
ract
iveindexofm3wi
threspecttom1
 Realdept hist heactualdepthoft heobj ectinthe
medium whi eappar
l entdepthisthev i
rtualdepthof
theobjectint hemedi um. Thedi f
ferencebet ween
Sol
uti
ons
realdepth and appar entdepthi s called verti
cal
)nas
a inθa=n1s
inθ1 di
spl
acement
.

1s
in35=1.
5si
nθ

θ=s
in-1 (1s
in35
1.
5 ) 48o
=22.

b)n1s
inθ1=n2s
inθ2

1.
5si 48= n2s
n22. in27
 Refr
act
iveindicesofmater
ial
scanbeexpr
essedi
n
1.
5si
n22.
48 ter
msofr ealandappar
entdepths.
n2=
si
n27

=1.
263

real
dept
h
Re
fra
cti
vei
nde
x,n=
appa
rentde
pth
s
inθa va
c
) =
v
s
inθ1 1
Condi
ti
onf
orUseoftheFormul
a:Thi
sfor
mul
aappl
i
es
s
in35 0×108
3.
= onl
ywhentheobj
ecti
sviewednor
mall
y.
v1
s
in22.
48

si 48×3.
n22. 0×108
v1=
sin35

0×108
=2.

n2s
inθ2
d)n3= ;
sinθ3

1.
263si
n27
⇒ n3= =1.
357
si
n25

1n3=1 2 n x2n3

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

s
in22.
48 si
n27
1n3= × =0.
9040
si
n27 si
n25

Exampl
es. bel
ow

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page58
 232

1. Acoininaglassjarfi
l
ledwi
thwaterappearst
obe
24.
0cm f
rom t
hesurf
aceofwat
er.Cal
cul
ate:

I
. Thehei
ghtoft
hewat
eri
nthej
ar,gi
ven t
hat
4
r
efr
act
ivei
ndexofwat
eri
s.
3
Ift
hespeedofl
i
ghti
nthemat
eri
ali 25x108 ms-1
s1.
I
I. Ver
ti
caldi
spl
acement cal
cul
ate:

Sol
uti
on a) Theappar
entdept
hoft
hemar
k

real
dept
h b) The vert
ical di
spl
acement of the mar
k
I
.Re
fra
cti
vei
nde
x,n= 8 -
1
appa
rentde
pth (speedofli
ghtinai
r=3.0x10 ms

4 re
al de
pth 5) Api ni
spl acedatthebott
om ofatal
lparal
lelsidedglass
=
3 24 j
arcont ainingat r
anspar
entl
iqui
dwhenvi ewednor mall
y
fr
om thet op,thepinappear
snearerthesur f
acet hanit
4 actual
l
yi s:
Re
alde
pth= ×24=32c
m.
3
Wi
tht
heai
dofdi
agr
am,
expl
aint
hisobser
vat
ion
Thi
sist
hehe
ightofwa
teri
nja
r.
6) Thetabl
ebelowshowsther
esul
tsobt
ainedwhensuchan
I
I.Ve
rti
cal
dis
pla
ceme
nt=r
eal
dept
h-a
ppa
rentde
pth exper
imentwascarri
edoutusi
ngvari
ousdepthsofthe
l
iqui
d
Ve
rti
cal
dis
pla
ceme
nt=( c
32-24)m =8c
m
Realdept
h 4.
0 6.
0 8.
0 10.
2 12.
8 14
(cm) .
0
2. Cal
culat
ethedisplacementandapparentdepthoft
he
obj
ectshowni nthef i
gurebelow assumingthatt
he Apparent 2.
4 3.
6 4.
8 6.
10 7.
32 8.
obj
ectis viewed normall
y and boundaries oft
he depth(cm) 4 6 8 54
mediaareparal
lel
.
a) Pl
otagr
aphofappar
entdept
hagai
nstr
ealdept
h

b) Usingthegr
aph,det
ermi
net
her
efr
act
ivei
ndexof
theli
quid

c) Whati
stheverti
caldi
splacementoft
hepi
nwhen
t
heappar
entdept
his1.22cm?

7) How longdoesi ttakeapulseofl ighttopassthrougha

gl
assbl ock15cm i
nlengt
h?(Refract
iveindexofgl
assis1.
5
Sol
uti
on andvelocit
yofli
ghtinai
ris3.0x108ms- 1)
r
ealde
pt h
a
ppa
rentde
pth=
Re
fra
cti
veindex,
n
Tot
alI
nter
nal
Ref
lect
ion
12 8 9
Tot
ala
ppa
rentde
pth= + + =20c
m
4 8 3
 Totalinternalreflecti
on referstot he compl et
e
3 5 2
bounci
ngof fofli
ghtatt heboundarybet weent wo
Ve
rti
cal
displa
c e
me nt mediai ntheopticallydensermedium.Thel awsof
=tot
alrea
l depth-tot
ala
ppa
rentde
pth ref
lect
ionareobey edintotali
nter
nalr
eflect
ion.

Ve
rti
cal
dis
pla
ceme
nt=( c
12+8+20)m-20c
m
=20c
m

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Exer
cise

1) Atankf ul
lofwaterappearst
obe0.5m deep.I
fthe
hei
ghtofwat eri
nthet anki
s1.0m,calcul
atethe
ref
ract
iveindexofwater
.

2) A gl assbl
ockoft hi
cken12cm i splacedona
mark drawn on a pl
ain paper
.The mar kis viewed
normall
yt hr
ough t
he glass.Cal cul
atethe appar ent Condi
ti
onsf
orTot
alI
nter
nalRef
lect
ion
depth ofthe mark hence the vert
icaldi
splacement
3
(
ref
ract
ivei
ndexofgl
ass=  Fori
nter
nal
ref
lect
iont
ooccur
:
2
a) Li
ghtmustbetravel
l
ingf
rom opt
ical
l
ydensert
o
3) Abeakerpl acedoveracoi ncontainsablockof
gl
assoft hi
ckeners12cm.overt hi
sblocki swat erof opt
ical
l
ylessdensemedi
um.
depth20cm.cal culatethever ti
caldi
spl
acementoft he
b) Theangl
eofinci
denceintheopt
ical
lydenser
coi
n and hence,i ts apparentdepthi fi tis viewed
normally.Assume t he boundar i
es oft he media are medi
um mustbegr
eatert
hant
hecr
iti
calangl
e
paral
lelandtakerefracti
veindicesofwaterandgl assto
4 3 Cr
it
icalAngl
e
be and r
espect
ivel
y
3 2
Cr
it
icalanglei
st heangl
eofinci
denceinopt
ical
l
ydenser
4) Amar konapaperisviewednor
mal l
ythrougha medium forwhi
cht heangleofr
efract
ioni
ntheopt
ical
ly
rect
angul
arbl
ockofat
ranspar
entmater
ialasshown l
essdensemedi um is900

Sol
uti
on

n1s
inθ1=n2s
inθ2

2.
4si
nc=1.
467s
in90

c=s
i (
n-1
1.
467s
2.
4
i
)
n90
66o
=37.

Rel
ati
onshi
pbet
weenCr
it
icalAngl
eandRef
ract
ive
I
ndex Exer
cise

 Consi
dera r ay of l
i
ghtst
ri
ki
ng a gl
ass-
air 1. What do you under
stand by t
he t
erm t
otali
nter
nal
i
nter
faceasshownbelow r
efl
ect
ion?

2. Statet he condi
ti
ons necessar
yfor t
otalunder
stand
refr
act
ion

3. Defi
ne cri
tical angl
e. Derive an expr
essi
on for t
he
rel
ati
onshi
pbetweencr i
ti
calangl
eandrefr
acti
vei
ndex

4. Thefi
gurebelowshowsapl anemirorat300 t
r ofaceofa
ri
ghtangled isoscel
es pr
ism ofr efr
acti
ve i
ndex 1.
50.
Fr
om Snel
l
’sl
aw, Complet
et hepathofli
ghtr
ayafterr
efl
ecti
on.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page60
 232

s
inc
 gna=
sn900
i

sn900
i
 Butang=1/gna=
s
inC
5. Cal
cul
atecr
it
icalangl
efordiamond-wat
eri
nter
face
 s
in900=1 (anw =1.
33,and =2.
46)
1
 ang=
si
nC

1
Thus
,re
fra
cti
vei
nde
x,n=
si
nC

Ef
fect
sofTot
alI
nter
nalRef
ract
ion

1.Mi
rage

 Mir
agereferstoopti
cali
ll
usi
onofani nv
ert
edpoolof
watert
hatiscausedbytotal
int
ernal
refl
ect
ionl
i
ght.
Examples
1.Thefi
gurebelow showsthepath ofa rayofli
ght
 Duri
ngahotdayai rneart
hegroundiswarmerand
passi
ngt
hrougharect
angul
arblockofPer
spexpl
acedin
ai
r ther
efor
ep hysi
call
ylessdensethanairawayf r
om
theground.Ther efor
eonahotdayt heref
racti
ve
i
ndexincreasesgraduall
yfr
om t
hegroundupwards.

 Ar ayofl i
ghttravel
l
inginairfr
om skytoground
undergoescont
inuousref
ract
ionandf
inal
l
yref
lect
ed
Cal
cul
atet
her
efr
act
ivei
ndexofPer
spex i
nternal
ly.

Sol
uti
on

1
n=
si
nc

1
n= =1.
467
si
n43

2.Arayofli
ghttr
avelsf
rom at
ranspar
entmedi
um i
nto  Miragesareal
sowi tnessedinver
ycoldregi
onsin
Per
spexasshowninthef
igur
e whichtheref
ract
iveindexincr
easesgr
aduall
yfr
om
thegroundupwards.Imagesappeari
nver
tedinthe
sky.

i
. Whi
choft
het
womedi
aisopt
ical
l
ydenser
?

Tr
anspar
entmat
eri
al

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

i
i
. Cal
cul
atet
hecr
it
icalangl
eC

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page62
 232

2.At
mospher
icRef
ract
ion I
V. I
nver
sionwi
thoutdevi
ati
on

 Thisisaphenomenoni nwhichl
ightraysfrom
the sun ar e ref
ract
ed and then r efl
ected
i
nternal
l
yt owar
dst heeart
h.Asaresult
,thesun
i
sseenev enaft
erithassetorbef
oreitri
ses.

Appl
i
cat
ionsofTot
alI
nter
nalRef
lect
ion

1.Per
iscope

 Li
ghti
sdev
iat
edt
hrough900byf
ir
stpr
ism bef
oret
he
secondpri
smsdev iat
esi
tfurt
hert ough900 i
hr nthe
opposi
tedi
rect
ion.Anupr
ightvi
rt
uali
mageisf
ormed

Tot
alI
nter
nalRef
lect
ionPr
isms

 Ri
ght
-angl
edi
soscel
esgl
assorPer
spexpr
isms
ar
eused.

I
. Tot
urnar
ayofl
i
ghtt
hrough900

Note:Pri
smsar eprefer
redtoplanemi r
rorsforusei
n
per
iscopesandotheropt
icali
nst
rumentbecause:

a) Mi
rr
orsabsor
bsomeoft
hei
nci
dentl
i
ght

b) Thesil
ver
ingonmi
rr
orscanbecomet
arni
shedand
peelof
f
I
I. Tot
urnar
ayt
hrough1800
c) Thi
ckmi
rr
orspr
oducemul
ti
plei
mages

2.Opt
icalFi
bre

 Anopt
icalf
ibr
eisat
hinf
lexi
blegl
assr
odofsmal
l
diameterintheorderof10-6m.Thecent
ralcaseof
theglassi scoatedwit
hgl assoflowerrefract
ive
i
ndex(cladding)
I
II
. I
nver
sionwi
thdevi
ati
on

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

 Ar ayofl i
ghtenteri
ng the f
ibre undergoes t
otal
i
nternalrefl
ect
ionsont heboundaryoft hehighand
l
owr efracti
vei
ndexglass.Thelightther
eforetr
avel
s
thr
ought heenti
relengthoft hef i
brewi t
houtany
gett
inglost.

Advant
agesofopt
icf
iber
soveror
dinar
ycabl
es.

-
theyhavehi
ghcar
ryi
ngcapaci
ty

-
theyar
ethi
nnerandl
i
ght
er

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page64
 232

UsesofOpt
icalFi
bre Revi
sionExer
cise

I
. Usedinmedi
cinet
ovi
ewi
nter
nalor
gansoft
he 1. Thedi
agr
am bel
owshowt
wopr
isms
body.

I
I. Used i ntel
ecommunicati
on wheret heyhave
higheradvantage t
han ordi
narycablessince
theyhave highercarryi
ng capaci
ty,theyare
thinnerandl
ight
er.

Gi
venthatthecri
ti
calangleoftheglassi
nbothpri
sms
Di
sper
sionofWhi
teLi
ght s 420 sket
i ch the paths of the two beams of
monochromati
cli
ghtuntiltheyl
eavethef
lasks.
 Di
spersi
onoflighti
st hespli
tt
ingofwhit
eli
ght
i
ntoits componentcol or
s..Whit
elightis a 2. Thefi
gurebel
owshowhowwhi
tel
i
ghtbehaveswheni
tis
mixt
ureofsev
encolors. i
nci
dentonaglasspr
ism.

 Thecomponent sofwhit
el i
ghttr
av elwi
thsame
vel
oci
tyi
nv acuum butt
heirvel
oci
tiesarenott
he
sameinothermedia.

I
. Det
ermine t
he cr
it
icalangl
e oft
he gl
ass
mater
ial

I
I. Det
erminet
her
efr
act
ivei
ndexoft
hegl
ass
mater
ial

3. The di agram below shows a t ransparentwatert

ank
containingwater.Anel ectri
cdampcover edwi
thashield
whichhasanar r
owsl itf
ixedatonenearoft het
ank.Al
ight
rayfrom theslitreachest hewatersurfaceatanangl
eof
0
42 asshownbel ow.
 Cause of Di
sper
sion of Whi
te Li
ght
: The
separ
ati
onofwhitel
ighti
ntoconsti
tuentcolours
i
s due tothei
rdiff
erentvel
ocit
ies in a given
tr
anspar
entmedium.

 Thevel
ocit
yofredishighestwhilethatofviol
et
i
stheleast.Redcolourhaslongestwavelength I
. Deter
minetheangleofr
efr
act
ionf
ort
her
ay
whi
levi
oletthel
eastwavelengt
h(v=λ f) showninthedi
agram

I
I. Deter
minetheangleofinci
denceforwhich
theangl
eofref
ract
i s900(
oni ref
ract
ivei
ndex
TheRai
nbow ofwater=1.
33)

 Rai
nbow i
sabow-
shapedcol
ourbandoft
he 4. Thedi
agr
am bel
ow showsar
ayofl
i
ghti
nci
dentont
he

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

v
isi
blespect
rum seeni
nt .I
hesky ti
sfor
med gl
ass–airi
nter
facefrom t
hei
nsideoft
hegl
ass.Theangl
e
when whi
teli
ghtfrom the sun isref
ract
ed, ofi
nci
denceI,i
sslightl
ysmal
lerthant
hecri
ti
calangl
eof
di
sper
sedandtot
all
yint
ernal
l
yr ef
lect
edbyrain gl
ass.
dr
ops.

St
ateandexplai
nwhatwoul
dbeobser
vedontherayi
f
adropofwaterwaspl
acedatt
hepoi
ntofi
nci
dence,o

Chapt
erFi
ve GASLAWS

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page66
 232

Speci
fi
cobj
ect
ives Cont
ent

Bytheendoft
hist
opi
c,t
hel
eanershoul
dbeabl
e 1. Boyl
e’
slaw,Char
les’l
aw,pr
essur
elaw,absol
ute
t
o: zer
o

a) st
atet
hegasl
awsf
orani
deal
gas 2. Kel
vi
nscal
eoft
emper
atur
e

b) v
eri
fyexper
iment
all
ythegasl
aws 3. Gasl
awsandki
net
ict
heor
yofgases

c) explai
nhow t
heabsolut
ezerotemper at
ure pv
4. Pr
obl
emsofgasl
aws(
incl
udi
ng =const
ant
)
may be obtai
ned fr
om t he pressure– T
temperat
ure and v ol
ume-t
emper at
ure
graphs

d) conver
tCel si
us scal
eto kel
vi
n scal
e of
temperat
ure

e) statethebasicassumpt
ionoft
heki
net
ic
theoryofgases

f
) expl
aint
hegasl
awsusi
ngt
heki
net
ict
heor
y
ofgases

g) sol
venumer
ical
probl
emsi
nvol
vi
nggasl
aws

I
ntr
oduct
ion Char
act
eri
sti
csofi
deal
gas

 To st
udy t
he behavi
ourofgases,pr
essur
e, I
. I
dealgascont
ainsi
dent
icalpar
ti
cl
esofnegl
i
gibl
e
vol
ume and t emperat
ure of the gas are vol
ume
considered.Achangei noneofthesevari
abl
es
causest heother
stochange.Thegasl
awsdealsII
. Thereis no int
ermol
ecul
arf
orces ofat
tract
ion
wit
hr elati
onshi
pbetweenthi
sparameter
s. betweenpar
ti
cles

I
II
. Molecul
esunder
goperfect
lyel
ast
iccol
l
isi
onwi t
h
 NBGaslawsapplyforidealgasesonl
y.Ani
deal
ot
hermolecul
esandwi
ththewal
lsoft
hecont
ainer
.
gasi
sonethatobey
st hegaslaws.

Relati
onshi
pbetweenKel
vi
n(Absol
ute)andCel
sius
ScalesofTemperat
ure TheGasLaws
 Absol
utezerotemper
aturerefer
stothel owest Char
les’
law
0
temperat
ureagascanf al
lto.Iti
s-273celcius
(0Kelvi
n).
Thekel
vi
nscaleoftemperatur
est ar
ts  Itstatest
hat
,“
thevol
umeofaf
ixedmassofagasi
s
2730c(
at- 0K)
. dir
ect
ly propor
ti
onal to t he absol
ute (
kel
vi
n)
temper
atur
eatconstantpr
essure”
.
o
 TheKel
vi
n(K)scal
eisr
elat
edt
otheCel
si
us(C)
by:  Mat
hemat
ical
l
y,Char
les’
lawcanbeexpr
essedas:

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Te
mpe
rat
urei
nke
lvi
n,TK(
K) V∝T(
atc
ons
tantpr
ess
ur)
e;

=Te
mpe
rat
urei
nce
lci
usTC(
, o
c
.)+273 ⇒V=kTwhe
rki
sac
ons
tantofpr
opor
tiona
lit
y

V
Note:Alltemperat
uresmustbeexpr
essedi
nkel
vi
n =c
ons
tant
T
anycal
culati
on.
Whe
reVi
sthevol
umea
ndTi
stheke
lvi
nte
mpe
rat
ure

V1 V2
The
ref
ore
, =
T1 T2

Where V1,V2 and T1,T2 ar

et heini
ti
alandf inal
val
uesofv olumeandt emper atur
erespect
ivel
y.

Exer
cise

1.Conver
tthef
oll
owi
ngt
emper
atur
etokel
vi
n

I
. 400C
-

I
I. 550C

2.Conver
tthef
oll
owi
ngval
uesoft
emper
atur
etodegr
ees
val
ues

I
. 45K

I
I. 300K

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page68
 232

Gr
aphi
cal
l
y,Char
les’l
aw canbeexpr
essed I
II
. Descr
ibehowt
hesetupcanbeusedt
over
if
yChar
les’
asshownbel
ow; l
aw.

 Thei
nit
iall
engthoftheai
rcol
umnist
akenandr
ecor
ded
aswel
lastheini
ti
alt
hermomet
err
eadi
ng.

 Thewaterbat
hisheat
edandnewhei
ght(col
umn)ofairi
s
takenandrecor
dedwithit
scorr
espondi
ngt emper
atur
e
readi
ng

 Thisisr epeatedseveralti
mesatsuit
abl
etemper
atur
e
i
nterval
st ogetsever
alpair
sofr
esul
ts

 A gr
aph ofvol
ume (
hei
ght
,h (
cm)
)agai
nstabsol
ute
t
emperat
urei
spl
ott
ed.

 I
tisast
rai
ghtl
i
newi
thposi
ti
vegr
adi
ent
.

 Thisshowsthatthevol
umei
sdi
rect
lypr
opor
ti
onalt
o
absolut
etemperatur
e.

3.Thevolumeofagasencl osedwithamovablepi
stonis300
3 o
cm whent hetemper
aturei
s2o C.Det er
minet
hetemperatur
e
atwhichthevolumeofthegasincr
easesto355cm3.(Assume
pressur
edoesnotchange)

Sol
uti
on
3 3
V1=300c m , T1=20o.C=(
m , V2=355c K
20+273)
=293K,T2=?

Sampl
equest
ionsonChar
les’
law V1 V2
=
T1 T2

3 3
300c
m 355cm
=
1.St
ateChar
les’
lawf
orani
deal
gas. 293K T2

The volume of a f ixed mass of a gas is dir

ect
ly 355c
3
m ×293K
proport
ionalt ot he absol
ute(kel
vi
n) temperat
ure at T2= 3
=346.
72K
300c
m
constantpressur
e.
I
nce
lsi
uss
cal
e,T2=346. 72o.C
72-293=53.
2.Theset-
upbelow showsanar rangementt
hatcanbe
usedtodeter
minetherelat
ionshi
pbet weent
emperat
ure Exer
cise
andvol
umeofagasatconst antpr
essure.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

1.I
nanexperi
menttofi
ndtherel
ati
onshi
pbetweenthevolume
andtemper
atureofamassofai ratconst
antpressur
e,the
fol
l
owingr
esult
swereobtai
ned:

Volume 31 3 3 3 4 43
(cm) 3 5 8 0

Temper
atur 0 2 4 6 8 100
e0c 0 0 0 0
I
.St
ateanyt
wousesofsul
phur
icaci
d.
Draw agraphtoshow ther el
ati
onshi
pbetweenvol
umeand
 Usedasapoi
ntert
ovol
umeoft
hegasont
hescal
e
temperat
ureanduset hegrapht ocalcul
atet
heincr
easei
n
 Usedasadr
yingagentf
ort
heai
r volumeofthegasperuni
tri
seintemperat
ure

 Usedt
otr
apai
r 2.Some r esult
s ofan experimentto st
udy t
he ef
fectof
temperatur
eonvol umeofafixedmassofagasataconstant
I
I.Whati
stheuseoft
hest
ir
rer
? pressur
earedisplayedi
nthet
ablebel
ow.

Tost
irt
hewat
erbat
hforuni
for
m di
str
ibut
ionofheat Volume 2 2 4
(cm)3 0 5 0
I
II
. St at
ethemeasur
ementt
hatusedt
obet
akeni
n
t
hisexper
iment
. Temper
atur 0 - 20
0
e(c) 136 5
 Temper
atur
e
Fi
l
lthemi
ssi
ngr
esul
ts
 Ai
rcol
umnhei
ghtwhi
chcor
respondst
ovol
ume
3.Thevolumeofagasenclosedwi
thamovablepist
oni s0.
02
3 0
m whent hetemper
aturei
s42 c.Det
ermi
nethetemper at
ure
atwhichthevol
umeofthegasincr
easest
o0.4cm3( assume
pressur
edoesnotchange)

4.Atubecontainsagasencl osedwi t
hat hreadofmercury.The
tube was placed hor
izontall
yi n a wat erbat h.The init
ial
0
temperat
ureoft hewaterwas20candt helengthofthegas
columnwas25cm.Det er
mi net het emperatur
eatwhi cht he
l
engthofai rcolumnwoul dbe20cm ( Assumet hepressure
doesnotchange)

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page70
 232

Pr
essur
elaw I
I.Descr
ibehowt
heset
-upcanbeusedt
over
if
ypr
essur
elaw

 Pressur
elawstat
esthat,“
thepr
essureofafixed  Thei
nit
ialt
emper
atur
eandpr
essur
ereadi
ngar
etaken
massofagasi sdir
ect
lypropor
ti
onaltothe andr
ecorded
absolut
e(kel
vi
n)temperat
ureataconstant  Thewat erbat
hisheatedgent
lyandsomemor epai
rs
volume”. ofpressureandtemperat
urer
eadingsaretakenand
recor
dedatsuitabl
etemperat
ureint
erval
s
 Mathemat
ical
l
y,pr
essur
elawcanbeexpr
essed
as:  Agr
aphofpr
essur
eagai
nstt
emper
atur
eispl
ott
ed.

P∝T(
K),a
tcons
tantt
empe
rat
ure  I
tisast
rai
ghtl
i
newi
thposi
ti
vegr
adi
ent
.

P  Thisshowsthatthepressur
eisdi
rect
lypr
opor
ti
onalt
o
=c
ons
tant(a
tcons
tantvol
ume
)
T(
k) absolut
etemperatur
e.

P1 P2 3.Agasi
ncont
ainerhaspr
essur 0x105Pawhent
eof3. he
t
her
efor
e, =
T1 T2 t
emper
atur
eis200c.Det
ermi
net
hepr
essur
eoft
hegaswhen
t
hetemper
atur
elower
edt
o-50c(
assumet
her
eisnochangei
n
Wher eP1,P2andT1,
T2arei
niti
alandf
inal
v
olume)
pressur
eandt emperat
urev
alues
respect
ivel
y 0×105Pa
T1=20o.C=293K, T2=-5o.C=268K, P1=3. ,P2
=?
 Gr
aphi
cal
l
y,pr
essur
elawcanbeexpr
essedas:
P1 P2
=
T1 T2

0×105Pa P2
3.
=
293K 268K

0×105Pa×268K
3.
P2= 744×105Pa
=2.
293

Exer
cise

1.80cm3ofhydr
ogengaswascollectedatatemperat
ureof15
o
Candnormalat
mospheri
cpressure.Deter
minethepressur
e
o
ofagaswhenthetemper
atur
eislower edto0 Catconstant
vol
ume.

2.I
nanexper i
mentt odeter
mi netheabsolut
ezerotemperat
ure
(0K),t
hepressureofagasatconst antvolumewasmeasur ed
asthetemperaturewasvariedgradual
ly.Thetabl
ebelow
showstheresultsobtai
ned.

Sampl
equest
ionsonPr
essur
elaw Pressur
e 75 77 80 82 85 88 9
(mmHg) 0 6 2 8 4 0 0

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

1.St
atepr
essur
elaw. Temper
at e0
ur 15 25 35 45 55 65 7
C 5
Thepr essur
eofafixedmassofagasisdi
rectl
y
proporti
onaltot
heabsolut
e(kel
vi
n)t
emperatur
eata I
. Pl
otagr
aphofpr
essur
eagai
nstt
emper
atur
e
constantvolume.
I
I. Det
ermi
net
heval
ueofabsol
utezer
ofr
om t
hegr
aph
2.Thesetupbel
owshowsanar r
angementusedto
determi
nether
elat
ionshi
pbetweentemperat
ureand I
II
. Givear
easonwhyt
hetemper
atur
einI
Iabovecoul
d
pressur
eofagasatconstantvol
ume. notbepr
acti
cal
l
yobtai
ned

Boy
le’
sLaw

 Boyl
e’
slawst at
esthat
,“t
hepressureofafi
xedmass
ofagasisinver
sel
yproport
ionalt
oitsvol
umeata
const
anttemperat
ure”
.

 Mat
hemat
ical
l
y,Boy
le’
slawcanbeexpr
essedas:
I.St
atethemeasur
ementt
hatneedt
obet
akeni
nthi
s
experi
ment 1
P∝ (
atc
ons
tantt
empe
rat
ure
)
V
a) Temper
atur
e
PV=c
ons
tant(
atc
ons
tantt
empe
rat
ure
)
b) Pr
essur
e
t
her
efor
e,P1V1= P2 V2

whe
reP1,P2a
ndV1,V2ar
eini
ti
alandf
inal
val
uesof
pr
essur
eandv
olumer
espect
ivel
y.

 Gr
aphi
cal
l
y,Boy
le’
slawcanbeexpr
essedas: Sampl
equest
ionsonPr
essur
elaw

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page72
 232

1.St
ateBoy
le’
slaw.

Thepressur
eofafi
xedmassofagasisi
nver
sel
ypr
opor
ti
onal
toi
tsvolumeataconst
antt
emper
atur
e.

2.Theset -upbel
owshowsanarrangementusedt
odet
ermi
ne
therelat
ionshi
pbetweenpr
essur
eandv ol
umeofagasat
constanttemperat
ure.

Descr
ibehowt
hesetupcanbeusedt
over
if
yBoy
le’
slaw.

 Witht
hetapopen,
airi
spumpedunti
ltheoilr
aisesasmal
l
butmeasur
abl
eheight
.Thetapi
sthenclosed.

 Thevalueofpressur
eandi
tscor
respondi
nghei
ghti
sread
andrecorded.

 Thi
sisrepeat
edtoobt
ainmorepai
rsofval
uesofpr
essur
e
andcor
respondi
nghei
ghts.

 Agraphofpr
essur
ever
susvol
umei
spl
ott
edusi
ngt
he
r
esult
.
 I
ftheexper i
mentonBoy l
e’sl
awisrepeatedat
di
fferentfixedt
emperaturesandt
her esult
s  I
tisasmoot
hcur
vewi
thnegat
ivei
nst
ant
aneousgr
adi
ent
pl
otted,isother
malcurv
esar eobt
ained.
 Thi
sshowst
hatt
hepr
essur
eisi
nver
sel
ypr
opor
ti
onalt
o
vol
ume.

3.Abubbl
eofai
rofv
ol 2cm3i
ume1. srel
easedatt
hebot
tom of
awaterdam 15m deep.Thet
emper at
ureofthewat s200c.
eri
Det
erminethevol
umeoft hebubbleasitemergesonthewat er
sur
facewherethepressur
eisthenormalatmospheri
cpressure
(
t er=1000kgm-3andnor
akeρofwat mal
atmospher
icpr
essur
e
0x105pa)
1.

Sol
uti
on

P1V1= P2V2

P1 = Patm +ρg
h

0×105pa+1000kg
1. 5×105pa
m-3×10Nkg-1×15m =2.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

0×105pa
P2 =1.

2c
V1 =1. m3

P1V
V2=?
,V2=
1

P2

5×105pa×1.
2. 2cm3
= 5
=3.
0cm3
1.
0×10pa

3.Anarr
owuni f
orm gl
asstubecontai
nsairenclosedbya10cm
thr
eadofmercury.Whenthetubeisheldvert
ical
,theaircol
umn
i
s600mm l ong.Whentit
ledsl
ight
ly,t
heaircolumni s679mm
l
ong.Thetemperatur
eisthesamei nbothcases(i.e.
temper
atur
eisconstant)

I
.Gi
vet
hereasonforthedi
ff
erencesi
nlengt
hoft
heai
rcol
umn
f
ort
hetwoposit
ions.

Whenver
ti
cal
,themer
cur
yexer
tspr
essur
eont
heai
r,r
educi
ng
t
hevol
ume.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page74
 232

I
I.Det
ermi
net
heat
mospher
icpr
essur
eatt
hepl ρof P1V1
ace( P2V2
=
mer
cur 6gcm-3)
y=13. T1(
K) T2(K)

Sol
uti
on 0×105Pa×1m
1.
3
06m3
P2 ×0.
=
100K 293K
3
0×105Pa×1m ×293K
1. 6
P2 = =4.
883×10Pa
100K×0.06m3

Exer
cise

P1V1=P2V2 1. Acontai nercarr

ies3000cm3ofoxygenatapr essureof
6 0
1.0x10 paat emp er
a t
ureo f20 cinacyl
inder
.Wh atisthe
P1=Patm +ρg m-3×10Nkg-1×0.
h; P1=Patm +13600kg 1m
vol
umeoft hegasi nthecylinderatthet
opoft hemount ai
n
=Patm +1.36×104pa
wherepr essureis0.8x106pa andt emperat
ureis-170c?
P2=Patm

Le
tthec
ros
s–s
ect
iona
lar
eaoft
het
ubebeA
Ki
net
icTheor
yofGases
V1=60Ac
m3, V2=67.
9Ac
m3
 Thekineti
ctheor
yofgasespr
oposesthatt
he
( moleculesofagasar
einaconti
nuousrandom
36×104pa
Patm +1. )×60Ac
m3=Patm ×67.
9Ac
m3 motion.
5
8.
16×10pa Basi
cAssumpt
ionsoft
heKi
net
icTheor
yofGases
Patm = 03×105pa
=1. .
7.
9
I
. Att
ract
ionbet
weent
hemol
ecul
esofagasi
s
Exer
cise
negl
igi
ble.
1.Acolumnofair26cm longistr
appedbymer cur
y
thr
ead5cm longwhenver t
ical
.Wheniti
splaced I
I. Thev
olumeoft
hemol
ecul
eoft
hegasi
szer
o.
hori
zontal
l
y,theai
rcolumnis28cm.Findthe
I
II
. Col
l
isi
onsbet
weent hemol
eculesandwi
ththe
atmospheri
cpressur
einmmHg.
wal
lsoft
hecont
ainerandper
fectl
yel
ast
ic.
2.Thetabl
ebelowshowst heresult
sobtai
nedi
nan
exper
imenttostudythevar
iat
ionofthevol
umeofa
fi
xedmasswi t
hpr essur
eatconstantt
emperat
ure.
Ki
net
icTheor
yofGasesandGasLaws
Pr
essur
e(cm Hg) 60 - 90 -
Boyl
e’
sLawandKi
net
icTheor
y
Vol
ume(
cm ) 3 36 80 - 40
 Whent emperat
ureisconstant,achangeinvolumeof
Fi
l
lint
hemi
ssi
ngr
esul
ts. gasresult
sinachangei nnumberofcol l
i
sionsper
uni
ttimebetweenmol ecul
esandbet weenmol ecul
es
3. Explai
nwhyabubbleincreasesinsizeandf
inal
l
y andwal l
sofcontai
ner.Asar esul
t,pr
essurechanges.
burstwheni
treachesthesurface.
Char
les’
LawandKi
net
icTheor
y

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
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 232

Gener
alGasEquat
ion(
Equat
ionofSt
ate)f
orI
deal  Whenpressur
eisconstantandtemper
atur
eisraised,
Gases t
hespeedofmoleculesri
sescausingt
hem tooccupy
alar
gervol
umeofthecont ai
ner
.
 I
tisobt
ainedbycombi
ninganyt
wooft
het
hree
gasl
aw. Pr
essur
eLawandKi
net
icTheor
y

PV  Achangei ntemper at
urechangest hekineticenergy
=c
ons
tant
T(K) andhencet hespeedofmol eculesofagasand
P1V1 P2V2 ther
efor
eifv ol
umeofgasi sconstant,pressure
The
ref
ore
, = changesasar esultchangeintemper ature.Pressur
e
T1(
K) T2(
K)
i
scausedbycol li
sionsbetweenmol eculesofthegas
whe
reP1,P2;
V1,V2 andT1T2ar
ethei
nit
ial
and andwiththewal l
soft hecontai
ner.
f
inal
val
uesofpr
essure,
volumeand Li
mit
ati
onofGasLaws
t
emperat
urer
espect
ivel
y.
 Gaslawsdonotappl
yinr
eal
gases.Realgases
l
iquef
ybefor
ethevol
umeofthegasreducestozer
o.
Exampl
e

I
nt hemanuf act
ureofoxygen1m3oft hegaspr oducedat-
0
173candnor malatmospheri
cpressureiscompr essed
3
i
nt oacyli
nderofvol
ume0. 06m andst oredata
temperat eof200c.Whati
ur sthepressur
eoft hegasi n
thecyli
nder?(Normalat
mospher i
cpressure,Patm =
1.0x105pa)

Sol
uti
on
3
0×105Pa
P1 =1. , P2=? , 06m3,
V1=1m ,V2=0.

o o
T1=-173.C=100K,,T2=20.C=293K,

Revi
sionExer
cise

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

1. Aballoonf i
ll
edwithorgangasavolumeof200cm3at
theearth’
ssur f
acewherethetemperatur
eis200C,
andthepr essure760mm ofmercury.Ifi
tisal
lowed
toascendt oaheightwherethetemperatur
eis00C
andthepr essure100mm ofmercury,cal
cul
atethe
volumeoft heball
oon.

2. Af i
xedmassofOxygenoccupi esavol
umeof0. 01m3
5 0
atapressur
eof1x10 Paandat emper at
ure0C.If
6
the pr
essur
ei si ncreased t
o 5 x10 pa and t he
temperat
ureisincreasedto250C.Whatvol umewil
l
thegasoccupy?

3. Anempt ybarometertubeofl ength90cm islowered

vert
ical
lywit
hi t
smout hdownwar dsint
oat ankof
water.Whatwillbethedepthatt hetopoft hetube
whent hewaterhasri
sen15cm i nsidethetube,gi
ven
thattheat
mospheri
cpr essur
ei s10m ofwat er
?

4. A handpumpsui t
ablef ori
nfl
ati
ngaf ootbal
lhasa
cyli
nderwhichis0.
24m inlengt
handani nt
ernalcr
oss
-secti
onalar
eaof5. 0x10-4m2.Toi nfl
atethefoot
ball
thepump handl ei spushed in and airispumped
throughaone-wayvalve.Thevalveopenst ol
etairi
n
tot heballwhent heairpressurei nthepumphas
reached150000pa.( Assumet heai rtemperatur
e
remainsconstant
}

I
. Ift
hepressur
einthepumpi sini
ti
all
y100000pa,
cal
cul
ate how f
art he pi
ston mustbe pushed
i
nwardsbefor
etheonewayval vesopens.

I
I. Whentheone-wayvalveopensthetotalpr
essur
e
i
nthecyli
nderwi
llbe150000pa.Whatf orcewi
ll
beexer
tedonthepist
onbyt heai
rinthecyli
nder
?

5. Thegraphi nfigurebel
owshowstherelati
onship
betweenthepr essureandt
emperatur
ef oranideal
gas.Uset hisinformat
ioni
nthef
iguretoanswer
questi
onst hatfoll
ow.

I
. St
atet
he uni
toft
he hor
izont
alaxi
s

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
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 232

quant
it
y.

I
I. Wri
teastatementoft
hegaslaw
r
epresent
edbytherel
ati
onshi
p.

6. Drawaxesandsket chagraphofpressur
e(p)against
reci
procalofvolume( 1/
v)forafi
xedmassofani deal
gasataconst anttemperat
ure.

7. Abal
loonisfi
ll
edwit
hairtoavolumeof200mlata
t
emperat
ureof293k.Det
ermi
nethevol
umewhenthe
t
emperat
ureri
sest
o353katthesamepressur
e

8. Show thatthe densi

tyofa f
ixed massofgasi s
dir
ect
ly pr
oport
ionaltothe pressur
e atconst
ant
temper
ature.

9. The pr essur
e of heli
um gas of vol ume 10cm3
decreases to one t
hir
d ofits or
igi
nalvalue ata
constanttemperat
ure.Deter
minethefinalvol
umeof
thegas.

Chapt
erSi
x WAVESI
I

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page78
 232

Speci
fi
cobj
ect
ives Cont
ent

Bytheendoft
hist
opi
cthel
ear
ner
sshoul
dbeabl
e 1. Propert
ies ofwav es incl
udi
ng sound waves:
to: Refract
ion,
dif
fracti
on,
inter
fer
ence(
exper
imental
tr
eatmentrequi
r ed)
a) Descr
ibeexper
iment
stoi
l
lust
rat
epr
oper
ti
es
ofwaves 2. Constr
uctiv
e i nt
erf
erence and destr
uct
ive
i
nter
ference(Qual
i
tat
ivetr
eat
mentonl
y)
b) Expl
ain const
ruct
ive and dest
ruct
ive
i
nter
ference 3. Stat
ionar
ywav es(qual
i
tat
iveand exper
iment
al
tr
eatmentrequi
red)
c) Descr
ibeexper
iment
stoi
l
lust
rat
est
ati
onar
y
waves

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
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 232

TheRi
ppl
eTank

 Ar i
pplet ankisusedt odemonst r
atepropert
ies
ofwav esinthelaboratory.Itconsi
stsofat ray
containingwater,apoi ntsour ceofli
ghtabove
thetray ,awhit
escreenpl acedunderneathanda
smallel ect
ri
cmot or(
vibrator).

 Thewav esar egeneratedbyanel ectri

cv i
brator
asripplesandt heytr
av elacrossthesur faceof
the shallow wat erint he tray.Cr est
s appear
bri
ghtwhi et
l roughsappeardar kwhent hewav e
i
si l
lumi nated.Not ethatcr estsactaspl ano-
convex l ens and ther
ef or
e conv erge rays on
them whi l
et r
oughsactaspl ano-concaveand
ther
eforedi vergerayi
ncidentont hem.

2.Ref
lect
ionofWaves

 Refl
ecti
onisthebounci ngoffofawav efrom an
 Abarattachedtovibr
at oducespl
orpr anewaves
obstacl
e.When r ef
lect
ion t
akes place,onlyt he
whi
le ci
rcularwavesar eproducedbyf i
xi
nga
di
recti
on of t he wave changes; Wavel ength,
smal
lballtothebar.
fr
equencyand speed ofthe wavedonotchange.
Refl
ecti
on ofwaves obeyst he laws ofr
eflecti
on
whichareappl
i
cabletotheref
lect
ionofli
ght
.
Pr
oper
ti
esofWav
es

1.Rect
il
inearPr
opagat
ion

 Rectil
i
nearpropagati
on i
st heproper
tyofthe
wav es to travel i
n a st rai
ght li
ne and
perpendi
cul
artothewavefront
s.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page80
 232

a)Pl
anewavesonast
rai
ghtr
efl
ect
oratanangl
e Afterr
efl
ect
ion,t
hewavesappeart
obedi v
ergi
ngf
rom a
pointbehi
nd themir
rorconvexref
l orhasa v
ect ir
tual
focus.

e)Ci
rcul
arwav
esonast
rai
ghtr
efl
ect
or

Pl
b) anewavesonast
rai
ghtr
efl
ect
oratanangl
eof
90o

f
)Ci
rcul
arwav
esonaconcav
eref
lect
or

c)Pl
anewav
esonaconcav
eref
lect
or

3.Ref
ract
ionofWaves

 Refracti
onofawav erefer
stochangei ndi rect
ionof
thewav e.Whenawav eisr efr
acted,i
tchangesi t
s
Af t
err ef
lect
ion,thewav esconv erget
oapoi nti
n speed, direct
ion and wavel ength but not i t
s
frontoft her ef
lect
ingsurf
ace.Hence,theconcav
e frequency.
refl
ectorhasar ealfocus
 Whenwat erwavescr ossint
oshal l
owr egionfrom a
d)Pl anewavesonaconv exref
lector deep region,the separat
ion between wave f ront
s
becomessmal leri
.e.thewavelengthdecreases

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
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 232

a) Wavef
ront
spar
all
elt
oboundar
y

 Dur
ingt
heni
ghtai
rcl
oset
ogr
oundi
scool
ert
han
thathigherabove.Soundwav
espr oducedcloseto
b) Wavef
ront
smeet
ingt
heboundar
yatanangl
e groundar eref
ract
eddownwardsbecausethewav e
front
sneart hegroundmoveslowerthanthoseon
theupperpar s.Thi
t si
swhysoundwavest ravelf
ar
f
rom sour
cedur
ingt
heni
ght
.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page82
 232

c) Refr
acti
onofst r
aightwat
erwavesonshal
l
ow Exer
cise
convexsur
face
Thef i
gurebel
ow showsplanewavest r
avel
l
inginashall
ow
regi
onofar i
ppl
etank.Theshall
ow regi
onisinci
dentona
deeperregi eof450asshown.
onatanangl

a) Descr
ibehowt
hewavesar
egener
atedi
nar
ippl
etank

d) Refr
acti
onofst r
aightwat
erwavesonshal
l
ow b) Complet
ethediagr
am toshowt
heappear
anceoft
hewave
concavesur
face fr
ont
sinthedeepregi
on

c) Whatproper
tyofwavesi
sil
l
ust
rat
edi
nthedi
agr
am you
havedr
awn?

4.Di
ff
ract
ionofWaves

 Di
ffr
act
ioni sthespreadingofawav easitgoes
ar
ound an obstacl
e ort hr
ough a smal
laperture.
Whendiff
racti
onoccursther
eisachangeindi
recti
on
butnoti
nvel
oci
ty,
frequencyorwavel
engt
h.

Ref
ract
ionofSoundWaves

 Dur
ing t
he day ai
rcl
ose t
o gr
ound i
s much
warmert hant heai
rhigherabove.Soundwav es
produced close tot he ground ar
er efr
acted
upwardsbecausewav efr
ontsnearthegr ound
mov efasterthanthoseont heupperpars.Thi
t s
i
s whysound waves ar
e nothear
dfarf
rom
sour
cedur
ingt
heday.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

 Whenaper t
ureiswiderthanthewav el hof 5.I
engt nter
ferenceofWaves
theplanewav e,thewav epassest hrough as
planewavesbutwhent hewidt
hoft heaperture  I
nterfer
enceistheint
eract
ionoftwoormorewaves
i
snearlyequalorl
essthanthewavel
engt hofthe of the same f r
equency emitt
ed fr
om coher
ent
planewave,t
hewav ef
rontsemergecircul
ar sources.Theresul
t ofint
erf
erencecanbeamu ch
l
ongerwav e,asmall
erwaveornowaveatall
.

 Interf
erence is an i mpor
t of t he princi
ple of
super posi
ti
onwhi chstat
esthat:“
theresul
tanteff
ect
oft wowavest ravel
li
ngatagi venpointinthesame
medi um i sthe vector sum of t heirrespect
ive
displacement”

Condi
ti
onf
ori
nter
fer
ence

Thewav cesmustbecoher
esour enti
.e.

I
. Havesamef
requencyorwavel
engt
h

I
I. Haveequalorcompar
abl
eampl
i
tudes

I
II
. Haveconst
antphasedi
ff
erence

Typesofi
nter
fer
ence
 Dif
fract
ionofsoundwavesisthereasonasto a)Constr
ucti
veint
erf
erence
whysoundf rom aloudspeakeri
nar oom i
s
heardroundacornerwit
houtthesourcebei
ng  Itoccurswhent hewav eamplit
udesrei
nforceeach
seen. otherbui
ldi
nga waveofevengr
eaterampl
it
ud e

Exer
cise

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page84
 232

1.Expl
ainwhydi
ff
ract
ionofl
i
ghtwavesi
snotacommon
phenomenon.

2.Givet
hedef
ini
ti
onoft
hef
oll
owi
ngt
ermsasconnect
ed
wit
hwaves

a) Wavel
engt
h

b) Fr
equency
b)Dest
ruct
ivei
nter
fer
ence
c) Wavef
ront
I
toccurswhenthewaveamplit
udeopposeeachot
her
3.Fi ve successive wave f
rontsinar i
ppl
et ank are r
esul
ti
nginwavesofr
educedampli
tude.
observed to spread overa dist
ance of6.4cm.Ift he
vibr
atorhasaf r
equencyof8Hz,det er
minethespeedof
thewaves

4.

a) Whati
sdi
ff
ract
ion?

b) Whatfact
orsdet
ermi
net
heext
entofdi
ff
ract
ion
t
hatoccur
s?

c) Descr
ibe an exper
imentthatcan be sett
o
i
ll
ustr
atet
hisphenomenon
 For thi
s case, t
he waves under
go compl
ete
5.Di
ffr
act
ion,r
efr
act
ionandref
lect
ionar
eal
lpr
oper
ti
esof dest
ructi
ver
esul
ti
ng i
n a pul
se (
a wav
e)ofzero
waveswhichoneoft
heseaff
ects: ampli
tude

a) Di
rect
ionbutnotspeed?

Speedanddi
rect
ionoft
raveloft
hewaves?

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

I
nter
fer
encei
nwat
er(
usi
ngar
ippl
etank) I
nter
fer
encei
nli
ght(
Young’
sdoubl
esl
i
texper
ience)

 Coherent water waves ar

e gener
ated by  Int
erf
erencei
nli
ghtisevidencethatli
ghti
sawav e.I
t
at
tachingtwosimi
larbal
l
sonthebari
ncont
act can be demonst r
ated by y oung’
s double sli
t
wit
hv i
brat
or. exper
imentwhosepi oneeristhePhy si
cistThomas
Young.

 Observati
on:Al
ternatedarkli
nes( i
nregionsof
destr
uctiv
ei nt
erfer
ence)and br i
ghtlines (
in  Amonochr omati
cli
ghtsour ceisusedi nt hedoubl
e
regi
onsofconstructi
veint
erf
erence)ar
eseenon sl
itexper
iment
.TheslitS1 dif
fractsl
ightfall
i
ngonit
thewhitescr
een. i
ll
uminati
ngbothsli
tS2andS 3infrontofit
.

 Aseri
esofalt
ernatebr
ight(
inregionsofconst
ruct
ive
i
nter
fer
ence)and dark( inregions ofdestruct
ive
I
nter
fer
encei
nsound i
nter
fer
ence)verti
calbands/f
ringesaref or
med on
t
hescreen.
 Coherentsour
cesofsoundcanbeobt ainedby
connecti
ng two i dent
ical loud speaker s i
n  The centr
al fri
nge is the br
ight
est (
regi
on of
paral
lelto an audio-frequency generatoras const
ruct
ivei
nterf
erence)
.
showbelow.

S
2O=S
3Oi
mpl
yingt
hatpa
thdi
ff
ere
ncei
sze
ro

S
3P-S2P=1λi
.
e.pa
thdi
ff
ere
nceonewa
ve.

1 1
 Al
ter
nat
e l
oud (
in r
egi
ons of const
ruct
ive S
3R-S1R= λi
.
e.pa
thdi
ff
ere
nce= wa
ve
2 2

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page86
 232

i
nter
ference)and softsound (i
nr egi
ons of Exampl
e
dest
ructi
veint
erf
erence)i
shear
dalongXY.
1.I
.Di
sti
ngui
shbet
weendi
ff
ract
ionandr
efr
act
ionofwaves
 Al
ongCOonl
yloudsoundi
shear
d
I
I.Thef
igur
esbel
owshowspl
anewavesappr
oachi
ngsl
i
ts
 I
fwav esfrom onespeakerar eexact
lyoutof
phasewit
ht hosef
rom theothersof
tsoundwi l
l
beheardal
ongCOi .
e.destr
ucti
veint
erf
erence

 Connecti
ng the speaker
s t o same audio
fr
equency gener
atormakes they sat
isf
ythe
condi
ti
onofbeingcoher
entsour
ces.

 Ifthefrequencyofthesignalisincreased,the
point
s of const r
uct
ive i nt
erf
erence ( l
oud
sounds)al ong XY wi
l
lbecome mor e closel
y
spacedandsamewayt ot hoseofdest ructi
ve
i
nterf
erence

Complet
ethedi
agr
amst
oshow t
hepat
ter
nacr
osst
he  A node is a poi
ntof zero displ
acementof a
sl
it
s. st
ati
onar
y wave whie an ant
l i
node i
s a poi
ntof
maxi
mum displ
acementofastati
onar
ywave.

II
I.Thefi
gurebel
ow showst woraysofmonochr
omat
ic
l
i
ghtinci
dentont
woadjacentsli
tsS2andS3.
2.St
ati
onar
ylongi
tudi
nal
wav
e

 Anexampleisasl
inkyspr
ingwi
thoneendf
ixedand
theot
herendmovedtoandfror
api
dlyi
nahorizont
al
dir
ect
ion.

Gi
veanexpressi
onf
ort
hewavel
engt
hoft
hel
i
ghti
n
t
ermsofd,
x,andy
 Anotherexampleiswhentwospeaker
sconnectedt
o
I
V.I nt hespacebelow,sket
chthei
nter
fer
encepat
ter
n same audio-frequencygener
atorar
e ar
ranged t
o
observed if whi t
e light was used inst
ead of faceeacho t
her
.
monochr omat
icl
ight

V.Expl
ainthevari
ati
onoff
requencyacr
osst
hepat
ter
n
di
spl
ayedin(I
V)above

VI.Gi
vet
hatthewavel
engthoft
hemonochromat
icli
ght
7
usedin(
II
I)abovei
s1.0x10 m,cal
cul
atei
tsfr
equency

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

(
speedofl
i
ghti 0x108ms-1)
s3.

2.Twoobser versPandQar est

ationed2.5km apart,each
equippedwithast ar
tergun.Qf i
resthegunandobser ves
Precor dthesound7. 75secondsaf terseei
ngthesmoke
fr
om t hegun.LaterPf i
rest
hegunandobser vesQr ecord
sound7. 25secondsaf t
erseei
ngt hesmokef rom t
hegun.
Determine;

a) Thespeedofsoundi
nai
r  Alt
ernat
eloud (
atant
inode)andsof
t(atnode)i
s
heardal
ongAB.
b) Thecomponentoft
hespeedoft
hewi
ndal
ongt
he
st
rai
ght  Diagram b)showsthewaveform f
ormed on t
he
screenofaCRO whenami cr
ophoneconnect
edto
li
nejoi
ningPandQ theCROi smovedal
ongAB.

St
ati
onar
y(St
andi
ng)Wav
es
Exer
cise
 I
tisdefinedasawav eformedwhent woequal
Thediagr
am belowshowsanarrangementt
hatcanbeusedt
o
pr
ogressive waves t r
avel
l
ing in opposit
e determinethespeedofsoundi
nair.
di
rect
ionsaresuper
posedoneachother.

Ty
pesofst
ati
onar
ywav
es

1. St
ati
onar
ytr
ansv
ersewav
e
A micr
ophoneconnect edt oaCRO wi t
hi t
st imebaseoni s
movedalongani magi
narylineABbet weent hewallandt he
 Anexampl ethiscasei sawav epr oducedwhen
l
oudspeaker.
j
er ki
ngast ri
ngf ixedatoneend.A t ransverse
wav et r
avelsal ongt hestri
ngt othef ixedend I . Use a di agram to explai
n whatis observed as the
and t hen r efl
ect ed back. The t wo wav es microphonei smovedf rom At
oB
travel
li
ngi nopposi tedir
ecti
onsalongt hest r
ing
thencombi ne/super posetof or
m ast ati
onary II
. I fthe frequency oft he sound emitt
ed by the loud
speaker is 1650 Hz and t he di
stance between a
transversewav e.
minimum andt henextmaxi mum is0.05m,calcul
atethe
vel
ocityofsoundair.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page88
 232

I
II
. Ifthe fr
equency of t
he vi
brat
ing loud speaker i
s
decr
easedwhathappenstothedi st
ancebetweent wo
adj
acentmaxi
mum?

Condi
ti
onsNecessar
yforFor
mat
ionofSt
ati
onar
y 5. Thesket
chgr aphshowstheresul
tsofanexper
imentt
o
Waves st
udydif
fract
ionpat
ter
nsusi
ngdoublesl
it
.

 Thetwoprogr
essivewavestrav
ell
ingi
nopposi
te
di
rect
ivet
oform astat
ionar
ywav emustbe:

I
. Samespeed

I
I. Samef
requency i
. Sketch an exper
iment
alsetup t
hatmaybeused t
o
obtainsuchapatter
n.
I
II
. Sameornear
lyequalampl
i
tudes
i
i
. Nameani
nst
rumentf
ormeasur
ingi
ntensi
ty

i
i
i. Expl
ainhow t
hepeaksl
abel
l
edA andB andt
roughs
Pr
oper
ti
esofaSt
ati
onar
yWave l
abel
edCareformed.

i
. A st ati
onar
y wave i s pr oduced by 6. Whatmeasurabl
equanti
tyisassoci
atedwithcolour
sof
super
posit
ion of two progressive waves l
ight
?
tr
avel
li
nginopposi
tedi
rect
ions. 7. Ci
rcul
arwat
erwavesgener
atedbyapoi ntsour
ceatt he
cent
reO oft
hepondar eobservedtohavet
hepat
ter
n
i
i
. The wave has nodes atpoints of zer
o
showni
ntheFi
g.Expl
ainthepat
tern.
di
spl
acementand antinodes atpoint
s of
maxi
mum displ
acement.

i
i
i. I
nt hewave,vi
brat
ionofpart
icl
esatpoi
nts
bet
weensuccessi
venodesi
sinphase.

i
v. Bet
ween successi
ve nodes par
ti
cl
es have
di
ff
erentampl
it
udesofvibr
ati
on.
8. I
nanexper i
menttoobservei
nter
ferenceofl
i
ghtwaves,a
v. Thedistancebet
weensuccessi
venodesor doubl
e sl
iti
s pl
acedcl
osetot
he source.
λ
ant
inodesis. i. St atethefunct
ionoft
hedoubl
esi l
t.
2
i
i
. Descr
ibewhati
sobser
vedont
hescr
een.
Di
ff
erences bet
ween st
ati
onar
y and pr
ogr
essi
ve

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

waves i
i
i. St
atewhati
sobser
vedont
hescr
eenwhen

St
ati
onar
ywaves Pr
ogr
essi
vewaves a) Thesl
i
tsepar
ati
onS1S2i
sreduced.

They do nottransf
er Theytr
ansferenergyfr
om b) Whiteli
ghti
susedi
nsteadofmonochromat
icsour
ce.
energyfr
om onepoint onepointtoanothersi
nce13.TheFig.showsanexper
imentalar
rangement
.S1andS2ar
e
to anot
her si
nce the the wave f or
ms mor enarr
owsli
ts.
wave forms do not thr
oughthemedium
move t hr
ough t he
medium

Thedi st
ancebetween Distance bet
ween
two successi
venodes successive t
roughs or
λ crestsisλ
oranti
nodesis
2

Vibr
ati
onsofpar ti
cles Phase of par
ti
cles near
at poi nt
s bet ween eachotheraredi
ff
erent
successi
ve nodes are St
atewhati
sobservedont
hescr
eenwhenthesour
cei
s:-
i
nphase i
) Monochromati
c (i
i) Whit
eli
ght

14.Thefigshowst
hedi
splacementofapract
iceinpr
ogressi
ve
The ampl
i
tudes of Theampl
i
tudeofanyt
wow ave i
nci
denton a boundar
y between deep and shal
low
regi
ons.
parti
cles between part
icl
es whi
ch are i
n
successive nodes is phasearet
hesame
dif
ferent

Revi
sionExer
cise

I
. Complet
ethediagr
am t
oshowwhatisobser
vedaf
ter
1. Name a pr
oper
tyofl
i
ghtt
hatshows i
tisa boundar
y.(
Assumenolossofener
gy)
.
t
ransver
sewave.
I
I. Expl
aint
heobser
vat
ioni
n(i
)above.
2. Inanexper i
mentusi ngarippletankt
hef requency,f
oft heelect
ri
cpul segeneratorwasr educedt oone
thir
dofi tsori
ginalval
ue.How doest henew wave
lengthcomparewi ththeinit
ialwavel
ength? Explain
youranswer.

3. Di st
ingui
shbet
weenst
ati
onar
yandpr
ogr
essi
ve
waves.

4. St
atethecondi
tionforami
nimum t
ooccuri
nan
i
nter
fer
encepat
tern.

Chapt
erSev
en CURRENTELECTRI
CITYI
I

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page90
 232

Obj
ect
ives Cont
ent

Bytheendoft
hist
opi
cthel
eaner
,shoul
dbeabl
e 1. Scal
ereadi
ngammet
er,
vol
tmet
er
to:
2. El
ect
ri
cci
rcui
ts:
cur
rent
,pot
ent
ial
dif
fer
ence
a) Def
inepot
ent
ial
dif
fer
enceandst
atei
tsuni
ts
3. Ohm’
slaw(
exper
iment
alt
reat
mentr
equi
red
b) Measurepot
ent
ialdi
ff
erenceandcur
renti
na
ci
rcui
t 4. Resistance t
ypes ofr
esi
stor
s,measur
ement
s of
resi
stanceunit
s
c) Ver
if
yOhm’
slaw
5. El
ect
romot
ivef
orce(
emf
)andi
nter
nalr
esi
stanceof
d) Def
iner
esi
stanceandst
atei
tsuni
t acel
l
(E=V+I r
)

e) Deter
mine exper
iment
all
yt he volt
age-
cur
rent 6. Resistor
sinser
iesandi
nparall
el
rel
ati
onshi
psforv
ari
ousconduct
ors
7. Problemsonohm’ slaw,r
esistor
sinser
iesandi
n
f) Defi
neemfandexplai
nint
ernalr
esist
anceofthe parall
el.
cel
l

g) Deri
vet hefor
mulaeforeff
ecti
ver
esi
stanceof
resi
stor
sinseri
esandinpar
all
el

h) Sol
venumer
icalpr
obl
emsi
nvol
vi
ngohm’
slaw
r
esi
stori
ntheser
iesandi
npar
all
el

El
ect
ri
cCur
rent ne It
I= n=
t e
 Elect
ri
c cur
rentr
efer
stot
he r
ate off
low of -19
NBc
har
geona
nel
ect
ron, 6×10 c
e=1. oul
omb
charge.

 The mov ement of char ged part

icles cal
l
ed
el
ectronsconsti
tutesanel ect
ri
cchargeandthe Exampl e
conducti
ngpatht hroughwhi chel
ectronsmov e
i
scalledanelect
riccircui
t. Calculat
etheamountofcharget hatpassest
hroughapoi
nti
na
ci
rcuitin3seconds,
ift
hecurrentinthecir
cui
tis0.
5A.
chang
e,Q
c
urr
ent
,I= Sol
uti
on
t
ime,t
Q=I
t
Q
I=
t
Q=0.
5A ×3s=1.
5C
 SIuni
tofel
ect
riccurr
entist
heamper
e(A)
aft
er
t amousphy
hef sici
stMari
eAmper
e.
Exer
cise
 Sub-
mul
ti
ples of t
he amper e mi
e ar l
li
-

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

amper
e(mA)andmi
cro-
amper
es(
μA) 1.Acur
rentof0.
08Apassesi
naci
rcui
tfor2.
5mi
nut
es.

IμA=1×10-6A
ImA=1×10-3A, I
. How much char
ge passes t
hrough a poi
nti
nthe
ci
rcui
t?

I
I.Calculat
et henumberofel
ect
ronspassi
ng t
hrough
Def
ini
ti
onoft
heamper
e thepointpersecond

 An amper
erefersto an el
ect
ri
c cur
r hat 2. Acur
entt rentof0.5Afl
owsinacir
cui
t.Determinet
hequant
it
y
fl
ows i
n a conduct
orwhen a charge of1 ofchargethatcr
ossesapoi
nti
n4mi nutes.
coul
ombfl
owsperunitt
ime

Tot
alchar
gepassi
ngt
hroughapoi
nti
naci
rcui
t Measur
ementofEl
ect
ri
cCur
rent
 Inel
f ect
ronspassthroughapointandt hateach  El
ect
ri
ccur
renti
smeasur nganammet
edusi er.
el
ectr
oncarri
esachar geethenthetotalcharge
Q passing thr
ough the poi
ntis equalt o ne
coul
ombsi .
e.Q=ne;butQ=It,⇒ne=It

Exampl
e

I
nmovingachar
geof30coulombsfr
om poi
ntB t
oA 150
j
oul
esofwor
kdonewhati
sthedeci
malpl
acebet
weenAand
B?

Sol
uti
on

W
p.
d=
Q

150
p.
d= =5V
30

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page92
 232

 Zer
oerror
,ifany,shoul
dber ect
if
iedusi
ngthe
adj
ust
mentscrewbeforeusi
ngtheammeter.
Def
ini
ti
onoft
hevol
t
 Theammet sconnect
eri edinseri
eswit
hother
 Avolti
sdef
inedastheener
gyneededtomoveone
component
s in the circui nce i
t si ti s an
coul
ombofchar
gefr
om onepoi
nttoanot
her
i
nstr
umentoflowr
esist
ance.

Measur
ementofPot
ent
ialDi
ff
erence

 Pot
ent
ial
dif
fer
encei
smeasur ngav
edusi olt
met
er.
 Anappropr
iat
escal
eshoul
dbechosentosafe
guar
dthecoil
oft
heammeterf
rom bl
owi
ngup

 Accur
acyofeachscal
eoftheammetermustbe
obser
vedwhenrecor
dingr
eadi
ngs.

Exer
cise

Givether
eadi
ngsshownbybot
hscal
esoft
heammet
er
 Zer
o err
or,i
fany,shoul
d be recti
fi
ed usi
ng t
he
below.
adj
ust
mentscr
ewbef
oreusingt
hev ol
tmet
er.

 Thevol
tmet sconnect
eri edinacross(i
nparal
lel
)the
component
sintheci
rcui ncei
tsi tisaninst
rumentof
hi
ghr
esi
stance.

 Anappropr
iat
escaleshouldbechosentosaf
eguar
d
thecoi
loft
hev ol
tmet
erfrom bl
owingup.

 Accur
acyofeachscaleofthevolt
met
ermustbe
obser
vedwhenr
ecor
dingr
eadi
ngs.

Exer
cise

Gi
vet
her
eadi
ngsshownbybot
hscal
esoft
hevol
tmet
erbel
ow.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Pot
ent
ial
Dif
fer
ence(
Vol
tage)

 Electr
icpot
ent
ialdi
ff
erencebetweentwopoi
nts
refer
st oworkdone( i
nj oul
es)inmovi
ngone
coulombofchargefr
om onepointtot
heot
her.

 TheSIuni
tofpot
ent
ial
dif
fer st
encei hevol
t(v)
.

 Thebat
ter
yisthesourceofpowerf
ormov
ing
char
get
hrought
heci
rcuit
.

workdoneW(
, i
njoul
es)
pot
ent
ial
dif
fer
enc
e=
Cha
ngemoved,
Q(i
ncoul
ombs)

W
p.
d=
Q

Cur
rentandv
olt
agei
nser
iesar
rangement

I
.Cur
renti
nser
iesar
rangement

 Consi
dert
heset
-upbel
ow:

e:Forbot
Not hthe ammet
erand vol
tmet
er,t
he
negativeter
minalisconnectedtonegat
ivetermi
nal
oft he batter
y and posi
tive t
ermi
nalt o posi
ti
ve
ter
mi nalofbat
ter
y.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page94
 232

Cur
rentandVol
tagei
nPar
all
elCi
rcui
tAr
rangement  Whent
heswi
tchi
scl
osed,
iti
sobser
vedt
hat
;

Re
a ngonA1=r
di ea ngonA2=r
di ea ngonA3=r
di ea ngonA3
di

I
.Cur
renti
nPar
all
el  Theref
ore,when components are connect
ed i n
ser
ies,same curr
entflows t
hrough each oft he
 Consi
dert
heci
rcui
tshownbel
ow: componentseveni
fthecomponent
sarenotident
ical

I
I.Vol
tagei
nSer
iesAr
rangement

 Consi
dert
heset
-upbel
ow

 Whent
heswi
tchi
scl
osed,
iti
sobser
vedt
hat
:

Re
a ngonA1 +r
di ea ngonA2
di

+r
ea ngonA3 =Re
di a ngonA4
di

 Theref
ore,thesum oft hecurrentsinpar
all
el
ci
rcuit
sisequalt othet ot
alcurr
ent.Thet
otal
curr
entflowing i
ntojuncti
onisequalt ot
otal
curr
entfl
owingout.  Whent
heswi
tchi
scl
osed,
iti
sobser
vedt
hat
:

Re
a ngonV1+r
di ea ngonV2+r
di ea ngonV3=r
di eadi
ngonV4

I
I.Vol
tagei
nPar
all
el  Theref
ore,when component s are connected i
n
ser
ies,the sum oft he vol
tage drop acr
oss t
he
 Consi
dert
hedi
agr
am bel
ow: componentsisequalt
othevoltagesuppl
y.

Exer
cise

1. Usi
ngt
hedi
agr
am bel
ow,
find:

I
. Thecurrentpassingt
hrough L1 i
nt hefi
gurebelow
gi
venthatammet erAr
eads1.2Aandcur r
entsthr
ough
L2andL3are0.
34Aand0.52Arespecti
vel
y.

I
I. V1V2 a
ndV3r
eadi
ngs
,gi
vent
hatV4r
eadsi
s3.
0V

 Whent
heswi
tchi
scl
osed,
iti
sobser
vedt
hat
:

Re
adi
ngonV1 =re
a ngonV2=r
di ea ngonV3
di
=rea
dingonV4

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

 Theref
ore, f or component s i n par al
lel
arr
angement,t he same vol
tage dr
ops acr
oss
eachoft hem (si
ncet hei
rtermi
nalsar
eatt he
sameelectri
cpotential
.

2. I
nthecir
cuitshownbelow,whati
sthep.
dacr
osst
he
bul
b,andtheswitchwhen:

I
. Theswi
tchi
sopen?

I
I. Theswi
tchi
scl
osed?

 From t
hegraph,v
olt
ageisdir
ect
lypr
oport
ional
tot
he
curr
entandthi
sisgraphi
cal
repr
esent
ati
onofOhm’s
l
aw.
3. Def
ineavol
t  Theslopeofthegraphgiesr
v esi
stance.
SIuni
tof
4. Twocell
s,A andB connect
edwi t
hpar
all
elar
ein resi
stanceistheohm(Ω)
ser
ieswi
thabulbasshownbelow.
∆V
=R
∆I

 Mul
ti
plesofanohm ar
e:

1ki
lohm (
1kΩ)=1000Ω(
103Ω)

I
. Copyt he di
agram to show wher
ethe ammet
er 1me
gohm ( 106Ω)
1MΩ)=1000000Ω(
should be connect
ed in or
derto measur
ethe
curr
entthr
oughcellA

I
I. Vol
tmetershould be connect
ed t
o measurethe Def
ini
ti
onofanohm
pot
ent
ialdi
ff
erenceacrossbothbul
bandcellB
 Anohm r ef
erst
ot heresi
stanceofaconduct
orwhen
acurr
entof1Aflowingthroughitcausesav
olt
age

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page96
 232

Ohm’
sLaw dr
opof1Vacr
ossi
tsends.

 Ohm’ slawr elatesthevoltageacr ossthe  Thereci

procal
ofr
esist
ancei
saquanti
tycal
led
conduct orandt hecurrentflowingthroughit
.It 1
conduct
ance,()whoseSIuni
t sΩ-1orSi
si emens(
S)
i
saf terthephy sici
stGeor geSimonOhm.I t R
statesthat,“t
hecur rentfl
owi ngthrougha
conduct ori
sdi rectl
yproportionaltothe
potentialdi
fferenceacr ossi t
sendspr ovi
ded Exampl
e
thattemper atureandot herphysicalproper
ti
es
arekeptconst ant”
.
Acurrentof6mAf l
owst hr
oughaconductorofresi
stance
4kΩ.Calcul
atet
hevol
tageacrosst
heconductor
.

Sol
uti
on

V=I
R
 Mat
hemat
ical
l
y,Ohm’
slawcanbeexpr
essedas:
V=6×10-3×4×103=24V
V∝I

V=RI
Exer
cise
 ∴V=IR,Wher eVist
hepotent
ial
diff
erence,Ii
s
thecur
rentandRaconstantofpropor
tional
i
ty 1. Calcul
atethecur r
entinmil
l-
amperesf
lowi
ngt
hrougha
-1
caledr
l esistance. conductorofconduct 2mΩ whena15vsour
ance0. cei
s
connectedtoi
t
 I
fsev
eral
val
uesofcur
rentandt
hei
r
2. I
nor dert
ost artacer tainlaw acurrentof36Amustf l
ow
cor
respondingval
uesofvolt
ageforni
chr
ome t
hroughthest artermot or
.Calcul
atet heresi
stancesoft
he
wir
eareobtainedandagraphofvolt
ageagai
nst motorgivent hatt hebat t
eryprovidesavol t
ageof12V
cur
rentplot
ted,ast
rai
ghtl
inethr
oughtheori
gin i
gnoretheint
er nalresi
stanceofthebat t
ery.
i
sobtained.

3. I
nanexperimenttoinvest
igat
etheV-
Irel
ati
onshi
pfor Fact
orsAf
fect
ingt
heResi
stanceofaMet
all
i
cConduct
or
aconduct
or,t
hefoll
owingresul
tswer
eobtai
ned
1.Temper
atur
e
P.
d(V) 2. 3. 4. 6.
0
0 0 0  Resi st
anceofamet al
licconduct
ori ncr easeswi t
h
temper at e.Thi
ur sisbecauseheat ingi ncreasesthe
Cur
rent
,I(
A) 1. 1. 2. 3.
0
vibrati
onofat omst herebyincr
easingt hecol l
isi
ons
0 5 0
percr oss-sect
ionarea.Theoppositiontot heflowof
Resi
stance, electronsthusincr
easesast emperatures.
R(Ω)
2.Lengt
hoft
heconduct
or
I
. Copyandcompl
etet
het
abl
e

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

I
I. Pl
otagr
aphofIagai
nstt
hevol
tage  Theresi
stanceofauni form conduct
orofagiv
en
mater
ial
isdirect
lypr
opor
tional
toitsl
engt
hi.
e.
I
II
. Det
ermi
net
her
esi
stanceoft
heconduct
or
R∝l
…….
.(1)
I
V. Commentont
henat
ureoft
heconduct
or
R=c
ons
tant×l
4. Twocell
s,eachof1.
5Var eusedt odr
iveacur
rent
thr
oughawireABofr
esi
stance90Ω R
=c
ons
tant
l
I
. Cal
cul
atet
hecur
renti
ntheci
rcui
t
3.Cr
oss-sect
ionar
eaoft
heconduct
or
I
I. Whatwoul d be t
he di
ff
erence,i
fany t
ot he
cur
rent
,ift
hetwocell
sar
econnectedi
npar
all
el?  Ther esi
stanceofamet all
i
cconduct orisinv
ersely
proporti
onalt oits cr
oss-sect
ionalarea,A. A
conductorwithlar
gercr
oss-sect
ionalar
eahasmany
Ohmi
candnon-ohmi cconduct
ors fr
ee el ect
rons t o conductor hence bet ter
conductivi
ty.
Ohmi
cconduct
ors
1
R∝ ………….
(2)
 Theseareconduct orswhich obeyOhm’ slaw A
and ther
eforevoltage dr
op acr oss them i s
1
dir
ect
lypropor
ti
onaltocurr
entthroughthem e.g. R=c
ons
tant×
A
metalconduct
orslikeni omeandel
chr ectrol
ytes
l
ikecoppersul
phate. RA=c
ons
tant

 Forohmicconduct
orsagraphofvol
tageagai
nst
cur
renti
sastrai
ghtli
net
hroughtheori
gin.
Resi
sti
vi
tyofamet
all
i
cconduct
or
Non-ohmi
cconduct
ors
 Resisti
vi
tyisther
esistanceofsampleofmat er
ialof
 Thesear
econductorswhichdonotobeyOhm’s unitlengt
ha nduni
tc r
o sssecti
onalareaatacert
ain
l
awandt her
eforet
heirr
esist
ancechangeswit
h temperature.
cur
rentf
low e.g.t
hermist
or,ther
mioni
cdiode,
 Combi ni
ngequati
ons(1)and( 2)above;
f
il
amentbul
b,semi
conduct
ordi
odeet
c.
l
 A gr
aph ofvol
tage agai
nstcurr
entf
ornon- R∝
A
ohmi
cconduct
orsisnotastr
aight
.
Pl
R=
A

RA
ρ= , whe
rei
sρi
sther
esi
sti
vit
yoft
hec
onduc
tor
l

 TheSIuni
tofr
esi
sti
vi
tyi
stheohm met
er(
Ωm)

Exampl
e

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page98
 232

Two met
ers ofa resist
ance wir
e,area ofcr
oss sect
ional
50mm2,
0. hasaresi
stanceof220Ω.Cal
cul
ate:

I
. Ther
esi
sti
vi
tyoft
hemet
al

I
I. Thelengt
hofthewirewhich,
connectedi
nparal
l
elwi
th
the2meterl
ength,
wi l
lgi
vearesi
stanceof2.
00Ω.

Sol
uti
on

l=2m, A=0. 0×10-5m2,

50mm2=5.

R=220Ω

RA
ρ=
l
El
ect
ri
cal
Resi
stance
0×10-5m2
220Ω ×5.
ρ= 50×10-3Ωm
=5.
 El
ectr
icalresi
stanceist
heopposi
ti
onoff
eredby 2m
aconductortotheflowofel
ect
ri
ccurr
ent
.

 Amat eri
alwit
hhighconductancehasverylow
electr
ical r
esist
ance e.g. copper
. El
ectr
ical
resist
anceismeasuredusinganohmmeter.

Exer
cise b)Li
ghtdependentr
esi
stor(
LDR)

1. Giventhatt
heresist
ivi
tyofni
chr
omei 1x10-6  I
s1. tsr esi
stance decr
eases wi
thi
ncr
eases i
nli
ght
m,whatlengthofnichromewi r
eofadiamet
er0.
42m i
s i
ntensi
ty.
neededt
omakear esi
storof20Ω?

2. TwowiresXandBaresucht hatther adi

usofYis
twi
cet
hatofYandthel
engt
hofYi stwicet hatofX.i
fthe
two ar
e of same mat eri
al, determine t he rat
io
r
esi
sta
nceofX
r
esi
sta
nceofY

Met
hodsofMeasur
ingResi
stance

Resi
stor
s 1.Vol
tmet
er-
ammet
ermet
hod

 These ar
e conductor
s speci
all
y designed t
o
off
erpart
icul
arr
esist
ancetothefl
owofel ect
ri
c
cur
rent
. The sy mbol of r esistor is

 Int
hismethod,
curr Ihr
entt oughtheresi
storRand
cor
respondi
ngvol
tageVacrossi
tareobtainedand

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Typesofr
esi
stor
s resi
stanceoft
heresi
stori
sdet
ermi
nedusi
ngt
he
V
expressi
onR= .
I

1.Fi
xedr
esi
stor
s Di
sadvant
agesofvol
tmet
erammet
ermet
hod

 These ar
eresi
stor
s desi
gned t
o gi
vef
ixed  I
tisnotaccur
atesi
ncev
olt
met
ert
akessomecur
rent
r
esi
stance e.
g.wi r
e wound r
esi
stor
s,car
bon and ther
efor
e notal
lcur
rentpasses t
hrough t
he
(
col
ourcode)resi
stor
setc. resi
stor

2.Themet
er-
bri
dgemet
hod

 I
nthismethodar esi
storofknownresi
stancei
sused
i
nthedeterminat
ionofresist
anceofanotherr
esi
stor
whoseresi
stanceisnotknown.

 Thef
igur
ebel
owshowsaset
-upofMet
erBr
idge.
2.
Var
iabl
eresi
stor
s

 Thesear
eresi
stor
swhoser
esi
stancecan be
v
ari
ed.Theyi
ncl
ude:

a)Rheost
at

 This i s two a ter

minal v
ari
abl
e r esi
stor
represent
ed byanyofthe sy
mbols below in
electr
ical
cir
cui
ts.  The br i
dge is balanced byadjusti
ng the v
ariabl
e
resi
storLunt ilthereisnogal vanometerdef
lecti
on
i
.e.pointeratzeromar k.Atbal
ancedstate:

X L1
=
R L2


b)
Pot
ent
iomet
er

 This is a t
hree ter
minal var
iabl
e r
esi
stor
repr
esent
edbythesy
mbol bel
ow.

3.Nonl
i
nearr
esi
stor
s

 These ar
er esi
stor
si n whi
ch currentfl
owi
ng
thr
oughthem doesnotchangelinearlywi
tht
he
vol
tageappl
i
ed.Theyincl
ude:

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page100
 232

a)Ther
mist
or

 Thisisat emperat
uredependentr
esist
orwhose
resi
stance decr eases wi th increase in
temperature.I
tsel
ectr
ical
symboli
sasbel ow.

Resi
stor
sNet
wor
ks a) Cal
cul
atet
heef
fect
iver
esi
stance

1.Resi
stor
sconnect
edi
npar
all
el Sol
uti
on

1 1 1 1 1
 Consi
dert
hreer
esi
st sR1R2 a
or ndR3 connect
ed = + + +
RT R1 R2 R3 R4
i
npar
all
elasshownbel
ow:
1 1 1 1 1
= + + + =0.
9333
RT 4 3 10 4

1
RT = =1.
074Ω
0.
9333

b) Thecur
rentt
hrought
he10Ωr
esi
stor

Sol
uti
on

Sincetheresist
orsareinparal
l
el,thevol
tagedr
op
acrosseachofthem i
sthesamei
.e.4.
5V.

I
T =I
1+I
2I
V
3 I=
R
VT V1 V2 V3
= + + 10
RT R1 R2 R3 I= =2.
222A
4.
5
ButVT=V1=V2=V3
2.Thefigurebel
ow shows3r esist
orsinser
iesconnect
edt
o
powersource.Acur
rentof1.
5Aflowthrought
hecir
cuit
.
(
forr
esi
stor
sinspa
ral
lel
)

1 1 1 1
= + +
RT R1 R2 R3

Fornr
esi
stor
sinpa
ral
lel

1 1 1 1
= + + Cal
cul
ate:
RT R1 R2 R3

a) Thet
otalr
esi
stance
 I
ftwo r
esi
st s R1a
or ndR2 ar
e connect
ed i
n
par
all
elthen t
he equi
val
entr
esi
stance RE i
s b) Thevol
tageacr
osst
hesour
ce
gi
venby
c) Thevol
tagedr
opacr
osseachr
esi
stor
1 1 1 R1+R
= + =
2

RE R1 R2 R1 R Sol
uti
on
2

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

R1R2 a
)RT=R1+R2+R3
RE=
R1+R2
RT=4+6+3=13Ω
2.Resi
stor
sconnect
edi
nser
ies
b)V=I
R

V=1.
5×13=19.
5V

c
) V4Ω =1.
5×4=6V

V6Ω =1.
5×6=9V

V3Ω =1.
5×3=4.
5V

VT=V1+V2+V3

I
TRT =I
1R1+I
2R2+I
3R3

ButI
T =I
1=I
2=I
3

RT=R1+R2+R3

Exer
cise

1.The fi
gure bel
ow shows f
ive r
esi
stor
s and a sour
ce of
vol
tageof6V.

Exampl
es a) Fi
ndt
heef
fect
iver
esi
stanceoft
heci
rcui
t

1.The circui
tdiagr
am inthe f
igur
e bel
ow shows 4 b) Cal
cul
atet
hecur
rentt
hrough
resi
stor
sconnect
edacrossa4.
5vsupply
3.Si
xresi
stor
sar
econnect
edi
naci
rcui
tasshowni
nthef
igur
e
bel
ow.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page102
 232

a) Cal
cul
atet
het
otalr
esi
stanceoft
heci
rcui
t

b) Thet
otalcur
renti
ntheci
rcui
t

c) Thecur
rentt
hrought
he3Ωr
esi
stor

d) Thecur
rentt
hrought
he8Ωr
esi
stor

3.Twor
esi
stor
sconnect
edi
npar
all
elasshownbel
ow Exper
iment
alDet
ermi
nat
ionofI
nter
nalResi
stanceand
Emf

Met
hod1

 Consi
dert
hesetupbel
ow:
a) Cal
cul
ate:

I
. t
hecur
rentt
hatpassest
hroughR1

I
I. Ter
minalp.
dacr
osst
hebat
ter
y

El
ectr
omot i
veForce( Emf
)andInternal Resist
ancer I fsev er
alv al
ues ofcur rentand theircor
responding
valuesofv olt
agearecol l
ectedandgr aphofv ol
tageV
 Electromoti
vef orce(E)ofthecel lreferstothe againstcurr
entIisplot
ted.Itisastr
aightl
i
neofnegat i
ve
potenti
aldi
fferenceacrossitst erminalswhen sl opecutti
ngthroughthevoltageaxi
swhenextrapolat
ed.
nochargeisflowingoutofi
ti.e.whent heci
rcui
t
i
sopen.

 Terminalvolt
age(V)i
st hevol
tagedropacross
theterminal
softhecellorbat
terywhencharge
i
sf lowing outofitand itisduet o ext
ernal
resi
stanceR.

 UsingtheequationE=V+I ra ndthereforeV=E-I r
,
thesl
opeoft hegraphgivestheint
ernalresi
stanceof
thecel
lwhilethev ol
tage-i
nter
ceptgivestheemf( E)
ofthecel
l.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

 Inter
nalresistance( r)r ef
erst ot heoppositi
on  Ifseveralval
uesofcurrentandt hei
rcorr
esponding
offer
edbyt hesour ceofel ectromot ivefor
ceto values of volt
age are collect
ed and graph of
theflowofthecur r
entt hati
tgener ates. 1
recipr
ocalofcurr
entagai nstRisplott
ed,astrai
ght
I
 Lost v ol
tage i s t he di f
ference bet ween 1
elect
romotivef orceandt erminalv olt
ageandi t l
inewi thposit
ivegr
adientwhi chpassesthrough
I
i
sduet ointernalresistance. axisisobtai
ned.

Rel
ationshipbetweenEmf( E)
,Ter
minalVol
tage(
V)
andInternalResi
stance(
r)

 Consi
deraresi
storRconnect
edi
nser
ieswi
tha
cel
lofi
nter
nalr
esist
ancer

 Theint
ernalresi
stanceoft
hecellrisconsider
ed
toconnectedinserieswi
ththeexternalr
esist
or,
R

 Thecur
rentf
lowi
ngt
hrought
heci
rcui
tisgi
ven
by:

e.
m.f
c
urr
ent=
tot
alr
esis
tanc
e

E
I=
R+r

E=I
(R+r
)

E=I
R+I
r⇒E=V+I
r

eI
 Wher R=t
ermi
nal
vol
tag
e,I
r=l
ostvol
tag
e

1 Exer
cise
 Thegr adi
entofthegr aphgives andt heref
ore
E
the elect
romoti
vef orce oft he cel
lcan be 1.Thetabl
ebelowshowsr
eadi
ngobtai
nedinanexperi
mentto
deter
mine the e.
m.f
,E and int
ernalresi
stance R of a
obtai
ned whilet he R- i
nter
cept gi
ves i
nter
nal
accumulat
or
resi
stanceofthecellr.
Ext
ernalr
esi
stance,R 0.
3 0.
3 2.
(Ω) 5 75
Exampl
es Cur
rent
,I(
A) 2.
5 1.
0 0.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page104
 232

1.Twodr ycel
lseachoft heint
ernalresi
stance0.05Ωand 5
connectedinseriesareusedt ooperateanel ect
ri
cbellof
resi
stance10Ω.Thewi ri
ngoftheci r
cuithasar esi
stance Reci
procalofcur
rent
,
of0.2Ω.I fthebellrequir
esacur rentof0. 2Ator i
ng,to 1
whatval uecan t hecombi ned emff allbeforethebell I
comest or i
ng?
a) Dr
awasui
tabl
eci
rcui
tusedt
ogett
heabover
esul
ts
Sol
uti
on
1
b) Pl
ott
hegr
aphof agai
nstR
I

c) Det
ermine t
he val
ues ofi
nter
nalr
esi
stance rand
el
ect
romoti
veforce

2.Theci
rcui
tinthef
igur
ebel
owshowsthecurr
entatjunct
ion
P.Findt
heamountanddi r
ecti
onoft
hecurrentthatpasses
E=I
( )
R+r thr
ought
hewi r
eW.

E=0.
2(( 05×2)=2.
2)+0.
10+0. 06V

2.Youar
epr
ovi
dewi
thr
esi
stor
sofval
ues4Ωand8Ω

a) Dr
awaci r
cuitdi
agram showi
ngther
esi
stor
sinser
ies
wi
theachotherandwithbatt
ery. 3.Thr
eer
esi
stor
sar
econnect
edasshownbel
ow

Cal
cul
ate:
b) Calcul
atet otalr esistance oft
he ci
rcui
t(assume
a)Thet
otalr
esi
stancei
ntheci
rcui
twhen:
negli
gibl
einternalresistance)
i
. 1i
S sopen
Sol
uti
on
i
i
. 2i
S scl
osed
RT=R1+R2
b)Thecur
rentt
hrougheachoft
her
esi
stor
swhen:
RT=4+8=12Ω
i
. 1i
S sopen

i
i. S1iscl
osed
c) Giventhatthebatteryhasanemfof6Vandi nt
ernal
resi
stanceof1.33Ω,cal
cul
atet
hecurr
entt
hrough c)Thepot
enti
aldi
ff
erenceacrosseachr
esi
storwhen S1 i
s
cl
osed
i
. 8Ωwhent hetwoar ei
nser
ies.
4.A batter
yofemf12V andi
nter
nalr
esi
stanceof0.
6Ω i
s
Sol
uti
on
connect
edasshownbel
ow
E
8Ω = I
I 4Ω =
R+r

6
= =0.
4501A
12+1.
33

i
i
. 4Ωr
esi
storwhent
het
woar
einpar
all
el. a)Cal
cul
atet
hecur
rentt
hrought
he3Ωr
esi
storwhenswi
tchi
s:

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
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 232

Sol
uti
on i
. Open

i
i
. Cl
osed

b)Findthetot
alpot
ent
ialdi
ff
erentacr
osst
he7Ω r
esi
stor
s
whenSisopen

5.Acel
lofemfof6.
0Vanddr
ivescur
rentof2.
0At
hroughR1
R1R2 whenswi
tchSi
sopen
RT = +r
R1+R2

4×8
RT = +1.
33=4Ω
4+8

V4Ω E-Vr
4Ω =
I =
R4Ω R4Ω

6-64×1.
33 Cal
cul
ate:
I
4Ω = =1.
001A
4 a) Thecur
rentt
hrought
he2Ωr
esi
stor

b) Thei
nter
nalr
esi
stanceoft
hecel
l

c) Thecur
rentthr
ougheachoft
her
esi
stor
swhent
he
swi
tchSiscl
osed

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page106
 232

The ci
rcui
t bel
ow can be used as a l
i
ght sensor
. 8. a) Int heci
rcui
tdiagr
am shown,
cal
cul
atet
heef
fect
ive
resi
stancebetweenYandZ.

a) Expl
ainhow itwor
ksascondit
ionschangef
rom
pi
tchdarknesst
obrightl
i
ght

b) Ift
heresi
stanceofLDRi ndimli s1x104Ω
ghti
b) Det
ermi
net
hecur
rentt
hrought
he3r
esi
stor
.
cal
cul
atet
hep.dacross1kΩresi
stor
c) Oneoft
he6resistor
shasal engt
hof1m andcr
oss-
secti
onal
-
5 2
ar
eaof5.0x10 m .Calcul
atether
esi
sti
vi
tyoft
hemateri
al.
Fur
therExer
cise
9. I
nthecircui
tdi
agram fi
veresi
storsareconnectedtoa
1. Astudentlear
ntthatabatter
yofeightdr
ycells bat
ter
yofemf .4V,andnegli
gibl
einter
nalresi
stance.
each1.5vhasatotalemfof12Vt hesameasacar Det
ermine:
batt
ery.Heconnectedinseri
eseightnewdr ybat
teri
esto
hi
scarbutfoundthattheycouldnotstar
ttheengine.Gi
ve
areasonfort
hisobservat
ion

2.
a)Youarerequi
redt
odet er
minetheresist
anceperuni
t
l
engthofani chr
omewi rex,youarepr ovi
dedwithd.c
powersuppl
y,anammeterandvol
tmeter
.
i
. Thet
otalr
esi
stanceoft
heci
rcui
t.
I
. Drawacircui
tdiagr
am t
oshowhowyouwoul
d
connectt
hecircui
t. i
i
. Thecurrentf
lowi
ngt
hrought 5
he5.
resi
stor
.
I
I. Descri
behowyouwouldusethecircui
tin(a)(i)
abovetodet
ermi
netheresi
stanceperunitlength i
i
i. Thepot
ent
ial
satpoi
ntsYandO.
ofx.
i
v. Thepot
ent
ialdi
ff
erencebet
weenYand
c) 2.Four5resi
stor
sareconnect
edt
oa10Vd.c.suppl
yas O.
showninthedi
agram bel
ow.
11.Ast udentwishest oinvesti
gatet herelat
ionshi
pbetween
currentandvoltagef oracer t
aindeviceX.I nthespace
provide,drawaci rcui
tdiagram includi
ngt wocell
s,
rheostat,ammet er,volt
meterandt hedeviceXthatwould
besui t
ableinobt aini
ngthedesi redresul
ts.

Cal
cul
ate;
- 12.Inthecir
cuitdiagram showni
nfi
gur
e7,theammeterhas
negli
gibl
eresistance.Whentheswi
tchSiscl
osed,
the
a. Theef
fect
iver
esi
stancei
ntheci
rcui
t. ammet erreads0. 13A.

b. Thecurr
entIf
oll
owi
ngi
ntheci
rcui
t(t
hrough
thesour
ce)

3.St
udythecircui
tdiagram bel
owanddet er
minet
he
pot
entialdropacrossthe3r esi
stor
.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
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 232

4. Stat
etwocondi
ti
onst
hatar
enecessar
yfora
conduct
ort
oobeyOhm’sl
aw.

5. a) Tworesistor
sR1andR2ar econnectedinseri
es
t
oa10Vbat ter
y.Thecur r
entflowi
ngthenis 0. 5A.
WhenR1onl
yi sconnectedtothebatt
erythecurr
ent
fl
owi
ngis0.8A.Calcul
atethe

i
. Val
ueofR2

i
i
. Cur
rentfl
owingwhenR1andR2ar
econnect
edi
n
par
all
elwit
hthesamebatter
.

6. Acurr
entof0.
08Apassesi
ncircui
tfor2.5minut
es.
Howmuchchargepassest
hroughapoi nti
nthe
ci
rcui
t?

7. Anammet er,avolt
meterandabulbareconnect
edi n
acircui
tsoast omeasur ethecur
rentf
lowi
ngandt he
potenti
aldif
ferenceacr
ossboth.Sketchasuit
able
ci
rcuitdi
agram forthearr
angement.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page108
 232

Chapt
erEi
ght HEATI
NGEFFECTOFANELECTRI
CCURRENT

Speci
fi
cobj
ect
ives Cont
ent

Byt
heendoft
hist
opi
cthel
eanershoul
dbeabl
eto: 1. Si
mpl
eexper
iment
sonheat
ingef
fect

a) Perf
orm anddescri
beexperi
mentstoi
l
lust
rat
e 2. Fact
orsaf
fect
ingel
ect
ri
cal
ener
gyW =VI
, t,
P=VI
heat
ingeff
ectofanelect
ri
ccur
rent
3. Heat
ingdevi
ces:electr
icalket
tl
e,el
ect
ri
cali
ron
b) Stat
ethef
act
orsaf
fect
ingheat
ingbyel
ect
ri
c box,
bulbf
il
ament,elect
richeat
er
curr
ent
4. Probl
ems on el
ect
ri
calener
gy and el
ect
ri
cal
c) Deri
vet heequati
onsf
orel
ect
ri
calener
gyand power
el
ectr
icalpower

d) Ident
ifydev
icesinwhichheat
ingef
fectofan
elect
ri
ccurr
enti
sappl
ied

e) Solv
e numericalproblems i
nvol
vi
ng el
ect
ri
cal
energyandel
ectr
icalpower

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

I
ntr
oduct
ion Fact
orsAf
fect
ingHeat
ingbyEl
ect
ri
cCur
rent

 When an el ectri
c cur r
entpasses t hrough a Theheatproducedbyaconduct orcar
ryingcurrent
conductor,itgener at
es heatener gy
.Thi sis dependson:
cal
ledt heheatingef f
ectofanel ectr
iccurr
ent
andi tisduet ot her esist
anceof f
eredbyt he a) Amountcur r
entpassingthr
oughtheconduct or
conductortot hecur r
ent.Heatingef f
ectofan
 The heatpr oduced i
s dir
ectl
ypr oport
ionaltot he
el
ectri
ccur r
entwasf i
rstinvest
i edbyJames
gat
squareofcur rentthrough t
heconduct orprovided
Joule,aManchest erbrewerinUK.
thatsameconduct ori
susedforthesamet i
me.i.
e

he
ate
ner
g 2
y∝I.
Demonst
rat
ingHeat
ingEf
fectofanEl
ect
ri
cCur
rent
Usi
ngaCoilofWi
re

 Theset-
upbelowcanbeusedt oexperi
mental
ly
demonstr
ateheati
ngef
fectofanel
ectr
iccur
rent
i
nthelaborat
ory
.

b)Resi
stanceoft
heconduct
or

 Heatpr oduced i
n a conductorcar
ryi
ng curr
enti s
 Precaut
ion:t
hecoi
lshoul
dbeful
l
yimmersedi
n di
rectl
ypr opor
ti
onalt
or esist
anceoft heconductor
waterbutshoul
dnottoucht
hebott
om orwal
l
s pr
o vi
ded cur
rent and t
ime ar
e const
an t
.
ofthebeaker
. he
atenergy∝R.

 Obser
vati
on:Iti
sobservedthatthetemperat
ure
ofwateri
ncreaseswit
hr esi
stance,cur
rentand
ti
me.

 Expl
anati
on:Elect
ri
calenergyis conver
ted to
heatenergyresul
ti
nginar isei
nt emperature.
The heat energy i
ncr
eases wit
hr esi
stance,
curr
entandti
me.

c) Ti
me f
or whi
ch cur
rent f
lows t
hrough Exampl
es

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page110
 232

conduct
or 1.Anelect
ri
cbulbr ated40W isoper
ati
ngon240V mai
ns.
Det
ermi
netheresi
stanceofi
tsf
il
ament.
 Keeping cur
rentand r
esi
stanceconstantheat
producedinaconductori
sdi r
ect
lypropor
tonal Sol
i uti
on
to t he t i
me f or which cur r
ent f l
ows.
V2
He
ate
ner
g .
y∝t P=
R

2402
40= ; ⇒R=1440Ω
R

2.Whenacurrentof2Af lowsinaresi
storf
or10minut
es,15KJ
ofelect
ri
calener gyisdissi
pat
ed. Deter
minethevolt
age
acr
ossther esi
stor.

Sol
uti
on

E=VI
t

15×1000=V×2×10×60

15000
Vv
dA` V= =12.
5V
1200

Q 3. Howmany100W el
ectr
icir
onscouldbesaf
elyconnect
ed
t
oa240Vmovi ngcir
cuitf
it
tedwit
ha13Af use?
El
ect
ri
cal
Ener
gy,
E
Sol
uti
on
 Consideracur rentIfl
owingthr
oughaconductor
ofr esistanceR f ora t met.I
i fa pot ent
ial P=VI
di
fference V dr ops acr
oss the ends ofthe
(
No.
ofi
rons
)x1000=VI
conduct or,t
hen;
13x240
W No.
ofi
rons=
V= 1000
Q
=3.
12=3i
rons
(
fromde
fini
ti
onofpot
ent
ial
dif
fer
enc
e)

E=W =VQ
Exer
cise
(
whe
reW i
sthee
lec
tri
cal
wor
kdonei
nmovi
ng

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

c
har
geQ) 1. Aheat erofresistanceR1i
sr atedPwat
ts,Vvoltswhil
e
v
anotherofresistanceR2israted2Pwatt
s, /2v
olts.
Thi
sise
lec
tri
cal
wor
kdonei
sconve
rte
dtohe
at
DetermineR1/R2

e
ner
gy,
E
2. Stat
eTHREEf
act
orswhi
chaf
fectheat
ingbyanel
ect
ri
c
char
geQ curr
ent
.
butc
urr
ent
,I= ; ⇒Q=I
t
ti
me ,
t
3. Whati
spowerasi
trel
atest
oel
ect
ri
calener
gy?
E=V(
It; ⇒E=VI
) t
4. Anelect
ri
calappl
i
ancei
srat
edas240V,200W.Whatdoes
S
inc
e, R(
V=I f
romohmsl
aw)
;el
ect
ric
ale
ner
gy thi
sinf
ormati
onmean?

c
ana
lsobee
xpr
ess
eda
s; 5. Anel
ect
ri
calheat
eri
slabel
l
ed120W,
240V.

E=VI
t=(
I I
R)t; 2
⇒E=IRT Cal
cul
ate;

a. Thecurr
entthroughtheheat
ingel
ement
whentheheaterison.
El
ect
ri
cal
Power
,PE
b. Theresi
stanceoft
heel
ementusedi
nthe
 Poweri
sther
ateofdoi
ngwel
l heat
er.

work 6. Anelectri
ctoyisr
ated100W,
240V.Calcul
atethe
powe
r= resi
stanceofthet
oywhenoperat
ingnor
mally.
ti
me

el
ect
ric
ale
ner
gy VI
t 7. Anelect
ricbul
bwi t
haf i
lamentofresi
stance480is
= = ;
ti
me t connect
edt oa240Vmai nssupply.Determi
netheener
gy
di
ssipat
edin2mi nutes.
⇒El
ect
ric
alpowe
r,PE=VI

S
inc
eV=I
R,e
lec
tri
cal
powe
rca
nal
sobee
xpr
ess
eda
s;

PE=(
I I
R)
2
PE=IR

V V
Orf
romI= ; P=V()
R R

V2
PE=
R

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page112
 232

El
ect
ri
cal
Dev
icesf
orLi
ght
ing 3. Acurrentof3.3Aispassedthroughar esi
storof25Ωf
or2
hour
scal cul
atetheel
ectri
calenergyconvertedt
oheat
a)Fi
l
amentl
amps ener
gyi n20mi nut
es.

4. Anelect
ri
ccur r
entir
onconsumes2. 9MJofenergyin1
hour10 minutes when conver
ted t
ot he mai
ns power
suppl
yof240v.Cal
culat
etheamountthrou

5.
 When curr
entflows through the fi
l
ament,it 6. ghthef
il
amentintheel
ectr
iciron
gl
owswhitehotandthereforepr
oducesli
ght.
7. I
nthecircui
tshowninthefigurebel
oweachbul
bisr
ated
 Thefi
l smadeoft
amenti ungst
enmet aldueto 6v,
3w,
i
tsver
yhighmelti
ngpoint(3400oC)
.

 Thebul
bi sf
il
ledwi thinertgasli
keargonand
ni
tr
ogentopr
ev entoxi
dationoft
hefil
ament.

b)Fl
uor
escentl
amps a) Calcul
atethecurr
entthr
ougheachbul
b,when
thebulbsarewor
kingnor
mall
y.

b) How many coulombs ofchar

ge pass i
n6
secondst
hrougheachbul
b?

c) Whatwoul
dt he ammet
erread when al
lthe
 Theyareeff
ici
entthanfi
lamentl
ampsbecause bul
bsar
eworki
ngnormal
l
y.
theylastmuch longerand hav
elow r
unning
d) Cal
cul
atet
heel
ect
ri
calpowerdel
i
ver
edbyt
he
cost
.
bat
ter
y.
 I
tconsist
soft hemercur
yvaporwhichproduces
8. Star
ti
ngfrom el
ectr
icalpower,
P,generat
edinaconduct
or
ul
tr
aviol
etradiat
ionwhenthel ampisswitched V2

on.The radiati he powderon t

on makes t he showthatP= , wherethesymbolsthei
rusualmeani
ngs.
R
i
nside of t he tube produce visi
ble li
ght
(
fl
uor
esce)
.

El
ect
ri
cal
Dev
icesf
orHeat
ing

A)Fuse

 Afuse i
s a shor
tlength ofwir
e of
mat
eri
alwit
hlow melt
ingpoint(
ti
nned

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

copper)
,which melts and breaks the
ci
rcuitwhencurr
entthroughitexceeds
acertai
nv al spr
ue.Thi otect
sel ectri
cal
appli
ancesandprev
entsfireoutbreaks.

Ot
herel
ect
ri
calheat
ingdevi
cesi
ncl
ude:

a) Radi
antel
ect
ri
cheat
er

b) El
ect
ri
cali
ronbox

c) El
ect
ri
cket
tl
e

d) Hotwi
reammet
er

a)

Revi
sionExer
cise

1. Atouchbulbiscall
ed2.
5v,0.
4A.Whati
sthepower
r
ati
ngofthebulb?

2. Anel
ect
ri
cbul
bisl
abel
ed50w,
240v,
cal
cul
ate

b) Ther
esi
stanceoft
hef
il
amentusedi
nthe
l
amp

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page114
 232

c) Thecur
rentt
hroughthef
il
amentwhent
he
bul
bworksnor
mal l
y

Chapt
erNi
ne ELECTROS
TATI
CSI
I

Speci
fi
cobj
ect
ives Cont
ent

Bytheendoft
hist
opi
cthel
eaner
,shoul
dbeabl
e 1. El
ect
ri
cfi
eldpat
ter
ns.
to:
2. Char
gedi
str
ibut
iononconduct
ors.
a) Sket
ch el
ect
ri
cfi
eld pat
ter
ns ar
ound char
ged
bodi
es. 3. Spher
ical
andpearshapedconduct
ors.

b) Descri
bechargedi
str
ibut
iononconduct
orsof 4. Act
ionatpoi
nts;
li
ght
ingar
rest
ors.
var
iousshapes.
5. Capaci
tance, unit of capacitance ( f
arad,
c) Def
inecapaci
tanceandst
atei
tsSIuni
t. micr
ofarad)
,fact
orsaf
fect
ingcapaci
tance.

d) Descri
be charging and dischar
ging of a 6. Appl
i
cat
ionsofcapaci
tor
s.
capacit
or (cal
culat
ion i
nvol
vi
ng curves not
7. Pr
obl
ems on
capaci
tor
s
requi
red).
1 1 1
(
usi
ngQ=CV,
CT =CI+C2, = + )
e) Statethefact
orsaffect
ingt
hecapaci
tanceofa CT C1 C2
paral
lelpl
atecapaci
tor

f
) St
atet
heappl
i
cat
ionsofcapaci
tor
s.

g) Sol
venumer
ical
probl
emsi
nvol
vi
ngcapaci
tor
s.

El
ect
ri
cFi
eld

 Anel ectr
icfi
eldr
eferstotheregi
onwherea
chargedbodyexperi
encesafor
ceofat
tract
ion
orrepulsi
on.

2. I
sol
atednegat
ivepoi
ntchar
ge

Di
rect
ionofanEl
ect
ri
cFi
eld Thefiel
dli
nesar
eradi
all
yinwar
dst
owar
dst
henegat
ive
char
ge
 Thedi r
ecti
onofanelectr
icf
ieldatapar
ti
cul
ar
pointisdefi
nedast
hedi r
ect
ioninwhi
chaunit
posit
ivechargei
sfr
eet omov ewhenpl
acedat
thatpoint
.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

3. Twoequalposi
ti
vepoi
ntchar
ge

El
ect
ri
cLi
neofFor
ce(
Elect
ri
cFi
eldLi
ne)

Thi
sisthepat
hal
ongwhichaunitposit
ivechar
ge
woul
dtendt
omoveint
heelect
ri
cfi
eld.

Pr
oper
ti
esofEl
ect
ri
cFi
eldLi
nes
4. Twoequalunl
i
kepoi
ntchar
ge
1. El
ect
ri
cfi
eldl
i
nesst
artat900f
rom t
heposi
ti
ve
char
geandendont
henegat
iv geat900.
echar

2. Theydonotcr
osseachot
her
.

3. Theytendt ocontr
actorexpandsot
hatt
hey
neveri
nter
esteachother

El
ect
ri
cFi
eldPat
ter
ns

. I
sol
atedposi
ti
vepoi
ntchar
ge

The fi
eldli
nes ar
eradi
all
y out
war
ds f
rom t
he
posi
ti
vechar
ge

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

5. Twounequalposi
ti
vepoi
ntchar
ge

 Nochargeisfoundontheout
sideofahollow
conduct
or.Forahol
lowconductor
,thecharge
resi
desontheoutsi
de.
6. Posi
ti
vepointchar
geandast
rai
ghtmet
alpl
ate Poi
ntAct
ion
havi
ngnegati
vecharge
 Pointact
ionref
erst
othefastl
ossorgainofchar
ge
atsharppoint
sduetohi
ghchargeconcentr
ati
onat
thepoint
s.

Demonst
rat
ionofPoi
ntAct
ion

 Ahighl
ychargedshar
ppoi
nti
spl
acedcl
oset
oa
Bunsenbur
nerflame.

7. Posi
ti
vepointchargeandunchar
gedr
ingpl
aced  I
tisobser
vedt
hatt
hef
lamei
sbl
ownaway
.
i
ntheelect
ri
cf i
eld

8. Twoparal
l
elmetalpl
ateshavi
ngopposi
te Expl
anat
ion
char
geandpl
acedclosetoget
her
.
 Burni
ngf lamecont ainspositi
veandnegat i
veions.
Theelect
ri
cf i
eldbet
weent
hem i
suni
for
mi.
e.f
iel
d Whent hesharp[pointisbroughtclosetotheflame,
l
inesequall
yplaced. negativeionsareat t
ractedtothesharppoint,whil
e
posit
iveionsarer epell
edawayf rom thesharppoint.
Astheposi ti
veionsar erepell
ed,theycreat
ean
“el
ectricwind”whichbl ows.

 I
ftheconduct
orisbr
oughtver
ycl
oset
othef
lame,
t
heflamespli
ts.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
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 232

Char
geDi
str
ibut
iononConduct
ors

 Dist
ri
buti
onofchargeonthesur
faceofa
ordependsont
conduct heshapeofthe
conduct
or.

 Forspheri
calconduct
or,char suni
gei for
mly
di
stri
butedonthesurface.

 Forpearshapedconduct
or,
chargei
s
concent
ratedatt
hesharppoint
.

 ForCuboi
danddiamondconduct
orshi
gh
char
gedensi
t satt
yi hever
ti
ces.

Exampl
e

1.Acandl
efl
ameisplacednearashar
ppointedpi
nconnect
ed
tot
hecapofanegat
ivel
ychargedel
ect
roscopeasshown
bel
ow.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Stat
eandexplai
nwhati
sobser
vedont
hel
eafoft
he
el
ect
roscope.

Sol
uti
on

Itisobser vedthattheleafcol
lapses.Thisisbecausethe
shar ppointionizestheairar
oundi t
.Thenegat i
veions
(electrons)movet ot heshar
pendt oionizethepositi
ve
char gedai ri
onswhi chareatt
ractedtothecap( orpositi
ve
airionsneut ral
izethenegati
vechargeont heshar ppoint)
.

1. I
selectr
icalf
iel
dst
rengt
hascal
arorvect
orquant
it
y?
Expl
ain
2.I
tisdangeroustocar
ryapoi
ntedumbr
ell
awheni
tis
rai
ning.Expl
ain. 2. Expl
ainhownegati
vel
ychar
gedpoi
ntededgeget
s
di
schargedbyi
tsel
f.
Sol
uti
on
3. I
tisnotadvi
sablet
otakeshel
terunderat
reewheni
tis
Thesharppointoftheumbr
ellaat
tract
schar
gereadil
yto
r
aini
ng.Expl
ain.
neutr
ali
zethechargeint
hecloudwhichmayelect
rocute
thepersonholdi
ngtheumbr
ella. 4. Whati
sthepur
poseoft
hespi
kesont
hel
i
ght
ningar
rest
or?

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Appl
i
cat
ionofPoi
ntAct
ion Capaci
tor
s

Poi
ntact
ioni
sappl
i
edi
nthewor
ki hel
ngoft i
ght
ing  Acapaci
tori
sadev
iceusedf
orst
ori
ngchar
ge.
ar
rest
or.
 Capaci
torsy
mbol
is

TypesofCapaci
tor
s

1. Papercapaci
tor
s

2. El
ect
rol
yti
ccapaci
tor
s

3. Var
iabl
eai
rcapaci
tor

4. Par
all
elpl
atecapaci
tor

Par
all
elPl
ateCapaci
tor

 Aparal
lelpl
atecapaci
torconsi
stsoft
womet al
Wor
kingMechani
sm oft
heLi
ght
eni
ngAr
rest
or pl
atesseparat
edbyani nsul
ati
ngmateri
alcal
l
ed
di
elect
ri
c.
 Whenanegat i
velychar
gedcloudpassesover
thearrest
ori
tinducesposi
ti
vechar
geont he
spikesandnegati
vechargeonthepl
ate.

 Thenegat
ivechar
geontheplat
eisimmedi
atel
y
di
schar
gedtothesur
roundi
ngground.

 Negati
veionsar
eattr
act
edt
othespikesandar
e
di
schargedbygi
vingupt
hei
rel
ect
rons.
Char
gingaCapaci
tor
 Atthesametime,positi
veionsar
erepel
led
upwardsfr
om thespikesandtheyneut
rali he  Theci
zet r
cui
tdi
agram bel
owshowsaset
-upf
or
negati
vechar
geont heclouds. char
gingacapaci
tor
.

Exer
cise

1. Thef i
g.showsahol l
ow negat i
velychar gedsphere
withmet aldi
skattachedtoani nsulatorplacedi
nside.
 Obser
vati
on:Whent heswit
chiscl
osedi sobser
ti ved
Statewhatwoul dhappent ot heleafofanunchar ged
el
ectroscopei ft
hemet aldiskwer ebr oughtnearthe t
hatmill
i
-ammeterreadingdecr
easeswhil
evol
tmeter
capofelectroscope.Givear easonf oryouranswer. r
eadingi
n cr
eases.

 Expl
anat
ion:Negat
ivechar
gef
lowf
rom t
henegat
ive

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page120
 232

terminal
oft hebatter
ytoplateBoft hecapaci
tor
.At
thesamer atenegat i
vechargefl
owf rom t
heplat
eA
ofcapacitort opositi
veter
mi nal
ofbat t
ery
.
Therefor
e, equal posi
tiv
eandnegat i
vecharges
appearont heplatesandopposet heflowof
elect
ronswhi chcauses.

 Thechargi
ngcurr
entdropst
ozer
owhent
he
capaci
tori
sful
l
ycharged.

 Pot
enti
aldi
ff
erenceacrosst
hecapaci
toral
so
dev
elopsdur
ingchar
ging.

Capaci
tance

 Capaci
tanceisdefi
nedasthechar
gest
oredi
na
capaci
torperuni
tvolt
age.

cha
nge,Q
c
apa
cit
anc
e,c=
vol
tag
e,V

Q
C=
V

 TheSIuni
tofcapaci
t st
ancei hef
arad(
F)

Def
ini
ti
onoft
hef
arad

 Afar
adisthecapaci
tanceofabodyi
fachar
ge
of1coul
ombr ai
sesi
tspotent
ial
by1vol
t

Submul
ti
plesoft
hef
arad

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Di
schar
gingaChar
gedCapaci
tor 1mi
crof
ar 1µF)=10-6F
ad(

 Theci
rcui
tfordi
schar
gingi
sasbel
ow. 1Nanof
arad( )=10-9F
1µf

1Pi
cof
ar 1pF)=10-12F
ad(

Fact
orsAffect
ingt
heCapaci
tanceofaPar
all
el-Pl
ate
Capaci
tor

1.Ar
eaofOver
lapoft
hePl
ates,
A
 Obser vat
ion:Iti
sobservedthatmi
ll
iammeter  Capaci
tanceisdir
ect
lypr
opor
ti
onal
tot
hear
eaof
readingdecreasesfrom maximum val
ueto over
lapoftheplat
es(C ∝A)
minimum.Thepoi nterdef
lect
sinopposi
te
directi
ontot hatdur
ingchargi
ng.  I
fthepositi
veplat
eisconnectedt
othecapofa
posit
ivel
ychar
gedelect
roscope,t
hediv
ergence
 Expl
anati
on:Dur i
ngdischargi
ngchargefl
owin oftheleafi
ncr
easesastheareaofoverl
ap
theopposi
tedi
r ecti
oni.e.f
rom pl
atetoAunt
il i
ncreases
thechar
geont heplatesiszero.
2.Di
stanceofSepar
ati
onoft
hePl
ates,
d
 Pot
enti
aldi
ff
erenceacrossthecapaci
tor
pr
acti
cal
lydi
minishest
ozero.  Capacit
anceisinver
selypropor
ti
onal
todi
stance
1
ofseparat
ionoftheplates(C∝ )
d

 Ift
heposi t
iveplat
eisconnect edtot
hecapofa
posit
ivel
ychargedel ect
roscope,t
hediv
ergence
oftheleafdecreasesast hedist
anceofthe
separati
onincreases.

3.Nat
ureoft
heDi
elect
ri
c Capaci
torNet
wor
ks

 Diel
ectr
icref
erstot
heinsul
ati
ngmat er
ial a)Capaci
tor
sinSer
ies
betweentheplat
esofaparal
lelpl
ate
capaci
tor.  Consi
dert
hecapaci
tor
sar
rangementbel
ow.

 Permi
tti
viyεoft
t heinsul
ati
ngmater
ial
isa
const
antdependentonthemedi
um between
thepl
ates.

 I
ftheplat
esareinavacuum,
theconst
antis
denot
edby ε
o(epsi
l
onnaught
)andit
svalue
i 85×10-12Fm-1
s8. Cha
rgeonC1=c
hang
eonC2=c
hang
eonC2=Q

Expr
essi
onf
orcapaci
tance butV=V1+V2+V3(
I.
e.c
ompone
ntsi
nse
rie
s)

A Q
C∝ a
ndV=
d C

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page122
 232

ɛ
A QT Q1 Q2 Q3
C= ,whe
reɛ=pe
rmi
tt
ivi
ty = + +
d CT C1 C2 C3

1 1 1 1
= + +
CT C1 C2 C3

whe
reCTi
sthec
ombi
nedc
apa
cit
anc
e

 Foronl
ytwoc
apa
cit
ors
,C1a
ndC2a
rei
nse
rie
s,

1 1 1 C1+C2
= + =
CT C1 C2 C1C2

Exampl
es C1C2
The
ref
oreCT =
1. Twopl atesofaparall
elpl
atecapaci t
orare1mm apart C1+C2
andeachhasanar eaof6cm2..
Giventhatt
hepotent
ial
B)Capaci
torsinParal
l
el
dif
ferencebetweentheplat
esi s90V, cal
cul
atet
he
chargest or
edinthecapaci
tor
.( Take  Considerthearr
angementbel
ow.
-12 -1
ɛ0=8.
85×10 Fm )

Sol
uti
on

ɛA
C=
d

85×10-12Fm-1×6×10-4m2
8.
C= 31×10-12F
=5.
1×10-3

Q=CV  Thep.dacrosseachofthecapaci
torsist
hesame
-12
31×10 ×90=4.
Q=5. 779×10 C -10 asthep.dacrossthesour
cesincetheyar
e
connect
edinparal
lel
.
2. Findtheseparati
ondistancebetweentwopl at
eifthe
capacitancebet
weent henis6×10-12Fandt he QT =Q1+Q2+Q3
enclosedareais3.0cm2(takeɛ0=8.8510-12Fm-1)
butQ=CV
Sol
uti
on
CTV=C1V1+C2V2+C3V3
ɛA
C=
d CT =C1+C2+C3
-12 -1 -4 2
85×10 Fm ×3×10 m
8.
6×10-12F=
d

85×10-12Fm-1×3×10-4m2
8.  Not
e:Tr
eatc
har
gea
scur
rents
inc
ecur
renti
s
-4
d= 425×1
=4. 0 m2
-12
6×10 F
t
her
ateoff
lowofc
har
ge.

Exer
cise
Exampl
es

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
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 232

1. Achargeof4x104cwasst or
edinapar al
l
elpl
ate 1. Threecapacit
orsofcapaci
tance3µF,4µFand6µFar
e
capacit
orwhenapotent
ialdi
ff
erenceof5Vwas connect
edt oapotenti
aldi
ffer
enceof24Vasshown
appli
edacrosst
hecapacit
or.Workoutthe bel
ow.
capacit
anceoft
hecapacit
or.

2. Thef i
gurebelow r
epresent
stwopar al
l
elplates
ofacapacitorseparat
edbyadi st
anced.Each
plat
ehasanar eaofAsquar eunits. Suggest
two adjust
mentst hatcanbemadeso ast o
reducetheeff
ecti
vecapacit
ance.

Fi
nd: Q3µF=C3µFV3µF;

a) Thecombi
nedcapaci
tance Q3µF=3µF×9.
0048V=27.
0144µC

Sol
uti
on Q4µF=C4µFV4µF;

1 1 1 1 Q4µF=4µF×9.
0048V=36.
0192µC
= + +
CT C1 C2 C3

1 1 1 1 Q2µF=C2µFV2µF;
= + +
CT 3 4 6
Q2µF=2µF×9.
0048V=18.
0096µC
1 4+3+2
= =0.
75
CT 12 Q5µF =C5µFV5µF ;

1
CT = =1.
333µF Q5µF =5µF×14.
9952V=74.
976µC
0.
75

b) Thet
otalchar
ge

Sol
uti
on Exer
cise

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page124
 232

Q=CV 1. I
ntheci
rcui
tbel
owC1=4µF,=C2=3µFandC3=1µF.Gi
ven
t
hatV=12V,calcul
ate:
Q= 1.
333µF ×24V=32µC

c) Thechar
geoneachcapaci
tor

Sol
uti
on

Sincethecapaci
torsar
einser
ies,
char
ge
oneachoft hem i
sthesameandisequalt
o
totalchar
gei.
e.32µC
I
. Thechar
geoneachcapaci
tor
d) Thevol
tageacr
osst
he4µFcapaci
tor
I
I. Thevol
tageacr
osseachcapaci
tor
Sol
uti
on
2. Intheci
rcui
tshownbel
owcal
cul
atet
hechar
geont
he
Q capaci
tor
V=
C

32µC
V= =8V
4µF

2. Fourcapacit
orsofcapaci
tance3µF,4µF,
2µF,
and
5µFarearrangedasshownbel ow.

Fi
nd: 3. Thefigurebel
ow Showspartofacir
cui
tcont
aini
ngt hr
ee
capaci
tors.Wri
tean expr
essi
on f
orCT. (Theef f
ecti
ve
capaci
tancebet
weenAandB. )

a) Thecombi
nedcapaci
tance
4. St
atet
hel
awofel
ect
rost
ati
cchar
ge.
Sol
uti
on
5. Thecapacit
orsintheci
rcui
tinFi
gur
ebel
ow ar
eident
ical
Capaci
tor
s3µF,4µFand2µFar ein andi
nit
ial
lyunchar
ged.
par
all
elandthei
rtot
alcapaci
tanceisi
n
ser
ieswit
hthe5µFcapaci t
or.

1 1 1
= +
CT C1+C2+C3 C4

1 1 1 14
= + =
CT 2+4+3 5 45

45
CT= =3.
214µF
14
6. Swit
chS1 i
sopenedandswitchS2 cl
osed.Det
ermi
net
he
b) Thet
otalchar
ge fi
nalr
eadi
ngofthevol
tmet
er,
V.

Sol
uti
on 7. I
ntheci
rcui
tdi
agr
am showni
nfi
gur
ebel
oweachcel
lhas

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

QT=CTVT an emfof1. 5 and int

ernalresi
st 5.The
ance of0.
-6
capaci
tanceofeachcapaci
tori
s1.4F.
QT=3.
214µF×24V=74.
976µCor74.
976×10 C

c) Thechar
geoneachcapaci
tor

Sol
uti
on

Vol
tageacross3µF,4µFand2µFist
he
samesincetheyareinpar
all
el;

QT
VT-V5µF=VT-
C5µF

74.
976µC
24- =9.
0048V
5µF

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page126
 232

2. A10µFcapaci tori
schargedtoap.
dof200Vand
i
solated.I
tisthenconnectedi
npar
all
elt
oa20µF
capacit
or.Fi
nd:

I
. Ther
esul
tantpot
ent
ialdi
ff
erence.

Sol
uti
on
8. Whent
heswi
tchsi
scl
oseddet
ermi
net
he:

a. Ammet
err
eadi
ng

b. Char
geoneachcapaci
tor

9. Threecapaci
tor
sof1.5F,2.
0 Fand3.
0 Far
e
connect
edinser
iest
op.
d.of12V.Fi
nd;
-
QT=Q10µF+Q20µF
I
. Thecombi
nedcapaci
tance.
10×10-6F×200V=(10×10-6)
V+( V
20×10-6)
I
I. The t ot
al char
ge st
ored i
n t
he
arr
angement
. 10×10-6F×200V
V= =10V
(10×10-6)
F+( F
20×10-6)
I
II
. Thechar
gei
neachcapaci
tor
.

10.Inthecir
cui
toft
hefi e3C1=2F,C2 =C3 =0.
gur 5F
andEi sa6Vbatter
y.Calcul
atet
het
otalchargeand I
I. Theener
gyst
oredbef
oreconnect
ion.
p.dacr
ossC1
Sol
uti
on
Ener
gySt
oredi
naCapaci
tor
1
Ene
rgys
tor
ed= CV2
2
 Char
gingacapacitorinvol
vesdoingwork
agai
nstrepulsi
onofthenegat i
vepl
atetomor
e 1 -6 2
Ene
rgys
tor
ed= ×10×10 F×(
200V)=0.
2J
el
ectr
onsf l
owingi
nandat t
racti
onofthe 2
posi
ti
veplateonelectronsfl
owingout.

 Thisworkisst
oredi
nfor
m ofpotent
ialener
gy.
I
II
. Theenergyi
nthet
wocapaci
tor
saf
ter
Theenergymaybeconver
tedtoheat,l
i
ghtor connect
ion.
otherf
orms.
Sol
uti
on
wor
kdone(Energ
yst
ore
dincapaci
tor)
=aver
agec
ha r
ge×potent
ial
dif
fer
e nc
e 1 1
Ene
rgys
tor
ed= CV2;
Ene
rgys
tor
ed= ( V2
C1+C2)
2 2
1
Ene
rgys
tor
ed= QV 1
2 Ene
rgys
tor
ed= ( 10 =1.
10×10-6+20×10-6) 5×10 J 2 -3

2
1 2
Ene
rgys
tor
ed= CV (
sinc
eQ=CV)
2
I
V. Theenergydi
ff
erencebet
weenI
IandII
I
Q2 Q
Ene
rgys
tor
ed= ( s
inc
eV= ) aboveandcommentonyouranswer
.
2C C
Sol
uti
on
 Agraphofchar
ge,Qagai
nstvol
t Vi
age, sa
st
rai
ghtl
inethr
ought
heorigi
n. (
0. 0015)J=0.
2-0. 1985J

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
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 232

Thee
ner
gyi
sconve
rte
dtohe
ata
ndl
ight

Exampl
es

1. A4µFcapacit
orischar
gedtoapotent
ial
di
ff
erenceof80V.Fi
ndtheener
gystor
edini
t

Sol
uti
on

1
Ene
rgys
tor
ed= CV2
2

1
Ene
rgys
tor
e 4×10-6F)×80V
d= (
2

Ene
rgys
tor
e 6×10-4J
d= 1.

Exer
cise

1. A10µFcapaci
tori
schar
gedbyan80Vsupplyandthen
connect
edacr
ossanunchar
ged20µFcapaci
tor
.Calcul
ate:

I
. Thef
inalp.
dacr
osseachcapaci
tor
.

I
I. Thef
inalchar
geoneach.

Theini
ti
alandf
inalener
gyst
oredbyt
he
capaci
tor
s.

2. A 2Fcapaci t
orischargedtoapot enti
alof200V,the
supplyisdi
sconnect
ed.Thecapaci
tori
st henconnect
edto
anotherunchar
gedcapacit
or.Thep.d.acrosstheparal
lel
arr
angementi s80V.Findthecapacit
anceoft hesecond

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page128
 232

capaci
tor
.

3. A5Fcapacit
orischar
gedtoap. dof200vandisolat
ed.It
i
sthenconnectedt
oanotherunchargedcapaci
torof10F.
Cal
cul
ate:

I
. Ther
esul
tantp.
d

I
I. Thechar
gei
neachcapaci
tor
.

4. Inanexperimenttost
udythevariati
onofcharge
storedoncapacit
orandthepotenti
aldi
ffer
ence
acrossit
,thefol
l
owingresul
tswer eobt
ained.

Char
geQ 0.
0 0.
1 0.
2 0.
3 0.
4 0.
56
(C) 8 6 4 2 0

p.
d(v) 2.
0 4.
0 6.
0 8.
0 10. 14.
0
0

5. Pl
otagr
aphofchar
geQ.agai
nstp.
d

6. Useyourgr
apht
odet
ermi
ne:
-

a. Capaci
tanceoft
hecapaci
tor
.

Ener
gyst
oredint
hecapaci
torwhent
hep.
d
acr
ossit
splat
eis10V.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
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 232

Appl
i
cat
ionsofCapaci
tor
s

1. Usedi
nsmootheningcir
cuit
st osmoot
hent
he
d.
cout
puti
nrect
if
icat
ionprocess

2. Usedi nr educti
onofspar ki
ngininduct
ioncoil
contact.Var i
abl
ecapaci t
orisused int ur
ning
ci
rcuitofar adi
or ecei
veri
nwhichi
tisconnected
i
npar al
leltoindi
cat or

3. Capaci
tor
sareusedindel
aycircui
tsdesi
gnedto
gi
veint
ermit
tentf
lowofcur
rentincari
ndicat
ors.

4. A capaci
tori
sincl
uded i
nfl
ash ci
rcui
tofa
camerainwhi
chi
tdi
schar
gesi
nst
antl
ytofl
ash.

Chapt
erTen QUANTI
TYOFHEAT

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
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 232

Speci
fi
cobj
ect
ives Cont
ent

Bytheendoft
hist
opi
c,t
hel
eanershoul
dbeabl
e 1. Heatcapaci
ty,
specif
icheatcapaci
tyuni
ts
to: (
experi
mentaltr
eatmentrequi
red)

a) Defi
neheatcapaci
tyandspeci
fi
cheat 2. Lat
entheatoffusion,l
atentheatofv
apori
zat
ion,
capaci
ty uni
ts(exper
imentaltr
eatmentnecessar
y)

b) Deter
mineexperi
mental
lyspeci
fi
cheat 3. Boi
l
ingandmel
ti
ng
capaci
tyofsol
idsandli
qui
ds
4. Pr
essur
ecooker
,ref
ri
ger
ator
c) Defi
nespeci
fi
clatentheatoff
usionand
speci
fi
clat
entheatofvapori
zat
ion 5. Pr
obl
emsonquant
it
yofheat

d) Deter
mineexperi
mentall
ythespeci
fi
clatent (
Q=MC∆θ, Q=Ml
)
heatoff
usionoficeandthespeci
fi
clatent
heatofvapor
izat
ionofst
eam.

e) Stat
ethef
act
orsaf
fect
ingmel
ti
ngpoi
ntand
boil
i
ngpoi
nt

f
) Expl
aint
hef unct
ioni
ngofapr
essur
ecooker
andaref
rigerat
or

g) Sol
vepr
obl
emsi
nvol
vi
ngqual
i
tyofheat
.

Def
ini
ti
onofHeat Sol
uti
on

 Heatisform ofener
gythatfl
owsfr
om onebody Q=C∆θ
toanotherduetotemperatur
edi
ff
erence
Q=600J
K-1×(
600-200)
betweenthem.
=24000J
Di
ff
erencesbet
weenHeatandTemper
atur
e

Heat Temper
atur
e
Exer
cise
Heati
saf orm ofenergy Temperatur
ei sthe
t
hatfl
owsf rom onebody degr
eeofhot nessor Anel ect
ri
calheat
err
ated240V,3Arai
sesthetemperat
ureof
t
oanotherduet o col
dnessofabody l
iqui
dXf om 25oCt
r o55oCin10minut
es.Calcul
atet
heheat
t
emperaturediff
erence measuredonsamescalecapaci
tyofl
iqui
dX.

PHYSI
CSFORTECHNOLOGYANDI
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 232

Measur
edi
njoul
es Measur
edi
nkel
vi
n Speci
fi
cHeatCapaci
ty,
c

HeatCapaci
ty,
C  Speci
ficheatcapaci
tyi
sthequanti
tyofheat
requi
redtorai
sethetemperat
ureofaunitmassofa
 Heatcapaci
tyr
efer
stot hequant
it
yofheat subst
ancebyonekel vi
n(K)
.
ener
gyrequi
redtorai
set het
emperat
ureofa
gi
venmassofasubst ancebyonekelvi
n. heate
ne r
gyabsorbed,
Q
S
pec
ifi
che
atc
apa
cit
y,c=
ma
ss,m ×temper
aturechange
,∆θ
heate
nergyabsorbe
d,Q
He
atc
apa
cit
y,C= Q
t
empera
turechange,
∆θ c=
m∆θ
Q
C= orQ=C∆θ Q=mc
∆θ
∆θ
 TheSIuni
tofspeci
ficheatcapaci
tyi
sthej
oul
eper
 TheSIuni
tofheati
sthej
oul
eperkel
vi
n(K-1)
J -
1-1
ki
logr
am perkel
vi
n(Jkg k )
 Not
e:Di
ff
erentmat
eri
alshavedi
ff
erentr
atesof
 Speci
fi
cheatcapaci
tycanal
sobeexpr
essedas:
heatabsor
pti
onandt
her
efor
edi
ff
erentheat
capaci
ti
es.

Exampl
es he
atca
paci
ty,
C
S
pec
ifi
che
atc
apa
cit
y,c=
mass
Calcul
atet hequant
it
yofheatrequi
redtorai
sethe
temperatureofamet albl
ockwithaheatcapaci
tyof He
atc
apa
cit
y,C=s
pec
ifi
che
atc
apa
cit
y,c×ma
ss,
m
-1 0 0
600JK f rom 20Cto60C.

 Note:Iftwosubstancesofthesamemassar e i
. St
atetheprecauti
onst
hatneedtobetakent
o
subjectedtothesameamountofheat ,
they mini
mizeheatlossest
othesurr
oundi
ngs
acquiredif
fer
enttemperatur
echangesbecause
Sol
uti
on
theyhavediff
erentspeci
ficheatcapaci
tiese.g.
thespecifi
cheatcapacit
yofi r
onis460Jkg k1.
-
1-
i
. Thecal
ori
met
ershoul
dbehi
ghl
ypol
i
shed.
Thismeanst hat1kgofironwouldtakeinor
i
i
. Thecal
ori
met
ershoul
dbeheavi
l
ylagged.
giveout460Jofheatwheni tstemperat
ur e
changesby1k. i
i
i. Thecal
ori
metershoul
dbecl
osedusi
ngan
i
nsul
ati
ngli
d(li
dmadeofapoorconduct
or)
.

Exampl
e

Ablockofcopperofmass10.0kgandspeci
fi
cheat
-
1- 1 0 0
capaci
ty460Jkg k coolsfr
om 80Cto40C.Findt
he
quanti
tyofheatgivenout
.

Sol
uti
on

Q=mc
∆θ

80-40)0C
kg-1K-1(
Q=10kg×460J

Q=184000J=184kJ Det
ermini
ngspecif
icheatcapaci
tyofl
i
qui
dsusi
ng
methodofmixt
ures

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page132
 232

Exer
cise

1.Ablockofmetalofmass1.5kgwhichi
ssuit
ably Exampl
e
i
nsulatedi
sheat
edf om 300Ct
r o500Cin8minut
esand20
secondsbyanel
ectri
cheatercoi
lrat
ed54watt
s.Fi
nd: Abl ockofmet alofmass0. 15kgat1000cwas
transferredtoacoppercal ori
met erofmass0. 4kg
a) Thequant
it
yofheatsuppl
i
edbyt
he containingal iquidofmass0. 0
8kgat20c.Thebl ockand
heat
er
thecal orimeterwi thit
scontent sevent ual
lyreacheda
b) Theheatcapaci
tyoft
hebl
ock commont emper at eof400c.Gi
ur vent hespecificheat
-
1-1
capaci tyofalumi nium 900Jkg k , andt hatofcopper
c) Thespeci
fi
cheatcapaci
ty -
1-1
400Jkg k , calculatethespecificheatcapaci tyofthe
2Findthefi
nalt
emperat
ureofwateri
faheat ersour
ce l
iquid.
0
r
ated50W heat
s100gwaterfr
om 25Ci n5mi nut
es
Sol
uti
on
(
speci
fi
cheatcapaci
tyofwat
eris4200Jkg-1k-1)
He
atl
ostbytheme
tal
bloc
k
=hea
tgaine
dbyli
qui
d+he a
tga
ine
dbyc
alor
ime
ter

Met
hodsofDet
ermi
ningSpeci
fi
cHeatCapaci
ti
es mBc
B∆θB =mLc
L∆θL+mcc
c∆θc

1.Met
hodofMi
xtur
es 0.
15×900×(
100-40)
=0.8 ×c
L×(
40-20)+0.
4×400×(
40-20)
 Inthi
smet hod,arel
ati
velyhotsubst
anceis
8100-3200
mixedwithar el
ati
vel
ycoldsubstance.Heat c
L= kg-1K-1
25J
=306.
16
energyi
stransferr
edfrom hotbodytocold
bodyunti
lthermalequil
ibri
um i
sestabli
shed

2.El
ect
ri
calmet
hod

Det
ermini
ngSpecif
icHeatCapaci
tyofaSol
i
dUsi
ng  I
nt hi
smethod,
electr
icheati
ngcoi
lsuppl
i
esthe
MethodofMi
xtures heatener
gywhichisabsorbedbyot
hersubst
ances.

Det
ermi
ningSpeci
fi
cHeatCapaci
tyofaMet
alBl
ock
Usi
ngEl
ectri
calMet
hod
Exampl
e
 Theset
-upt
hatcanbeusedi
nthi
scasei
sasshown
Al aggedcoppercal
ori
meterofmass0. 50kgcontains
bel
ow.
0.
4kgofwat erat250C.Ametall
icsoli
dofmass1. 2kgis
0
tr
ansf er
redfr
om anovenat350Ct ot
hecal or
imeteranda
steadytemperat eof500Ci
ur sreachedbyt hewateraft
er
sti
r r
ing.

i
i
. Calcul
atethespeci
ficheatcapacit
yofthe
mat er
ialoft
hesoli
d( Speci
fi
cheatcapaci
tyof
-1-1
copperis 400Jkg k andt er4200Jkg-
hatofwat
1-1
k)

Sol
uti
on

He
atl
ostbythes
olid
=hea
tgaine
dbyt hewa
ter+he
atg
aine
dbyc
alor
ime
ter

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
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 232

msc
s∆θs=mwc
w∆θw +mcc
c∆θc

2×c
1. s×(350-50)
=0.4×4200×( 50-25)+0.
5×400×(
50-25)

42000+5000
c
s= =130. kg-1K-1
56J
360

Pr
ecaut
ions ChangeofSt
ate

I
. Themetalblockmustbehi
ghl
ypol
i
shedand  Changeofasubst
ancefrom sol
idtol
iqui
d,f
rom
heavi
l
ylagged. l
iqui
dtogasortherever
seinvol
veschangeofst
ate.

II
. Thet woholesshouldbefi
l
ledwit
hahighoi
l Lat
entHeat
toi
mpr ovet
hermalcont
actwit
htheheat
er
andthermometer
.  Lat entheatref
erst
oamountofheatrequi
redt
o
changest at
eofasubst
ancewi
thoutchangein
Exampl
e temperature

Amet alblockofmass0. 5kgisheatedel

ectr
ical
ly.I
fthe  Iti
stheheatener
gyabsor
bedorgi
venoutdur
ing
voltmeterreads20V, t
heammet er4Aandthe changeofstat
e.
temperatureoftheblockri
sesfom 250Ct
r o950Ci n8
mi nut
es,calcul
atet
hespecifi
cheatcapaci
tyofthemet al Lat
entHeatofFusi
on
block.
 Latentheatoff
usionref
erstot
heamountheat
Sol
uti
on requir
edtochangethestat
eofasubstancefr
om
He
atsuppl
iedbythehea
ter
soli
dtoliqui
dwithoutt
emperat
urechange.
=hea
ta bs
orbedbythemetal
bloc
k
 Note:Whenali
qui
dchangest
osol
i
dst
ate,
lat
ent
t=mbc
VI b∆θ heatoff
usioni
sgi
venout.

20×4×( 5kg×c
8×60)J=0. b×( K-1
95-25) Speci
fi
cLat
entHeatofFusi
on,l
f

38400  Speci
fi
clat
entheatf
usionrefer
stothequant
it
yof
c
b= =1097. kg-1K-1
14J
35
heatr
equir
edtochangeaunitmassofthesubstance
fr
om sol
idtol
iqui
dwithoutchangei
ntemperatur
e.

Det
erminingSpeci
fi
cHeatCapaci
tyofLi
qui
dUsi
ng Q
f= , ⇒Q=ml
l
El
ectr
icalMethod m f

 TheSIunitofspeci
fi
clatentheatoff
usi
oni
sthe
Exampl
e -
1
j
ouleperkil
ogram (
Jkg ).
Inanexper imenttodet erminethespecifi
cheatcapaci ty
ofwat eranelectr
icalheaterwasused.I fthevoltmeter Lat
entHeatofVapor
izat
ion
readingwas24Vandt hatofammet er2.0A, calcul
atethe
specifi
cheatcapaci t
yofwat erift
hetemper atureofa  Latentheatofvapor
izat
ionrefer
stotheheat
massof1. 5kgofwat erina0. 4kgcoppercalorimeterrose requir
edtochangethestateofasubstancefr
om
0
by6caf ter13.
5mi nutes. l
iquidtogaswithoutchangeintemperat
ure.

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
ON.
Page134
 232

Speci
fi
clat
entheatofvapor
izat
ion,

 Speci
fi
clat
entheatofvapori
zati
onisthequanti
tyof
heatr
equir
edtochangeauni tmassofasubstance
fr
om l
iqui
dtovaporwithoutchangei
nt emper
ature.

Q
V = , ⇒Q=ml
l V
TheSI m
uni
tof
Sol
uti
on speci
ficl
atentheatofvapor
izat
ioni
sthej
oul
e
-1
perki
logr
am (Jkg ) .
He
atsuppl
iedbyt
hehea
ter
=hea
tg ai
nedbywat
er+he a
tga
ine
dbyc
alor
ime
ter

t=mWc
VI W∆θW +mcc
c∆θc

24×2×(
13. 5×c
5×60)=1. W ×6+0.
4×400×6

38880-960
c
W= kg-1K-1
33J
=4213.
9

Exer
cise

Amanwant edtohaveawar m bat hat350c.Hehad4.0kg

0
ofwaterinabasi nat90c. whatmassofcol dwaterat
230cmusthehaveaddedt othehotwat ertoobtai
nhis
choiceofbath.Neglectheatl
ossesandt akespeci
ficheat
capacit
yofwat eras4200Jkg-1k-1

PHYSI
CSFORTECHNOLOGYANDI
NNOVATI
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 232

Exampl
es

1.Cal
culat
et heamountofheatr equir
edtoconver t4kgof
0 0
i
ceat-10ct ol i
quidat5c( speci
fi
cheatcapaci
tyofwat er
s4200Jkg-1k-1,
i speci
fi
cheatcapacit ce=2100Jkg-1k-1,
yofi
speci
fi
clatentheatoffusionofi 000Jkg-1k-1)
ce=340,

Sol
uti
on

Q=micec
ic
e∆θice+micel
fi
ce
+ mwaterc
wat
er∆θwater

Q=4×2100×(
10-0)+4×340000+4×4200×(
5-0)

Q=84000+1360000+84000

Q=1528000Jor1.
528MJ

2.Aket t
leratedat4.0kW cont ai
ning2.0kgofwat erisl
eft
switchedon.Howl ongwi l
littaket hewat ertoboi
ldryi
n
thekettl
eiftheini
ti
altemper atureofwat eris200c
(specif
icheatcapacityofwat eris4200Jkg1k-1speci
-
fi
c
6 -
1
l
atentheatofvaporizati
onofwat eris2.26x10Jkg . a. Fr
om t
hegr
aph,
det
ermi
net
heboi
l
ingpoi
ntoft
hel
i
qui
d.

Sol
uti
on b. I)Determinetheheatgivenoutbyt heheaterbetweent
he
ti
mest =0.5minutesandt =5.
0 mi nutes.
Q=Pt= mwaterc
wat
er∆θwater+ml
Vwa
ter I
I)From thegraphdet er
minethetemper at
urechange
betweent het
imest =0.5minutesandt=5.0minutes.
26×106
100-20)+2×2.
4000×t=2×4200×(
II
I)Hencedet
ermi
net
hespeci
fi
cheatcapaci
tyoft
he
672000+4520000
t= =1298s l
i
quid
4000
c. 1.8gofvaporwascoll
ectedfrom t
heli
quidbetweent
he
ti
mest=6.8minut
esandt =7.
3mi nut
es.Deter
minethe
Exer
cise speci
fi
clat
entheatofvapori
zati
onoftheli
quid.

1. Calcul
atetheamountoft hermalener gyrequi r
edto
0 0
change5gofi ceat- 10ct ost eam at100c( speci
fi
c
Fact
orsaf
f ect
ingmel
ti
ngandboi
l
ingpoi
nts
heatcapacit
yofi 10Jg-1k-1,
ce2. specif
iclatentheatof
fusi
onofice336Jg-1, specif
icl atentheatof Theyaret
wo:
vapori
zati
onofst eam 2260Jg-1, speci
fi
cheatcapaci ty
ofwater4.2Jg-1k-1) I. Pr essure

2. Acoppercal ori
meterofmass60gcont ains100gof II. I mpur i
ti
es
0 0
oilat20c.api eceoficeofmass28gat10ci s
addedt otheoi l.Whatmassofi cewi l
lbel eftwhen Ef
fectofpressureonmel ti
ngpoi
nt
thet emper at
ureofthecalor
imet erandi tscontent
s
0  I ncr
easeinpr essur
elowerst
hemel
ti
ngpoi
ntofa
willbe10C?Speci f
icheatcapacityofcopper=0. 4Jg-
1- substance.
k1,specifi
cheatcapaci t
yofoil2.4Jg-1k-1,
specif
ic
l
at entheatoff usi
onofice336Jg-1)  Consi dertheset-upbel
ow.

PHYSI
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3. Dryst eam i spassedi ntoawel llaggedal umi ni

um
calorimet erofmass400gcont aining1. 2kgofi ceat
0oC.Themi xtur
eiswel lsti
rredandst eam suppl ycut
offwhent hetemper atureoft hecal orimet erandi t
s
cont entsr eaches53oC.Negl ectingheatl osses,
deter minet hespecificlatentheatofvapor i
zati
onof
wat erif400gofst eam i sfoundt ohavecondensedt o
-1
wat er.(speci f
iclat
entheatoff usionofi ce336Jg ,
speci f
icheatcapaci t
yofwat er4. 2Jg-1k-1,speci
fic
-
1 -1
heatcapaci tyofalumi ni
um is900Jkg K.

4. a)St atetwof actorsthataffectt

heboil
ingpoi nt
ofal iquid.b)100gofal iquidatatemperatureof
10Ci spour edintoawel llaggedcal
ori
met er.An  I
tisobser
vedt
hatthewi
recut
sit
swayt
hrought
he
electricheaterrated50W isusedt oheatthe
i
ceblock,
butl
eavesi
tasonepi
ece.
li
quid.Thegr aphinthefigurebelowshowst he
variati
onoft het emperatureoftheli
qui
dwi th
ti
me.

Expl
anat
ion Evapor
ati
on

 Thewir
eexertspr
essur
eont hei
cebeneathi
t  Evaporat
ionistheprocessbywhichal
iqui
dchanges
andt
herefor
emakesitmeltatatemper
ature toagas.Itoccursatallt
emperat
ures(
i.
e.hasno
l
owerthanit
smelti
ngpoint
. fi
xedtemperature)

 Thewat erf
ormedbymel tedicefl
owsoverthe Factor
saffectingrateofevapor ati
on
wireandi mmedi atel
ysoli
difi
esandgivesout
l
atentheatoff usiontocopperwi rewhi tuses 1.Temper
chi ature
tomel ticebelowit.
Rateofevapor at
ionincreaseswi t
ht emperaturesince
 Not e:Copperwi r
eisusedi ntheexperi
ment i
ncreaseintemper atureincreaseskineti
cenergyof
because: mo l
eculesan dtheref
o r
e surfacemo lecul
es easi
l
y
escape.
I. I tisagoodconduct orofheat.
2.Surf
acear ea
I
I. I thashi gherthermalconductivi
tyt
hanother
met als.  Rat eofevapor ati
oni ncreaseswi t
hsurfacearea
becausemanymol eculesareexposedwhensur f
ace
Appl
icati
onsofef fectsofpressureonmel t
ingpoi
nt areaislarge.Wat erinbasin(a)evaporatesfaster
ofi
ce thantheonei n(b)i nthefigur
ebel ow.

I
. Iti
sappliedoniceskatinginwhichweightof
theskateractsonicethroughthi
nbladesof
skates.Thi
smel tst
heicet ofor
m afil
m of
wateronwhi chtheskatersli
des. 3.Dr
aught

I
I. I
tcanbeusedi
njoi
ningt
woi
cecubesunder  Draughti
ncr
easesrateofevaporat
ionsinceit
pr
essur
e. sweepsawayevaporati
ngmol ecul
escleari
ngaway

PHYSI
CSFORTECHNOLOGYANDI
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 232

Ef
fectofi
mpur
it
iesonmel
ti
ngpoi
nt formoremolecul
estoescape.Thi
siswhycl
othes
dryfast
eronawindyday.
 Impur
iti
eslowert
hemel
ti
ngpoi
ntsof
subst
ances. 4.Humi
dit
y

 Thisisappli
edinhumi
dareasduri
ngwint
eri
n  I
ncr
easeinhumidi
tylower
srat
eofevaporat
ion.Thi
s
whichsalti
sspreadonroadsandpat
hsto i
swhyclothest
akelongtodr
yonahumi dday.
preventf
reezi
ng.
Ef
fect
sofEvapor
ati
on
Ef
fectofi
mpur
it
iesonboi
l
ingpoi
nt
I
. Onefeelscoldwhenmet hylat
edspiri
tisappli
ed
 I
mpuriti
esrai
setheboil
i
ngpointsofsubst
ances. onhisheadaf t
ershavi
ng.Thisisbecausethe
Thisi
swhysaltedfoodcooksfast
erthan evapor
atingspir
itget
slatentheatofvapori
zati
on
unsal
tedone. fr
om hisbody.

 Consi
dert
heset
-upbel
ow. I
I. Thinl ayeroffrostfor
msar oundt heout si
deofa
testtubewal l
swhenai ri
sbl ownt hr
ough
met hyl at
edspiri
t.i
nthetube.Thisi sbecausethe
evapor at
ingspir
itobt
ainslatentheatof
vapor izati
onfrom wall
soft hetubecr eati
nga
coolingef fect
.

Di
ff
erencebet
weenEvapor
ati
onandBoi
l
ing
 Whent hel
iqui
dsar eheatedt
oboili
ng,t
he
boil
ingpoi
ntofsaltsolut
ioni
sobservedtobe Evapor
ati
on Boi
l
ing
highert
hanthatofdisti
ll
edwater
.Thisi
s
becauseimpuri
ti
esr ai
setheboil
i
ngpointofa I
ttakesplaceatal
l I
ttakesplaceatf
ixed
l
iquid. t
emper at
ures. t
emper at
ure.

Ef
fectofPr
essur
eonBoi
l
ingPoi
nt I
ttakespl
aceonthe I
ttakesplacet
hroughout
l
i
quidsurf
aceonl
y. t
heliqui
d.
 Decreasei
npressur
elower
stheboil
i
ngpoi
ntof
ali
quidwhil
eincr
easeinpr
essur
erai
sest
he Decreasi
ng Decreasi
ngatmospher
ic
boi
li
ngpoint. atmosphericpr
essur
e pressur
elowerst
he
i
ncreasestherateof boil
ingpoi
nt.
 Thi
siswhyfoodt
akeslongert
ocookathigh evaporat
ion.
al
ti
tudest
hanatl
owal t
it
udesbecausepr
essur
e
athi
ghalt
it
udei
shigher.

 Itisalsothereasonast owhywaterinasufuri
a Appl
i
cat
ionsofCool
i
ngbyEvapor
ati
on
closedwi t
hal i
dboi lsfast
ert
hantheclosed
one.Thest eam pr essurei
naclosedsufur
iais 1. Sweat
ing
higherandt hi
sr ai
sest heboi
l
ingpointofwater
 Evaporat
ingsweatabsorbsl
atentheatof
maki ngittoboilfaster.
vapori
zati
onfrom bodyandt
hereforecreat
inga
 cooli
ngeffect
.

2. Cool
i
ngofWat
eri
naPor
ousPot

 Wat
erseepi
ngoutoft
hepott
hroughpor
es

PHYSI
CSFORTECHNOLOGYANDI
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 232

evapor
atesf
rom sur
faceofpot
,cr
eat
ingacool
i
ng
eff
ect.

1. TheRef
ri
ger
ator 3 .4. St
atet
wofact
orst
hatwoul
drai
set
heboi
l
ingpoi
ntof
0
watert
oabove100C

5. a)Stat
ewhati
smeantbyt
het
erm speci
fi
clat
entheat
ofvapor
izat
ion

b)Inanexperi
menttodeter
minethespecif
iclat
entheatof
0
vapori
zat
ionofwater
,st
eam at100cwaspassedi ntowater
contai
nedinawell
-l
aggedcoppercalor
imeter
.Thefoll
owing
measurementsweremade:

 Massofcal
ori
met
er =50g

 I
nit
ialmassofwat
er =70g

 Fi
nalmassofcal
ori
met
er+wat
er+condensedst
eam
=123g

 Fi
nalt
emper
atur
eofmi
xt e=300C
ur

(
Speci
ficheatcapaci
tyofwater=4200Jkg-1Kandspeci
fi
c
heatcapaci
tyforcopper=390Jkg-1K-1)Det
ermi
net
he:

I
. Massofcondensedst
eam

 Itconsi
stsofahighlyvolati
l
eli
quid(Freon) I
I. Heatgai
nedbyt
hecal
ori
met
erandwat
er
whichtakeslat
entheatofvaporizat
ionfrom
content
s( f
ood)inrefr
igerat
orandevaporates. I
II
. GiventhatLi
sthespeci
fi
clat
entheatofevapor
ati
on
ofsteam
 Thepumpr emovesthevapori
ntol
owercoil
i
. Wri
teanexpr
essi
onf
ort
heheatgi
venoutby
outsi
dethecabi
netwhereiti
scompressedand
st
eam
changedtoli
qui
dform.
i
i
. Det
ermi
net
heval
ueofL.
 Theevaporat
orandcondenserpipesarehi
ghly
coi
ledtoincr
easesurfaceareaf
orabsorpt
ion 6.Aheat ingel
ementr ated2.5KW isusedt orai
sethe
andlossofheatrespecti
vel
y. temperatureof3.
0kgofwat erthrough500C.Calcul
atet
he
ti
mer equiredt
oaffectthis.(Specif
icheatcapaci
tyofwat
eri
s
 Thepipesar
emadeofgoodconduct orofheat 4200J/ kg/K)
(copper
)toi
ncr
easeheatconduct
ivi
ty.
7.St
atet
wof
act
orst
hataf
fectt
hemel
ti
ngpoi
ntofi
ce.
 Thecopperf
insenhanceheatlosstot
he
8.St 0gat1000ci
eam ofmass3. spassesi
ntowaterofmass
sur
roundi
ngatthecondenserpipe. 0
400gat10c.Thef i
nalt
emperat
ureoft
hemixt
ureisT.The

PHYSI
CSFORTECHNOLOGYANDI
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 232

Revi
sionExer
cise contai
nerabsorbsnegl i
gibl
eheat .(
Specif
iclatentheatof
vapori
zat
ionofst eam=2260kJ/ kg,speci
fi
cheatcapaci t
yof
1.Anel ectr
icheaterrated6000W i susedtoheat1kgof wat er=4200Jk-1)
iceini
tial
lyat- 100cunt i
lallthemasst ur
nst o st
eam.
Giventhatlatentheatoffusionofice=334kJkg-1,specifi
c I
. Der i
veanexpr essionfortheheatlostbythesteam as
heatcapaci tyofi ce= 2, -
1 -
1
260J kg K ,speci f
ic heat itcond ensestow aterattemperat
u r
e T.
-1 -1
capaci
tyofwat er=4,200Jkg K andl atentheatof
II
. Der i
veanexpr essionfortheheatgainedbyt hewat
er.
vapori
zationofwat er=2,260KJkg-1 K -1,calculatet he
minimum t i
mer equi
redforthisacti
vi
ty. I
II
. Det erminet hevalueofT.

2a)Expl ai
nwhyabur nfr
om thesteam ofboili
ngwateri
s 9.Acantogetherwi
thst ir
reroftotalheatcapaci
ty60j/
k
mor eseverethanthatofwateri
tself
? contai
ns200gofwat 0
erat10 c.dr yst 0
eam at100ci spassedin
whil
ethewaterissti
rreduntilthewholereachesat emperatur
e
b)Anener gysavingstovewhenbur ni
ngsteadil
yhasan
of300cCal
culatet
hemassofst eam condensed.
effi
ciencyof60%.Thest ovemelts0.03kgoficeat00ci
n
180seconds.Cal cul
ate;
- 10.Animmersionheat erwhichtakesacur r
entof3Af r
om 240V
mainsr
aisedthet
emperat
ur er300ct
eof10kgofwat o
I
. Thepowerr
ati
ngoft
hest
ove. 0
50c.Howl ongdi
ditt
ake?
I
I. Theheatener
gywast
edbyt
hest
ove.
11.100gofboi l
i
ngwat erarepouredi nt
oamet alvessel
weighing800gatat emperat eof200ci
ur fthef i
nal
0
temper at
ureis50c.Whati sthespecificheatcapacit
yof
3Ani mmer si
onheaterrated90W i
splacedinaliquidof themet al?(Specif
icHeatcapacityofwat er4.2x
mass2kg.Whent heheat eri
sswit
chedonf or15mi nutes, 103J/kg/k)
0 0
thetemperatur
eoftheliquidri
sesfr
om 20Ct o30C.
Determinethespeci
fi
cheatcapacityoftheli
quid. 12.0.02kgofi ceand0. 01kgofwater00car einacont ai
ner
.
0
Steam at100ci spassedinuntilal
ltheiceisjustmelted.
Howmuchwat eri
snowi nthecont ai
ner?

13.I
nadomest i
coil-
fi
redboi
ler
,0.5kgofwat erfl
owst hrough
theboilereverysecond.Thewat erent
erstheboilerata
temper atureof300candleavesatat emperat eof700c,
ur re
-ent
eringt heboil
ersaft
erfl
owi ngaroundtheradiatorsat
300c.3. 0x107Jofheatisgivent othewaterbyeach
kil
ogram ofoi lburnt
.Thespeci f
icheatcapacit
yofwat eris
-
1 -1
4200Jkg K.

I
. Uset hei
nformati
onabovetocalculat
ethe
energyabsorbedbythewatereverysecondasi
t
passesthroughtheboi
ler
.

I
I. Usethesamei nf
ormati
onabovetocal
cul
atet
he
massofoilwhichwouldneedtobebur
nti
norder
topr
ovidethi
senergy.

14.Youarepr ovidedwi t
htwobeaker s.Thefirstbeaker
0
contai
nshotwat erat70c.Thesecondbeaker
contai
nscol dwat erat200c.Themassofhotwat eri
s
thr
icethatofcoldwat er.Thecont ent
sofbot h
beakersaremi xed.Whati sthetemper at
ureofthe

PHYSI
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 232

mi
xtur
e?

15.Calcul
atetheheatevol
vedwhen100gofcopperar
e
0 0
cooledfr
om 90ct o10c.(Speci
fi
cHeatCapaci
tyof
Copper=390J/ Kg/
k).

16.An-immersi
onheaterr
ated150wi splacedinali
qui
d
ofmass5kg.Whent heheateri
sswi t
chedonf or25
minutes,t
hetemperat
ureoftheli
quidri
sesf r
om 20-
0
270c.Det er
minethespecif
icheatcapacit
yoftheli
qui
d.
(Assumenoheatlosses)

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