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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 = 13600kgm−3
Specific heat capacity of water = 4200 J kg−1 K−1
Page 1 of 33
Plastic lid
Rubber stands
Heating element
1. (a) State Charles’s law. (01 mark)(b) Describe an experiment to verify Charles’s law. (06 marks)(c) At 27oC a fixed mass of gas occupies a volume of 120 cm3. Find the
volume of the gas if it is cooled to 7oC , at constant pressure.(02 marks)
(d) The diagram below shows a shiny silver surfaced electric kettle being used to heat water from a mains supply.
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)
(e) A kettle rated at 1600 W, 240 V is used to heat 2000 cm3 water initially at 25o C to its boiling point. If the specific heat capacity of the material of the kettle is 120 J kg−1 K−1, find how much time is taken for the water to reach its boiling point. (04 marks)
Page 2 of 33
A
B
C
2. (a) State Snell’s law of refraction. (01 mark)(b) Describe an experiment to determine the refractive index of glass using a
glass block. (06 marks)
(c)
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 .
(i) Find the angle yo . (04 marks)(ii) 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)
(d) 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)
3. (a) State the law of charges. (01 mark)(b) Describe with the aid of suitable diagrams how a gold leaf electroscope
can be charged negatively by induction. (05 marks
Page 3 of 33
AS
3 Ω 6 Ω1 Ω
1.5V,r=1Ω
1.5V,r=1Ω
(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)
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 33
P
Bale of cotton
piston
A B
Effort
4. (a) (i) Define moment of a force. (01 mark)(ii) State the principle of moments. (01 mark)
(b)
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 2m2. If the pressure exerted on the cotton bale is 4200 N m−2, find the minimum effort required to exert this pressure when the system is in equilibrium. (04 marks)
(c) (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)
(d) A spring balance reads 3.56 N when a metal cube of side 2.0cm 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 33
Velocity (ms-1)
5
10
15
20
0 10 25 35 Time(s)
5. (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.
(i) Find the force acting on the body for the first 10 seconds of its motion.(03 marks)
(ii) 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. (02 marks)
(d) In a game of pool, a white ball of mass 120 g moving with a velocity of 5 m s−1 hits a stationary red ball of mass 150 g .
After the collision, the white ball moves with a velocity of 1m s−1 in the opposite direction. Find;
(i) The velocity of the red ball after the collision. (03 marks)(ii) State two possible energy losses during the collision. (02 marks)
Page 6 of 33
10
0 0.2 0.4 0.6 Time (µs)
Displacement (m
)
6. (a) State three differences between sound and radio waves. (03 marks)
(b) Describe an experiment to show that sound requires a physical medium for its transmission. (05 marks)
(c) (i) What is an echo? (01 mark)
(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) .
(Note : 1 μs=1.0 ×10−6 s¿
The graph above shows the wave profile of a radio wave. Find;
(i) The amplitude of the wave. (01 mark)(ii) The frequency of the wave. (03 marks)
7. (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 33
(c) 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)
(d) (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)
8. (a) (i) Define the term half-life as applied to a radioactive material. (01 mark)
(ii) In 168 seconds, the activity of Thorium falls to one-eighth of its original value. Determine its half – life.
(03 marks)(b) (i) With the aid of a well labeled diagram, explain how X –
rays are produced. (06 marks)(ii) State two uses of X- rays. (02 marks)
Page 8 of 33
Radioactive material
(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
MARKING GUIDE
Page 9 of 33
Thermometer
Conc. Sulphuric acid
Capillary tube
Gas column
Stirrer
mm scale
water
Heat
QUESTION 1
(a)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.
(01 mark)
(b) An experiment to verify Charles’s law.
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.
The values of T and Volume V are recorded in a suitable table.
(06 marks)
Page 10 of 33
V(mm3) T(o C)
A graph of Volume, V and temperature, T is plotted.
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)T 1=27+273=300 K ,V 1=120 cm3
T 2=7+273=280 K ,V 2=?
applyingV 1
T 1=
V 2
T 2
120300
=V 2
280
V 2=112 cm3
(02 marks)
(d)(i) The heating element is placed at the bottom of the
Page 11 of 33
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.
(01 mark)
(01 mark)
(01 mark)
(e)h eat supplied byheating element=heat gained by water+material of kettle
power × time=(mw cw+mk ck ) ∆ θ
1600 ×t=(V w 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=84301600
s=395.16 s
(04 marks)
QUESTION 2
(a) Snell’s law states that: (01 mark)
Page 12 of 33
H
O
i
A B
CDG
r
P2
P3
P4
EP1
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.
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.
The glass block is replaced on its outline and optical pins P1 and P2 stuck on the line EO,
(06 marks)
Page 13 of 33
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∧60o.
The results are recorded in a suitable table including values of sin i∧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:
The refractive index of the glass block is determined as the slope of the straight line graph obtained.
Page 14 of 33
A
B
C
(c)
(i)
ηasin ia=ηg sin ig
1×sin 20=1.6 sin x
x=sin−1( sin201.6 )=12.34o
Now:
y=70o−x
y=70o−12.34o
y=57.66o
(04 marks)
(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. Light must be travelling from an optically dense medium to a less optically dense medium.
2. The angle of incidence must be greater than the angle of refraction of the dense medium.
(d)With the aid of a ray diagram, find the location of
(02 marks)
(03 marks)
Page 15 of 33
the image of an object, 2 cm tall that is placed 10 cm in front of a concave lens of focal length
15 cm.
QUESTION 3
Page 16 of 33
Electron flow
Electron flow from the earth via earth connection
(a)
The law of charges states that:
Unlike charges attract, like charges repel each other,
(01 mark)
(b) Charging a gold leaf electroscope negatively by induction.
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,
With the positively charged rod still in position, the brass cap is momentary earthed.
Electrons flow from the earth via the earth connection to neutralize the charge on the plate and leaf and the leaf collapses.
(05 marks)
Page 17 of 33
+¿
+¿
+¿+¿
+¿
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)
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
(d) (i) Polarization is the formation of hydrogen gas at the positive electrode . The hydrogen bubbles insulate the positive
(04 marks)
(02 marks)
Page 18 of 33
AS
3 Ω 6 Ω1 Ω
electrode from the electrolyte and this increases the internal resistance of the simple cell,
(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)
(e)
resistance∈¿ , Rp=3 × 63+6
=189
=2Ω
total externalresistance , R=1+2=3 Ω
total e . m. f , E=1V
total internal resistance ,r=1 ×11+1
=12=0.5 Ω
E=I ( R+r )
1=I (3+0.5 )
I= 13.5
=0.286 A is the ammeter reading.
(03 marks)
QUESTION 4
(a)
Page 19 of 33
P
Bale of cotton
piston
A B
Effort
(i) The moment of a force about any point is the product of the force and the distance perpendicular to its line of action 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)
(01 mark)
(b)
Force on ram, F=pA=4200 ×2=8400 N
Total downward force at end A of the rod
F A=weight of cotton bale−force exerted on piston
F A=(10000−8400 ) N=1600 N
Taking moments about P
F A ×0.5=effort × 1.5
effort=F A ×0.5
1.5=1600× 0.5
1.5=533.3 N
(04 marks)
(c) (i) An experiment to demonstrate that pressure in liquids changes with depth.
(04 marks)
Page 20 of 33
Tall can 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)
Atmospheric pressure at water surface:
PA=75
100×13600 ×10=102000 N m−2
Pressure due to water column above sea diver
PW=25× 1200 ×10=300000 N m2
(03 marks)
Page 21 of 33
Total pressure on diver:
P=PA +PW=102000+300000=402000 N m−2
(d)
Volume of metal cube , V = 2×2×21000000
=8.0 ×10−6 m3
weight of displaced mercury , W l=V ρH g
W l=8.0 ×10−6 ×13600 ×10=1.088 N
weight of meatal cube∈air ,W a=3.56 N
The reading of the spring balance when the cube is completely immersed in mercury is:
W =W a−W l
W =3.56−1.088=2.472 N
(03 marks)
QUESTION 5
(a) Acceleration is the rate of change of velocity with time.(01 mark)
Page 22 of 33
Velocity (ms-1)
5
10
15
20
0 10 25 35 Time(s)
(b)
(i)acceleration , a1=
v−ut
=20−510
=1510
=1.5 ms−2
Force , F1=m a1=5×1.5=7.5 N
(03 marks)
(ii)deceleration , a2=
v−ut
=0−2010
=−2010
=−2.0 ms−2
decelerating force , F2=ma2=5×2.0=10.0 N
total distance , d=12
×10 × 20=100 m
power , p= Fdt
=10×10010
=100 W
(05 marks)
(c) Distinguish between elastic and inelastic collisions.
Elastic collision Inelastic collision
(02 marks)
Page 23 of 33
k.e is conserved k.e is mot conserved
Bodies do not stick together after collision.
Bodies stick together after collision.
(d) (i)
mw uw+mr ur=mw vw+mr vr
120×5+150× 0=120 (−1 )+150vr
600+120=150 vr
vr=4.8 ms−1.
(03arks)
(ii)
Two possible energy losses during the collision.
1. Sound2. Heat3. Energy loss due to deformation (change of shape).
(02 marks)
QUESTION 6
(a) Three differences between sound and radio waves.(03 marks)
Page 24 of 33
Electric bell
Bell jar
To vacuum pump
To battery
Gong
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.
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.
This shows that sound requires a physical medium for its transmission.
(05 marks)
Page 25 of 33
10
0 0.2 0.4 0.6 Time (µs)
Displacement (m
)
(c) (i) An echo is reflected sound. (01 mark)
(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)
velocity , v=2 ×depthtime
depth= velocity × time2
depth=1400 ×0.122
=84 m
(03 marks)
(d) .
(Note : 1 μs=1.0 ×10−6 s¿
(i)
The amplitude of the wave is 10 m
(01 mark)
Page 26 of 33
(ii)
time , t= 34
× period
period , T=43
×0.6 ×10−6 s=8.0 ×10−5 s
frequency , f = 1T
= 18.0 ×10−5 =1.25× 104 Hz
(03 marks)
QUESTION 7
(a) Magnetic saturation occurs when all the dipoles in the (01 mark)
Page 27 of 33
N SX
Soft iron ring
N S
different domains have been made to face in the same direction and the material has achieved its maximum magnetic strength.
(b)
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.
(03 marks)
(c) magnetic field patterns resulting from two straight conductors placed vertically near each other carrying a current in:
(i) the same direction.
(01mark)
(ii) (ii) opposite directions. (01 mark)
Page 28 of 33
X
Soft iron ring
F
F
B1 B2
N SP
Q R
S
C1 C2
Coil
Carbon brushes
Half-split commutatorsK
(d)(i)
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.
(05 marks)
(ii) Two ways in which the power generated by the d.c. motor may be increased.
1. Increasing the number of turns of the coil.2. Strengthening yhe magnetic field.
(02 marks)
Any 2 pts.
Page 29 of 33
3. Increasing area of the coil.4. By winding the coil on a soft magnetic material.5. By increasing the current flowing in the coil.
(e)
I=I S+ I G
I S=15−0.025=14.975 A
Potential difference:
I S RS=IG RG
14.975 RS=0.025 ×4
RS=0.025 ×414.975
=0 .00668 Ω.
(03 marks)
QUESTION 8
(a)(01 mark)
Page 30 of 33
Cooling fins
X-rays
Tungsten target
High Voltage
Filament Vacuum
Glass tube
low Voltage
Focusing cup
Cathode rays
(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 one-eighth of its original value. Determine its half
18 ( M o )=M o( 1
2 )Tt
18 ( M o )=M o( 1
2 )168
t
3=168t
half −life , t=56 seconds
(03 marks)
(b) (i)
The filament cathode is heated using a low voltage supply.
The cathode emits electrons by process of thermionic emission.
(06 marks)
Page 31 of 33
Radioactive material
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. Used in radiography (X-ray photography).2. Used in detection of cracks or flaws in metal plates.3. Used in radiotherapy(treatment of cancerous tissue)4. Used to study crystals.5. Used in security measures to detect hidden firearms or
illegal materials.
(02 marks)
(c)
(i)
A is beta radiation.
B are gamma rays.
C is alpha radiation
(03marks)
(ii) The alpha particles will be stopped by the paper. Both gamma and beta radiations will penetrate through the paper.
(01 mark)
Page 32 of 33
END
Page 33 of 33
