Physics Nov\Dec 2009

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Question 2 Part I: Candidates were required to answer any five questions from this part.
A projectile is launched with a velocity of 42 ms-1 at an angle of 30o to the horizontal. If the time of flight is 4.2 seconds, calculate the maximum attainable height. [g = 10 ms-2]

ANSWER

The projectile takes 2.1 seconds to reach the maximum height H.
H = (U sinθ) t – ½ qt2

=   (42 sin 30) (2.1) – ½ x 10 x (2.1)2
=    44.1   –   22.05
=    22.05m

            OR

H =  U2 sin2θ
              2g
=   422 x ( sin 30)2
2 x 10

= 22.05m

Question 3 Part I: Candidates were required to answer any five questions from this part.
A ray of lights is incident on an air-glass boundary at an angle θ. If the angle between the partially reflected ray and the refracted ray is 90o, calculate angle θ given that the refractive index of glass is 1.54.

ANSWER

Diagram

n =  sin θ                    =    sin θ    = tan θ
Sin (90 – θ)          cos θ

1.54   =   tan θ
θ   =    57o

Question 4 Part I: Candidates were required to answer any five questions from this part.
State: two applications of electrolysis in an industry; one application of electrolysis in a school laboratory

ANSWER

(a)    In industry we use electrolysis for;
(i)     electroplating;
(ii)   extraction of some metals (e.g. aluminum);

production of some gases (e.g. hydrogen);
purification of some metals (eg. copper).

(b) In the school laboratory, we use electrolysis to:

calibrate ammeter;
electrolyze acidified water.

Question 5 Part I: Candidates were required to answer any five questions from this part.
Explain the effect of temperature on the surface tension of a liquid.

ANSWER

The higher/lower the temperature of a liquid the lower/higher its surface tension.

This is because at higher/lower temperatures the molecules, including those at the surface, have gained/lost more energy and become more/less mobile thereby decreasing/increasing the forces at the liquid surface.

Question 6 Part I: Candidates were required to answer any five questions from this part.
Define: elastic limit: yield point.

ANSWER

Elastic Limit:
The minimum stress/force applied on an elastic material beyond which a permanent deformation occurs.

Yield Point
The point on a stress/strain or load/extension graph beyond which the elasticmaterial becomes plastic.

Question 7 Part I: Candidates were required to answer any five questions from this part.
When a steady current of 1.57 A is passed through a copper voltameter for 20.00 minutes, a mass of 6.25 x 10-4 kg of copper is deposited at the cathode. Calculate the relative atomic mass of copper. [Faraday’s constant = 96500 C]

ANSWER

M1 = M2
Q1 Q2

M1 = 0.625 x 10-4
2 x 96500 1.57 X 20 x 60

M1 = 0.064g (or 6.4 x 10-2) kg
∴ Relative atomic mass = 64

Question 8 Part I: Candidates were required to answer any five questions from this part.
What is an electron “gun”? Name one equipment in which an electron gun is an essential component.

ANSWER

a)        An “electron gun” is a device that produces a narrow beam of high speed/velocity
            electrons.
e.g
(b)        –           Cathode ray tube
            –           X-ray tube
            –           Cathode ray oscilloscope
            [Accept any other correct equipment].

Question 9 Part I: Candidates were required to answer any five questions from this part.
Explain threshold frequency with respect to photoemission.

ANSWER

When radiation of appropriate frequency falls on a metal surface, electrons are ejected from the surface. The minimum frequency of a radiation that must be incident on the metal surface to liberate/set free electrons from the surface is the threshold frequency.

Question 10 Part I: Candidates were required to answer any five questions from this part.
Electrons from a hot cathode are accelerated from rest onto a target in a vacuum tube maintained at a potential difference of 180 volts. Calculate the velocity of the electrons on reaching the target. [Mass of electron = 9.0 x 10-31 kg; charge of an electron = 1.6 x 10-19 C].

ANSWER

Let the velocity of the electron at the target be u

½ mu2 = e V

or u2 = 2eV
m

     =   2 x 1 . 6 x 10-19 x 180
                        9.0 x 10-31

u   =   8 x 10⁶ ms‑1

Question 11
Part II: Candidates were required to answer any five questions from this part.

(a)        State the energy transformations that take place when:
            (i)         an electric current is used to drive a crane;
            (ii)        an electric bulb is switched on.

(b) Explain why roads are banked at the bends.

A bucket filled with water is tied to a piece of string and whirled in a vertical plane as illustrated in the diagram above.

            (i)         Identify the forces F1 and F2.
            (ii)        At which position X or Y will the bucket be heavier? Explain.
            (iii)       Give the reason why the water does not pour out at X.

(d) An electric car of mass 600 kg and efficiency 75% moves up a hill at a constant speed in 80 seconds. If the hill is 80m long and 5 m high and the car operates on a 36 volts battery, calculate the current supplied by the battery.
[g = 10 ms_2]

ANSWER

(a)        (i)         Electric energy →   mechanical energy   →   potential energy

   OR   Electrical energy → mechanical energy

(ii) Electric energy → heat energy → light energy

(b) At unbanked roads the frictional force required to keep a vehicle in a circular
path is not sufficient. Banking provides an additional force due to the normal reaction on the wheels to prevent the vehicle from slipping.

The bucket is heavier at position Y.

This is due to the direction of the centripetal force acting.

At position X, weight = mv2 – mg
r

At position Y, weight = mv2     + mg
                                              r
(iii)       The weight acts in the opposite direction to the tension.
The weight of the water is less than the centrifugal force.

Efficiency = Power output

Power input

But Power output = force x distance
Time

Po = 600 x 10 x 5.0
80

= 375 W = 75 P1
100

Power input = 100 x 375 = 500W
75

                      P1   =   IV
                      I   = P1 V   =  500    =   13.9A
                                               36

Question 12 Part II: Candidates were required to answer any five questions from this part.
(a) (i) Define latent heat of fusion. (ii) Diagram The diagram above illustrates part of the set-up for determining the specific latent heat of vaporization of water. State the purpose of the curved tube P. (b) Explain how pure water can be made to boil at a temperature below 100o C. (c) The resistance in the element of a platinum resistance thermometer is 9.60 Ω at 0.oC, 12.10 Ω at 100oC and 10.20Ω room temperature. Calculate the room temperature on the scale of the resistance thermometer. (d) A body of mass 2 kg at 100 oC is dropped gently into a mixture of ice and water at 0oC. Calculate the mass of ice that melted at 0oC. Specific latent heat of fusion of ice = 3.3 x 105 J kg-1

Question 12
Part II: Candidates were required to answer any five questions from this part.

(a) (i) Define latent heat of fusion.
(ii) Diagram

The diagram above illustrates part of the set-up for determining the specific latent heat of vaporization of water. State the purpose of the curved tube P.

(b) Explain how pure water can be made to boil at a temperature below 100o C.

(c) The resistance in the element of a platinum resistance thermometer is 9.60 Ω at 0.oC, 12.10 Ω at 100oC and 10.20Ω room temperature. Calculate the room temperature on the scale of the resistance thermometer.

(d) A body of mass 2 kg at 100 oC is dropped gently into a mixture of ice and water at 0oC. Calculate the mass of ice that melted at 0oC.

Specific latent heat of fusion of ice = 3.3 x 105 J kg-1

ANSWER

(a)        (i)         This is the amount of heat required to change a substance from the solid
state to the liquid state without a change in/at constant temperature.
(ii)        It is to ensure that only dry steam is used.

(b) Pure water boils at a temperature below 100oC when its s.v.p is less than normal atmospheric pressure.

To achieve this:

Fill a round bottom flask to ¾ of its volume.
Insert a thermometer through the stopper and heat the flask to about 80oC.
Invert the flask and place it under flowing tap water so as to reduce the pressure in the flask as a result of condensation of vapour.

θ   =   10.2 – 9.6 x 100
                      12.1 – 9.6

            = 0.6 x 100
                       2.5

            =   24. 0o C

Heat lost by hot body = Heat gained by ice

MbCb Δθ = MiLf
2 x 2200 x (100 – 0) = Mi x 330000

Mi = 2 x 2200 x 100 = 4/3
330000

= 1.33 kg.

Question 14 Part II: Candidates were required to answer any five questions from this part.
State: three ways by which the e.m.f. of an a.c. generator can be increased; how an a.c. generator may be modified to produce direct current. (i) Define the root mean square value of an alternating current. Explain the term resonant frequency as it relates to an L-C-R series circuit. A particle of charge -6 μC released from rest in a uniform electric field E moves a distance of 5cm. If its kinetic energy is 4 x 10-3 J, calculate the: work done by the electric field; potential at its initial position; magnitude of the electric field intensity.

Question 14
Part II: Candidates were required to answer any five questions from this part.

State:
1. three ways by which the e.m.f. of an a.c. generator can be increased;
2. how an a.c. generator may be modified to produce direct current.

(i) Define the root mean square value of an alternating current.
1. Explain the term resonant frequency as it relates to an

L-C-R series circuit.

A particle of charge -6 μC released from rest in a uniform electric field E moves a distance of 5cm. If its kinetic energy is 4 x 10-3 J, calculate the:
1. work done by the electric field;
2. potential at its initial position;
3. magnitude of the electric field intensity.

ANSWER

(a) (i) E.m.f. of an a.c. generator can be increased by:

increasing the number of turns in the coil of the armature;
using larger magnets/increasing strengths of the magnetic field;
increasing the speed of rotation of the armature;
the core should be made of soft iron.

Accept any other valid example.

By replacing the slip rings with split rings (commutator)

(b)(i)   The root mean square value is the direct current that would dissipate the same
amount of energy in a resistor as is dissipated by the actual alternating current.
                                         OR
The root mean square value is that steady current that dissipates the same energy or quantity of heat in the same time in the same resistance.

        (ii)    Resonance frequency fo is the frequency at which minimum impedance occurs.
                                    OR
                 Resonance frequency fo is the frequency at which XL = Xc .
                                    OR

Resonance frequency fo is the frequency when maximum current is obtained from such a circuit.

(c)(i) Work done = kinetic energy = 4.0 x 10-3 J

(ii) W = QV

V = W/Q = 4 x 10-3
6 x 10-6

V = 0.667 X x 10-3 = 667V

(iii) E = V/d

= 667
0.05

= 13333.4 NC-1

Question 15 Part II: Candidates were required to answer any five questions from this part.
(a) (i) A radioactive substance emits α, β and γ radiations. Arrange them in order of decreasing; (A) penetrating power; (B) ionizing power. (ii) With the aid of a labelled diagram, explain how the radiations in (i) above are affected by a magnetic field placed perpendicularly to the source. (b) State one factor each that is responsible for the: (i) occurrence of photoemission; (ii) number of electrons emitted during photoemission; (iii) energy of photoelectrons. (c) A photon has a frequency of 5.02 x 1014 Hz. Calculate the energy of the photon in: (i) joules; (ii) electron volts. [h = 6.6 x 10-34 Js; 1eV = 1.6 x 10-19 J]

Question 15
Part II: Candidates were required to answer any five questions from this part.

(a) (i)   A radioactive substance emits α, β and γ radiations. Arrange them in order of decreasing;
            (A) penetrating power;
(B)   ionizing power.
     (ii)   With the aid of a labelled diagram, explain how the radiations in (i) above are affected
by a magnetic field placed perpendicularly to the source.
(b)        State one factor each that is responsible for the:
            (i)         occurrence of photoemission;
            (ii)        number of electrons emitted during photoemission;
            (iii)       energy of photoelectrons.
(c)        A photon has a frequency of 5.02 x 1014 Hz.
            Calculate the energy of the photon in:
            (i)         joules;
            (ii)        electron volts.
                        [h = 6.6 x 10-34 Js; 1eV = 1.6 x 10-19 J]

ANSWER

(a) (i) (A) gamma rays → beta rays → alpha particles.

                        (B)       alpha particles → beta rays → gamma rays.

            Accept:    γ → β → α   and
                            α → β → γ
       (ii)

β ray is strongly deflected (because it is negatively charged.

1. α particle is deflected but not as strong as the β – beta (because it is massive)
2. γ rays is not affected (because it has neither charge nor mass).

(i) –     The energy/frequency of incident radiation must be greater than thework function/threshold frequency of the metal
(ii) –   the intensity of the radiation
(iii)- Energy/frequency of incident radiation/ work function of the metal irradiated

     (c) E = hf
= 6.6 x 10-34 x 5.02 x 1014

= 3.31 x 10-19 J
1 e V = 1.6 X 10-19 J
∴ 3.31 x 10-19 J = 2.07 e V

Physics Nov\Dec 2009

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