CE Phy 2006 (Paper 1)
1.
(a)
v
3 × 108
Wavelength λ = f = 1.2 x 109 = 0.25 m
(b)
Time lag = (6.5)(5 × 10-6) = 3.25 x 10-5 s
1
Distance = 2 (3 x 108)(3.25 × 10-5) = 4875 m ( ≈ 4880 m)
2
(a)
(i)&(ii)
X
on top of the
radar station
Reflection line at the base of plane,
with angle of incidence = angle of
reflection
(b)
(c)
3.
(a)
Cooling the exhaust gas reduces much of the IR radiation, making the
IR sensors harder to detect the plane.
Quiet engines
Paint the plane in black (visually blended in with the background)
Distance = area under v-t graph
1
= 2 (9.25 - 8.90)(2.95 - 0) = 0.516 m
(b)
(c)
PE loss = mgh = (0.154)(10)(0.516) = 0.795 J
From graph, the final speed = 2.95 m s-1
1
1
Maximum KE = 2 mv2 = 2 (0.154)(2.952) = 0.670 J
4.
(d)
There is energy loss due to air resistance.
(a)
(i)
normal reaction
friction
weight
1
s = ut + 2 at2
(ii)
1
5 = (0)(2) + 2 a(22)
a = 2.5 m s-2
Net force = ma = (10)(2.5) = 25 N
(b)
P
5.
(a)
Q
R
S
L is a concave / diverging lens
because the image formed is erect and diminished.
(b)
6.
(c)
Larger angle of view
(a)
Whole tank of water can be heated evenly by convection.
(OR To prevent overheating in case the level of water is too low.)
(i)
Energy required E = mcΔT
= (24.1 x 1)(4200)(65 - 15) = 5.06 x 106 J
(b)
(ii)
Assuming 100% efficiency,
E
Power P = t = VI
5.06 x 106
24.3 x 60 = (220)I
(c)
*(7)
I = 15.8 A
Since the current in the wires is large, using thicker wires can reduce
the resistance and thus effectively reduce the power loss in the wires.
Use the electromagnet to lift up one end of the chain of iron clips.
Count the number of clips that the electromagnet can hold just before the
chain falls.
Repeat the above steps by varying the number of turns in the coil and
recording the corresponding changes in the number of clips that the
electromagnet can hold just before the chain falls.
In every trial, keep the current constant.
8.
(a)
(b)
(c)
(d)
β radiation and γ radiation
It prevents the film from exposing to ordinary light, which would give
similar results as if it is exposed to radiations.
(i)
John was definitely exposed to γ radiation because
*
the region on the film beneath the 5 mm lead sheet was
blackened
*
γ radiation can penetrate lead sheets and the region on
the film beneath the lead sheet was blackened.
Mary was definitely exposed to β radiation because
*
the region on the film beneath the 5 aluminium sheet
was blackened but that beneath the lead sheet was not
*
the regions beneath the 1mm and 3 mm aluminium
sheets were blackened to different degrees.
(ii)
Ken was exposed to higher dose of radiation than Mary.
It can destroy or damage living cells.
(any one)
It can lead to cancer.
It can change the DNA structure.
9.
(a)
(i)
(1)
(b)
10.
(a)
(b)
(c)
(e)
From the graph, N ≈ 7.5 when v2 = 0
As thickness of each box is 1 m, stopping distance ≈ 7.5 m
By v2 - u2 = 2as,
0 – 272 = 2a(7.5)
a ≈ 48.6
Average resistive force = ma = 1600 × 48.6 = 77800 N
(2)
From the graph, N ≈ 7.5 when v2 = 0
The minimum number of boxes required = 8
(ii)
A straight line graph with a slope with a smaller value and a
small intercept on the vertical axis
Concrete is more rigid, which will make the collision time shorter. A
larger force will be exerted on the car during collision.
When the molecules of the refrigerant with greater kinetic energy leave
the liquid refrigerant, the average kinetic energy of molecules in the
liquid refrigerant decreases. The temperature of the refrigerant drops.
The coiled tube increases the contact surface area with air (and the
fins).
It helps to condense/collect water from the air.
Decreases
Any Two:
*
The coil in Part B of the dehumidifier is painted black to radiate
heat.
*
The air fan creates a flow of air current.
*
*
(e)
11.
(a)
The metal fins and/or the coiled tube carry heat away by
conduction.
The holes in Part B of the dehumidifier allow a flow of air
current.
(i)
(ii)
E = mLv = (1.5)(2.26 × 106) = 3.39 x 106 J
Mass of the air in the room = volume × density
= 400 ×1.3 = 520
Energy released by water vapour = Energy gained by
surrounding air
3390000 = 520 (1030) (ΔT)
ΔT =6.33 ˚C
(i)
The brightness of bulb remains unchanged. It reveals that the
voltage across the bulb does not change when the resistance of
the variable resistor changes.
(ii)
A
(iii)
Resistance of R = 5 Ω
3
V
Total resistance = I = 3 = 1 Ω
1
1
1
+
=
5
RB
1
RB = 1.25 Ω
*(b)
(c)
When the resistance of R is reduced, the total current will increase.
The voltage drop due to internal resistance of the dry cells increases.
The voltage across the bulb decreases.
As a result the brightness of the bulb decreases.
Mary’s answer is incorrect.
*
The bulb is short-circuited.
*
No current passes through the bulb.
*
The voltage across the bulb is zero.