REDS POT
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SECTION I: MEASUREMENT Topic 1
Measuremen t
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SECTION II: NEWTONIAN MECHANICS Topic 2 Topic3
Topic4
TopicS
Topic6
Topic 7 Topic8
Kinematics D ynamics F orces
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Work, Energy and Power
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Motion in a Circle
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Gravitational Field
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0sci11at ion s
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SECTION III: THERMAL PHYSICS Topic 9
Topic 11
Topic 13
Topic 14
Topic 15 Topic 16
Topic 17
Topic 19
Topic 20
49
65 78
87
98
Wave Motion
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126
S uperpos1tlon
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EI ectric Fields
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Current of Electrici ty D. C. Circuits
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Electromagnetism
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Electromagnetic Induction Alternating Currents
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SECTION VI: MODERN PHYSICS Topic 18
36
I 09
SECTION V: ELECTRICITY AND MAGNETISM Topic 12
2I
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Thermal Physics
SECTION IV: WAVES Topic 10
5
Quantum Physics
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Lasers and Semiconductors Nuclear Physics
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139
I 57
175
189
208 223 235
244
257
266
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l
Topic
7.
Measurement
Freq uently Exami ned
Which pair contains one vecto r and one scalar quantity?
A B
force : kinetic energy
C
momentu m : velocity
D
displacement : accel e ration
C
the Young mod u l u s
D
weight
Units in
Vectors have directions while scalars do not. A: both q uantities are vecto rs
= kg ms-2 B = kg mss
Units in C =
force is a vector while energy is a scalar
kg ms2 m
Units in D = kg ms-
C : both q uantities are vecto rs
D:
� x m3 x ms- 1 = kg ms- 1
k
m
1
H elping concept!'
B:
x area
Helping concepnUn its in A =
powe r : speed
Questions
2
x
2 2 m = kg ms-
2
both q uantities are scalars
8.
What is the ratio
A 10-3 c 10-12
11.
1 .um ? . 1 Gm
B 10-9
D
Express the joule in terms of the base units of mass, length and time, the kg, m and s.
10-1s
H elping concept'!' 1pm 1 Gm
1x10-5 m = 10 - 1s 1x109 m
The u n it of work , the joule, may be defined as the work done when the point of application of a force 1 newton is moved a distance of 1 metre in the direction of the force.
A
kg m-1 s 2
C
kg m 2 s -1
H elping concepn=>
W = F x d [ W] = [F][d] J = [ma][d] = kg m s-2 m = kg m 2 s -2 =>
9.
Which experimental tech nique reduces the sys tematic error of the q uantity being investigated?
A
adjusting an ammeter to remove its zero er ro r before measu ring a cu rrent
B
measuring several i nternodal distance on a stan ding wave to find the mean internodal distance
C
measuring the diameter of a wire repeatedly and calculating the average
D
timing a large n u mber of oscillations to find a period
12.
The resistance of an electrical component is mea sured. The following meter readings are obtained.
0
H e lping concepnA zero error is a systematic error.
What is the resistance?
2.5n
A c 10.
Which quantity has different u n its from the other t h ree?
A B 1000
density
x volume x velocity
rate of change of momentum
Pb�sics 'Jlf/ci
with HELPS
B
2500Q
D
2.7n
2700n
Helping concepnR
= '{_I
=
200 mV 0.48 A
2_5n
page 7
Topic l
Measurement
Frequently Exa m i ned
1 3. What is the reading shown on this milliammeter? 4
A
C
6
1 .0 mm 1 .038 mm
Ques!tions
1 .00 mm 1 .04 mm
B D
H e lping c,onceptl-'
Mean measurement = 1 · 02 x 4 + 1 · 01 = 1 .02 mm 5 Value of d= 1 .02 - (-0.02) = 1 .04 mm
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A
C
2.35 mA 3.4 mA
B D
2.7 mA 3.7 mA
1 6. A student makes measurements from which i;he
calculates the speed of sound as 327.66 m s-1 • She estimates that her result is accurate to ± �1%. Which of the following gives her result expres !led to the appropriate number of significant figu ms?
H e lping c,onceptl-'
Each interval is 4 - 2 = 0.4 mA. 5 For 3.5 interval, its value is 3.5 x 0.4 = 1 .4 mA.
Measured reading is 2 + 1 .4 = 3.4 mA.
1 4. Errors in measurement may be either systematic
or random. Which of the following involves random error? A not allowing for zero error on a moving-coil voltmeter B not subtracting background count rate when determining the count rate from a radioactive source C stopping a stopwatch at the end of a race D using the value of g as 1 0 N kg- 1 when cal culating weight from mass
A C
327.7 ms-1 330 ms-1
328 ms- 1 300 ms-1
B D
H e lping c,oncept:r
Value of error = 327.66 x 3% = 9.83 ms- 1 "'1 O ms- 1 (1 sig. fig.) Speed of sound = 330 ms- 1 (rounded to the same t uns place as the error)
1 7. A metal sphere of radius r is dropped into a tank
of water. As it sinks at speed v, it experiences a drag force F given by F krv, where k is a con stant. What are the SI units of k? kg m 2s- 1 B kg m- 2s- 2 A =
H e l p i n g c,onceptl-'
Random error arises when measured value is higher or lower than its true value.
C
kg m - 1 s- 1
D
kg ms- 2
Helping c,onceptl-'
15. A micrometer, reading to ±0.01 mm, gives the fol lowing results when used to measure the diam eter d of a uniform wire : 1 .02 m m 1 .02 m m 1 .01 m m 1 .02 mm 1 .02 mm When the wire is removed and the jaws are closed, a reading of -0.02 mm is obta in ed. Which of the following gives the value of d with a precision appropriate to the micrometer? 1000
Pb"9sics
'fflc i with HELPS
F = krv =>
[k ] =
=>
k
= f_ rv
2 n = kg ms- = kg m- l s -1 [r][v] m -ms 1
1 8. An Olympic athlete of mass 80 kg competes in a
1 00 m race.
page 8
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Topic 1
Measurement
Frequently Exa m i ned
What is the best estimate of his mean kinetic en ergy during the race? A B 4 x 1 03 J 4 x 1 02 J c D 4 x 1 05 J 4 x 1 04 J H e lping concept"l-'
Estimate time to complete 1 00 m = 1 O s Average speed = 1 OO m = 1 O ms- 1 10s Average K.E. = lmv 2 = 1(80)(1 0) 2 = 4 x 1 0 3 J 2 2
Questions
Which properties apply to ra ndom errors? B P, , 02 , R2 A P, , O,, R2 D P2 , 0, , R, C P2 , 02 , R, H e l p i n g concept"l-'
systematic error random error P2 Pi 01 02 R1 R2
2 1 . Argon and neon are monatomic gases. One mole 1 9. In a simple electrical circuit, the current in a re
sistor is measured as (2.50 ± 0.05) mA. The resis tor is marked as having a value of 4.7 n ± 2%. If these values were used to calculate the power dissipated in the resistor, what would be the per centage uncertainty in the value obtained? A 2% B 4% c 6% D 8% H e l p i n g concept"l-'
of argon has mass 40 g and one mole of neon has mass 20 g. What is the ratio no. of atoms in 1 mole of argon ?· no. of atoms in 1 mole of neon A always 1 B always 2 C 1 , only if both gases are at the same tem perature and pressure D 2, only if both gases are at the same tem perature and pressure
Power P i s related t o current I i n the resistor of resis tance R by P = 12R
The number of atoms in one mole of any monatomic gas is always the Avogadro number (6.02 x 1 02 3).
= o.o5 2% I 2.50 R = 4. 7 n ± 2% => ti.R = 2% R . . Percentage uncertainty in power P is thus
22. An i nstrument g i ves a n u m e rical read ing of
Given I = 2.50 ± 0.05 mA =>
M
=
= 2 M + ti.R = 2(2%) + 2% = 6% I P R
jj.p
20. When comparing systematic and random errors,
the following pairs of properties of errors in an experimental measurement may be contrasted : P, : error can possibly be eliminated P2 : error cannot possibly be eliminated o,: error is of constant sign and magnitude 02 : error is of varying sign and magnitude R, : error will be reduced by averaging repeated measu rements R2 : error will not be reduced by averaging re peated measurements
1000
Pb�sics
rflctt with HELPS
Hel p i n g concept"l-'
0.001 60 ± 0.00005.
Which statement is correct? A The actual uncertainty is 5. B The fractional uncertainty is 5 x 1 0-s. C The fractional uncertainty is 5 . 16 D The percentage uncertainty is 3%. H e lping concept"l-'
Fractional uncertainty = o.oooo5 = __§___ = 0.031 25 0.001 60 1 60 Percentage uncertainty = fractional uncertainty x 1 00% = 3.1 25% ,,,3 3 page 9
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Topic 2
Kinematics
Frequently Exa m i ned
The horizontal component of the projectile's velocity is zero. C The kinetic energy of the projectile is zero. D The momentum of the projectile is zero. B
H elping concep�
Questions
What is the length of side X of the vector dia gram? B Ft A F C
at
u + at
D
H elping concep�
At the highest point Q, the vertical component of speed is zero. The horizontal component of speed is non zero. This means kinetic energy and momentum at Q is non-zero. There is no horizontal acceleration in projectile mo tion.
From the vector diagram, X = V-U
For constant acceleration,
v = u+ at v - u = at X = at
1 5. Which graph represents the motion of a car that
is travelling along a straight road with a uniformly increasing speed?
L l::: c "12: 'I�: A
acceleration
acceleration
8
o
o
time
0
time
O
time
0
time
0
H elping concep�
A uniformly increasing speed means the air is accel erating at a constant value ( v = u + at).
1 7. The diagram shows a velocity-time graph for a
car.
12
v I m s-' 10 8
6 4
2
0
I/
v
.. v
/
0
v
v""
2
3
/
v
tis
4
What is the distance travelled between time and t= 4 s? A 2.5 m B 3.0 m D 28 m C 20 m
t= O
Helping concep�
Distance travelled = area under the graph 1 6. An object has an initial velocity
u.
It is subjected to a constant force F for t seconds, causing a constant acceleration a. The force is not in the same direction as the initial velocity. A vector diagram is drawn to find the final velocity
= � (2+12)(4) = 28 m
v.
u
1 000
P{msics
rflci with HELPS
1 8. A man stands on the edge of a cliff. He throws a
stone upwards with a velocity of 19.6 ms-1 at time t= 0. The stone reaches the top of its trajectory after 2.00 s and then falls towards the botton of the cliff. Air resistance is negligible. Which row shows the correct velocity v and accel eration a of the stone at different times? page 25
Topic 2 A
Kinematics
tis
1 .00 B 2.00 c 3.00 D 5.00
v / ms- 1 a / ms-2 9.81 9.81 0 0 9.81 -9.81 -9.81 -29.4
H e lping conce:pn-
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Upward velocity of + 1 9.6 ms-1 means the convention is upwards direction is taken to be positive. Acceleration due to gravity will be -9.81 ms- 2, at all times. At f = 3.00 s, v = u+at = 1 9.6 + (-9.81)(3) = -9.81 ms- 1 At f= 5.00 S, v = u+at = 1 9 .6 + (-9.81)(5) = -29.4 ms- 1
Frequently Exa m i ned
A
Quei;tions
B
a
--_,t.... 0o!--
c
H elping conce:pn-
The ball will eventually reach terminal velocity from a tall building. This means there will be no net force and hence no net acceleration, as time is large.
t of the of a vehicle moving along a
2 1 . The graph shows the variation with time
displacement straight line.
t of the velocity v of a bouncing ball, released from rest. Downward velocities are taken as positive.
s
1 9. The graph shows the variation with time
s
At which time does the ball reach its maximum height after bouncing? 00�1'-tH--t-��;---t-�-+-t-....��� .. D
During which time interval does the acceleration of the vehicle have its greatest numerical value? H elping conce:pn-
H elping conce:pn-
At maximum height, the velocity of the ball is zero.
The gradients for
B,
C and
D
are constant, i.e.
constant. Hence a= 0 for all except A.
ds dt
is
20. A tennis ball is released from rest at the top of a
tall building. Which graph best represents the variation with time t of the acceleration a of the ball as it falls, assuming that the effects of air resistance are ap preciable?
1 000 Pb;gsics
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22. The diagram shows a ball which has been thrown
and is being acted on by air resistance. Which labelled arrow shows the direction of the resultant force on the ball when it is at the posi tion shown? page 26
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Topic 2
Kinematics
-------�
path of ball
/
\
it,'
--
24. A train travelling at 2.0 ms-• passes through a
ball ---
.,.,.. - -�
B
D
station. The variation with time t of the speed v of the train after leaving the station is shown below. vims · '
c
12 10
H elping conceph'
8
At the highest point of projectile, there is a resistive force acting opposite to the motion of ball. The force is to the left. There is also a gravitational force, i.e. the weight of ball .
6 4 2 0
resistive force
raw
Questions
Frequently Exam i ned
� �l t
I/ 0
/
10
/
20
/
30
1/
40
50
tis
What i s the speed of the train when it i s 1 50 m from the station? B 8.0 ms·• A 6.0 ms·• D 1 2 ms·• C 1 0 ms·•
weight
H elping conceph'
Acceleration of train = gradient of graph
4-2
23. A motorcycle stunt-rider moving horizontally takes
10-0 = 0.2 ms·2
off from a point 1 .25 m· above the ground, landing 1 0 m away as shown.
1 25 m
re- - - - - - - - - a
Using v 2 = u2 + 2as, v 2 2 2 + 2(0 . 2)(1 50) v 8.0 ms· ' =
---
·····�W////7////7/////ff//)//,_; .
10m
What was the speed at take-off? A 5 ms·• B 1 0 ms·• C 1 5 ms-1 D 20 ms·• H elping conceph'
Take off speed consists only of horizontal component of speed. 1 0 - - - - -(1 ) u s -t t = - =
=
·
25. The velocity of an object during the first five sec
onds of its motion is shown on the graph.
velocity/ms_
,
15
To find time taken t, s
ut + .!.at 2 2 -1 .25 0 + .!_ ( -9.81)t 2 2 t 0.50 s =
10
=
=
Subs!. t into (1 ): u ..!.9.. 20 ms- 1 0.5 =
5
=
1000 Pb�sics rflci with HELPS
0 0
2
3
4
time/
5
s
page 2 7
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Topic 3 Dynamics
A B C
D
Frequently Exa m i ned
It increases uniformly with respect to time. It is constant but not zero. It is proportional to the displacement from a fixed point. It is proportional to the velocity.
H elping conce:pnStatement A is incorrect as the resultant force acting on the mass is constant since the acceleration is constant ( uniform ) .
Statement C is incorrect as it would imply that the resultant force changes with the displacement from the fixed point, as in the case of S.H.M. Statement D is incorrect as it would imply that the resultant force changes with time.
1 2. Which graph best shows the variation with time of the momentum of a body accelerated by a conslant force?
A
c
lo
�� IL:= t= � 0
B
0
0
time
D
. time
O
time
0
O
atom can be stationary after the collision.
H elping conce:ptl-By definition , for an elastic collision, the relative speed of approach between two atoms equals their relative speed of separation. Also total kinetic energy of the two atoms is conserved.
m and travel v, are moving towards each other.
1 4. Two similar spheres, each of mass ling with speed
G
G
The spheres have a head-on elastic collision . Which statement is correct?
A The spheres stick together on impact. B The total kinetic energy after impact is C
mv2 .
The total kinetic energy before impact is zero.
D The total momentum before impact is
2mv.
H elping conce:ptl-Kinetic energy is conserved for elastic collision.
2 = mv 2 2 ..!.mv 2 2 + ..!.mv
time
H elping conce:pn-
A body accelerated by a constant force has constant acceleration. Its velocity increases linearly with time. Hence, its momentum also increases linearly with time.
1 3. In perfectly elastic collisions between two atoms,
1 5. The diagram shows two trolleys, X and Y, about to collide and gives the momentum of each trolley before the collision. 20 N s
12 Ns
A the initial speed of one atom will be the same
After the collision, the directions of motion of both trolleys are reversed and the magnitude of the momentum of X is then 2 Ns.
B
What is the magnitude of the corresponding mo mentum of Y?
it is always true to say that
C
1000
D whatever their initial states of motion, neither
c
�
Questions
as the final speed of the other atom. the relative speed of approach between the two atoms equals their relative speed of sepa ration. the total momentum must be conserved, but a small amount of the total kinetic energy may be lost in the collision.
Pb;9sics
?rlci with H E LPS
A C
6 Ns 1 0 Ns
B D
8 Ns 30 Ns page 39
Topic 3
Dynamics
Questions
Frequently Exa m i ned
H elping conce:pt¥
H elping conce:pt¥
Since linear momentum is conserved, 20 - 1 2 = -2 + momentum of Y momentum of Y = 1 0 Ns
Change in momentum = fF dt = area under the curve = (2)(2) + i (2 + 6)(4) = 20 kg ms -1
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1 6. Two railway trucks of masses m and 3m move
towards each other in opposite directions with speeds 2 v and v respectively. These trucks col lide and stick together. What is the speed of the trucks after the collision? v v B 2 A 4 Sv D c v 4
H elping conce:pt¥
This is a perfectly inelastic collision. � m 2v '..:.::../ �
�own, V1
a
v
��
m(2v) + 3m(-v) = (m + 3m)v 1 -mv = 4mv 1 V1 = - 41 V
1 8. A wooden block of mass 0.60 kg is on a rough
horizontal surface. A force of 1 2 N is applied to the block and it accelerates at 4.0 ms-2 • wooden block
4.0 m s-2 � "-
) 12 N
What is the magnitude of the frictional force <act ing on the block? B 9.6 N A 2 .4 N D 16 N C 14 N H elping conce:pt¥
Net force = ma 1 2 - F = (0.60)(4) F = 9.6 N
4.0 m s-2 friction , F
j
0.60 kg
1 9. Which graph best· shows the variation with time of
1 7. The graph shows how the force acting o n a body
varies with time. force/N 6
t
A momentum
- - - -l - - - - - 1' I I I I I I I 4 _ _ _ _ _. _ _ _ _ _ I I I I I I I I -
I
21----.r-- I I I I
0o
the momentum of a body accelerated by a C<)n stant force?
- - - -
I - - -i - -
2
I I I I 4
C --
6
Assuming that the body is moving in a straight line, by how much does its momentum change? B 36 kg ms-1 A 40 kg ms-1 D 1 6 kg ms-1 C 20 kg ms-1
1 000 Pb;gsics 'Yflci with HELPS
O� t�
momentum
0
0
time/s
./
time
�
0
H elping conce:pt¥
time
B
momentum
O
D momentum
t
/
�
t---
0
•
time
l_______.
O 0
·
time
Net force is the rate of change of momentum, which is the gradient of momentum vs time graph. Since net force is constant, gradient is also constaint and not equal to zero. page �10
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Topic 3
Dynamics
Frequently Exam ined
20. A body, initially at rest, explodes into two masses
What is the ratio � ? V2
A
<
60 cm s - 1
M, and M2 that move apart with speeds v, and v2 respectively.
40
Questions
m c s-1
+ - M%1
What is the speed of the masses after impact? B 20 cm s- 1 A 1 0 cm s-1 D 50 cm s-1 C 40 cm s-1
B
H e lp in g concept¥
Let the mass be m. H e l p i n g concept¥ ---. +
c·· .'M1 )i----• ·-.Y
v1
By conservation of momentum,
0 = M1 v1 + M2 (-v2 ) � = M2 V 2 M1
1 r-;:::;-i 60 cm s -
L...'..'.:._J �
�+
1 40 cm s -r-;;;-,
��
before collision
alter collision
m(60) + m(-40) = (m + m)v 20m = 2mv v = 1 0 cm s - 1
23. A force of 54 N pushes two touching blocks of 2 1 . A ball of mass 2 kg travelling at 8 ms- 1 strikes a
ball of mass 4 kg travelling at 2 ms-1·• Both balls are moving along the same straight line as shown. 8 ms- 1 ---
2 ms- 1 --
v ---
8 8
After collision, both balls move at the same veloc ity v. What is the magnitude of the velocity v? A 4 ms-1 B 5 ms- 1 C 6 ms - 1 D 8 ms-1 H e lping concept¥
mass 6.0 kg and 2.0 kg along a flat surface. The frictional force between the blocks and the sur face is 6.0 N. 54 N
What is the magnitude of the resultant force on the 6.0 kg mass? B 36 N A 12 N C 45 N D 48 N H e lping concept¥
Net force on (6.0 + 2.0) kg = 54 N - 6.0 N 48 N =
For inelastic collision, momentum is still conserved. momentum before collision = momentum after collision m,u, + m2 u2 = (m, + m2 )v (2)(8) + (4)(2) = (2 + 4)v v = 4 ms -1
Net acceleration, a = f_ m 48 8 = 6.0 ms -2 Resultant force on 6.0 kg mass, F = ma = (6)(6) = 36 N
22. Two equal masses travel towards each other on
a frictionless air track at speeds of 60 cm s- 1 and 40 cm s-1 • They stick together on impact.
1 000
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1ff!ct:t with H ELPS
page 4 1
Topic 3 Dynamics
Freq uently Exa m i ned
Y held sta tionary and connected by an extended elastic cord. The mass of X is twice that of Y.
24. The diagram shows two trolleys X and
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pin
Ques1:ions
Y , of masses m and 3m re spectively, are accelerated along a smooth hori zontal surface by a force F applied to block X are shown.
26. Two blocks X and
cork
The trolleys are released at the same instant. They move towards each other and stick together on impact. Just before the collision, the speed of X is 20 cm s- 1 • What is the speed of Y after the collision? A zero B 5 cm s-1 C 7 cm s- 1 D 1 0 cm. s- 1
H elping concept!'
What is the magnitude of the force exerted by bli0ck X on block Y during this acceleration?
A c
By conservation of momentum,
F 4 F 2
B D
F 3 3F 4
H elping concept!' initial momentum before release = momentum after collision Using F = ma , 0 = (mx + mv )v V=O acceleration, a = __E_ 4m Considering block Y , ------
Fxv
2.0 N
25.
=
F ) = 3F 3m( 4 4m -
27. The diagram shows a situation just before a head 2.0 N
A ruler of length 0.30 m is pivoted at its centre. Equal and opposite forces of magnitude 2.0 N are applied to the ends of the ruler, creating a couple as shown. What is the magnitude of the torque of the couple on the ruler when it is in the position shown? A 0.23 Nm B 0.39 Nm C 0.46 Nm D 0.60 Nm
H elping concept!'
on collision. A lorry of mass 20000 kg is travelll i ng at 20.0 ms- 1 towards a car of mass 900 kg trav elling at 30.0 ms-1 towards the lorry. 20.0 ms -•
-
30.0 ms - •
mass of lorry 2 0 000 kg
What is the magnitude of the total momentum? A 373 kNs B 427 kNs C 3600 kNs D 441 0 kNs
Couple = foece x perpendicular distance between 2 equal forces H elping concept!' = 2.0 x (0.3 sin 50° ) Momentum is a vector and is given by mass x veloc:ity. = 0.46 Nm Momentum of lorry = (2 x 1 0 4 )(20) 5 = 4 x 1 0 Ns 1000
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page 42
Topic 4 Forces 1 1 . A ruler of length 0.30 m is pivoted at its centre.
Equal and opposite forces of magnitude 2.0 N are applied to the ends of the ruler, creating a couple as shown.
1 3. A cylindrical block of wood has a cross-sectional
area A and weight W. It is totally immersed in water with its axis vertical.
The block experiences pressures P, and top and bottom surfaces respectively.
2.0 N
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C
B
2.0 N
What is the magnitude of the torque of the couple on the ruler when it is in the position shown? A 0.46 Nm 0.64 Nm D 0.79 Nm C 0.72 Nm
H elping conceptl'
The torque produced by a couple = force x perpendicular distance between forces = 2.0 x 0.30sin50° = 0.46 Nm
1 2. A car with front-wheel drive accelerates in the di
rection shown.
(Pb - Pi )A + W ( Pb - Pi )A
B
Which diagram best shows the direction of the total force exerted by the road on the front wheels?
c
..
D
H elping concept¥-
A/ . .lli . .
I
D
H elping conceptl'
(Pb - Pt ) (Pb - pi )A - W
Upthrust is a force, and acts upwards. Pressure, p = .
. Upthrust = (Pb - P1 )A
�
�
force, F = px A
1 4. A surface is bombarded by particles, each of mass
m, which have velocity v normal to the surface. On average, n particles strike unit area of the surface each second and rebound elasticall�1. What is the pressure on the surface? A nmv 2nmv ..:!. nmv 2 ..:!. nmv 2 c D 3 2
H elping conceptl'
r
B
pb at: its
Which of the following expressions is equal to the upthrust on the block?
A
A
Questions
Freq uently Exa m i ned
B
Momentum of a particle due to v is mv. After impact, the momentum is -mv. :. Change in momentum of a particle on impact = mv - (-mv) = 2mv Force per unit area = (momentum change per second per unit area) = (number of particles strike the surface per second per unit area)(2mv) = 2nmv Hence, pressure on the surface = 2nmv.
normal force
force
force on wheel by road
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1 5. A child drinks a liquid of density
p througl1 a vertical straw. Atmospheric pressure is p0 and the child is ca pable of lowering the pressure at the top of the straw by 1 0%. The acceleration of free fall is g.
page 52
Topic 4
Frequently Exa m i ned
Forces
Questions
What is the maximum length of straw that would enable the child to drink the liquid?
A
_f?_E_
1 0pg
B
c
&
D
pg
9po 1 0pg 1 0p0 pg
pressure
B
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Difference in pressure = hpf where h is the maximum length of the straw.
(0 1)Po = hpg ·
=>
D
H elping conce:ptl'
Ahpg + p1 A P2 A - p1A
Upthrust ( U) is a force.
h = _&_ 1 0pg
U = pressure difference x
= ( P2 - p , ) A
1 6. A uniform beam in a roof structure has a weight of 1 80 N. It is supported in two places X and Y,
a distance 3.0 m apart. A load is placed on the beam a distance of 0.80 m from X. The support provided by Y is 220 N.
x
area
1 8. A force F is applied to a beam at a distance d
�F
from a povot. The force acts at an angle line perpendicular to the beam.
220 N 3.0 m
x
pivot
y
0.80 m be am load
1 80 N
What is the value of the load? A 270 N B 490 N D 830 N C 520 N
11
L be the weight
L(0.80) + 1 80 x 1 .5 220 x 3 L 488 N :::: 490 N
e
to a
I
d
Which combination will cause the largest turning effect about the pivot? F
H elping conce:ptl'
Taking moments about point X and let of load.
p2
Which of the following is an expression for the upthrust on the block?
H elping conce:ptl'
:.
=
A large B large c small D small
d large large small large
e
large small large small
Helping conce:ptl'
=
=
p, height h and horizontal surface area A is immersed in a liquid. The pressures of the liquid at the liquid at the upper and lower surfaces are p1 and p2 re spectively.
Turning effect of F is F cos O(d). For moment to be large, F and d must be large. In order for cosO to be large, o must be small.
1 7. A solid block of material of density
1 000 Pb-;gsics
rf!c<t with H E LPS
p at a depth h in a liquid by the formula p = hpg.
1 9. The hydrostatic pressure
of density
p
is given
Which equation, or principle of physics, is used in the derivation of this formula? page 53
Topic 4
Forces
Ques;tions
Freq uently Exa m i ned
A density = mass + volume B potential energy = mgh C atmospheric pressure decreases with height D density increases with depth
force/N 550 500
y
H elping concept!'
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By definition, pressure = force area weight area =
mg A
(density = mass + volume used)
pVg A
pAhg A
= phg
1 0.0 1 2.0 extension/mm
What is the total work done in stretching the sample from zero extension to 1 2.0 mm? Simplify the calculation by treating the region XY as a straight line. A 3.30 J B 3.55 J c 3.60 J D 6.60 J H elping concept!'
Work done = ..!F
2
·
x
= area under the graph
20. A lump of ice floats in water as shown.
= ..!(500)(1 o x 1 o-3) + ..!(500 + 550)(2 x 1 0 3)
2
2
= 3.55 J
ice
water
Which statement is correct? A The lump of ice floats because the area of its lower su rface is larger than the area of its upper surface. B The pressure difference between the lower and the upper surfaces of the lump of ice give rise to an upthrust equal to its weight. C The ice has a greater density than the water. D The mass of water displaced by the ice is equal to the upthrust.
H elping concept!'
The upthrust acting on a floating object is the pressure difference between the lower and upper surface.
22. A rigid uniform bar of length
zontally at its mid-point.
2.4 m is pivoted hori
0.8 m
200 N
0.8 m
6
£
6
300 N
Weights are hung from two points of the bar as shown in the diagram. To maintain horizontal equi librium, a couple is applied to the bar. What is the torque and direction of this couple? A 40 Nm clockwise B 40 Nm anticlockwise c 80 Nm clockwise D 80 Nm anticlockwise H elping concept!'
2 1 . The graph shows the behaviour of a sample of a
metal when it is stretched until it starts to undergo plastic deformation.
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with H E LPS
-·-- ---1 200 N
0. 8 m
<
..6
08m
0.4 m
.
l
300 N
•
page 54
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Topic 1 1 A c
Superposition
!!..
4
3 11' 4
rad rad
Freq uently Exa m i ned
rad
B
!!..
D
11' rad
2
H elping concept!'
Wavelength of wave, Arp
J
=
speed = 340 = 0.78 m frequency 500
Questions
Helping concept!'
A.D X=a where x = fringe separation, A. = wavelength of light, D = distance between slits and screen, a = slit separation.
Ax
Using - = 2 11' J '
Arp 0. 1 7 = => Arp = !!... rad 2 211' 0.78 where tJ.rp = phase difference, Ax = distance apart.
1 7.
The diagram shows a long rope fixed at one end. The other end is moved up and down, setting up a stationary wave.
vibratandion down up
0
c
B
..!. 11' rad 4
D
� 11' rad 4
Light of frequency 6.0 x 1 014 Hz passes through a d iffraction g rating with 4 . 0 x 1 0 3 l i n es per centimetre. What is the angle between the two third-order dif· fraction maxima? B 2 3° A 1 2° D 74° c 37°
Helping concept!'
d sin8 = n 4 4.0 x 1 0 3 lines cm- 1 = 4.0 x 1 0 5 lines m- 1 '1
What is the phase difference between the oscil lations at X and Y? A
1 9.
= � = 3.0 x 1 os14 = 5.0 x 1 0 -1 f 6.0 x 1 0
A. sine = n = d
m
(3)(5.o x 1 0-7 ) = 0.60 1 ) ( 4.0 x 1 0 5
=>
e=
36.9°
The angle between 2 maxima is 2(} = 2(36.9°) = 74 °
H e lping concept!'
Particles of rope within two nodes are of the same phase. Hence, phase difference is zero. 20. The graph represents a stationary wave at two
different times. displacement
1 8.
A two-slit arrangement is set up to produce inter ference fringes on a screen. The fringes are too close together for convenient observation when a monochromatic source of violet light is used. In which way would it be possible to increase the separation of the fringes? A Decrease the distance between the screen and the slits. B Increase the distance between the two slits. C Increase the width of each slit. D Use a monochromatic source of red light.
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distance along the wave
What does the distance XV represent? A half the amplitude B half the frequency C half the period D half the wavelength page 1 43
Topic 1 1
Superposition
H elping conce;pn-
XY is the distance between adjacent antinodes of the stationary wave which is half the wavelength.
Frequently Exa m i ned
QuesHons
between the screen and the slits, or decreasing the distance between the slits.
23. A stationary sound wave is set up in a pipe using 2 1 . A double-slit light interference experiment is set
up.
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U nder which conditions will the separation of bright fringes be greatest?
A B c D
distance distance from wavelength between slits slits to screen of source small large short small large long large small short large small long
an oscillator of frequency 440 Hz. The extent of the vibration of the molecules in the pipe is illus trated. A centimetre scale is shown alongside the pipe.
lllli�l Ii �b i;kli I �I l l �i i �1 Pi � i i � ii 'J; ii 'J,' 1 1 I� I I� ii�� cm 1
1
What is the speed of sound in the pipe? B 328 ms-1 A 1 76 ms- 1 D 352 ms- 1 C 337 ms- 1
H e lping conce;pn-
H elping conce;pn-
A = ax => x = Ad d a where A = wavelength of light, a = separation of two slits, x = fringe separation, d = distance from slits and screen.
Distance between 2 nodes 40 cm Length of wavelength = 2 x 40 = 80 cm
For x to be large, be small.
A
=
Speed of wave, v = fA = ( 440)(80 x 1 0-2 ) = 352 ms- 1
and d have to be large and a to
24. The diagram represents a stationary wave cin a
stretched string.
x
22. Coherent monochromatic light illuminates two nar
row parallel slits and the interference pattern that results is observed on a screen some distance beyond the slits.
Which modification increases the separation be tween the dark lines of the interference pattern? A decreasing the distance between the screen and the slits B increasing the distance between the slits C using monochromatic light of higher frequency D using monochromatic light of longer wavelength
H e lping conce;pn-
The separation between the dark fringes is given by the formula y = AD . This can be increased by using a
light of a longer wavelength, increasing the distance
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What is represented by point P and by the ler n gth x? A B c D
point P anti node anti node node node
length x one wavelength two wavelengths one wavelength two wavelengths
H e lping conce;pn-
p is a node that is permenantly at rest in a stationary wave. Length x is equivalent to one wavelength. page 1 44
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Topic 1 1
Superposition
F requently Exa m i ned
25. The frequency of the fundamental mode of trans
verse vibration of a stretched wire 1 000 mm long is 256 Hz. When the wire is shortened to 400 mm at the same tension, what is the fundamental fre quency?
D
1 62 Hz 41 6 Hz
A
C
3 1 2 Hz 640 Hz
B
H e l p i n g conceptl-
The frequency of the fundamental mode of a stretched wire is
where
/
f = _!_ fI 21 'J µ
is the length of the wire,
T is the tension in the wire,
is the mass per unit length of the string. Since T and µ are held constant, JI
Given f = 256 Hz when
At
/
.!
f
oc
/
= 1 000 mm.
I
_f_ = 1 000 256
400
=>
f = 640 Hz
26. Monochromatic light is incident on a diffraction
grating and a diffraction pattern is observed. Which line of the table gives the effect of replac ing the grating with one that has more lines per metre? no. of orders of angle between 1 st and diffraction visible 2nd orders of diffraction A decreases decreases B decreases increases c increases decreases D increases increases
H e lping conceptl-
For a diffraction grating, d sinO = n..t where d= distance between 2 lines in a grating, n = orders of light, o = angle between the zero order and nth or der. For a grating with more lines per metre, d is smaller. 1 000
P�sics
n will be smaller as ..t is constant and sin O = 1 for maximum value. Since n is smaller for 0 � O � 90° , this means the or· ders are more spread out with larger angle between one another.
27. W h e n m o n oc h ro m atic l i g h t of wavelength
5.0 x 1 0-1 m is incident normally on a plane dif fraction grating, the second-order diffraction lines are formed at angles of 30° to the normal to the grating. What is the number of lines per millimetre of the grating? A c
D
250 1 000
B
H el p i n g conceptl-
= 400 mm, frequency is calculated to be
'ff!c'I with HELPS
Questions
500 2000
When monochromatic light passes through a plane dif fraction grating, bright or principal maxima are obtained when d sinO = n..t where d = spacing of lines, A. = wavelength of light, n = order of bright image. Given
0 = 30°, n = 2, A = 5.0 x 1 0-1 m.
1 :. d = 2(5.0 x 1 0- ) 2 x 1 0-a m sin 30° Hence, number of lines per millimetre of the grating is .! x 1 0-3 = .! x 1 06 x 1 0-3 = 500 2 d
28. A diffraction grating is ruled with 600 lines per
millimetre. When monochromatic light falls nor mally on the g rating, the first-order d iffracted beams are observed on the far side of the grating each making an angle of 1 5° with the normal to the grating. What is the frequency of the light? A B
c D
1 .2 x 1 0 1 3 4.7 x 1 013 1 .9 x 1 014 7.0 x 1 0 1 4
Hz Hz Hz Hz page 1 45
Topic 1 1
Superposition
Frequently Exa m i ned
30. A narrow beam of monochromatic light fall�1 at
H elping conceptJ'
The angle of diffraction, On , for nth order diffraction pattern of the light is given by d sinOn = nA. where d = spacing between adjacent slits in diffraction grating =
1 0-3 600
m,
When monochromatic light passes through a plane dif fraction grating, bright or principal maxima are obtained from d sinO = n). where d= spacing of lines, A = wavelength of light, n = order of bright image. Given n = 3 occurs at 45°.
81 = 1 5°.
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---
1
1 0-3 . sin 1 5 0 -- 4. 3 x 1 0 -1 600
--
m
Hence, frequency of light, f = £. f 3 x 1 08 4.3 x 1 0 7 = 7.0 x 1 0 14 Hz
-
normal incidence on a diffraction grating. Third order diffracted beams are formed at angles of 45° to the original direction. What is the highest order of diffracted beam 1Pro duced by this grating? B 4th A 3rd C 5th D 6th
H elping conceptJ'
n = 1 for first-order diffraction beam,
d sin 01 _ . . ,,., -_ -
Questions
-
. . d sin45° = 3J.
29. In the diagram, T represents a transmitter of mi
crowaves and P represents a metal plate.
�
� = .!sin 45° d 3
. . sin e = n(�) = � sin 45 ° = � d 3 3 v2 Since maximum angle allowed is 900, the highest order of diffracted beam is thus given by n = truncated integer of 3./2 = 4
i.e. highest order nmax = 4.
T
detector
p
The detector is connected to a galvanometer. The distance TP is much greater than the wavelength of the microwaves. As the detector is moved between T and P, what happens to the galvanometer reading? A B
C D
It decreases steadily. It reaches a maximum at P. It reaches a maximum midway between T and P.
It increases and decreases regularly.
3 1 . Under which of the following sets of conditions
will the separation of the bright fringes of a double slit interference pattern be greatest? distance distance from wavelength between slits slits to screen of source short A small small B short small large long small c large D short large small
H elping conceptJ'
H elping conceptJ'
Since TP is much greater than the wavelength of the microwaves, stationary waves are formed along TP. The galvanometer reading is low at nodes and high at antinodes.
slits
df
bright ' "· n g e s
__ __ __
_ ·A
1 .s
_-_-i
j j I Yn 1 Yn+ 1 screen
L
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page 1 46
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Topic l 2
A)(-8
Frequently Exa m i ned
Electric Fields
i
D
+
-----
- • <-0
Helping conce;pn A positive charge moves along the electric field lines from a high potential to low potential, and work is done only by moving parallel to field line.
'' ' ' '
C
Questions
�PE = F · x
o • - - - - - - - - - - - - - - - - - - • +o
point charge is surrounded symmetrically by six identical charges at distance r as shown in the diagram.
38. A
H elping concept!Ee
��
- - - - - - - - - - - - - - -." °•
E
I I I i i I I I I I ·- - - + Oc
----------·
+O e
The vector sum of EA and Ee is in the same direction of E8• This is because of the square arrangement of 3 charges. Hence, resultant electric field is in the direction of A.
0
0
How much work is done by the forces of electro· static repulsion when the point charge at the centre is removed to infinity? A
c
zero 60 2 4 m:0 r
B
60 4 m:0r
D
60 4 m:o r 2
Helping conce;pt!-
The potential of the point charge due to individual small positive charge, placed at a point L inside a uniform electric field, experiences a force of magnitude F.
37. A
·1- -�
_
,_
x
j
:
:s:;_JM
_________________
electric field lines
The charge is moved from point L to point M . What is the change in the potential energy of the charge? A
gain of Fx
B
C
loss of Fx
D
1 000 P/msics
rf!c'J
with HELPS
gain of F � loss of F Jx 2 + y 2
charges surrounding it is U = _st_ _ Since the six 4 m:0 r charges are identical, the resultant potential of the point 2 charge is 6U or 60 . Hence, the work done on 4 m:0 r 2 moving the point charge to infinity is 60 4 J1T.0 r
positive charge and a negative charge of equal magnitude are placed a short distance apart.
39. A
Which diagram best represents the associated electric field? page 1 67
Topic l 2
Electric Fields
Frequently Exa m i ned
Questions
The separation of the plates is d. The droplet is observed to be stationary when the upper pliate is at potential + V and the lower at potential - V. A
For this to occur, the weight of the droplet is equal in magnitude to A
c
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B
Vq
2Vq
B
d
Vd q
d
2Vq
D
d
H el p i n g conceptl'
The electric field acting on the oil droplet is E = _g�. d
The electric force acting upwards on the oil droplet is F = qE = 2qV . d
Since the charge is stationary, this electric force bal ances the gravitational force acting on the oil droplet 2 V whose weight is thus equal numerically to
c
�
41 .
H e l p i n g conceptl'
An oil drop of mass m, carrying a charge q, is in the region between two horizontal plates. When the potential difference between the upper and lower plates is V, the drop is stationary. The potential difference is then increased to 2 V. What is the initial upward acceleration of the drop?
The concept of the electric field is represented in dia gram by lines of force, which form a convenient way of visualising electric field patterns. The lines of force are always originated from a positive charge and ended on a negative charge; the direction of which is the direction that a positive charge would move when placed in the electric field. Hence, diagram 8 best represents the electric field pattern of the two unlike charges.
A c
g
2g
B
2qV m
-g
2qV
D
m
H e l p i n g conceptl'
Initially, when the oil drop was stationary, upward force due to electric field is qE -- qV ' where d is the disd
lance between the two plates. This balances the weig1ht qV of the oil drops, mg. Thus, mg = - . d
40.
An oil droplet has a charge - q and is situated between two parallel horizontal metal plates as shown in the diagram.
·I
1 000 rb�sics
•
-q
Jlf/c<j- with HELPs
When the potential difference was increased to 2 V, 2 11� the upward force acting on the oil drop is thus
�
By Newton's 2nd law, 2qV mg = ma, where a is tile -
d
-V
upward acceleration of the oil drop. Thus, a = 2qV md
- g.
page 1 �18
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Topic 1 4
O. C. Circuits
Frequently Exa m i ned
For variable resistor = 1 .0 kD, the effective circuit is as follow:
� n %n �1 n <
Now,
<
49. __
__
XZY
gives
the combinations in order of increasing resistance. -- -·-- ···-·--·
Effective resistance of AB = (-1- + 1 ) -1 1 .0 kD 1 .0 kD = 0.50 kD Using potential divider method, the value of voltmeter is 1 .0 kD x 12 V = 8 V ( 1 .0 + o .5) kn
and hence, only
Questions
___
,,,
_ _ _ ··--�----..-
The diagram shows three resistors of resistances 4 D, 1 O n and 6 n connected in series. A poten tial difference of 1 0 V is maintained across them, with point a being earthed. 10V
.-------c ..
---� ------.
Which graph represents the change in potential along the resistor network? 48.
Three resistors of resistance 1 D, 2 D and 3 D respectively arc used to make the combinations X, Y and Z shown in the diagrams. x
y
z
2n
20
3U
Q -0 Q
A
A
XYZ
B
XZY
c
YXZ
D
ZXY
2 10
c
O ·. p
.
Which of the following gives the combinations in order of increasing resistance?
1:
<
c
<
2c 10
=
�- D 3
= -0-
R
QJ
2 8 0 a.
5
Q
R
S
<
2c 10 0 a.
2 0
s
R
-5 -a
H elping
concept¥
I I------.
Y:
3D
For
'E
2
a.
10V �---� I · · · · · · ·
For
Iii
s
0
Q
1211•1 2 -+ 4
<
2
D
H e l p i n g concepfy
F or X:
a
B
Z:
PL
%n
� = � = -01n
1 000 Pb;gsics
2n
rflci with HELPS
30
potential at point Q,
V0
=
-1 0 0.5 A 4+10+6 Potential across the 4 D resistor, Vj,o = (4 )( 0.5 ) 2 V => Vp
Current,
I
=c
·· ·
0V
-- =
=
=
2V page 205
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Topic 1 5
Electromagnetism
Frequently Exa m i ned
plotting compass is placed near a solenoid. When there is no current in the solenoid, the com pass needle points due north as shown.
23. A
CD y
x
When there is a current from X to Y, the magnetic field of the solenoid at the compass is equal in magnitude to the Earth's magnetic field at that point. In which direction does the plotting compass set?
H e lp i n g conceptl'
The magnetic field set up by current in the coil is as follows.
�DBOf &:a. x
y
Since Bh, magnetic field experienced by the compass, is numerically equals to e the Earth's magnetic field, the resultant field is B as indicated in the diagram. Deflection of needle in compass is thus best illustrated by diagram A .
Questions
Tho hot air is directed between the poles of a strong magnet, as shown. What happens to tho ions? They arc deflected A towards the north pole N . B towards the south pole S. C downwards. D upwards. H e l p i n g conceptl'
The motion of the positively charged ions forms the current flow /, through the magnetic field B as shown in the diagram below.
The magnetic force acting on tho ions F is thus up wards as shown. Hence, tho ions are deflected up wards.
beam of electrons, travelling horizontally towards the right, passes between two horizontal charged parallel metal plates.
25. A
beam of
+
+
+
+
electrons
•.
24.
Hot air from a hair-dryer contains many positively charged ions. The motion of these ions consti tutes an electric current.
There is a vertical electric field between the plates. There is no deflection of tho beam because of tho presence of a uniform magnetic field in the region botwc.e n tho plates. In which direction must this magnetic field be? A out of the paper B into tho paper C vertically upwards D vertically downwards H e l p i n g concept�
Tho electric force on the electrons is vertically upwards by looking at the upper positive plate. This means that tho magnetic force on the electrons must be vertically downwards, so that there is no deflection of the elec trons, i.e. the electrons will travel in a straight line. Using Fleming's left hand rule, the magnetic field must be into the page.
1000
Pb;gsics
'fflc i with HELPS
page 2 1 5
Topic 1 6 Electromagnetic Induction 0. 1 0 A
2.0 A
�"�o co�, primary
secondary
Frequently Exam i ned
Questions
The induced e.m.f. E in the coil is thus given by E
t = d¢ = -r/> o 2 ff cos( 2 ff )
T T dt Note: A negative cosine function which is best repre sented in graph D .
Which set of values in the table is correct?
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A B c D
no. of turns on no. of turns on p . d . across load / V primary coil secondary coil 12 300 6000 12 6000 300 300 6000 4800 4800 6000 300
1 7. In a laboratory experiment to test a transformer, a student used the circuit shown in the diagram to take measurements.
H e l p i n g concept!'
� = NP = VP Ip N S vs
Two of the original entries in the student's results table are missi ng as shown below.
Since I, > /P , this transformer is a stop-down transformer with the number of turns on the primary coil being larger than that on the secondary coil and VP > v,
VP I V /P I mA NP turns Vs I V Is I mA Ns turn:Sl ?
2.0
240
Hence, only (B) satisfies both conditions.
?
50
50
J
Assuming the transformer was 1 00% efficient, what are the missing resu lts?
NP turns Vs I V
1 6. The magnetic flux ¢ through a coil varies with time t as shown in the diagram.
2 50 480 1 250
A B c D
6000 9.6 1 .0 9.6
H e l p i n g concep-C-K-
E�
E�. [\[\/\ _ lf\/\. EIAAI
Which graph best represents the variation with of the e.m.f. E induced in the coil?
A
C
B
J ,
Assume the transformer to be tested is an ideal one, then
G iven VP = 240 V,
N,
t
H e l p in g concept!'
From the given graph of ¢, we may define ¢ by the equation
50 turns,
I, = 50 mA, /P 2.0 mA.
D
t
=
=
. . 240 = �
NP
= 50 25 vs 50 2 Vs = 9.6 V and NP = 1 250 turns =
;
2 t ) ¢ = r/J o sin(
1 000
Pb;9sics
rJ!lci
with HELPS
page 228
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Topic l 7
Alternating Currents
Frequently Exa m i ned
Questions
The peak value of the sinusoidal waveform = 1 22°0
Which graph shows the variation with time of the power dissipated in the resistor?
= 6.00 v The root-mean-square value of the applied p.d. 6.00 J2
A
=
time/s
= 4.24 v B time/s
The graph shows how the potential difference across an alternating supply varies with time.
21 .
p.d.N
c
110
·
-uo] - - A
·
o-r--..-�..--+-,.--.�+--.-�r-+-..--,.�+----.--,-1--r--
·
D
B
c
Let R be the resistance of the resistor and if we rep resent the current by I = 10 sin(2nft), the power dissi pated in the resistor is then given by
1 56 78 1 56 78
power, P = 1 2 R = 18 R sin 2 (2nft) = � 18 R [ 1 - cos(4nft)]
H el p i n g oonuptl
Period,
T
=
=
25 ms 25 x 1 0-3 s
Frequency, f Vnns
22.
=
o J2 V
=
=
;
=
i.e. variation of power as function of time has a fre quency double that of the current function.
40 Hz
1J2 1 0 78 V =
23.
An alternating current flows through a resistor. The variation with time of this current is shown.
•
1 000
Plmsics
time/s
H el p i n g oonuptl-
f / Hz Vnns / V
40 40 400 400
-
D
What are the frequency f and the root-mean-square potential difference vrms of the a.c. supply?
�ru/\ : 1� /
time/s
'ff!c'I with H E LPS
The variation with time of the current through, and of the potential difference across, a resistor are shown below. potential difference
current
time/s
page 241
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Topic l 8
Frequently Exa m i ned
Quantum Physics
I 1 11�
x
1 2.
Which statement is true for emission line X of the spectrum? A It has the longest wavelength and is at the ultra-violet end of the spectrum. B It has the highest frequency and is at the ultra violet end of the spectrum. C It has the lowest frequency and is the red end of the spectrum. D It has the shortest wavelength and is at the red end of the spectrum.
An atom makes a transition from a state of energy E2 to one of lower energy E, .
Which of the following gives the wavelength of the radiation emitted, in terms of the Planck con stant h and the speed of light e?
E2 - E 1 he he E2 - E,
A c
H e lp i n g concepn-
Since
The emission line spectrum from left to right is in the increasing wavelength. The higher the wavelength, the smaller the frequency of the emission spectrum. Hence, the left-most line represents the ultra-violet emission while the right-most line is associated with the red light emission. Therefore, emission line X has the highest frequency (thus shortest wavelength) and is at the ultraviolet end of the spectrum.
f = �).
radiation,
1 3.
is the frequency of the radiation emitted. where ;. is the wavelength of the emitted
The diagram shows the first five energy levels of an atom. -----Es ------ E.
Which graph shows how the energy E of a pho ton of light is related to its wavelength ;. ?
-------- E 3 ---E2
A
'L::_ o E� 0
0
.
A
Photon energy
E = hf = he).
A
Photons of electromagnetic radiation are emitted when an electron falls from one energy level to a lower level. Which transition corresponds to a photon with the greatest wavelength? A B D
in the usual notation.
E2 E5
E4
Es
to to to to
E,
E,
E
3
E4
H e l p i n g concepn-
Greatest wavelength means smallest energy transitions
. . E oc �).
P{msics
.
------ E ,
c
H e l p i n g concepn-
1 000
D
he he E2 E, e h(E2 - E1 )
---
E2 - E1 = hf f
c
B
H e l p i n g concepn-
where
11.
Questions
between levels, using
rtlci with HELPS
E hf = he). . =
page 24 7
Topic l 8
F requently Exa m i ned
Quantum Physics
c
11 I
30.
The diagram shows five energy . levels of the hy drogen atom, labelled in the unit of electron-volt.
I 11
D
eV 0
- 1 . 5 ----
The greater the difference in the energy levels during the transition, the higher the frequency of the line rep resented in the spectrum since E hf. Hence, there should be three lines of high frequency and two lines of comparatively lower frequencies in the spectrum.
- 3 .4
=
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---
- 0 . 9 ----
H e l p i n g concept¥
-
29.
Questions
----
1 3 . 6 ----
Wl:lich statement is correct? A An atom in the level -3.4 eV can change levels by emitting photons of energy 1 .9 eV, 2.5 eV, 3.4 eV and 1 0.2 eV. B An atom in the level -3.4 eV can emit a photon of wavelength 650 nm to arrive in the level -1 .5 eV. C An electron with energy 1 0.2 eV colliding with an atom in level -1 3.6 eV can move it to the level -3.4 eV by losing all of its kinetic en ergy. D Most hydrogen atoms will be found in the level with zero energy.
A photocell is connected in a series circuit with a variable d.c. power supply and a sensitive amme ter. The photocell is illuminated with ultra-violet ra diation and photoelectrons are emitted. The elec tromotive force (e.m.f.) of the supply is then re duced and reversed and a graph is plotted of current against e.m.f. as shown. cu rren t
H e l p i n g concept¥ e . m.f.
Which graph is obtained if the experiment is re peated with a lower intensity of the same ultra violet source? current
�-- - A
L---- B original graph _i:...---- C
�---- 0 e.m.f.
Option A: An atom in the level -3.4 eV can change levels by e m itti n g photons of e n e rgy 1 0.2 eV and absorbing photons of energy 1 .9 eV, 2.5 eV and 3.4 eV. Option B: An atom in the level -3.4 eV can absorb a photon of wavelength 650 nm to arrive in the level -1 .5 eV. Option D: Most hydrogen atoms will be found in the level with - 1 3.6 eV. This is the lowest energy level. If the electrons are at the zero energy level, these elec trons are at the highest energy level in the diagram shown above.
H e l p i n g concept¥
Intensity of light source is proportional to the current detected. The same ultra-violet source means the stopping po tential of the emitted photoelectrons remains u n changed.
1 000
Pb�sics
mc'J with HELPS
31.
An electron of energy E is incident on the left hand side of a potential barrier of energy U. The energy U is greater than E. Which diagram represents the wave function of the electron to the right of the barrier? page 253
Topic 1 9 A B C D
Lasers and Semiconductors
Atoms in solids are much closer together than those in gases. Atoms in solids are much denser than those in gases. Solids are bettor electrical conductors than gases. Solids are not fluids but gases are fluids.
H elping conceptl-'
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Fact. The denser material (solid) causes the energy levels to band together.
1 4.
Freq uently Exa m i ned
Tho depletion region at p-n junction arises be cause of A diffusion of charges from p-type to n-type semiconductor and vice-versa. B absence of charges. C presence of tho external battery. D reverse bias of tho battery.
depletion layer potential barrier A decrease decrease B decrease increase c increase decrease D increase increase H e lping concepn-
ln theory, a forward-biased p-n junction will incmase tho diffusion current flowing through it. This will hap pen when the depiction layer and potential barrier decrease.
1 7.
H elping conceptl-'
Charges move from p-type to n-type semiconductor and vice versa. In the process, an electric field is set up. This prevents further movement of the charge carriers.
1 5.
What is meant by spontaneous emission of elec trons in solids? A Electrons being emitted by the solids through photoelectric effect when irradiated with e lec tromagnetic radiation. B Incident electrons colliding with electrorn:; in solids, and releasing double the number of incident electrons. C Excited electrons going back to lower encirgy states immediately by releasing energy. D Electrons in solids arc emitted without any external stimulus through radiation.
H e lping concept:�
Metals arc better electrical conductors than semi conductors because A there are more charge carriers in metals. B of the overlap in valence and conduction bands in metals. C the density of metals is higher than semicon ductor. D there is no doping in metals.
H elping concept:�
The overlapping of valence and conduction band in metals allows the charge carriers (i.e. electrons) to move freely into the conduction state.
1 6.
Q uestions
When a p-n junction is forward biased, what will happen to its depletion layer and potential bar rier?
1000
Pb�sics
'fflc i with HELPS
Electrons de-excited immediately to a lower ene•rgy state by releasing energy. All matters in the univc�rse tend to lower energy state wherever possible.
1 8.
What is meant by stimulated omission of electrons in lasers? A Electrons are de-excited to lower energy state by incoming photons. The energy and freque11cy of the incoming and emitted photons are the same. B Electrons are de-excited to lower energy state by incoming photons. The energy of the incom ing and emitted photons is the same but with different frequency. C Electrons are excited to higher energy state by incoming photons. These electrons will stay in the meta-stable state. D Electrons are excited to higher energy state by incoming photons. These electrons will 1Jn dergo spontaneous emission. page 260
Topic 1 9 31 .
Q uestions
In the diagram below, the symbols + + + and represent the majority carriers in the p-type and n-type sides of a p-n junction. Which pair of diagrams illustrates how a p-n junc tion acts as a rectifier?
A
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F req uently Exa m i ned
Lasers and Semiconductors
8
c
D
+
voltage source
n
p
volta ge source
+
voltage source
v
o l ta g e
p
+ �- + + + + + + + + +
r +
sou rce
+ + + + + + + + + + + --
n
p
+ + + + + + + + + + +
n
t
conventional
l
c u rrent
r
p
+
no cu rrent
n -=::J
conventional
c" rrenl
+ + + +
p
+ + + +
no cu rrent
n
l
c u rrent
j
rn rr e n l
conventional +
p
+ + + +
no current
n
-::::i
p n
+ + + + + t + + + + +
conventional
l_ +
L_
+ + + +
__ _
p
no cu rrent
n
H elping co-ncepn-
For forward-bias of rectifier, the p-typc j unction is aligned to the positive polarity of the battery. In the reverse-bias (where there is no current flow), the p-type junction is connected to the negative polar ity of the battery.
1 000
Pb�sics
rf!ci with HELPS
page 265
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Topic 20 31 .
Freq uently Exam i ned
Nuclear Physics
How many up quarks and down quarks must a proton contain?
Each of the nuclei below is accelerated from rest through the same potential difference. Which one completes the acceleration with the lowest speed? A
�H
c
� Li
A
� He � Be
B D
B c D
H elping concept¥
up quarks down quarks 3 0 1 1 2 1 1 2
Helping concept¥
Potential energy = kinetic energy e V = i- mv 2 V=
Charge on proton is +e. By simple addition,
l:V
2 + (- -e ) = e 2(-e) 3 3
The speed v is the lowest if the ratio of (�) is the m lowest.
� Li
has the smallest ratio of (�) of (�) . m 7 34.
32.
Q uestions
A thin gold foil is bombarded with shown.
a
- particles
as
What is the decay constant of a radioactive sub stance? the constant of proportionality in the equation relating the rate of decay of the substance to the number of undecayed nuclei B the number of disintegrations of the substance occurring in one half-life of the substance C the number of disintegrations of the substance occurring per second D the average time taken for half the nuclei ini tially present in the substance to decay A
incident a-particles
gold foil
The results of this experiment provide information about the A binding energy of a gold nucleus. B energy levels of electrons in gold atoms. C size of a gold nucleus. D structure of a gold nucleus.
H elping concept¥ dN dt
=
_
.J..N
=>
le
dN
= __Jj1._ N
dN where (dt) is the rate of decay of substance and N
is the number of undecayed nuclei.
33.
Most of the a - particles pass straight through and very few get deflected, showing the size of the nucleus to be very small.
Protons and neutrons are thought to consist of smaller particles called quarks. The 'up' quark has a charge of has a charge of - � e, where
charge ( + 1 .6 x 1 0-1 9 C). 1 000
Helping concept¥
Pb�sics
rf/cCJ wit h H ELPs
�e; a 'down' quark e is the elementary
35.
A radioactive isotope has a decay constant ), and a molar mass M. Taking the Avogadro constant to be L, what is the activity of a sample of mass m of this isotope? page 2 73
Topic 20
F req uently Exa m i ned
Nuclear Physics
56.
H elping concept¥
The experiment shows that most electrons passed through the gold foil without change in direction.
Isotope Y forms after two successive decays. Ea1ch decay can be either alpha-emission or beta-emis sion. Four other isotopes, P, Q, R and S, are shown on the diagram.
neutron number
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55.
The binding energy (B.E.) per nucleon is a useful measure of the stability of a nucleus. Which graph shows how the B.E. per nucleon varies with the nucleon number (A)? A
B.E. per nucleon in MeV
s : 2
/
O +--r--.�..--.----..� 0
40
1 20
200
nucleon number (A)
B
B.E. per nucleon in MeV
6
6
2
... 0 -'-�--.r-��....� 1 20 200 0 40
n ucleon number (A)
B . E . per nucleon
�
0 -+---r--ir--r-....,..----.-� 0 40 200 1 20
nucleon number (A)
B . E . per
8
nucleon in MeV
6
:
... 0 -'-�--.r-�--r-....� 1 20 200 0 40
nucleon
number (A)
H e lping concept¥
The B.E. per nucleon will increase as nucleon number increases from 1 to a peak at 8.5 MeV at approxi mately nucleon number of 56 before decreasing in a linear trend.
1 000
Pb�sics
1 38 - .. -- -
-- --->-------+----
Q
. · - ---- -
R
1 37 - ---�--------·, ----j--...------r--- •
p
___,._ . __,�-...---
1 36 -+----t---� --
__
-· 1 35 - --
- -- ·
..
.
'
y
. .. --
-......,.---.,- -
--
-t -t--�---t--1 34 +-----
1 33 +----+---.,--t---r----;-�-r�-, 96 97 93 90 91 92 95 94
proton
number
How many of the isotopes P, Q, R and S could be the initial isotope that, after two successive decays, became Y? A B 2 c 3 D 4 To find the original isotope before decay, isotope Y has to be added either with alpha or beta emission. There are 4 possible isotopes with the following pro cesses: Process 1 : Y + alpha + alpha y
in MeV
D
1 40 -.-----.--,---,--- , 1 39 -
H elping concept¥
4
c
Questons
---------- ------ ----
rf!ctt with HELPS
neutron 1 35 proton 93
alpha alpha total 2 1 39 2 2 2 97
Process 2: Y + alpha + beta y
neutron 1 35 proton 93
alpha beta total 2 0 1 37 2 94 --1
Process 3: Y + beta + alpha y
neutron 1 35 proton 93
beta alpha total 0 2 1 37 -1 2 94
Process 4: Y + beta + beta y
neutron 1 35 proton 93
beta beta total 0 0 1 35 -1 -1 91 page 280
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:di!�
1 000
v
Topic 1 Measure me n t I.
6.
1 1. 1 6.
B D
B
2.
7.
c
A
1 3. 1 8.
c
27.
31.
A
32.
36.
A
37.
c
47.
46. 51. 56. 61.
D B
8.
c
26.
B
3.
1 7. 22.
41.
ANSWE RS
1 2.
A
21.
B B
Ph ysics M CQ
42. 52. 57.
c
D
c
D D D
c B
D
23 . 28.
6.
1 1. 1 6.
D D B
2.
7. 1 2.
c
1 7.
21.
A
22.
26.
c
27.
31. 36. 41. 46.
B
c
D D
32. 37. 42.
Topic 3 Dynam i cs I.
6.
1 1. 1 6. 21. 26. 31. 36. 41.
B
A
B D B
D D A B
43. 48.
c
42.
Topic 4 Forces I.
c
2.
6.
c
7.
1 1.
A
1 2.
A B D B
B D
44. 49.
3.
D
4.
8. 1 3. 1 8. 23 .
A
D D D
D
40.
c
45.
c
50.
B
c
D
5.
A
[)
1 0.
1 4.
A
1 5.
1 9.
[)
20.
9.
24.
c
34.
38.
A
39.
43.
A
44.
28.
B
c
B
B
60.
33.
B B B B B B
A
30. 35.
D
29.
8.
D
D
c
c
3.
B D
A
55.
28.
23 .
c
29.
59.
A
32.
c
A
22.
3 7.
25.
24.
58.
1 3.
B
c
54.
1 8.
A B
20.
A
c
27.
1 5.
c
53.
1 7.
c
c
1 9.
39.
A
A D
1 4.
c
1 2.
A
D
1 0.
38.
B
A D
c
5.
A
34.
2. 7.
B
A
9.
A
B
c
c
[)
4.
33.
Topic 2 Kin emat i cs I.
B D
with � elps
4. 9. 1 4.
B D
D D
25.
[) A [)
c
30.
A
35.
c
40.
c
c
45.
B
[)
5.
[)
B B
1 9.
A
24.
A
29.
A
1 0. 1 5. 20.
25.
30.
D
c
B c A A B
33.
c
34.
c
35.
38.
B
39.
A
40.
[)
5.
c
1 0.
c
43.
D D
4.
8.
c
9.
13.
c
1 4.
3.
B B
1 5.
A page 284
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