Download 1473066326.

535/1
PHYSICS
Paper one
AUGUST 2016
2 hours 15 minutes
GAYAZA HIGH SCHOOL
MOCK EXAMINATIONS
PHYSICS
Paper one
2 hours 15 minutes.
Instructions to candidates:
Attempt FIVE questions only.
Start each question on a fresh page.
The following values may be useful
Density of water
= 1000 𝑘𝑔𝑚−3
Density of mercury
= 13600 𝑘𝑔𝑚−3
Specific heat capacity of water
= 4200 𝐽𝑘𝑔−1 𝐾 −1
Page 1 of 34
State Charles’s law.
(01 mark)
Describe an experiment to verify Charles’s law.
(06 marks)
At 27𝑜 𝐶 a fixed mass of gas occupies a volume of 120 𝑐𝑚3 . Find the
volume of the gas if it is cooled to 7𝑜 𝐶 , at constant pressure.
(02 marks)
The diagram below shows a shiny silver surfaced electric kettle being used to
heat water from a mains supply.
1. (a)
(b)
(c)
(d)
Plastic lid
Rubber
stands
Heating
element
Briefly explain how each of the following features of the kettle
enhances its efficiency.
(i)
The heating element is situated at the bottom of the kettle.
(01 mark)
(ii)
The kettle is made of shiny silvered surface.
(01 mark)
(iii)
The lid is made of plastic material.
(01 mark)
3
(e) A kettle rated at 1600 W, 240 V is used to heat 2000 𝑐𝑚 water
initially at 25𝑜 𝐶 to its boiling point. If the specific heat capacity of the
material of the kettle is 120 𝐽𝑘𝑔−1 𝐾 −1 , find how much time is taken for
the water to reach its boiling point.
(04 marks)
Page 2 of 34
2. (a)
(b)
State Snell’s law of refraction.
(01 mark)
Describe an experiment to determine the refractive index of glass using a
glass block.
(06 marks)
(c)
B
70𝑜
20𝑜
𝑥𝑜
𝑦𝑜
C
A
𝑧
(i)
(ii)
(d)
3. (a)
(b)
𝑜
The diagram above represents a ray of light travelling from air as it enters
and emerges from the prism ABC of refractive index, 1.6 .
Find the angle 𝑦 𝑜 .
(04 marks)
State the conditions necessary for the ray of light to behave as shown in the
diagram when it is incident on the face BC of the prism.
(02 marks)
Use the graphical method to find the location of the image of an object,
2 cm tall that is placed 10 cm in front of a concave lens of focal length
15 cm.
(03 marks)
State the law of charges.
(01 mark)
Describe with the aid of suitable diagrams how a gold leaf electroscope
can be charged negatively by induction.
(05 marks
Page 3 of 34
(c)
++
++
++
++
A sharp pin is placed on the cap of the gold leaf electroscope as shown in the
diagram above. A positively charged rod is held next to the sharp end of the pin.
Draw the diagram and use it to explain what happens to the electroscope and the
charged rod.
(04 marks)
(d)
(i)
Explain what is meant by polarization as supplied to a simple
cell.
(02 marks)
(ii) State how polarization can be minimized in a simple cell.
(01 mark)
(e)
1.5V,r=1Ω
1.5V,r=1Ω
3Ω
1Ω
A
6Ω
S
The figure above shows two cells each of e.m.f. 1.5 V and internal
resistance of 1 Ω, connected to three resistors and a switch, S. Find the
reading of the ammeter, A when the switch is closed.
(03 marks)
Page 4 of 34
4. (a)
(i)
(ii)
Define moment of a force.
State the principle of moments.
(01 mark)
(01 mark)
(b)
Bale of cotton
piston
A
P
B
Effor
t
The diagram above represents a system used to compress a bale of cotton of
weight 10000 N. The rigid rod AB is 2 m long and is pivoted at a point P,
0.5 m from end A. At end A of the rod is hinged a piston arm at the opposite
end of which is a ram of area 2 𝑚 2 . If the pressure exerted on the cotton bale
is 4200 𝑁𝑚−2 , find the minimum effort required to exert this pressure when
the system is in equilibrium.
(04 marks)
(c)
(d)
(i)
Describe an experiment to demonstrate that pressure in liquids
changes with depth.
(04 marks)
(ii)
Find the pressure exerted on a deep sea diver who is 25 m
below the surface of the sea whose waters are of density
1200 𝑘𝑔 𝑚3 , assuming the atmospheric pressure is 75 cmHg.
(03 marks)
A spring balance reads 3.56 𝑁 when a metal cube of side 2.0 𝑐𝑚 is
suspended in air from the spring. Find the reading of the spring
balance when the metal cube is completely submerged in mercury.
(03 marks)
Page 5 of 34
(a)
Define the following term acceleration in relation to motion in a
straight line.
(01 mark)
(b)
The graph below represents the velocity-time graph of the motion of a
body of mass 5 𝑘𝑔.
Velocity (ms-1)
5.
20
15
10
5
0
(i)
(ii)
10
25
35
Find the force acting on the body for the first 10 seconds of its motion.
(03 marks)
Find the power generated by the retarding force in the last 10 seconds
of the body’s motion.
(05 marks)
(c)
Distinguish between elastic and inelastic collisions.
(d)
In a game of pool, a white ball of mass 120 𝑔 moving with a
velocity of 5 𝑚𝑠 −1 hits a stationary red ball of mass 150 𝑔.
After the collision, the white ball moves with a velocity of
1 𝑚𝑠 −1 in the opposite direction. Find;
(i)
(ii)
Time(s)
The velocity of the red ball after the collision.
State two possible energy losses during the collision.
(02 marks)
(03 marks)
(02 marks)
Page 6 of 34
(a)
State three differences between sound and radio waves.
(b)
Describe an experiment to show that sound requires a physical medium
for its transmission.
(05 marks)
(c)
(i)
What is an echo?
(ii)
The captain of ship transmitted an ultrasonic sound to the
bottom of the sea-bed in order to detect coral reefs. The echo
was detected 0.12 𝑠 later after transmission. If the speed of
sound in water is 1400 𝑚𝑠 −1 , find the depth of the coral reefs
from the ship.
(03 marks)
(d)
(03 marks)
(01 mark)
.
Displacement (m)
6.
10
0
0.2
0.4
0.6
Time (µs)
(Note : 1𝜇𝑠 = 1.0 × 10−6 𝑠)
The graph above shows the wave profile of a radio wave. Find;
(i)
(ii)
7.
The amplitude of the wave.
The frequency of the wave.
(01 mark)
(03 marks)
(a)
What is meant by magnetic saturation?
(01 mark)
(b)
With the aid of a diagram, explain what is meant by magnetic
shielding.
(03 marks)
Page 7 of 34
(c)
(d)
8.
Draw a diagram to show the magnetic field patterns resulting
from two straight conductors placed vertically near each other
carrying a current in:
(i)
the same direction.
(01 mark)
(ii)
opposite directions.
(01 mark)
(i)
Draw a labeled diagram to show the essential parts of a
simple d.c. motor and explain how it works.
(05 marks)
(ii)
State any two ways in which the power generated by the
d.c. motor may be increased.
(02 marks)
(e)
A moving coil galvanometer has a coil of resistance 4 Ω and
gives a full scale deflection of 25 mA. Find the value of the
resistance required to convert to an ammeter which reads 15 A
at full-scale deflection.
(03 marks)
(a)
(i)
(ii)
(b)
(i)
(ii)
Define the term half-life as applied to a radioactive
material.
(01 mark)
In 168 seconds, the activity of Thorium falls to oneeighth of its original value. Determine its half – life.
(03 marks)
With the aid of a well labeled diagram, explain how X –
rays are produced.
(06 marks)
State two uses of X- rays.
(02 marks)
Page 8 of 34
(c)
+
𝐴
Radioactive
material
𝐵
𝐶
−
A radioactive material emits radiations which are directed
between oppositely charged metal plates as shown in the
diagram above.
(i)
Name the radiations labeled A,B and C.
(03 marks)
(ii) What happens when the radioactive material is completely
covered with an ordinary sheet of paper?
(01 mark)
END
Page 9 of 34
MARKING GUIDE
QUESTION 1
(a)
(01 mark)
Charles’s states that:
The volume of a fixed mass of a gas in directly
proportional to the thermodynamic temperature, provided
the pressure is constant.
(b)
An experiment to verify Charles’s law.
(06 marks)
Thermometer
Stirrer
Conc. Sulphuric acid
Capillary tube
mm scale
Gas column
water
Heat
A column of a fixed mass of gas is trapped by an index of
concentrated sulphuric acid in a capillary tube sealed at one end.
The tube is tied to a mm scale and placed in a beaker containing
water and a thermometer.
The initial temperature, T of the water and the volume, V of the
gas column are measured and recorded.
The water is heated and after suitable intervals of time, the
temperature of the water, T and the corresponding volume, V of
the gas column are measured and recorded.
Page 10 of 34
The values of T and Volume V are recorded in a suitable table.
V(𝑚𝑚3 )
T(o C)
A graph of Volume, V and temperature, T is plotted.
V[mm3)
-273oC
T(oC)
The straight line graph of constant gradient verifies that the
volume of a fixed mass of a gas is directly proportional to the
thermodynamic temperature provide the pressure is constant.
(c)
(02 marks)
𝑇1 = 27 + 273 = 300 𝐾,
𝑉1 = 120 𝑐𝑚
𝑇2 = 7 + 273 = 280 𝐾,
𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔
𝑉1
𝑇1
=
3
𝑉2 =?
𝑉2
𝑇2
120
𝑉2
=
300 280
𝑉2 = 112 𝑐𝑚3
Page 11 of 34
(d)
(i) The heating element is placed at the bottom of the
kettle in order to facilitate the heat transfer by
convection in the water.
(ii) The kettle has a shiny silvered surface which
minimizes heat loss to the surroundings by radiation.
(iii)The lid is made of plastic material to minimize heat
loss by conduction. In addition, it reduces heat loss by
evaporation.
(e)
ℎ𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝑏𝑦 ℎ𝑒𝑎𝑡𝑖𝑛𝑔 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
= ℎ𝑒𝑎𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑤𝑎𝑡𝑒𝑟 + 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑜𝑓 𝑘𝑒𝑡𝑡𝑙𝑒
(01 mark)
(01 mark)
(01 mark)
(04 marks)
𝑝𝑜𝑤𝑒𝑟 × 𝑡𝑖𝑚𝑒 = (𝑚𝑤 𝑐𝑤 + 𝑚𝑘 𝑐𝑘 )∆𝜃
1600 × 𝑡 = (𝑉𝑤 𝜌𝑤 𝑐𝑤 + 𝑚𝑘 𝑐𝑘 )∆𝜃
1600 × 𝑡 = (2.0 × 10−3 𝑚3 × 1000 × 4200 + 0.25 × 120)∆𝜃
1600 × 𝑡 = (8400 + 30)(100 − 25)
𝑡𝑖𝑚𝑒, 𝑡 =
8430
𝑠 = 395.16 𝑠
1600
Page 12 of 34
QUESTION 2
(a)
Snell’s law states that:
(01 mark)
The ratio of the sine of the angle of incidence to the sine of
the angle of refraction is a constant for a given pair of
optical media.
(b)
Describe an experiment to determine the refractive index
of glass using a glass block.
(06 marks)
E
P1
P2
A
i
O
B
r
D
G
C
P3
P4
H
A glass block is placed on a white sheet of paper fixed on a soft
board with thumb pins.
The out line of the glass block ABCD is traced out on the white
sheet of paper.
The glass block is then removed and a normal to face AB is
drawn at point O, about 2 cm from A.
A line EO making an angle 𝑖 = 10𝑜 with the normal at O is
drawn to meet face AB at O.
Page 13 of 34
The glass block is replaced on its outline and optical pins P1 and
P2 stuck on the line EO,
Pins P3 and P4 are then stuck into the soft board such that they
are in line with the images of pins P1 and P2 when viewed
through the glass block from the opposite face CD,
The four pins and the glass block are then removed from the
paper.
A line joining the marks left by P3 and P4 is drawn to meet CD
at G.
Point G is also joined to O and the angle of refraction, r is
measured and recorded.
The procedures are repeated for 𝑖 = 20𝑜 , 30𝑜 , 40𝑜, , 50𝑜 𝑎𝑛𝑑 60𝑜 .
The results are recorded in a suitable table including values of
sin 𝑖 𝑎𝑛𝑑 sin 𝑟 as shown below:
i (o)
r (o)
sin 𝑖
sin 𝑟
A graph of sin 𝑖 against sin 𝑟 is plotted as below:
sin 𝑖
sin 𝑟
The refractive index of the glass block is determined as the slope
of the straight line graph obtained.
Page 14 of 34
(c)
(04 marks)
B
70𝑜
20𝑜
𝑥𝑜
𝑦𝑜
C
A
𝑧
𝑜
(i)
𝜂𝑎 sin 𝑖𝑎 = 𝜂𝑔 sin 𝑖𝑔
1 × sin 20 = 1.6 sin 𝑥
𝑥 = 𝑠𝑖𝑛−1 (
sin 20
) = 12.34𝑜
1.6
Now:
𝑦 = 70𝑜 − 𝑥
𝑦 = 70𝑜 − 12.34𝑜
𝑦 = 57.66𝑜
(ii)
The conditions necessary for the ray of light to behave as shown
in the diagram when it is incident on the face BC of the prism
are:
1.
2.
(02 marks)
Light must be travelling from an optically dense
medium to a less optically dense medium.
The angle of incidence must be greater than the angle
of refraction of the dense medium.
Page 15 of 34
(d) With the aid of a ray diagram, find the location of
the image of an object,
2 cm tall that is placed 10
cm in front of a concave lens of focal length
15 cm.
(03 marks)
Page 16 of 34
QUESTION 3
(01 mark)
(a)
The law of charges states that:
Unlike charges attract, like charges repel each other,
(b)
Charging a gold leaf electroscope negatively by induction.
(05 marks)
+
+
+
+
+
− −
−
−
Electron flow
+
+
A positively charged rod is brought near the cap of the uncharged
GLE.
This cause electrons to flow from the brass plate and the leaf
towards the brass cap,
+
+
+
+
+
+
+
+
Electron flow from the
earth via earth connection
With the positively charged rod still in position, the brass cap is
momentary earthed.
Page 17 of 34
Electrons flow from the earth via the earth connection to
neutralize the charge on the plate and leaf and the leaf collapses.
−
− − −
−
−
The earth connection is broken and the positively charged rod is
also removed.
The GLE will be left with a net negative charge , since it has
gained electrons from the earth.
(c)
++
++
++
++
+
(04 marks)
+
−−
−−
+
+
The positively charged rod induces a negative charge at the sharp
end of the pin and a positive charge at its opposite end as well as
the brass p[ate and leaf of the GLE.
Point action then takes place at the sharp end of the pin, leading
to ionization of the air around it.
The negative ions are attracted to the positively charged rod and
neutralize it.
The positively charged ions are attracted to the sharp point and
GLE causing leaf divergence. The GLE becomes positively
charged
Page 18 of 34
(d) (i)
Polarization is the formation of hydrogen gas at the
positive electrode . The hydrogen bubbles insulate the positive
electrode from the electrolyte and this increases the internal
resistance of the simple cell,
(02 marks)
(ii) Polarization in a simple cell can be minimized by
introducing a depolarizing agent like manganese (iv) oxide that
oxidizes the hydrogen to water .
(01 mark)
(03 marks)
(e)
3Ω
1Ω
S
A
𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙, 𝑅𝑝 =
6Ω
3 × 6 18
=
= 2Ω
3+6
9
𝑡𝑜𝑡𝑎𝑙 𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑅 = 1 + 2 = 3Ω
𝑡𝑜𝑡𝑎𝑙 𝑒. 𝑚. 𝑓, 𝐸 = 1𝑉
𝑡𝑜𝑡𝑎𝑙 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 , 𝑟 =
1×1 1
= = 0.5Ω
1+1 2
𝐸 = 𝐼(𝑅 + 𝑟)
1 = 𝐼(3 + 0.5)
𝐼=
1
3.5
= 0.286 𝐴 is the ammeter reading.
Page 19 of 34
QUESTION 4
(a)
(i)
The moment of a force about any point is the product of
the force and the distance perpendicular to its line of action
(01 mark)
measured from that point.
(ii)
The principle of moments states that:
For any system of forces acting on a rigid body, the sum of the
anticlockwise moments about any point is equal to the sum of
clockwise moments about the same point provided the body is in
a state of equilibrium.
(01 mark)
(04 marks)
(b)
Bale of cotton
piston
A
P
B
Effort
Force on ram , F = pA = 4200 × 2 = 8400 N
Total downward force at end A of the rod
𝐹𝐴 = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑜𝑡𝑡𝑜𝑛 𝑏𝑎𝑙𝑒 − 𝑓𝑜𝑟𝑐𝑒 𝑒𝑥𝑒𝑟𝑡𝑒𝑑 𝑜𝑛 𝑝𝑖𝑠𝑡𝑜𝑛
𝐹𝐴 = (10000 − 8400)𝑁 = 1600 𝑁
Taking moments about P
𝐹𝐴 × 0.5 = 𝑒𝑓𝑓𝑜𝑟𝑡 × 1.5
𝑒𝑓𝑓𝑜𝑟𝑡 =
𝐹𝐴 × 0.5 1600 × 0.5
=
= 533.3 𝑁
1.5
1.5
Page 20 of 34
(c)
(i)
An experiment to demonstrate that pressure in
liquids changes with depth.
Tall can
(04 marks)
A
B
C
Three identical holes A,B and C are made at different depth of a
tall can.
With the holes plugged, the can is filled with water from an
overhead tap.
The plugs are then simultaneously removed from the holes to let
out water through them.
Observations show that the water jet from the lowest hole C,
travels furthest and fastest from the can, followed by that form
bole B, and lastly by that from hole A.
This shows that pressure in liquids increase with depth.
(ii)
(03 marks)
Atmospheric pressure at water surface:
𝑃𝐴 =
75
× 13600 × 10 = 102000 𝑁𝑚−2
100
Page 21 of 34
Pressure due to water column above sea diver
𝑃𝑊 = 25 × 1200 × 10 = 300000 𝑁𝑚2
Total pressure on diver:
𝑃 = 𝑃𝐴 + 𝑃𝑊 = 102000 + 300000 = 402000 𝑁𝑚−2
(03 marks)
(d)
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑚𝑒𝑡𝑎𝑙 𝑐𝑢𝑏𝑒, 𝑉 =
2×2×2
= 8.0 × 10−6 𝑚3
1000000
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 𝑚𝑒𝑟𝑐𝑢𝑟𝑦, 𝑊𝑙 = 𝑉𝜌𝐻 𝑔
𝑊𝑙 = 8.0 × 10−6 × 13600 × 10 = 1.088 𝑁
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑚𝑒𝑎𝑡𝑎𝑙 𝑐𝑢𝑏𝑒 𝑖𝑛 𝑎𝑖𝑟, 𝑊𝑎 = 3.56 𝑁
The reading of the spring balance when the cube is completely
immersed in mercury is:
𝑊 = 𝑊𝑎 − 𝑊𝑙
𝑊 = 3.56 − 1.088 = 2.472 𝑁
Page 22 of 34
QUESTION 5
(a)
Acceleration is the rate of change of velocity with time.
(01 mark)
(03 marks)
Velocity (ms-1)
(b)
20
15
10
5
0
10
25
35
(i)
𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛, 𝑎1 =
Time(s)
𝑣 − 𝑢 20 − 5 15
=
=
= 1.5 𝑚𝑠 −2
𝑡
10
10
𝐹𝑜𝑟𝑐𝑒, 𝐹1 = 𝑚𝑎1 = 5 × 1.5 = 7.5 𝑁
(ii)
(05 marks)
𝑣 − 𝑢 0 − 20
20
𝑑𝑒𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛, 𝑎2 =
=
=−
= −2.0 𝑚𝑠 −2
𝑡
10
10
𝑑𝑒𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒, 𝐹2 = 𝑚𝑎2 = 5 × 2.0 = 10.0 𝑁
𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑑 =
1
× 10 × 20 = 100 𝑚
2
𝑝𝑜𝑤𝑒𝑟, 𝑝 =
𝐹𝑑 10 × 100
=
= 100 𝑊
𝑡
10
Page 23 of 34
(c)
Distinguish between elastic and inelastic collisions.
Elastic collision
Inelastic collision
k.e is conserved
k.e is mot conserved
Bodies do not stick together
after collision.
Bodies stick together after
collision.
(d)
(i)
(02 marks)
(03 arks)
𝑚𝑤 𝑢𝑤 + 𝑚𝑟 𝑢𝑟 = 𝑚𝑤 𝑣𝑤 + 𝑚𝑟 𝑣𝑟
120 × 5 + 150 × 0 = 120(−1) + 150𝑣𝑟
600 + 120 = 150𝑣𝑟
𝑣𝑟 = 4.8 𝑚𝑠 −1 .
(02 marks)
(ii)
Two possible energy losses during the collision.
1. Sound
2. Heat
3. Energy loss due to deformation (change of shape).
Page 24 of 34
QUESTION 6
(a)
Three differences between sound and radio waves.
Sound waves
Radio waves
They are longitudinal waves
They are transverse waves
They are mechanical waves
requiring a physical medium
for their transmission.
(cannot travel through a
vacuum)
They are electromagnetic
waves and do not require a
physical medium for their
transmission. (can travel
through a vacuum)
They are slower than radio
waves travelling at about
330 ms-1 in air
They are faster than sound
waves travelling at the speed
of light.
(b)
An experiment to show that sound requires a physical
medium for its transmission.
(03 marks)
(05 marks)
To battery
Electric bell
Gong
Bell jar
To vacuum
pump
An electric bell connected to a battery is hung in a bell jar.
The bell jar is connected to a vacuum pump.
When the electric bell is switched on, sound can be heard.
When the air is gradually by the pump, the sound decreases and
eventually dies away when the jar is completely evacuated, even
when the hammer is seen hitting the gong.
Page 25 of 34
This shows that sound requires a physical medium for its
transmission.
(c)
(i)
An echo is reflected sound.
(01 mark)
(ii)
The captain of ship transmitted an ultrasonic sound to the (03 marks)
bottom of the sea-bed in order to detect
coral reefs. The echo
was detected 0.12 𝑠
later after transmission. If the speed of
sound in water is 1400 𝑚𝑠 −1 , find the depth of the coral
reefs
from the ship.
(03 marks)
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑣 =
2 × 𝑑𝑒𝑝𝑡ℎ
𝑡𝑖𝑚𝑒
𝑑𝑒𝑝𝑡ℎ =
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 × 𝑡𝑖𝑚𝑒
2
𝑑𝑒𝑝𝑡ℎ =
1400 × 0.12
= 84 𝑚
2
.
Displacement (m)
(d)
10
0
0.2
0.4
0.6
Time (µs)
(Note : 1𝜇𝑠 = 1.0 × 10−6 𝑠)
Page 26 of 34
(i)
(01 mark)
The amplitude of the wave is 10 m
(ii)
(03 marks)
3
𝑡𝑖𝑚𝑒 , 𝑡 = × 𝑝𝑒𝑟𝑖𝑜𝑑
4
𝑝𝑒𝑟𝑖𝑜𝑑, 𝑇 =
4
× 0.6 × 10−6 𝑠 = 8.0 × 10−5 𝑠
3
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦, 𝑓 =
1
1
=
= 1.25 × 104 𝐻𝑧
𝑇 8.0 × 10−5
Page 27 of 34
QUESTION 7
(a)
Magnetic saturation occurs when all the dipoles in the
different domains have been made to face in the same
direction and the material has achieved its maximum
magnetic strength.
(01 mark)
(03 marks)
(b)
Soft iron ring
N
X
X
S
Magnetic shielding is when magnetic field lines are
prevented from reaching a certain region.
Magnetic shielding is achieved by placing a soft iron ring
in the path of a magnetic field such that the field lines are
concentrated to pass through the ring, preventing them to
reach region X found inside the ring.
(c)
magnetic field patterns resulting from two straight
conductors placed vertically near each other carrying a
current in:
(i)
the same direction.
(01mark)
Page 28 of 34
(ii)
(ii)
opposite directions.
(01 mark)
(05 marks)
Coil
(d)(i)
F
Q
F
P
N
B1
C1
K
S
C2
B2
R
S
Carbon brushes
Half-split
commutators
When switch K is closed, current flows through the coil situated
in the magnetic field of the two pole pieces.
Each of the sections PQ and RS of the coil experiences a force
whose direction is given by Flemming’s left hand rule.
The two forces form a couple causing the coil to rotate.
When the coil reaches the vertical position, the commutators
break contact with the carbon brushes. At this point no current
flows through the coil and no forces act on the coil. The coil rolls
over by virtue of its momentum.
After the vertical position, the commutators interchange contact
with the carbon brushes and the direction of the current in the coil
is reversed. This ensures that the direction of rotation of the coil is
maintained.
Page 29 of 34
(ii)
Two ways in which the power generated by the
d.c. motor may be increased.
1.
2.
3.
4.
5.
(02 marks)
Any 2 pts.
Increasing the number of turns of the coil.
Strengthening yhe magnetic field.
Increasing area of the coil.
By winding the coil on a soft magnetic material.
By increasing the current flowing in the coil.
(e)
(03 marks)
𝐼 = 𝐼𝑆 + 𝐼𝐺
𝐼𝑆 = 15 − 0.025 = 14.975 𝐴
Potential difference:
𝐼𝑆 𝑅𝑆 = 𝐼𝐺 𝑅𝐺
14.975𝑅𝑆 = 0.025 × 4
𝑅𝑆 =
0.025×4
14.975
= 0.00668 Ω.
Page 30 of 34
QUESTION 8
(01 mark)
(a)
(i)
The half-life of a radioactive substance is the time taken
for the substance to decay to half of its original amount.
(ii)
In 168 seconds, the activity of Thorium falls to oneeighth of its original value. Determine its half
1
1
(𝑀𝑜 ) = 𝑀𝑜 ( )
8
2
1
1
(𝑀𝑜 ) = 𝑀𝑜 ( )
8
2
(03 marks)
𝑇⁄
𝑡
168⁄
𝑡
3 = 168⁄𝑡
ℎ𝑎𝑙𝑓 − 𝑙𝑖𝑓𝑒, 𝑡 = 56 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
(b)
(i)
(06 marks)
High Voltage
+
−
Vacuum
Glass tube
Filament
Cathode rays
low Voltage
Focusing cup
Cooling fins
Tungsten
target
X-rays
Page 31 of 34
The filament cathode is heated using a low voltage supply.
The cathode emits electrons by process of thermionic emission.
The electrons are accelerated towards the tungsten target
embedded in the anode by a high voltage.
When the cathode rays (electrons) strike the target, X- rays are
produced.
(ii)
Two uses of X- rays(any two)
1.
2.
3.
4.
5.
(02 marks)
Used in radiography (X-ray photography).
Used in detection of cracks or flaws in metal plates.
Used in radiotherapy(treatment of cancerous tissue)
Used to study crystals.
Used in security measures to detect hidden firearms or
illegal materials.
(03 marks)
(c)
+
𝐴
Radioactive
material
𝐵
𝐶
(i)
−
A is beta radiation.
B are gamma rays.
C is alpha radiation
Page 32 of 34
(ii)
The alpha particles will be stopped by the paper.
Both gamma and beta radiations will penetrate
through the paper.
(01 mark)
Page 33 of 34
END
Page 34 of 34