Tung Wah Group of Hospitals Kap Yan Directors' College
2014-2015
Second Term Test
Physics
/73
Form 5 Time allowed: 1.5 hours Total marks: 73
Date: 31 Mar 2015 LPK
Name : __________________________________ Class: ____________ Class no.: _________
Answer ALL questions.
Section A : Structured-type Questions (48 marks)
1. (a) A kettle is rated at 1 kW, 220 V. Calculate the working resistance of the kettle. (2 marks)
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(b) When connected to a 220 V supply, it takes 3 minutes for the water in the kettle to reach boiling point.
Calculate how much energy has been supplied. (2 marks)
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(c) Different countries supply mains electricity at different voltages. Many hotels now offer a choice of voltage supplies as shown in the photograph.
220 V 110 V
(i) By mistake, the kettle is connected to the 110 V supply. Assuming that the working resistance of the kettle does not change, calculate the time it would take for the same amount of water to reach boiling point. (3 marks)
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P.1

(ii) Explain what might happen if a kettle designed to operate at 110 V is connected to a 220 V supply.
(2 marks)
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2. (a) In the circuit shown in Figure 2.1, a 12 V battery of negligible internal resistance is connected with a thermistor R and a resistor of resistance 120

. The graph below shows the variation of the thermistor's resistance with temperature.
Figure 2.1 Variation of thermistor’s resistance with temperature
(i) Find the resistance of the thermistor R at 20 o C. (1 mark)
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(ii) What is the potential difference V
AB
across A and B at 20 o C? (2 marks)
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P.2

(b) Wendy wants to confirm the above calculation by measuring V
AB
using a voltmeter of about 1 k
 resistance. She finds that the reading registered is slightly different from the value found in (a) despite making careful measurements. Explain why this is so.
Suggest how the accuracy of the measurement could be improved. (3 marks)
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(c) (i) The potential difference V
AB
is used to drive an electronic switch connected across AB to turn on a fan if the temperature rises above a certain value such that
V
AB
is 6.0 V or above. Using the information provided in the graph, find the minimum temperature needed to keep the fan on. Show your working. (2 marks)
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(ii) Without using additional components, complete the new circuit diagram below to illustrate how the circuit can be modified to turn on a heating device when the temperature falls below a certain value. Explain the action of the circuit.
No calculation is required. (3 marks)
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P.3

3. In the circuit shown in Figure 3, resistors R
1
and R
2
represent the heating elements in a heater using mains supply. Both resistors are immersed in water. heater
S
~ 220 V a.c.
Figure 3
R
2
R
1 water
The heater can be operated in two modes, namely, heating and keeping warm, and it is controlled by the switch S. The power consumed by the heater in the heating mode is 550 W and in the mode of keeping warm is 88 W. The mains voltage is 220 V a.c.
(a) In which mode is the heater operating when switch S is open? (1 mark)
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(b) Find the resistance of R
1
. (2 marks)
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(c) When switch S is closed, calculate the current passing through resistor R
2
. (3 marks)
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4. A beam of electron moving horizontally with an initial velocity of 2 x 10 7 m s -1 enters a vertical electric field set up by two parallel square metal plates, each of side 10 cm. The plate separation is 5 cm and a potential difference of 250 V is applied across the plates. Given that the charge and mass of an electron are
- 1.6 x 10 -19 C and 9.11 x 10 -31 kg respectively.
P.4

(a) Which plate, the upper plate or the lower plate, is positively charged? (1 mark)
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(b) Calculate the time lapsed as the electrons move in the electric field. (2 marks)
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(c) Calculate the acceleration of the electrons. (2 marks)
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(d) Calculate the vertical deflection
 y of the electrons in the electric field. (2 marks)
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(e) The electron travels at an angle

with respect to the horizontal when it leaves the electric field.
Calculate the angle

. (2 marks)
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5. The figure below shows an electric kettle and the 3-pin plug connected with it.
B
A fuse
C
P.5

(a) (i) State the colours of the coatings of wires A and C. (1 mark)
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(ii) The earth pin on the plug is longer than the other pins. Briefly explain why. (2 marks)
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(b) A student suggests that the switch S can also be connected to the neutral wire in terms of safety. Do you agree? (2 marks)
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(c) The rating of the kettle is ‘220 V, 1500 W’. What is the cost of electricity if it is switching on for 40 minutes? (Given: 1 kWh of electricity costs $1.1) (2 marks)
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6. The figure below shows a solenoid of length 80 cm. The solenoid has 500 turns and it carries a current of
50 mA.
(a) State the polarity of the magnetic field at R (1 mark)
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(b) Find the magnitude of the magnetic field at the centre O of the solenoid. (2 marks)
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P.6

(c) What is the major assumption in (b)? (1 mark)
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(d) What would happen to the magnetic field strength in the solenoid if a bar made of the following material is inserted into the solenoid?
(i) an iron bar (1 mark)
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(ii) a copper bar (1 mark)
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SECTION B: Multiple Choice Questions (25 marks)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A
B
C
D
1. In the circuit shown, all the light bulbs are identical. When the switch is open, A
1
reads I and the power dissipated by L
1
is P . What are the ammeter reading A
1
and the power dissipated by L
1
when the switch is closed?
Ammeter reading A
1
Power dissipated by L
1
A. I 2 P
B. I 4 P
C. 2 I 2 P
D. 2 I 4 P
2. What is the voltmeter reading in the circuit shown?
A. 0 V B. 4 V C. 6 V D. 12 V
3. A 100 W lamp connected to the 230 V mains is replaced by a lamp which has twice the resistance. The power of the new lamp is
A. 25W B. 50W C. 200W D. 400W
4. A resistor of resistance R is connected to three identical cells as shown. Each cell has an internal resistance r and an e.m.f. V . What is the current flowing through the resistor?
A.
3 V r

R
B.
3 r
3 V

R
C.
3 V r
3

R
D.
3 r
V

R
P.7

5. Two pupils Peter and Mary are asked to find out the value of an unknown resistance. They each use the same equipment – a d.c. supply of negligible internal resistance, an ammeter and a voltmeter. Peter uses circuit P and Mary uses circuit M.
Peter obtained the following readings: voltmeter reading V = 5 V ammeter reading A = 1 mA
What are Mary’s results?
Reading on V Reading on A
A. more than 5V more than 1mA
B. more than 5V less than 1mA
C. less than 5V more than 1mA
D. less than 5V less than 1mA
6. A fuse will melt and cut off the current if there is a short circuit between
(1) the neutral and the earth wire.
(2) the live and the neutral wire.
(3) the live and the earth wire.
A. (1) only B. (3) only C. (1) and (2) only D. (2) and (3) only
7. Three resistors are connected to a 6 V battery as shown.
6 V
Which of the following statements are correct?
(1) The power dissipated in the two 10

resistors are the same.
(2) The total power dissipated in the resistors is 3.6 W.
(3) The total power dissipated is larger if all resistors are connected in series. 10 
A. (1) and (2) only B. (1) and (3) only
C. (2) and (3) only D. (1), (2) and (3)
8. Which of the following results can prove that an object is a magnet?
20 
10 
A. The object is not attracted by an electromagnet.
B. A bar magnet is attracted by the object.
C. A bar magnet is repelled by the object.
D. The object is attracted by an electromagnet.
9. Which of the following figures show(s) the patterns of iron filings produced by magnets correctly?
A. (1) and (2) B. (1) and (3) C. (1) and (4) D. (3) and (5)
10. A motor lifts a load of 10 N vertically upwards at a steady speed of 2 m s
–1
. The voltage applied to the motor is 200 V and the current drawn is 0.5 A. Find the efficiency of the motor.
A. 0.2% B. 5% C. 20% D. 50%
P.8

11. Which of the following configurations has the smallest equivalent resistance?
12. Three identical conducting balls X, Y and Z are suspended separately from insulating threads. Ball X has a net charge of 1.0 x 10 -8 C while balls Y and Z are neutral. Balls X and Y are brought into contact and then separated. Then ball Y is brought momentarily into contact with ball Z and then separated. When balls X and Z are placed 1 m apart, the electric force between them is
A. 1.0 x 10 -7 N B. 1.1 x 10 -7 N C. 2.2 x 10 -7 N D. 9.0 x 10 -7 N
13. An artificial pacemaker is a device used to regulate the beating of the heart by emitting electrical pulses. A typical artificial pacemaker can produce potential difference of 10 mV and pulses of 16 mA current. If the duration of each pulse is 0.5 ms and the pacemaker generates 100 pulses each minute, find the total charge delivered by the pacemaker each minute.
A. 8 x 10 -6 C B. 8 x 10 -4 C C. 0.016
C D. 160 C
14. A 6 V 24 W lamp is to be operated from a 12 V supply. What is the value of the resistor R to be connected in series with the lamp to allow it to work at its rated value?
A. 2

B. 1.5

C. 3

D. 4

15. When the switch S in the circuit is closed, all the lamps light up. But after the circuit has been used several times, only L
1
lights up. Which of the following possibilities would account for this?
(1) There is a short circuit across L
4
.
(2) The filament of L
2
is burnt out.
(3) The filament of both L
2
and L
4
are burnt out.
A. (1) only B. (3) only
C. (1) and (2) only D. (2) and (3) only
16. A cell of e.m.f. E and a hearing aid are connected in series to a rheostat which controls the current in the circuit. When the rheostat is set to 300

, the current is 2.5 mA. On increasing the volume of the hearing aid, the resistance of the rheostat is decreased to 100

and the current is increased to 5.0 mA. If the hearing aid can be regarded as a resistor of constant resistance and the internal resistance of the cell is negligible, what is the value of E?
A. 1.00 V B. 1.25 V C. 1.50 V D. 1.75 V
17. A current-carrying long solenoid of N turns has a cross-sectional area A and length L. Which of the following sets of values would increase both the magnetic field strength and the magnetic flux at the middle of the solenoid?
Length Cross-sectional area Number of turns
A.
L A 2N
B.
2L A N
C.
L 2A N
D.
2L A 2N
P.9

18. The figure shows the cross-sections of two long, parallel straight wires R and S in vacuum. The current in
R is 2 I flowing into the paper while that in S is 4 I flowing into the paper. Find the resultant magnetic field at X.
A.
μ
0
I
2
π r
(downward) B.
μ
0
I
2
π r
(upward)
C.
μ
0
I
π r
(downward) D.
μ
0
I
π r
(upward)
19. The graph shows the I-V relationship of an electric component.
Which of the following deductions is/are correct?
(1) When the applied voltage is 1 V, the component has a resistance of 1

.
(2) When the applied voltage exceeds 2V, the component has a constant resistance.
(3) When the applied voltage exceeds 2V, the component is non-ohmic.
A. (1) only B. (3) only
C. (1) and (2) only D. (2) and (3) only
20. The diagram shows a rice-cooker being tested for safety. The cooker has a metal body. Clip C
1
is connected to the metal body of the cooker. Suppose the insulation of the live wire cable is broken and the live wire touches the metal case. The bulb will light up when C
2
is connected to
(1) the earth pin of the plug.
(2) the live pin of the plug.
(3) the neutral pin of the plug.
A. (1) and (2) only B. (1) and (3) only
C. (2) and (3) only D. (1), (2) and (3)
P.10

List of data, formulae acceleration due to gravity charge of electron electron rest mass permittivity of free space permeability of free space
A1. E = mc

T
 g = 9.81 m s e = 1.60

10 m e
= 9.11

0
0
10
= 8.85

10
= 4
 
10 energy transfer during heating and cooling


2

7
(close to the Earth)

19 C

31 kg

12 C
H m
2

1
N

1 m

2
D1. F =
Q
1
Q
2
4
π 
0 r
2
Coulomb’s law
A2. E = l
 m
A3. pV = nRT
A4. pV =
1
3
Nm c
2 energy transfer during change of state equation of state for an ideal gas
D2.
E =
Q
4 π 
0 r
2
D3. V =
Q
4
π 
0 r kinetic theory equation D4. E =
V d electric field strength due to a point charge electric potential due to a point charge electric field between parallel plates (numerically)
3 RT
A5. E
K
=
2 N
A molecular kinetic energy D5. I = nAvQ general current flow equation
D6. R =
 l
A resistance and resistivity
B1. F = m
 v
 t
=
 p
 t force D7. R = R
1
+ R
2 resistors in series
B2. moment = F
 d moment of a force D8.
1
R
=
1
R
1
1
+
R
2 resistors in parallel
B3. E
P
= mgh gravitational potential energy
D9. P = IV = I 2 R power in a circuit
B4. E
K
=
1
2 mv
2
B5. P = Fv =
W t
B7. F =
Gm
1 m
2 r
2 kinetic energy mechanical power
Newton’s law of gravitation
D10. F = BQv sin

D11. F = BIl sin

D13. B =

0
I
2 π r
D14. B =

0
NI l force on a moving charge in a magnetic field force on a current-carrying conductor in a magnetic field magnetic field due to a long straight wire magnetic field inside a long solenoid
P.11
End of Paper

F5 Physics Second Term Test (14-15) Marking Scheme
1. (a) R = V 2 / P (1M for formula R = V 2 /P)
= 220 2 / 1000 = 48.4

(1A)
(b) E = Pt = 1000 x 3 x 60 (1M)
= 180000 J (1.8 x 10 5 J) (1A) (3000 J scores 1 mark)
(c) (i) Since P = V 2 / R (1M for P proportional to V 2 )
Power is reduced to P/4 (1M for P/4 OR P = 1000/4 = 250W)
Energy required (E = Pt) is unchanged
Thus if P becomes P/4, time t required becomes 4t, i.e. 4 x 3 = 12 minutes.
(1A for 12 minutes, no mark for just 4 times the time)
(ii) Current will increase (or power increases). (1M)
The kettle may be damaged / the fuse may melt / the circuit breaker becomes off / the heating element may burn out (1M) (no mark for short circuit or explosion)
2. (a) (i) 100

(1A)
(ii) V
AB
= 120 x 12 / (120 + 100) (1M)
= 6.55 V (1A)
3. (a) Keeping warm
(b)
(c)
R
1

V
2
P

220
2
88
= 550

Total current I o

P
0
V

550
220
= 2.5 A
Current in R
1
, I
1

220
550
= 0.4 A
Current in R
2
, I
2
= 2.5 − 0.4 = 2.1 A
Or Power to R
2
, 550 W – 88 W = 462
W
Current in R
2
, I
2

P
2
V

462
220
= 2.1 A
1A
1M
1A
1M
1M
1A
1
2
3

4. (a) Upper plate (1A)
(b) Time elapsed = s / v x
= 0.1 / 2 x 10
7
= 5 x 10
-9
s (1M+1A)
(c) a = F/m = qE / m = eV/md = 1.6 x 10
-19
x 250 / (9.11 x 10
-31
)(0.05) = 8.78 x 10
14
m s
-2
.
(1M for qE/m+1A)
(d) Vertical deflection required = 0.5 at
2
= 0.5 x 8.78 x 10
14
m s
-2
x (5 x 10
-9
)
2
= 0.011 m
(1M+1A)
(e) v y
= u + at = 0 + 8.78 x 10
14
(5 x 10
-9
) = 4.39 x 10
6
m s
-1
. (1M)

= tan -1 v y
/ v x
= tan -1 4.39 x 10 6 m s -1 / 2 x 10 7 = 12.4
o (1A)
5. (a) (i) The coating of wire A is blue, while that of wire C is brown. (1A)
(ii) The earth pin is longer such that the neutral and live pin-holes are closed only when the earth pin is first plugged into the earth pin-hole (2A). OR
The earth pin is longer so that it is first connected before the other two pins are inserted.
This ensures that any leakage of current will flow to the ground. (2A)
(b) The student is wrong (1A). The switch should always be connected to the live wire such that the iron is at zero potential when the switch is off (1A). OR The live wire will still be at high potential if the switch connected to the neutral wire is open.
(c) Cost of electricity = 1.5 x (40/60) x 1.1 = $1.1 (1A)
(iii) From P

V
2
R
, the power of the kettle is halved when the resistance is doubled. (1M)
From t

E
P
, the time required to heat up the water is doubled, that is 262 .
5

2

525 s . (1A)
6. (a) North pole at R.
(b) B =

0
NI l
(1M)
=
(4 π 
10

7
)(500)(0.0
5)
0.8
= 3.93

10
–5
T (1A)
(c) Assume the length of the solenoid is much larger than its radius. (1A)
(d) (i) Increase (1A)
(ii) Unchanged (1A)
MC 1-5 D C B B B 6-10 D A C B C 11-15 A B B A A 16-20 A A A B D
MC explanations
(1A)
1. When the switch is open, current flows through L
3
and L
1
only. Hence,
I

V
2 R
, and P

I
2
R
When the switch is closed, current flows through L
1
only since L
3
and L
2
are both shorted.
I '

V
R

2 I , and P '

I '
2
R

( 2 I )
2
R

4 I
2
R

4 P
2. Since the voltmeter has high resistance, current will not pass through the voltmeter.
Required p.d. = p.d. across 100

= 6 V (Each 100

will take up half of the battery’s voltage.)

3. P

V
2
R
, so if resistance is doubled, power is halved.
4. Total emf is 3V, total internal resistance is 3r.
5. For Mary, the voltmeter is measuring the cell’s emf. But Peter’s voltmeter only measures the p.d. across the resistor. So Mary’s voltmeter reading is higher.
In circuit P, the ammeter reads the total current and is therefore larger than that in circuit M. In circuit Q, the voltmeter reads the total p.d. (battery’s emf) and is so larger than the p.d. across R only in circuit P.
10. Power output = Fv = 10

2 = 20 W
Power input = VI = 200

0.5 = 100 W
Efficiency =
20
100

100 % = 20%
12. After touching each other, balls X and Y will have equal amount of charge 1 x 10
-8
/2 =0.5 x 10
-8
C.
After touching each other, balls Y and Z will have equal amount of charge 0.25 x 10
-8
C.
So force between balls X and Z =
1
4

0
0 .
5

10

8 
0 .
25

10

8
1
2

1 .
12

10

7
N
13. For each pulse, charge flow = Q = It = 16 x 10
-3
x 0.5 x 10
-3
= 8 x 10
-6
C
In each minute, the total charge delivered = 100 x 8 x 10
-6
= 8 x 10
-4
C
14. P.d. across resistor = 12 – 6 = 6V
Current I = P/V = 24/6 = 4A
R = V/I = 6/4 = 1.5

16. The cell, the hearing aid and the rheostat are connected in series. Thus
E = 2.5 x 10
-3
x (300 + R)
E = 5 x 10
-3
x (100 + R)
R = 100

, and E = 1V
18. B =

0
( 2
2
 r
I )


0
2

( 4 I )
( 4 r )


0
I
2
 r
20. The circuit can be redrawn as follows: