SOLUTIONS TO PAST YEAR PAPERS
2002
-
20 1 2
TOPICAL APPROACH
•
A true step by step guide to questions
• • •
10 year series Detailed answers that guarantees the mark Yearly question analysis table provided for Papers 1 to 3
•
Descriptive explanation to understand the Physics phenomena, concept and theory.
FAIROZ AHAMED BookOne
ABOUT THIS GUIDE BOOK:
T h e a n s w e r s i n t h i s b o o k , f o r m o st pa rt of it w i l l g o b e y o n d w h at i s n e c e s sa ry t o g et t h e d e s e rv e d ma r k . T h i s i s d e l i b e rate l y d o n e s o f o r t h e b e n ef i t of st u d e nts
w h o m i g ht f i n d r e ca l l i n g and ret ra c i n g c o n c e pt s d iff i c u lt . T h e a n s w e rs a r e w r itte n i n p o i nt f o r m f o r t h e p u r p o s e o f c l a r ity l i ke t h i s l i st a n d at t i m e s a mat h e mat i ca l a p p r oac h i s g i v e n . S t u d e nts s h o u l d b e a b l e t o w r it e t h e m d e s c r i pt i v e l y i n pa rag ra p h f o r m a s w e l l , e s p e c ia l l y w h e n r e q u i re d t o d o s o b y t h e q u e st i o n . I n a way, t h i s b o o k att e m pt s t o a n s w e r t h e q u e st i o n l i k e g o i n g aft e r a f l y w i t h a baz o o ka . T h e Baz o o ka d o e s its j o b , b ut its e x p e n s i v e . O n ce y o u have l ea r nt h o w
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t o a p p r oa c h t h e q u e st i o n s , y o u may b r i n g a b o ut y o u r o w n f l y s watt e r t o t h e t e st ! I n q u e st i o n s w h e re m u l t i p l e a p p roac h e s a r e p o s s i b l e , t h e r e l e va n t m et h o d s of a n s w e r i n g a r e c l ea r l y stat e d . A y ea r l y pap e r q u e st i o n mat r i x i s p ro v i d e d t o f i n d t h e c o r r e s p o n d i n g to p i c of t h e q u e st i o n . V i s it t h e w e b s it e www. c i k g u-a h a m e d . c o m f o r e d it i o n s , P D F a n d v i d e o n ot e s etc FAIROZ AHAMED BIN PEER MOHAMMED BIN SHEIKH MOHAMMED BIN ADAM SAHIB WWW.CIKGU-AHAMED.COM
TYPES OF QUESTIONS
S i m p l e r e ca l l of d ef i n it i o n s a n d e q u at i o n s . S u b st i t ut i n g r e l e va n t va l u e s i nto
e q uat i o n s . G ra p h i ca l r e p r e s e ntat i o n of e q uat i o n s a n d w h e n pa ra m ete r s i n t h e e q u at i o n a r e m o d if i e d . M u lt i p l e s c e n a r i o q u e sti o n s ( u s ua l l y t w o ) . I n t h i s ty p e y o u m i g ht e n d u p g ett i n g two e q uat i o n s w it h u n k n o w n s t hat ca n c e l o ut b y s i m p l e a d d i t i o n o r d i v i s i o n of t h e e q uat i o n s . Q u e st i o n s t hat t e st y o u o n g e n e ra l u n d e rsta n d i n g of t h e p h y s i ca l w o r l d , s u c h as t h o s e on c o n s e r vat i o n of m o m e nt u m , e n e rg y, mas s et c . Oth e rs !
Abbreviations used
Their meanings Re m e m b e r & Reca l l
R & R
S k etc h e d i n 30
3D
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SID E TOP
As s e e n from the t o p
NlL
N e wto n ' s 1st l a w
N2L
N e w t o n ' s 2"d l a w
N3L
N ewto n ' s 3rd l a w
FLHR
F l e m i n g ' s l eft h a n d r u l e
FRHR
F l e m i n g ' s r i g ht h a n d r u l e
RHGR
R i g ht h a n d g r i p r u l e
COG
C e n t re of g ra v ity
COM
C e n t r e of m a s s
LL
B
Lenz's law
G Field
G r a v itat i o n a l f i e l d
E Field
E l e ct r i c f i e l d
Field I H Field
M a g n et i c f i e l d
PCOM
P r i n c i p l e of c o n s e rv at i o n of m om e n t u m
PCO E
A A AND
DATA
A s s e e n f ro m t h e s i d e
ORM
P r i n c i p l e of c o n s e rv at i o n of e ne r g y
A =
Speed of light in free space
-
c
Permeability of free space
--+
µ0
Permittivity of free space
-
£0
=
=
3.00 4 'If
x
x
10-7 H m-1 x
8.85
(1/(36'If) ) x
Elementary charge
-
e
1. 60
The Planck constant
-
h
6. 63
Unified atomic mass constant
-
Rest mass of electron
-
me
9.11
Rest mass of proton
-
m
1.67
Molar gas constant
--+
p
u
R
1. 66
=
8.
108 m s-1
x
x
10-12 F m-1
x
10-9 F m-1
1 o-19 c 1 o-34 Js 1 o-27 kg
x
x
10-31 kg
10-27 kg
31 J K-1 moz-1
the Avogad ro constant
---+
The Boltzmann constant
---+
Gravitational constant
---+
Acceleration of free fall
---+
Uniformly accelerated motion
---+
NA
G
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---+
Hyd rostatic pressure
---+
Gravitational potential
---+
Displacement of particle in s.h.m.
---+
Velocity of particle in s.h.m.
---+
Mean kinetic energy of a molecule of an id eal gas
---+
Resistors in series
---+
Resistors in parallel
---+
electric potential
---+
alternating current/voltage
---+
transmission coefficient
---+
s
6. 02
1. 38
6. 67
9.81
g
F or mulae
Work d one on/by a gas
=
k
=
=
v2
ut
u2
---+
d ecay constant
---+
-
2
+
2 as
Gm
</J
r
X O O
sin co s
E
3
kT
R
RI
x
V
v
±w 2
1
R
j
Vx � +
x
A
(J) t
(J) t
2 - x
R2
1 RI
+
Q
v
T
2
at
pg h
p
x
-
m s- 2 1
+
+ ....
1 R2
4 'If£0r ex:
x0
sin wt
exp(-2kd)
w h ere k
radioactive d ecay
x
p.J v
W
x
1023 mo[-I 10-23 J K-I 10-11 N m2 kg 2
x
=
B'lf'm( U
E)
h2
x0 exp(-A,t)
0. 693
+ ....
No. Theme I 1
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3
4
5
.
age
MEASUREMENT
1.
2
op1c
M e a s u r e m e nt
1
NE WTONIAN MECHANICS
2.
K i n e m at i c s
3.
D y n a m i cs
4.
Fo rces
5.
W o r k , E n e rg y & P o w e r
6.
M ot i o n i n a C i rc l e
7.
G r a v itat i o n a l F i e l d
8.
O s c i l l at i o n s
12
33
60
85
98
113 142
THERMAL PHYSICS
9.
T h e r m a l P h y s i cs
174
WAVES
10 .
W a v e M ot i o n
11 .
S u p e r p o s it i o n
218
229
ELECTRICITY & MAGNETISM
12 .
E l e ct r i c F i e l d
13.
C u r r e nt of E l e ct r i c ity
14 .
D . C . C i rc u its
15 .
E l e ct ro m a g n et i s m
16.
E l e ct ro m a g n et i c I n d u ct i o n
17 .
A lt e r n at i n g C u r r e nts
261 28 8
3 02
3 24 3 48
3 70
I
No. Theme I Topic 6
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I
MODERN PHYSICS 1 8.
Q u a nt u m P h y s i c s
20.
N u cl e a r P h y s i cs
19.
7
Page
L a s e rs & S e m i co n d u ct o r s
381 404
411
OTHERS 2 1.
22.
D ata A n a l y s i s Design & Pla n n i n g
441 478
1. MEASU REMENT
8.
N06. 1 .2 -
PAPER 1
D
Note : The choice of instrument depends on range and resolution desired . I n a way it is common sense to use an i nstrument that can best accommodate the dimension to be measured. w - vernier caliper preferred over the half meter rule x - micrometer screw gauge preferred 9.
N07. 1 . 1 - C
minimum
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,---"----..
1 0.
11.
volume = lOO cm x 2.5 cm x 0.5 cm = 1500cm3
Note : It is a meter rule, therefore the length has to be at least 1 00 cm. That narrows the options. The other dimensions need just a bit of imagination. N07. 1 .2 - B
the equation or formula :
(�J 1 A )2 = x ( A )2 = !_ x ( A ) = v;J x ( 1
P = K/2
�P
K
in units :
N08. 1 .2 -
/2
L
L
/
V I
L
Note : Simplify to the simplest of units that you know. Since its got to do with electricity recall equations for voltage, cu rrent, power and resistance.
D
tonnes speed
A.
B.
c.
1 2.......-"----. ........, KEofbusonthe = - x 2000 x 50 = 2,500,000 2 freeway power of light bulb = SW to lOOW {energy saving to the filament ) temperature of hot oven = 120° = 120 + 273.15 = 393 .15K Area of circle
(-t5 )2 x 0 .2 = 0 .03929 m3
,------A-
D.
volume of tyre = re x
o
Note : The diameter and width of the tyre was just an estimate but yielded a good result. Do note that the tyre looks more like a ''torus" (donut) and not a cyli nder which means its got to be less than our calculation with its center part removed . And the temperature given in option C was room temperature. 4
05m
0.2 m
© BOOK ONE PTE LTD
PAPER 1
2. KIN EMATICS
9.
N0? . 1 .3 - A
we know ,and taking t as positive : v = u + at = v - 9.8 1 x t
/
Linear relationship With negative gradient
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1 0.
11.
Note : Gradient of sit graph gives velocity, while gradient of v/t graphs gives acceleration. In this case a = g In this motion a = g = 9.81 and is constant throughout the motion of the ball. This implies that the gradient of the v/t graph is negative and is constant. Note that when velocity becomes zero, that is when & where the ball is at its maximum height.
NOB. 1 . 1 - A Note : Definition of velocity as rate of change of d isplacement with respect to time. Distance is a scalar so B & C is not correct for velocity D is not the appropriate definition for velocity, though implied
NOB. 1 .4 - C
AO
t I I I I I I I I I I
+
. . . BQ.}f cQ
i8m
Analysing motion from Ato B: s= u =O V= a = 9 . 81 t=l using : v = u + at v = 0 + 9.8 1 x l = 9.81
Note : This is a two stage s, u , v, a, t analysis as we have two journeys here. Two difference accelerations hence two different jou rneys. And we are taking down as positive The final velocity acquired in the motion from A to B becomes the initial velocity during its motion from B to C . We are assuming that the deceleration on the ball is u niform for the duration of the "stopping" and that is why the question asks for average acceleration.
Analysing motion from B to C : s = 8m u = 9.81 v=O a= t= using : v 2 = u 2 + 2as 02 = 9.8! 2 + 2 x a x 8m 2 a = 9 路 81 = 6014.75 m/2 Is 2 x 8m = 6 x l03 2
J-:
漏 BOOK ONE PTE LTD
15
2. K I N E MATICS
PAPER 3
c. ii.
max GPE that should
energy lost to air resistance
have been reached
- GPE that was reached
initial KE of ball
initial KE of ball
initial KE that should
=
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=
_
- GPE that was reached max GPE to converted have ------initial KE of ball
initial KE of ball - GPE that was reached initial KE of ball
lxmx252 - mxgx18.75 m(lx252 - gx18.75) 12 2 -21 xmx252 -xmx25 1 252 - x18.75 l2 x252 =0.4 114 =0.4 1 =
-X
g
- �2�-----
d. i.
tangent drawn on graph at v= I 0,
( 0, 21.5) (1.6, 0) 21.-5 -0 =-13.4375 = - 13 m%2 = 0 1.6 101% &
two extream points on tangent are : acceleration at v=
ii.
101% =
gradient of v/t graph at v=
+
t
v
s
taking t as + ve :
L FY = ma
mg + FR = m x a w
Also shown on graph next page!
( 0.350x (-13.4375 - (-9. 8 1)) = - 1.269625 = - 1.27
FR = ma - mg = m a - g ) =
air resistance is
1.27
N
N downwards.
Note : Draw the FBD to get a visual on the direction of the various vectors We took up as positive hence the negative on the air resistance is expected. This is one way of making sure that the values obtained validates our u nderstanding.
28
© BOOK ONE PTE LTD
PAPER 3
1 2. ELECTRIC FIELD
ii .
cons i d eri ng we i g h t
Ignori ng we i g h t
at the point of exit
at the point of exit
v tan (J = -1'... vx 6 . 1 53 x l0 -3 x !l + 1 .006 x l0 -7
6 . 1 53 x l0 -3 x !l tan 4.3 = q
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m
e.
=
3 .9 x l0 6
m_
_____
tan 43 x 3 .9 x l0 6 ----6 . 1 5 3 x 10 -3
tan 4.3 =
m
-------
/ kg
m
i.
""'/!°:::7:::' .
FE
t
{ /kg
!l = 4.77 x 10 1 C
9__ = 4 .77 x l0 7 C/
m
m
3 .9 x l0 6 q tan 43 x 3 .9 x l0 6 - 1 .006 x l0 -7 = 6 . 1 5 3 x l0 -3
So the weight was really insignificant!
� Phospho-luminance screen or detector Vertical and perpendicular to plates
v
1�
'""'" (<po<) w"°' • U off V=O
0
�arget (spot) when v is on V=3000V
d
M ount a p h osp h o- l u m i nance screen or some f rom of a d etector th at can i ll u m i nate the po i nt w h ere t h e e l ectron stream stri kes, perpen d i cu l ar to t h e p l ates, at a d i stance ' d ' from t h e center of th e p l ates. C on d uct two experi ments, one wi t h th e vo ltage at zero and th e ot h er w i t h t h e vo ltage of th e p l ates turned up. In both cases, mark t h e i mpact po i nts on th e screen . Th i s d i mens i on g i ves th e h e i g ht ' h '. App ly s i mp l e tri gonometry to get th eta, t h e ang l e of d efl ecti on. O ne may c h oose to i gnore t h e we i g ht as i t h as b een foun d to b e i ns i gn i fi cant.
h tan O = d N ote : Exp l a i n th e set up and t h e mat h b e h i n d t h i s We h ave to assume th at t h e tri ang l e set up d ue to th e d ef l ecti on starts at t h e center of t h e p l ates.
278
© BOOK ONE PTE LTD
1 3. CURRENT OF ELECTRICITY
1.
PAPER 1 , MCQ
N02. 1 .21 - D
For a thermistor T t, R ,!,
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2.
3.
For an LDR R ,!, Cd t , ( intensity )
Note : Recall the characteristics of these semiconductor devices.
N02. 1 .22 - A
y
Side view
x
y
Top view
)
•
x
1x1a' an/s
time taken for a 1 l=Q 1 cm trace to be drawn 1 x 104 t now we find how much charges arri es during this time 50µ = n x e x � 04 1 1 -x 50µ 50µ x -1 x 104 = 3.l25 x 1010 n = --�1x�l=O-4 = 1 .6 x l0- 19 e
-
C )
N03 . 1 . 1 9 - B
x
100 = 230 x / I = lOO = 0.4347 230
V = IR 230 = lOO x i 230 2 R = 230 = 529 100
y
60 = 230 x / 60 = 0.2608 1=230
60 x / 230 = 230 2 R = 230 = 881 . 6 60
P = Vl
© BOOK ONE PTE LTD
OR
y2 P = Vl = R V =constant if Pt
It
R ,!,
289
1 3. CURRENT OF ELECTRICITY
5.
PAPER 1 , MCQ
as V is t, I t, but T t, :. R t, which results in I :. Resistance should be seen increasing for non-ohmic conductors V=IR R= VI
N04. 1 .22 - B
.J...
v
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0
6.
7.
I
Note : Resistance is a function of V & I and not the gradient at that point. If you wish to see it in terms of gradient, then think of it as gradient of the line OX, where 0 is the origin and X is the point on the VI graph . That gradient should increase if R is to increase.
N04. 1 .23 - D
EMF = pd across r + p d across R EMF = V, + VR EMF = Ir + IR E = 3xr+3x1 = 3r + 3 <1 1!] E=2xr+2x2=2r+4 <I � m - (%x �) E -[% x E] = 3r + 3 -[% x (2r + 4)] _!_E2 = 3r+3 - 3r - 6 E=6
Note : The internal resistance is in series with the load , so the sum of the potential drops across them should equal to the E M F
_
N05. 1 .22 - A R
R= p Al whene &Aareconstant R= pA xi so as l t ,R t linearly if Ais J.. in steps, pA tin steps (gradient of RI graph) Note : This is more of a math question. Addition of two linear functions as one approaches a constant value. The graphs for the individual lengths can be superimposed .
0 © BOOK ONE PTE LTD
d 291
PAPER 1 , MCQ
13. CURRENT OF ELECTRICITY
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1 6.
1 7.
1 8.
N 1 0. 1 .27 - C
E = V, +VR E = IR, + IRR exptl E = 6x R, + 6 x l .2 E = 6R, + 7 .2 <l [!] expt 2 E = 5 x R, + 5 x 1 .6 E = 5R, + 8 <l rn 0 = 6R, + 7.2 - ( 5R, + 8 ) R, -0.8 = 0 R, = 0.80
1-
I I I
I I
: I I
L
- -------------, I I I
-
E
r
: I I
E
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
I A --+--...°""
R
B
A
c
Note : You will get a pair of simultaneous equation from the two experiments. This is a common question type.
E = 5 x 0.8 + 8 = 12V N 1 0. 1 .28 - C
V = IR R = V = _i_ = l .714 = 1 .70 I 3.5
Note : Resistance is a function of V & I and not the gradient at that point. If you wish to see it in terms of gradient, then think of it as gradient of the l ine OX, where 0 is the origin and X is the poi nt on the VI graph.
N 1 1 . 1 .25 - B Current. A It's a measure of the rate of flow of electric charge Q through a given cross-section of the conductor. E . M . F. V Between the terminals of a voltage sou rce is the electrical energy dissipated per unit charge in brining that charge from one terminal to the other via the external circuit. This is equal to the sum of the potential drops across the components in the circuit
WD = W emf = Q D
WD emf = Q
Stick to the known and agreed definitions of the terms.
pd across R = I x R = QxR t Š BOOK ONE PTE LTD
295
PAPER 3
1 8. QUANTUM PHYSICS
a. For electromagnetic radiation of wavelength , ;., , which exhibits particle behaviour, it will have an associated momentu m, p, given by
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p = Ah = mP vP where : mP = mass of particle vP = velocity of particle
b.
Note : Recall deBroglie's principle.
l0-34 = 2 .7625x 10-24 : . Petectron = !!_A = 62.63x .4 x l0-10 = 2 .76x l0-24 Ns the particle is an electron p= h =mv
A
ee
h =mv
This is the velocity this particle with that mass (me) should have to display the same diffraction pattern as that of a wave of wavelength A.
A ee h ve = - Ame 1 : . KEetectron = 2 x me x v� = _!_2 x me x
h )1 (Ame
lossin EPE = gain in KE U = KEelectron
()
Particle is accelerated in an E-field by electrodes that are held at a potential difference of V
(
)
6 .63 x 10_34 2 1 v - -1- x !!_ 2 x 2me e A 2x9 . l lx l0 21 x l .6x l0 19 2 .4 x 10-10 = 26 . 17798 = 26 .2 V Note : Recall 'Electric Field ' An electron travels from a region of lower potential to a region of higher potential opposite to the electric field direction.
© BOOK ONE PTE LTD
401
PAPER 1
20. NUCLEAR PHYSICS
1.
N02. 1 .30
-
C
Note : Most a-particles should pass through, evidence that most of the atom is empty space. Therefore at 0°, a much higher cou nt should be observed relative to what is obtained at other angles. There should be no other peaks at any other angle. Decreasing cou nt as angle is increased. Some will be repelled or bou nce back. Therefore, some cou nt is registered even at very large angles. 2.
N03. 1 .29
-
D
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a-
3.
particle = helium nucleus (positively charged) f3 - particle = electron (negatively charged) y - radiation = a photon (of EM radiation)
N03. 1 .30
-
D
jR & R I
we know :
A = AN Method 2
Method 1
using :
wiriting known values : fior X : }.,x = ln 2 24 ln 2 fior Y : l 16 "Y
_ _
:. total activity Ax + Ay ln 2 8 ln 2 = A e- 24 x48 + A e- 16 x4 =
[ =A e o
o
o
ln 2 - -x 48 24
+e
ln 2 - -x 48
16
)
48 years corresponds to : 24x n = 48 � n = 48 = 2half lives for X 24 16x n = 48 � n = 48 = 3half lives for Y 16 :. total activity = Ax + Ay
3 0 = A0 (0.25 +0.125) = -A 8
(�J + A0 (�J ((H + (HJ = A0 ( + _!_ ) = A° 4 8 8
= A0 =
Note : Total activity recorded would be that due to that of X and Y. Method 2 will work even when n is not a whole nu mber.
412
A.
_!_
�
© BOOK ONE PTE LTD
PAPER 2
20. N UCLEAR PHYSICS
a. i. Isotopes Atoms of the same element having the same number of electrons and protons but different number of neutrons [or] Atoms of the same element that has different mass number (Nucleon number)
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ii. Fission It is the disintegration of a heavy nucleus into lighter nuclei of comparable size with the release of energy and some other particles. The daughter nuclei possess a greater binding energy per nucleon therefore being more stable) . (higher mass defect) iii. Half Life The half-life for a given radioisotope is the time taken for half the radioactive nuclei in any sample to u ndergo radioactive decay. b. Method 1 of 3
using A = A0 e-.A.t to find time taken to go from A0 � � x
Note : Using the equation directly. Here, the time t will measure the duration .
= A e .A.t x o -1 = e Ao
-
-At
x x- 1 = e-.A.t ln x-1 = ln e--'1 -lx ln x = -A,t x ln e
ln x = A-t ln x t=A, A0 . A0 �:. to find time taken to go from 20 1 40 M =t� -t� 140
20
A,,
20
A_ 1 40
t,.,,_ 20
t� 140
( )
ln 140 = ? x l08 ( In 7) = l .965 1 x l09 = ln l40 - ln 20 = _!_(ln 140 - ln 20) = -ln12 20 ln 2 A, A, A, 7 x l08 = l .97 x l09 yrs
422
© BOOK ONE PTE LTD
DATA ANALYS IS
PAPER 2
e.
we know : P = Vl 75m = 0.5 x l I = 75m = 0.15 A 0.5
I
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To the external_______, circuit or load
Note : A current draw of 0 . 1 5 A at a potential difference of 0.5 V results in a power of 75 mW. The maximum current drawn is 0. 1 5 A.
i. P = 1 50 mW
&
p.d = 1 .0 v
power in circuit = P = VI = lx 0.15 = 150mW ii. P = 1 50 mW
&
Note : P . D is doubled to 1 .0 V hence 2 cells arranged in series Current drawn is 0 . 1 5 A in the branch so that each cell gives a power of 75 mV. Therefore, total power in circuit is: OR
power in circuit = P = 0.15 x 0.5 + 0.15 x 0.5 = 0.15 x 0.5 x 2 = 150mW
p.d = 0.5 v Note : P . D is 0.5 V. therefore across the terminals only 0.5 V should be seen. Current drawn is 0. 1 5 A in each branch so that each cell gives a power of 75 mV. Therefore, total power in circuit is stil l the same at 1 50 mW
ii. P = 300 mW
&
p.d = 1 .0 v Note : P . D is doubled to 1 .0 V hence 2 cells arranged in series Current drawn is 0 . 1 5 A in the branch so that each cell gives a power of 75 mV. Therefore, total power in circuit is:
power in circuit = P = (0.15 x 0.5 ) x 4 = 300mW 452
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DATA ANALYS IS
d.
PAPER 2
4.5
i
'
I
the point : x = O, 0.5 y =4+l9x 25 i .e. (0, 4.38)
l - ·
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4.0
·-
·-
'
w
-·
'
3.5
'
I
'
m
3.0
'
-
i.
I I
.
1 0.0
5.0
0
'
1 5.0
the point : x = 25.0 , 0.5 y =2.5 + 10.5 x 25 i .e . (25.0 , 2.7 1)
I�/ II I
2.5
470
-
'
20.0 xi cm 25.0
with room temperature = 20° C X
(}
5.0
77
(}E
14.04305 = 4.041
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PAPER 2
PLANN I N G
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Pu rpose and approach The purpose of this experiment layout and procedure is to find the efficiency of the solar panel model being provided. The efficiency of the solar panel water heater can be measu red by computing how much of the incident light energy is harnessed and converted into thermal energy in the flowing water. The incident power can be calculated using equation 2 and the power gained by the water can be calculated using equation 4.
Additional apparatus needed and their purpose 2 thermometers. One to measure the water temperature at the inlet and the other at the outlet A large beaker (high capacity) . To find the volume of water from the outlet, that got heated up. 1 stopwatch 1 power supply unit with variable voltage selector for the lamp. It should display voltage and current. Otherwise would need a voltmeter and an ammeter for voltage and current measurement respectively. A retort stand . To hold the lamp at fixed height, centered above the panel A measuring tape. To measu re the height of the lamp from the panel's surface. A piece of paper with a grid (of l ines) draw on it that fits the panel 's su rface An electronic balance. To measure the mass of water collected from the outlet.
The experiment setup
lamp
power supply
solar panel
ďż˝ / water outlet
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water
thermometer
/ water inlet
479
PLANN I N G
PAPER 2
Obtaining the frequency
.
: T ------•••!. ·-·····! 1-·····-····-·-····-···
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.
A
B
c
D
The above trace (or similar) should be observed in the oscilloscope when positioned as shown in the diagram and the l ig ht beam of the photogate closer to the minimum point but not at it. 'A' represents the wire cutting the l ight on its way down. 'B' is registered when the wire is on its way back up after reaching the minimum poi nt. 'C' is registered on its way down again and 'D' on its way u p again . A & C are registered when the wire is in the same state of vibration and hence the duration between these two dips in the signal would be the time taken for one complete wave. Using the time base scale, measure the d u ration between A & C or B & D to give the period of the wave. Use equation 2 to compute the frequency.
Analysis By plotting the graph of lg f vs lg µ as per equation 1 , a straight line is expected. The graph has to be linear for the equation to be valid.
Equations
the governing equation : f = kµ n lg f = n lg µ + lg k lg f = lg k + n lg µ where n & k are constants f =frequency of that mode µ= mass per unit lenght of wire plating lg f aginst lg µ will result in a straight line with gradient n and vertical intercept lg k frequency : frequency = f
484
=
1
1 = period .
T
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