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    Quiz 6 of 16

    2016 Physics WAEC Theory Past Questions

    WAEC: PHYSICS 2016 Physics WAEC Theory Past Questions

    Responses

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    (1) (a) State the dimension of impulse.

    Solution

    MLT-1

    (b) State the dimension of acceleration

    Solution

    LT-2

    (c) State the dimension of work

    Solution

    ML2 T-2

    (2) A projectile is fired with a velocity of 20 ms-1 at an angle of 40° to horizontal. Determine the components of the velocity of the projectile at its maximum height.

    Solution

    Vx = U cos θ = 20 cos 40° = 15.32ms-1
    Vy = 0

    (3) State three different materials that can be used to demonstrate Brownian motion.

    Solution

    – smoke particles
    – carbon particles
    – dust particles
    – pollen particles
    – chalk particles
    – potassium permanganate crystals

    (4) An electron enters perpendicularly into a uniform magnetic field which has a flux density of 0.12T . This results in a magnetic force of 9.6 × 10-12N on the electron. Calculate the speed of the electron as its enters the magnetic field.
    (e = 1.6 ×10-19 C ).

    Solution

    F= evB sin θ

    9.6 × 10-12 = 1.6 × 10-19 × v × 0.12× sin 90o

    v = \( \frac{9.6 \: \times \: 10^{-12}}{1.6 \: \times \: 10^{-19} \: \times \: 0.12\: \times \: 1} \)

    v  = 5.0 × 108ms-1

    (5) List three uses of rockets.

    Solution

    – space exploration
    – space travel
    – warfare
    – fireworks/entertainment
    – launching of (artificial) satellites(into orbits)

    (6) (a) What is doping ?

    Solution

    Doping is the introduction of impurity (atoms) into a semiconductor.

     

    (b) Explain how doping improves the conductivity of a semiconductor.

    Solution

    Doping reduces the energy gap between the conductor band and the valence band so that charges of minimum energy can move into the conduction band.

    The diagram above illustrates a cathode ray tube.
    Identify the components X, Y, and Z

    Solution

    X – filament/heater/cathode
    Y – anode
    Z – plate(s)/deflector(s)

    (8) (a) Explain the term net force.

    Solution

    Net force is the effective force resulting from the actions of a system of forces on a body.

    OR

    Net force is an unbalanced force that produces an acceleration of a body.

     

    (b) Define the principle of conservation of linear momentum and state one example of it.

    Solution

    In an isolated/closed system of colliding bodies, the total linear momentum in a fixed direction remains unchanged/constant.

    Example
    – rocket propulsion;
    – recoil of gun
    – colliding trolleys.

     

    (c) A ball of mass 200g released from a height of 2.0 m hits a horizontal floor and rebounds to a height of 1.8 m. Calculate the impulse received by the floor.
    (g = 10 ms-2)

    Solution

    v21 = u21 + 2gh1= 0 + 2 × 10 × 2

    v21 = 40

    v1 = √40

    v1 = 6.325ms-1

    v22 = u22 – 2gh2

    0  = u22 – 2 × 10 × 1.8

    u2= 36

    u = √36

    u = 6.000ms-1

    Impulse = change in momentum

    = mv1 -(-mu2) = m(v1 + u2)

    =0.2(6.325 + 6.000)

    = 2.46Ns or 2.46kgm/s

     

    (d) (i) A body of mass 20g performs a simple harmonic motion at a frequency of  5 Hz . At a distance of 10 cm from the mean position, its velocity is 200 cms-1. Calculate its maximum displacement from the mean position;
    (g = 10ms-2 ; π = 3.14)

    Solution

    v =  \( \scriptsize \omega \sqrt{r^2 \: – \: y^2} \)

    = \( \scriptsize 2 \pi f \sqrt{r^2 \: – \: y^2} \)

    2 = \( \scriptsize 2 \:  \times \: 3.14 \: \times \: 5 \: \times \:   \sqrt{r^2 \: – \: 0.1^2} \)

    r = 0.12m

     

    (d) (ii) A body of mass 20g performs a simple harmonic motion at a frequency of 5 Hz . At a distance of 10 cm from the mean position, its velocity is 200 cms-1.

    Calculate its maximum velocity;
    (g = 10ms-2 ; π = 3.14)

    Solution

    Vmax = ωr = 2πfr

    = 2 × 3.14 × 5 × 0.12

    = 3.77ms-1

     

    (iii) A body of mass 20g performs a simple harmonic motion at a frequency of 5 Hz. At a distance of 10 cm from the mean position, its velocity is 200 cms-1. Calculate its maximum potential energy.
    (g = 10ms-2 ; π = 3.14)

    Solution

    P.E = \( \frac{1}{2} \scriptsize mw^2r^2 \)

    P.E = \( \frac{1}{2} \scriptsize m(2 \pi f)^2r^2 \)

    P.E = \( \frac{1}{2} \scriptsize \: \times \:  0.02  \: \times \:   (2 \: \times \: 3.14 \: \times \: 5)^2\: \times \: (0.12)^2 \)

    = 1.42 × 10-1 J

    (9) (a)Explain the term:

    (i) thermal equilibrium;

    Solution

    Thermal equilibrium is a condition in which two bodies in contact have no net flow of heat between them

     

    (ii) Explain the term: fundamental interval.

    Solution

    Fundamental interval is the difference in temperature between the lower fixed point(temperature of melting ice at standard atmospheric pressure) and the upper fixed point(temperature of steam at standard atmospheric pressure).

     

    (b) List two uses of hydraulic press.

    Solution

    – lifting of objects;
    – compressing metal sheets;
    – compressing soft materials into bales;
    – compressing paper in printing industries.

     

    (c) Name the material used to reset the steel index in the Six’s maximum and minimum thermometer.

    Solution

    Magnet

     

    (d) (i) A nursing mother prepared her baby’s milk mixture at 85o C, in a feeding bottle. In order to cool it to 40o C, she immersed the bottle in an aluminium bowl of heat capacity 90 J K -1 containing 500g of water at 26o C. If the mass of the mixture is 300g, calculate the specific heat capacity of the mixture.
    [Neglect the heat losses and heat capacity of the bottle;
    specific heat capacity of water = 4200 J kg-1 K-1]

    Solution

    Heat lost by mixture = Heat gained by water + Heat gained by aluminium

    Mm Cm∆θm = Mw Cw∆θw + Ca∆θa

    0.3 × Cm × (85 – 40) = 0.5 × 4200 × (40 – 26) + 90 (40 – 26)

    13.5Cm = 29400 + 1260

    Cm = 2271.1JK-1K-1

     

    (ii) A nursing mother prepared her baby’s milk mixture at 85o C, in a feeding bottle. In order to cool it to 40o C, she immersed the bottle in an aluminium bowl of heat capacity 90 J K -1 containing 500g of water at 26o C. If the mass of the mixture is 300g

    (*) Name two ways through which the bottle losses heat
    (**) Name two industrial processes in which heat exchange is used

    Solution

    (*)
    – Conduction
    – Convection

    (**)
    – incubator;
    – cooling by radiators;
    – generation of electricity;
    – refrigeration;
    – air – conditioning;
    – sewage treatment

    (10) (a) Define critical angle.

    Solution

    Critical angle is the angle of incidence in the denser medium for which the angle of refraction in the less dense medium is 90o.

     

    (b) How are anti-nodes created in a stationary wave?

    Solution

    Anti-nodes are created when an incident and its reflected waves are superposed and 180o out of phase.

     

    (c) The angle of minimum deviation of an equilateral triangular glass prism is 46.2o. Calculate the refractive index of the glass.

    Solution

    n = \( \frac{sin \frac{A \: + \: Dm}{2}}{sin \frac{A}{2}} \)

    n = \( \frac{sin \frac{60 \: + \:46.2}{2}}{sin \frac{60}{2}} \)

    n = \( \frac{sin 53.1}{sin 30} \)

    n = 1.6

     

    (d) An illuminated object is placed in front of a concave mirror and the position of a screen is adjusted in front of the mirror but no image is obtained on the screen. Give two possible reasons for this observation.

    Solution

    No image is obtained on the screen because

    – Object, screen and mirror are not co-axial;
    – Screen is located at a distance less than/greater than actual image distance;
    – Object is placed at focal point of mirror;
    – Object is placed between the focus and the pole.

     

    (e) (i) An illuminated object is placed at a distance of 75cm from a converging lens of focal length of 30cm. Determine the image distance.

    Solution

    \( \frac{1}{u} \: + \: \frac{1}{v} = \frac{1}{f}  \) \( \frac{1}{75} \: + \: \frac{1}{v} = \frac{1}{30}  \)

    v = 50cm

     

    (ii) An illuminated object is placed at a distance of 75cm from a converging lens of focal length of 30cm.
    If the lens is replaced by another converging lens, the object has to be moved 25 cm further away to have its sharp image on the screen. Determine the focal length of the second lens.

    \( \frac{1}{u} \: + \: \frac{1}{v} = \frac{1}{f}  \)

    u = 75 + 25 = 100

    \( \frac{1}{100} \: + \: \frac{1}{50} = \frac{1}{f}  \)

    f = 33.3cm

    (11) (a) Explain briefly dielectric strength.

    Solution

    Dielectric strength is the maximum potential gradient/electric field strength an insulator can withstand without breaking down. Measured in Vmm-1/Vm-1

     

    (b) (i) An electromagnetic wave has its wavelength shorter than those of radiowaves and microwaves but longer than that of visible light. Identify the wave.

    Solution

    – Infra-red wave

     

    (b) (ii) An electromagnetic wave has its wavelength shorter than those of radiowaves and microwaves but longer than that of visible light. Name one suitable detector for the wave.

    Solution

    – Infra-red camera;
    – Photographic film;
    – Bolometer;
    – Thermopile.

     

    (b) (iii) An electromagnetic wave has its wavelength shorter than those of radiowaves and microwaves but longer than that of visible light. Name one source of the wave

    Solution

    -Hot bodies;
    – Sun.

     

    (c) An oil drop carrying a charge of 1.0×10-19 C is found to remain at rest in a uniform electric field of intensity 1200 N C -1. Calculate the weight of the oil drop.

    Solution

    E =  \( \frac{F}{q} = \frac{W}{q}\)

    OR

    W = mF = Eq

    = 1200 × 1.0 × 10-19

    = 1.2 × 10-16 N

     

    (d) (i) An RLC series circuit consists of a 100Ὠ resistor,0.05 H inductor, and a 25 μF capacitor. A220 V, 50 Hz mains voltage is applied across the circuit. Calculate the impedance.

    Solution

    XL = 2Ï€fL

    = 2 × 3.14 × 50 × 0.05

    = 15.7Ὠ

    XC = \( \frac{1}{2 \pi f C} \)

    XC = \( \frac{1}{2 \: \times \: 3.14 \: \times \: 50 \: \times \: 25 \: \times \: 10^{-6}} \)

    = 127.4Ὠ

    i. Z = \( \scriptsize \sqrt{R^2 \: + \: (X_L \: -\: X_C)^2} \)

    Z = \( \scriptsize \sqrt{100^2 \: + \: (12.74 \: -\: 15.7)^2} \)

    Z = \( \scriptsize 149.9 \Omega \)

    (ii) An RLC series circuit consists of a 100Ὠ resistor, 0.05 H inductor and a 25 μF capacitor. A220 V, 50 Hz mains voltage is applied across the circuit. Calculate the current.

    Solution

    I = \( \frac{V}{Z} \)

    I = \( \frac{220}{149.9} \)

    I = 1.47A

    (12) (a) (i) Explain the following term;
    mass defect;

    Solution

    Mass defect is the difference between the mass of a nucleus and the total mass of its nucleons.

    – expressed in atomic mass unit ;
    – total mass of nucleons – mass of nucleons;
    – it is a measure of binding energy

    (ii) Explain the following terms:

    binding energy of a nucleus.

    Solution

    Binding energy is the work done/energy needed to separate the nucleus.
    -measured in eV/J or ev or J.

     

    (b) (i) Assuming the wave nature of an electron, what is the effect of decreasing the speed of a photoelectron on its;
    (*) wavelength?
    (**) energy?

    Solution

    (*) wavelength: wavelength increases
    (**) energy: energy decreases

     

    (ii) A particle of mass 4.4 × 10-23 kg moves with a velocity of 105 ms-1, calculate its wavelength.
    (h = 6.6 × 10-34 Js)

    Solution

    λ = \( \frac{h}{mv} \)

    λ = \( \frac{6.6 \: \times \: 10^{-34}}{4.4 \: \times \: 10^{-23} \: \times \: 10^5} \)

    λ = 1.5 × 10-16m

     

    (12) (c)

    The diagram above shows part of a radioactive decay series. Use it to answer the following questions;
    (i) Name a pair of isotopes.

    Solution

    21284Po and 21684Po

     

    (ii) Name the isotope with which the series starts.

    Solution

    23290Th

     

    (iii) Write down a nuclear equation for two examples of each of
    (*) alpha decay;
    (**) beta decay.

    Solution

    [*]
    21284Po → 20882Pb + 42 He + Energy
    23290 Th → 22888 Ra + 42 He + Energy

    [**]
    21283 Bi → 21284Po + 0-1 e + energy
    22488 Ra → 22489 Ac + 0-1 e + energy

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