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F5 Physics First Term Exam (10-11) Marking Scheme
1.(a) (i) Temperature of the gas = 273 + 6 = 279 K
(ii) By pV = nRT,
1.8  10 5  0.037
pV
 2.87 mol
n=
=
8.31  279
RT
 There are 2.87 mol of gas in the bag.
(iii) Since V is constant, by the pressure law,
p1
p
= 2
T2
T1
T
(273  18 )
Pressure in bag = p1  2
= 1.8  10 5 
= 1.88  105 Pa
(273  6)
T1
(b) Since the bag lengthen the time of impact,
the force on the driver is reduced for the same momentum change.
(c) The number of molecules inside the bag does not change.
By general gas law,
p1V1
pV
= 2 2
T1
T2
Final temperature =
p2V2
2.3  10 5  0.03
 279 = 289 K
 T1 =
p1V1
1.8  10 5  0.037
F
,
A
the force = pA = 100  1000  1  104 = 10 N
(b) Mass of mercury column = volume  density
= 1  10–4  0.1  13 400
= 0.134 kg
Since the mercury column remains stationary, net force = 0
Force due to atmospheric pressure + weight of mercury column
= force due to pressure of trapped gas
10 + 0.134  10 = p  1  10–4
p = 1.13  105 Pa
 The pressure of the trapped gas is 1.13  105 Pa.
(c) Since the volume of the gas is kept constant, by the pressure law,
p1
p
= 2
T2
T1
T
273  57
p2 = p1  2 = 1.13  10 5 
= 1.29  105 Pa
273  17
T1
Increase in force due to trapped gas = increase in weight of mercury
p  A = mg
(1.29  105 – 1.13  105)  10–4 = m  10
m = 0.16 kg
 0.16 kg of mercury should be added.
2.(a) By p =
3.(a) Sound waves cannot travel through a vacuum.
(b) Sound waves can be transmitted up the wire suspending the bell to the
outside. / Air is not completely removed, a weak sound is still transmitted
through the bell jar to outside.
(c) (i) No. It is because the longitudinal sound wave is converted into
electrical signal. The CRO just displays how the electrical signal
varies with time and is actually not a wave.
(ii) No,
because there is no energy transferred across the screen.
(Or other reasonable explanations)
(iii) Period  6.5  0.1  103 = 6.5  104 s
1A
1M
1A
1M
1A
1A
1A
1M
1M
1A
1M
1A
1M
1M
1A
1M
1M
1M
1A
1A
2A
1A
2A
1A
1A
1A
Frequency =
1
T
1M
1
=
= 1540 Hz
6.5  10  4
(iv) The sounds produced by the bell and the guitar have different
quality,
the waveforms displayed on the CRO will be different.
1A
1A
1A
Q
4.(a) E =
1A
4π 0 r
where Q is the charge of the sphere, r is the radius of the sphere, ε0 is the
permittivity of free space.
2
1A
(b)
(i) By E =
Q
4π 0 r 2
,
Q = 4πε0r 2E= 4π(8.85  10–12)(0.2)2(3  106)= 1.3345  10–5 C  1.33  10–5 C
The maximum charge stored on the sphere is 1.33  10–5 C.
Q
(ii) V =
4π 0 r
=
1.33  10 5
4 π (8.85  10
12
)(0.2)
= 6.00  105 V
1M+1A
1M
1A
The max. electric potential at the surface of the sphere is 6.00  105 V.
5.(a)
(b)
(Correct shape of electric field lines.)
(Correct direction of electric field lines.)
By conservation of energy,
gain in KE = loss in PE
= qV
= (1.6  10–19)(1900) = 3.04  10–16 J
By KE 
1A
1A
1M
1
mv 2 ,
2
speed of electron, v =
1M
2  3.04  10 16
2  KE
=
= 2.58  107 m s–1
m
9.1 10 31
1A
(c)
Magnitude of the electric field strength =
V 450
=
= 15 000 V m–1
d 0.03
1M+1A
(d)
1A
(e)
(f)
19
F eE 1.6  10  15000


 2.63  1015 ms 2
31
m m
9.1  10
1 2 1
0.3
y  at   2.63  1015  (
) 2  0.178 m
7
2
2
2.58  10
a
1-5 B C B B C
6-10 D A C C C
11-15 D B D D C
1M+1A
1A for y=(1/2)at2, 1A for
time of flight, 1A for
answer
16-20 D D C A B
21-25 B D C A C
26 D
Explanations to mc
2. Pressure, volume and temperature are the same for both gases. Thus the number of moles is also the same.
For statement 2, since the temperatures are the same, so the average KE is the same for both gases.
Hydrogen being lighter, has higher rms speed.
For statement 3, hydrogen moves faster, so hydrogen molecules collide more frequently with the container
wall.
3. Use
p1V1
pV
= 2 2.
T1
T2
4. Use c rms 
3RT

5. TX > TY.
15. P.d. at Q = 2.5 = 2.5 (v/f) = 2.5 x 340/850 = 1 m
16. When f  2f,   (1/2) , i.e.  = 0.2 m
P.d. = 1 m = 5
18. All charges flow to earth finally.
19. Non-metals can only be charged by friction. Charge sharing and induction requires free electrons.
20. E = V/d = 60/0.02 = 3000 V/m
F = qE = 5 x 3000 = 15000 N
21. Statement (1) is obvious. (Imagine a positive charge is placed at X)
Statement (2) is also obvious.
 Q   Q 
 + 
 = 0
Potential at X = 
4

r
4

r
0  
0 

Electric field at X is the smallest. (Try some numbers to get answer)
22. It is not 100 % sure about the charges on C and D. It is because attraction may happen when one is charged
and one is uncharged and one is charged and the other is neutral.
23. The force and acceleration depends only on the direction of electric field but not dependent on the direction
of initial velocity.
26. Initially,
Po V = nRTo
Finally, LHS, PV = n’RTo
RHS
PoV = n’’RT
Since no gas is escaped,
n’ + n” = 2n
PV PV 2 P0V


RT0 RT
RT0
=> P = 2 Po T / (T + To)