2016 WAEC Essay May/June Answers

2016 WAEC Essay May/June Answers

1. i. dimension of impulse

Impulse = force x time

        = mass x acceleration x time = kg x ms^-2 x s

        = M x LT^-2 x T = MLT^-1

ii. dimension of acceleration

        acceleration = ms^-2 =LT^-2

iii. dimension of work

        work = force x distance = kgms^-2 x m =ML²T^-2

2. Given: u=20m/s   ϴ=40⁰

V_x = u cos ϴ = 20 cos 40 = 15.32 m/s

V_y = 0

3. Materials that can be used to demonstrate  Brownian motion: smoke particles, dust particles ,pollen grains , chalk particles ,potassium permanganate crystal, carbon particles

4. Given: q = e = 1.6 x 10^-19 c    B = 0.12 T   F = 9.6 x 10^-12 N 

ϴ = 90⁰     V = ?    F = qvBsin ϴ

V = ( F ) / (qBsin ϴ) = (9.6 x 10^-12) / (1.6 x 10^-19 x 0.12 x sin 90)

        = 5.0 x 10⁸ m/s

5. Uses of rockets : for space exploration, for space travel , for warfare,  for fireworks , for launching of artificial satellites into orbits

6a. Doping is the introduction of impurity atoms into a semiconductor

6b. Doping increases the number of charge carriers or electron holes thereby increasing the conductivity of the semiconductor

7.  X= Filament or heater or cathode Y= Anode

 Z =  plates or deflectors

8a. Net force is the effective force or resultant force from the actions of forces on a body. It causes the body to accelerate.

8b. The principle of conservation of linear momentum states that in an isolated or closed system of colliding bodies, the total linear momentum in a fixed direction remains constant

Example of the principle of conservation of linear momentum:

The recoil of a gun, rocket propulsion, colliding trolleys

8c. Given: m = 200g = 200/1000 = 0.2kg   h₁ = 2m   h₂ = 1.8m

 g = 10m/s²   impulse = ?

V₁² = u₁² + 2gh₁ = 0 + 2 x 10 x 2     v₁² = 40   v₁ = 6.325 m/s

V₂² = u₂² – 2gh₂   0 = u₂² – 2 x 10 x 1.8   u₂² = 36   u₂ = 6 m/s

Impulse = change in momentum = mv₁ – ( - mu₂ ) = mv₁ + mu₂

= m ( v₁ + u₂ ) = 0.2 (6.325 + 6 ) = 0.2 x 12.234 = 2.46 Ns

8d. Given: m = 20g = 20/1000 = 0.02kg   f = 5H_z 

y = 10cm = 10/100 = 0.1m v = 200 cm/s = 200/100 = 2 m/s

8dii.  V_{max =} wr = 2πfr = 2 x 3.14 x 5 x 0.12 =3.77 m/s

8diii. P.E = ½ mw²r² = ½ m(2πf)²r²

= 0.5 x 0.02 x ( 2 x3.14 x 5 )² x 0.12² = 0.14 joule = 1.4 x 10^-1 J

9a. i. Thermal equilibrium is a condition in which two bodies in contact have no net flow of heat between them

ii. Fundamental interval is the difference in temperature of melting ice at standard atmospheric pressure and the upper fixed point, i.e., temperature of steam at standard atmospheric pressure

9b. uses of the hydraulic press are:

Compressing metal sheet, compressing soft materials into bales, compressing paper in printing industries, lifting of objects

9c. The material used to reset the steel index in the six’s maximum and minimum thermometer is Magnet

9d. Given: m_m = mass of mixture = 300g = 300/1000 = 0.3kg

c_m = ?

ϴ_m = temperature change of mixture = 85⁰C – 40⁰C = 45⁰C

m_w = mass of water = 500g = 500/1000 = 0.5kg

c_w = specific heat capacity of water = 4200 jkg^-1k^-1

ϴ_w = 40⁰C – 26⁰C = 14⁰C

C_a  = heat capacity of aluminium = 90 J/k

ϴ_a =temperature change of aluminium bowl=40⁰C – 26⁰C = 14⁰C

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

m_m c_m ϴ_m = m_w c_w ϴ_w + C_a ϴ_a

c_m = ( m_w c_w ϴ_w + C_a ϴ_a ) / (m_m  ϴ_m )

c_m = ( 0.5 x 4200 x 14 ) + ( 90 x 14 )

        (0.3 x 45)

c_m = 2271.1 Jkg^-1k^-1

ii. α. Two ways through which the bottle losses heat : through conduction, through convection

β.   Industrial processes in which heat exchanger is used : incubation, refrigeration, air conditioning, sewage treatment, cooling by radiators, generation of electricity

10. a. Critical angle is defined as the angle of incidence in the denser medium for which the angle of refraction in the less dense medium is 90⁰

10. b. How antinodes are created in a stationary wave : antinodes are created in a stationary wave when an incident wave and its reflected waves are superposed and 180⁰ out of phase

10. c. Given : D_m = 46.2  A = 600  A/2 = 300  η = ?

η = (sin ((A + D_m )/ 2 )) ⁄ (sin (A/2)) = (sin ((60+46.2)/2))/(sin(30))

= 1.599

10d.   possible reasons why no image is obtained on the screen when an illuminated object is placed in front of a concave mirror and the position of a screen is adjusted in front of the mirror: - object, screen and mirror are not coaxial – object is placed at focal point of mirror – object is placed between the focus and the pole –screen is located at a distance less than or greater than actual image distance

10ei. Given: u = 75cm  v =?  f = 30cm  

1/u  + 1/v  = 1/f      1/v = 1/f  - 1/u    1/v = 1/30 -  1/75

 1/v = 5-2 / 150    1/v = 3/150  = 1/ 50 

V= 50cm

10eii. Given: u = 75 + 25 = 100cm f=?  V= 50cm   1/f = 1/u + 1/v

1/f = 1/100 + 1/50            1/f = 1+2  / 100          1/f = 3/100          
  f= 100/3 = 33.333cm

11a. Dielectric strength is the maximum potential gradient or electric field strength an insulator can withstand without breaking down. It is measured in V/m

11bi. The wave that has its wavelength shorter than those of radio wave and microwave but longer than visible light is infra-red wave

11bii. Suitable detector for infra red wave are: infra red camera, thermopile, bolometer, photographic film

11biii. Source of infra red are: hot bodies, sun

11c. Given: q=1.0 x 10^-19 C E= 1200 N/C  W=F=?   E=F/q

F=Eq = 1200 x 1.0 x 10^-19 = 1.2 x 10^-16 N

11d. Given : R=100Ω L=0.05H C=25µF= 25 x 10^-6 F  V= 220v  F=50H_z  Z = ?  I = ?  π=3.14

Z² = R² + (X_L – X_C)²      X_L = 2πFL = 2 x  3.14 x 50 x 0.05 = 15.7Ω

X_C = 1 / 2πFC = 1 / (2 x 3.14 x 50 25 x 10^-6) = 127.389Ω

11di.    Z² = R² + (X_L – X_C)²      = 100² + ( 15.7² – 127.389²)

Z² = 22474.433

Z = 150Ω

11dii.  V = IZ      I = V/Z   220/150   =  1.47A

12ai. Mass defect represents the mass of the energy binding the nucleus, and is the difference between the mass of a nucleus and the sum of the masses of nucleons of which it is composed . It is expressed in atomic mass unit ( a.m.u) . mass defect = total mass of nucleus – mass of nucleons

12aii. Binding energy is the work done or energy needed to separate the nucleons in  a nucleus  . It is measured in eV or J(joules)

12biα. Assuming wave nature of an electron, the effect of decreasing the speed of a photoelectron on its wavelength: wavelength increases

12biβ. Assuming wave nature of an electron, the effect of decreasing the speed of a photoelectron on its energy: energy decreases

12bii. M = 4.4 x 10^-23kg  V= 105 m/s  h = 6.6 x 10^-34 Js     λ = ?

λ = h / mV   =  (6.6 x 10 ^-34)/(4.4 x 10 ^-23 x 10⁵ )

λ = 1.5 x 10^-16 m

12c.