H2 PHYSICS COMMON TEST 1 SOLUTIONS:
MCQ
1.
1
2

Answer is B.
2. Since n
 d d sin
 sin

  n

As sin
 
1 , n
 d

Maximum order n

1

10

3
500
600

10

9

3 .
33
Hence maximum order is 3.
Including the central maxima, the total number of lines that can be seen = 2(3) + 1 =7
Answer is C.
3. Path difference

6

4

2 m

2
1
 
2

As the sources are antiphase, path difference of 2
 means the meet at P and result in destructive interference.
Since both waves have approximately the same amplitude,
Resultant amplitude = A
–
A = 0
Answer is A.

4. The two radio sources will interfere and form an interference pattern.
For the interference pattern,
Fringe separation


D
 a
3 .
0

10

2
15

2000

4 .
0 m
Period of fluctuations

4 .
0

0 .
044 s
90
Answer is A.
5. Net work done on gas = - Area under graph.



1
[( 3 .
231
2

221 .
9 J

1 .
012 )

10
5
][( 22 .
40

20 .
40 )

10

3
]
By First Law of Thermodynamics,

U
 
Q
 
W
Since the ideal gas went through a full cycle,

U

0
Hence,
0


Q

Q


(

221 .
9 )
221 .
9 J
Heat supplied by gas =
 
Q
 
221 .
9 J
Answer is C.
6. Statement A is wrong because more energy is needed to do work against the atmospheric pressure and also to break the bonds between molecules and not to increase random kinetic energy.
Statement B is wrong because internal energy consists only of random kinetic energy and not bulk kinetic energy.
Statement C is wrong because the statement is only true for an ideal gas.
Increase in internal energy could be due to increase in potential energy due to intermolecular forces and not increase in random kinetic energy. In that case, the temperature would not have increased.
Statement D is correct because for an ideal gas undergoing adiabatic process, net heat absorbed is zero and thus, work done on the gas translates to an increase in internal energy.
Answer is D.

7. The total energy required to melt the object
= heat required to raise temperature to melting point and heat required to melt the object.

Mc

T

Ml
Since rate of heat absorbed is H .
Time required

Mc

T
H

Ml

M ( c

T
H
 l )
Answer is C.
8. For an ideal gas, pV = nRT.
Hence, when pressure is increased 3 times, temperature will increase 3 times as well.
When temperature increases 3 times, the kinetic energy will increase 3 times and hence the root mean square speed will increase 3 times.
Answer is D.
9. Taking vector sum of E
A
, E
B
, E
C and E
D
, the resultant E acts to the left.
As there are equal amount of positive and negative charges at the same distance away from the centre, the resultant potential at the centre is thus zero.
(A)

Q
E
C
E
A
(B)
+Q

(D)
Q
E
B
E
D
+Q
(C)
Answer is C.

10.
P
Q
1
Q
2
The electric potential at P due to charges Q
1
and Q
2
is the algebraic sum of the electric potentials due to individual charges at point P. Hence, the electric potential caused by Q
2
at P is (100
– 70) = 30 V.
The electric field strength at P due to charges Q
1
and Q
2
is the vector sum of the electric field strengths due to individual charges at point P. A possible value of the electric field strengths caused by Q
2
at P is 50
2 
30
2 = 40 V m -1 .
Answer is A.
11. A Electric potential at a point is the work done in bringing unit positive charge from infinity to the point.
B Electric potential gradient is a scalar.
D Unit potential gradient exists between any two points, if one joule of work is done in transporting one coulomb of charge over one metre.
Answer is C.
12.
Since
E

E
  dv dr

150

89
 
10
3
 

4690 V cm -1
The direction of electric field intensity is the direction of decreasing potential, the electric field intensity points towards P.
Answer is C.

13. The resistance is constant when small p.d is applied. At higher p.d, temperature increases , the resistance increases and Ohm’s law is not obeyed, corresponding to the I-V characteristics of filament lamp.
Answer is B.
14. A The magnitude of current through both sections is the same
. (Kirchoff’s
First Law)
B The resistivity remains constant throughout because material is unchanged.
C Resistance =
 l/A. Since width of narrower section is half, then A of narrower section is half, & the resistance per unit length is thus twice.
D Since the relative lengths of the sections are not given, there is no way to verify statement D.
Answer is C.
15.
To operate normally, the current in the lamp, I = P/V = 0.04 A.
Thus the current in the whole circuit should also be 0.04 A.
E = V + IR = 6 + 0.04 (200) = 6 + 8 = 14 V.
Answer is D.
16. Q = I t => Q = (E/R) t
=> Q

E t

A
L
[using
[using E = I R]
R


L
]
A
Answer is D.
17. Effective Resistance across RS

1
1
PD across PQ

200
15

15
10

40

65

15 V

1
50
PD across QR

10
15

10

40

65

10 V

40

PD across RS

40
15

10

40

65

40 V
As Q is connected to earth, potential at Q is 0 V.
As P is of higher potential than Q, potential at P is 15 + 0 = 15 V.
As R is of lower potential than P, potential at R is 0 – 10 = -10 V.
As S is of lower potential than R, potential at S is -10 – 40 = -50 V.

Common misconception:
If the potential difference is 65 V, some thinks that it should be the higher potential being 65 V and the lower potential being 0 V.
Another common problem is reading the drop in potential incorrectly. This is the issue addressed in option C.
Answer is D.
18. The decrease in temperature will cause an increase in the resistance of the thermistor while the increase in light intensity will cause a drop in the resistance of the LDR. As a result, the p.d. across the thermistor will increase while the p.d. across the LDR should decrease.
The answer cannot be B because the current would have passed through the diode instead of the LDR and hence, the changes in the resistance of the
LDR will not have a significant effect on the potential difference reading V
2
.
Answer is D.
19. Equivalent circuit for Option A:
P
2.0 Ω 4.0 Ω
Q
2.0 Ω
3.0 Ω
1.0 Ω 2.0 Ω
1
Effective resistance

2 .
2

1
2 .
0

4 .
0

2 .
0

1
1
1
3 .
0

1 .
0
1

2 .
0

Equivalent circuit for Option B:
P
2.0 Ω
3.0 Ω
1.0 Ω 2.0 Ω
S
4.0 Ω
2.0 Ω
1
Effective resistance
1
2 .
0

2 .
0

4 .
0

1
1
1
3 .
0

1 .
0
1

2 .
0
Equivalent circuit for Option C:
2.0 Ω 4.0 Ω 2.0 Ω
3.0 Ω
S
1.0 Ω 2.0 Ω
Q
1
Effective resistance
2 .
0

1
4 .
0

2 .
0

1
3 .
0

1 .
0
1

2 .
0

1 .
6


1 .
3


Equivalent circuit for Option D:
2.0 Ω 2.0 Ω 4.0 Ω
1.0 Ω 3.0 Ω
R
2.0 Ω
Q
Effective resistance
1
2 .
0

1 .
0

1
1
1

1 .
2

1
3 .
0

2 .
0
1

2 .
0

4 .
0
Common Problem:
Unable to redraw the circuit into one which is visually more easy to decipher format because not being able to go back to the fundamental of what is in series and what is in parallel when deciding the relationship between resistors.
Answer is A.
20. When the filament of bulb 2 breaks, there is no more connection across bulb
2. As a result, the section becomes an open circuit.
The combined resistance of bulb 2 and 4 is lower than the resistance of bulb
4 alone. Hence, the overall resistance of the circuit increases.
Due to an overall increase in resistance, the overall current decreases.
The current passing through bulb 1 and 3 will decrease, hence brightness decrease for both bulbs.
As the potential across bulb 4 increases, since the resistance is higher as compared to the effective resistance of bulb 2 and 4, the current flowing through it should be higher. Hence, the brightness of bulb 4 increases.
Common Problem:
Some may think that with one less bulb, there should be less resistance and hence a larger current for all.
Some may think that since the effective resistance of the entire circuit increases, there should be less current flowing through all resistors and hence brightness of bulb 4 should decrease.
Answer is D.

21. Applying right hand grip rule. Compass needle will be pointing in direction of magnetic field at that point.
Answer is C.
22. As the circular motion is caused by the magnetic force,
Bqv
 mr

2
Bqr

Bq

 m
 mr

2
Bq
T

 m (
2

T
2
 m
Bq
)
Hence, T is independent of v.
Answer is B.
23. Initially,
In the vertical direction, s
 ut

1
2 at
2 s

1
2 at
2
, sin ce u

0
When the speed is doubled, t '

As
1 t
2
F

Bqv .
When the speed is doubled, the magnetic force is doubled and hence, the acceleration is doubled. s '

1
2
( 2 a )(
1
2 t ) 2 
1
2 s
Hence, deflection is halved.
Answer is A.
24. The electric field causes it to deflect rightwards and the magnetic field causes it to deflect downwards, according to Fleming’s left hand rule.
Answer is D.

25. Applying right hand grip rule,
The magnetic field created by the vertical wires cancels out each other. The magnetic fields created by the horizontal wires reinforce each other creating a resultant field into the paper.
Answer is D.
Structured questions
1(a) (i) It acts as a point source to ensure the light reaching the double slits are coherent and in phase. [1]
(ii) The width of the single slit, S should be approximately equal to that
(iii) of the wavelength of the light. [1] x

2 .
5

D
 d
10

3
4

620

10

9 d

0 .
80
[1] d

7 .
9

10

4 m [1]
(iv) The interference fringe pattern will disappear. Only a diffracted light pattern from the remaining slit will form on the screen. [1]
(V) The fringes will appear closer together. [1]
1(b) (i) d sin sin n

 d



 n n


 d
1

600

10

3
650

10

9 n<2.6
[1]
The maximum value for n is 2, hence the 3 rd order maximum does not exist. [1]
(ii) sin n
0

 n
0 d
(

0
0


 n
(

0 d n v

0
)
0
 d


1 )
 v n v
 v d
 v n
0 n v


 v

0
  v
2

1

3

600
[1]
400

400

2 [1]

sin

 
2

46 .
1


600

10

9
10

2
[1]

600
2(a) (i) At the mentioned temperature, the material is boiling and heat supplied is used to increase the potential energy of the particles as well as to do work against the environment as the material expands.
[1]
The heat energy supplied does not goes towards increasing the random kinetic energy of the particles, which is directly proportional to the temperature of the object. [1]
Hence, the temperature remains constant.
(ii) Power supplied, P

VI

2 .
00 ( 0 .
500 )

1 .
00 W
Total energy supplied to boil the material,
E

Pt

1 .
00 ( 49625

47787 )

1840 J [1]
Assuming no energy lost to the environment,
Total energy supplied to boil the material = total energy absorbed to boil 1.50 kg of the material. ml v
1 .
50 l
 v
1840

J
1840 l v

1230 J kg

1 [1]
(b) (i) The reason is because there is heat lost to the environment and not all the heat supplied has been used to boil the material. [1]
(ii) From the first experiment,
1 .
50 l v
 h

1840 J _ (1)
From the second experiment,
3 .
60 l v
 h

2 .
50 ( 0 .
700 )( 49625

3 .
60 l v
 h

3220 _ (2) [1]
Using (2)
– (1), [1]
2 .
10 l v

1380 l v

657 J kg

1 [1]
47787 )
(c) (i) A real gas behaves like an ideal gas under high temperature and low pressure. [1]
The reason is because at high temperature, the particles are moving so fast that intermolecular forces are negligible and at low pressure, the particles are too far apart for intermolecular forces to be significant. [1]

(ii) Applying the equation of state, pV
 nRT nR
 pV
T
As nR are constant in this case, pV
 nRT
( pV
) before

( pV
) after
T T
Since pressure has remained constant,
(
V
T
) before

(
V
T
) after
[1]
T after

V after
V before
T before

0 .
5 V before
( 400 )

V before
200 K [1]
(iii) Number of particles present,
N

N
 pV
N
A
[1]
RT
1 .
01

10
5
( 1 .
00
8 .
31 ( 200 )

10

3
)
( 6 .
02

10
23
)

3 .
66

10
22
[1]
3 (a) (i) Electric field strength
Electrostatic force
Acceleration a

F
F m


E
 eE
V d


1 .
6
6 .
45
0 x 10

0 x 10

19

3
 x 7500
7 .
5
 x 10
1 .
2
3
Vm x 10

1

15
.
N
1 .
2 x 10

15
9 .
11 x 10

31

1 .
32 x 10
15 ms

2
.
[0.5]
[0.5]
[1]
Direction of a is vertically downwards towards the positive plate. [1]
(ii)
Let the initial speed be u .
Horizontal motion:
Time taken to travel 25 mm t

25 x 10

3 u cos 15 o
….(1) [0.5]
Vertical motion:
Vertical displacement = 0, time taken = t as given in equation (1) [0.5]
0
 u y t

1
2 at
2
0
 u sin 15 o t

1
2 at
2 [0.5]

u

1
( 1 .
32 x 10
15
2
25 x 10

3
)( sin 15 o cos 15 o
)

[1]
8 .
12 x 10
6 ms

1
.
[0.5]
(b) (i) Initial electric PE





4
Q
1
Q
2
 o r
1 .
6
4


8

10
.
85

19



10

12
2


1 .
6
1 .
0


10
10

9

19


4 .
6

10

19
J [ 1 ]
(ii) By law of conservation of momentum,
Total initial momentum = Total final momentum
0 = m p v p
－ m
 v

0 = (1.67

10 -27 )(2.0

10 4 ) － (4

1.67

10 -27 ) v

[1] v  = 5.0

10 3 m s
－
1 [1]
(iii) Total KE

1
2 m p v p
2 
1
2 m
 v

2


1

1 .
67

10

27

2 .
0
2
4 .
2

10

19
J

10
4
2


1
2

4

1 .
67

10

27

5 .
0

10
3
2

[1]
(iv) Initial total energy = final total energy
KE
1
+ PE
1
= KE
2
+ PE
2
0 + 4.6

10
0.4

10
－
19 = 4.2

10
－
19 =


1 .
6
－
19 +
Q
1
Q
2
4
 o r

10
4



19
8 .
85


2

10

1 .
6

12 r


10

19
 r = 1.2

10
－
8 m [1]
[1]
4(a) The electromotive force of a cell is the amount of energy converted from other forms to electrical energy when a unit charge passes between the two terminals [1] while the potential difference between two points in an electric circuit is the amount of electrical energy converted to other forms of energy when a unit charge passes between the two points.[1]
(b) (i) There is internal resistance of the cell. [1]
The resistance of the ammeter is not zero. [1]
[1]

The resistance of the voltmeter is not infinite. [1]
(ii) Resistance of the wire =
V 9.80
=
I 4.00 ×10 -3
= 2.45k
Ω [1]
The method is inappropriate as the current used in the calculation is not the current through the resistive wire only but includes the resistance through the voltmeter. [1]
(iii) Given that the resistance of the wire is 5 k

, and the effective resistance of the wire and voltmeter connected in parallel is 2.45 k

,
R × R voltmeter = 2.45k
Ω
R + R voltmeter
5R voltmeter = 2.45k
Ω
5 + R voltmeter

R voltmeter
= 4.80k
Ω
[2]
(iv) The ammeter should be connected in series with the resistive wire so as to measure the current in the wire only.[1]
The voltmeter should be connected across the ammeter and the wire to measure the p.d across the wire since the pd across the ammeter is negligible as the resistance of the ammeter is much lower than that of the resistive wire. [1]
5(a) (i) At balance point,
V
XJ
 emf of 1 .
5 V cell

1 .
5 V
R
[1]
E = 10.0 V
V
A
[1]
V

XJ l
XJ l
XY
V
XY
[1]

l

XJ
V
XJ
V
XY l
XY
V
XY

R
XY

R
XY
R var iable
 r
E driver cell
Hence,
V
XY

1 .
40
1 .
40

2 .
00

0 .
70
( 6 .
00 )

2 .
05 V [1] l
XJ

1 .
50
2 .
05
( 1 .
00 )

0 .
732 m [1]
(ii) At balance point, the potential across XJ along the wire and the e.m.f of the 1.50 V cell is the same. [1]
Since the positive terminal of the cell is at the same potential as point
X, the potential difference across the galvanometer is thus zero. [1]
With zero potential difference, no current flows.
(iii) A practical voltmeter does not have infinite resistance and thus draws current from the circuit. As a measuring device should not disrupt the physical quantity that it is measuring, a practical voltmeter has failed to do that. [1]
However, the potentiometer does not draw current at balance point
(as shown by zero deflection in the galvanometer) and thus, it does not disrupt the circuit and can measure the potential difference accurately. [1]
(iv) Before balance point is reached, current still flows through the galvanometer. Without the 5.00 Ω resistor, the current flowing through the galvanometer could be significant and thus spoil the galvanometer.
(b) At balance point,
V
XJ
 p .
d .
across 5 .
00
 resistor [1] l

XJ
V
XJ
V
XY l
XY
From part (a)(i),
V
XY

2 .
05 V [1] p .
d .
across 5 .
00
 resistor

R
R
5 .
00
5 .
00
 r
E
1 .
50

5 .
5 .
00
00

0 .
50
( 1 .
50 )

1 .
36 V [1] l
XJ

1 .
36
2 .
05
( 1 .
00 )

0 .
663 m [1]

6(a) (i) The magnetic flux density is the force experienced by a conductor of unit length, placed perpendicular to the magnetic field and carrying unit current.
(ii) The tesla is equivalent to a force of 1 N experienced by a conductor of 1 m long, placed perpendicular to the magnetic field and carrying a current of 1 A.
(b) (i) x x x x
F
B x x x x x x
I
30 o x x x x x x
W=mg x x x x g
(ii) For vertical equilibrium,
F
B cos 30
  mg [1]
BIl cos 30
  mg
I

Bl mg cos 30


5 .
0

10

3
1 .
5

10

3 
0

9 .
81
.
50 cos 30


76 A [1]
(c) (i)
(ii)
B
 
0 nI

4
F

BIl [1]
F

3 .
1

10

3



10

7 
700

3 .
5
3 .
5

5 .
0

10

2 

3 .
1

10

3
T
5 .
4

10

4
N
[1]
[1]
(iii) For frame to remain in horizontal position,
Total clockwise moment = total anticlockwise moment
F
B l
QC
 mgl [1] l
5 .
4

10

4
( 15 .
0

0 .
083 m [1]

10

2
)

0 .
10

10

3 
9 .
81 l [1]
(d) For circular motion,
Bqv r

 mv
Bq mv
2
[1] r

1 .
67

10

27 
1

10
5
0 .
18

1 .
6

10

19

5 .
8

10

3 m [1]
Since r

5 .
8

10

3 m , the protons will emerge from the side AB. [1]