Download 2008-SRJC-PH-H2-J2-CT-w-soln

H2 PHYSICS COMMON TEST 1 SOLUTIONS:
MCQ
1.
1

2
Answer is B.
2. Since d sin   n
d sin 
n

d
As sin   1, n 

1
 10 3
 3.33
Maximum order n  500
600  10 9
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.
2
  2
1
As the sources are antiphase, path difference of 2 means the meet at P and
result in destructive interference.
3. Path difference  6  4  2 m 
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,
D 3.0  102  2000
Fringe separation 

 4.0 m
a
15
4 .0
 0.044 s
Period of fluctuations 
90
Answer is A.
5. Net work done on gas = - Area under graph.
1
  [(3.231  1.012)  105 ][( 22.40  20.40)  10 3 ]
2
 221.9 J
By First Law of Thermodynamics,
U  Q  W
Since the ideal gas went through a full cycle,
U  0
Hence,
0  Q  (221.9)
Q  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.
McT  Ml M (cT  l )

Time required 
H
H
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 EA, EB, EC and ED,
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
(B)
+Q
EA
EC
EB
ED
Q
(D)
Answer is C.
+Q
(C)
10.
P
Q1
Q2
The electric potential at P due to charges Q1 and Q2 is the algebraic sum of
the electric potentials due to individual charges at point P. Hence, the electric
potential caused by Q2 at P is (100 – 70) = 30 V.
The electric field strength at P due to charges Q1 and Q2 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 Q2 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   103
 32  19 
 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=It
=>
=>
Q = (E/R) t
A
Q  Et
L
[using E = I R]
[using R 
L
]
A
Answer is D.
17. Effective Resistance across RS 
PD across PQ 
1
1
1

200 50
 40 
15
 65  15V
15  10  40
10
 65  10V
15  10  40
40
 65  40V
PD across RS 
15  10  40
PD across QR 
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 V2.
Answer is D.
19. Equivalent circuit for Option A:
P
2.0 Ω
2.0 Ω
4.0 Ω
Q
3.0 Ω
1.0 Ω
Effective resistance
2.0 Ω
1
1

2 . 0  4 . 0 2 .0 
 2 .2 
1
1
1
1

3 .0 1 .0  2 . 0
Equivalent circuit for Option B:
P
2.0 Ω
S
4.0 Ω
2.0 Ω
3.0 Ω
1.0 Ω
Effective resistance
2.0 Ω
1
1
1

1
2 .0 2 .0  4 .0 
1
1

3 .0 1 .0  2 . 0
 1 .6 
Equivalent circuit for Option C:
2.0 Ω 4.0 Ω 2.0 Ω
3.0 Ω
Q
S
1.0 Ω
Effective resistance
2.0 Ω
1
1
1
1


2.0  4.0  2.0 3.0 1.0  2.0
 1.3 
Equivalent circuit for Option D:
2.0 Ω 2.0 Ω 4.0 Ω
1.0 Ω
R
3.0 Ω
Q
2.0 Ω
Effective resistance
1
1

2 .0 1 .0 
 1 .2 
1
1
1
1

3 .0 2 .0  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   mr 2
Bq  m
2
)
T
2m
T 
Bq
Hence, T is independent of v.
Bq  m(
Answer is B.
23. Initially,
In the vertical direction,
1
s  ut  at 2
2
1
s  at 2 , sin ce u  0
2
When the speed is doubled,
1
t ' t
2
As F  Bqv .
When the speed is doubled, the magnetic force is doubled and hence, the
acceleration is doubled.
1
1
1
s'  (2a)( t ) 2  s
2
2
2
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)
(ii)
(iii)
(iv)
(V)
1(b)
(i)
It acts as a point source to ensure the light reaching the double slits
are coherent and in phase. [1]
The width of the single slit, S should be approximately equal to that
of the wavelength of the light. [1]
D
x
d
2.5  103 620  109  0.80
[1]

4
d
d  7.9  104 m [1]
The interference fringe pattern will disappear. Only a diffracted light
pattern from the remaining slit will form on the screen. [1]
The fringes will appear closer together. [1]
d sin   n
n
sin  
1
d
d
10 3
n 
 600  650  10 9
n<2.6 [1]
The maximum value for n is 2, hence the 3rd order maximum does not
exist. [1]
(ii)
n0 0 nv v

d
d
n0 0 (n0  1)v

d
d
n0 (0  v )  v
sin  
v
400
 2 [1]
0  v 600  400
nv  2  1  3 [1]
n0 

2  600  10 9  600
10 2
  46.1 [1]
sin  
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)  1.00W
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.
mlv  1840 J
1.50l v  1840
lv  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,
1.50lv  h  1840 J _ (1)
From the second experiment,
3.60lv  h  2.50(0.700)(49625  47787)
3.60lv  h  3220 _ (2) [1]
Using (2) – (1), [1]
2.10lv  1380
l v  657 J kg 1 [1]
(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
pV
T
As nR are constant in this case,
pV  nRT
nR 
pV
pV
)before  (
) after
T
T
Since pressure has remained constant,
V
V
( )before  ( ) after [1]
T
T
Vafter
0.5Vbefore
Tafter 
Tbefore 
(400)  200 K [1]
Vbefore
Vbefore
(
(iii)
3(a)
(i)
Number of particles present,
pV
N
N A [1]
RT
1.01  105 (1.00  103 )
N
(6.02  1023 )  3.66  1022 [1]
8.31(200)
V
45  0
 7.5 x10 3 Vm 1 .
Electric field strength E  
3
d 6.0 x10
Electrostatic force F  eE  1.6 x10 19 x7500  1.2 x10 15 N
F 1.2 x10 15
Acceleration a 

 1.32 x1015 ms 2 .
31
m 9.11x10
Direction of a is vertically downwards towards the positive plate.
[0.5]
[0.5]
[1]
[1]
(ii)
Let the initial speed be u.
Horizontal motion:
Time taken to travel 25 mm
t
25 x10 3
u cos15 o
….(1)
[0.5]
Vertical motion:
Vertical displacement = 0, time taken = t as given in equation (1) [0.5]
0  uy t 
1 2
at
2
0  u sin 15o t 
1 2
at
2
[0.5]
u
1
25 x10 3
(1.32 x1015 )(
)
2
sin 15 o cos15 o
[1]
 8.12x106 ms 1.
(b)
(i)
[0.5]

Initial electric PE
Q1Q2
4o r
 1.6  10  2  1.6  10 

4 8.85  10 1.0  10 
19
19
12
9
 4.6  1019 J
(ii)
[1]
By law of conservation of momentum,
Total initial momentum = Total final momentum
0 = mpvp－mv
0 = (1.6710-27)(2.0104)－(41.6710-27) v [1]
v= 5.0103 m s－1

(iii) Total KE
[1]
1
1
2
2
m p v p  m v
2
2
2
2
1
1

1.67  10 27 2.0  104   4  1.67  10 27 5.0  103 
2
2
19
 4.2  10 J
[1]

(iv)
Initial total energy = final total energy
KE1 + PE1 = KE2 + PE2
－
－
0 + 4.610 19 = 4.210 19 + Q1Q2
[1]
4 o r
0.410－19 =
－8
r = 1.210
 1.6 10  21.6 10 
r
4 8.85 10
19
19
12
m
[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]
(ii)
The resistance of the voltmeter is not infinite. [1]
V
9.80
= 2.45kΩ
Resistance of the wire = =
[1]
I 4.00 ×10 -3
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 × Rvoltmeter
= 2.45kΩ
R + Rvoltmeter
5Rvoltmeter
= 2.45kΩ
5 + Rvoltmeter
 Rvoltmeter = 4.80kΩ
[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]
E = 10.0 V
[1]
R
A
V
5(a)
(i)
At balance point,
VXJ  emf of 1.5V cell  1.5V [1]
l
VXJ  XJ VXY [1]
l XY
l XJ 
VXJ
l XY
VXY
VXY 
RXY
Hence,
RXY
Edriver cell
 Rvar iable  r
1.40
(6.00)  2.05V [1]
1.40  2.00  0.70
1.50

(1.00)  0.732 m [1]
2.05
VXY 
l XJ
(b)
(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.
At balance point,
VXJ  p.d . across 5.00  resistor [1]
V
l XJ  XJ l XY
VXY
From part (a)(i),
VXY  2.05V [1]
R5.00
5.00
p.d . across 5.00  resistor 
E1.50 
(1.50)  1.36V [1]
R5.00  r
5.00  0.50
1.36
l XJ 
(1.00)  0.663 m [1]
2.05
6(a)
(i)
(ii)
(b)
The magnetic flux density is the force experienced by a conductor of
unit length, placed perpendicular to the magnetic field and carrying
unit current.
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.
FB
(i)
x
x
x
x
x
x
x
x
I
x
x
o
x 30 x
x
x
x
x
x
x
x
x
g
W=mg
(ii)
For vertical equilibrium,
FB cos 30  mg [1]
BIl cos 30  mg
I
(c)
(i)
(ii)
(iii)
mg
5.0 103  9.81

 76 A [1]
Bl cos 30 1.5 103  0.50 cos 30
B  0 nI  4  107  700  3.5  3.1  103 T [1]
F  BIl [1]
F  3.1  103  3.5  5.0  102  5.4  104 N [1]
For frame to remain in horizontal position,
Total clockwise moment = total anticlockwise moment
FBlQC  mgl [1]
5.4  104 (15.0  102 )  0.10  103  9.81l [1]
l  0.083 m [1]
(d)
For circular motion,
mv2
[1]
Bqv 
r
mv 1.67  1027  1 105
r

 5.8  103 m [1]
Bq
0.18  1.6  1019
Since r  5.8  10 3 m , the protons will emerge from the side AB. [1]