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Physics
WAEC / WASSCE 2016
Atomic, Quantum, Nuclear Physics & Electronics
The phenomenon by which two light atomic nuclear combine to form a heavy nuclide with the release of energy is known as
A
radioactivity
B
nuclear fusion
C
nuclear fission
D
chain reaction
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #45691)
Physics
WAEC / WASSCE 2016
Current Electricity & Electric Circuits
Let \(\Delta\)x be the uncertainty in the measurements of position and \(\Delta\)p the uncertainty in measurement of momentum. The uncertainty principle relation is given as
A
\(\Delta\)x . \(\Delta\)p = h
B
\(\Delta\)x . \(\Delta\)p \(\leq\) = h
C
\(\Delta\)x . \(\Delta\)p \(\geq\) = h
D
\(\Delta\)x . \(\Delta\)p > h
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #45693)
Physics
UTME / JAMB 2016
Current Electricity & Electric Circuits
Solution
The frequency of beats produced when two tunning forks of frequencies 258Hz and 270Hz are sounded close to each other
A
6Hz
B
264Hz
C
528Hz
D
12Hz
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Beat = f2 - f1 = 270 - 258 = 12Hz
Source: Official Exam / Past Question Archive
Physics
WAEC / WASSCE 2016
Newton's Laws, Momentum & Collisions
Solution
State the dimension of; (a) impulse; (ii) acceleration; (iii) work
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Quality
Dimensions
a) Impulse
b) Acceleration
c) Work
MLT\(^{-1}\)
LT\(^{-2}\)
ML\(^2\)T\(^{-2}\)
Source: Myschool Past Questions Archive (Record #52943)
Physics
WAEC / WASSCE 2016
Linear Motion, Projectiles & Kinematics
Solution
A projectile is fired with a velocity of 20 ms\(^{-1}\) at an angle of 40^\circ to the horizontal. Determine the components of the velocity of the projectile at its maximum height.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
The components of velocity at maximum height are:
Vy = 0
Vx = UCos \(\theta\) = 20 Cos 40^\circ
=15.32ms\(^{-1}\)
Source: Myschool Past Questions Archive (Record #52946)
Physics
WAEC / WASSCE 2016
Current Electricity & Electric Circuits
Solution
State three different materials that can be used to demonstrate Brownian motion.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Materials used to demonstrate Brownian motion are: Pollen grains, dust particles, smoke particles, potassium permanganate cystals, carbon particles, etc.
Source: Myschool Past Questions Archive (Record #52947)
Physics
WAEC / WASSCE 2016
Magnetism, Electromagnetism & AC Circuits
Solution
An electron enters perpendicularly into a uniform magnetic field which has a flux density of 0.12 T This results in a magnetic force of 9.6 x 10\(^{-2}\) N on the electron. Calculate the speed of the electron as it enters the magnetic field. (e =1.6 x \(10^{19}\) C)
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
F = evB sin \(\theta\)
V = \(\frac{F}{evB sin \theta} = \frac{9.6 \times 10^{12}}{1.6 \times 10^{-19} \times 0.12 sin 90^o}\)
= \(\frac{9.6 \times 10^{-12}}{1.6 x 10^{-19} \times 0.12 \times 1}\)
= 5.0 x 10\(^8 ms{-1}\)
Source: Myschool Past Questions Archive (Record #52948)
Physics
WAEC / WASSCE 2016
Current Electricity & Electric Circuits
Solution
List three uses of rockets.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Uses rockets are:
— for space travel
— for space exploration
— for warfare
— for launching of artificial satellites into orbits
— for entertainment, that is fireworks
Source: Myschool Past Questions Archive (Record #52949)
Physics
WAEC / WASSCE 2016
Current Electricity & Electric Circuits
Solution
(a) What is doping?
(b) Explain how doping improves the conductivity of a semiconductor
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Doping is the introduction of impurity atoms into a semicondoctor
(b) Doping increases the conductivity of a semicondoctor by increasing the number of chargers or electron or holes.
Source: Myschool Past Questions Archive (Record #52950)
Physics
WAEC / WASSCE 2016
Atomic, Quantum, Nuclear Physics & Electronics
Solution
The diagram above illustrates a cathode ray tube. Identify the components X, Y, and Z.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
The cathode ray tube components are:
X = Filament or heater or cathode
Y = Anode
Z = Plates or deflectors.
Source: Myschool Past Questions Archive (Record #52951)
Physics
WAEC / WASSCE 2016
Newton's Laws, Momentum & Collisions
Solution
(a) Explain the term net force.
(b) Define the principle of conservation of linear momentum and state one example of it.
(c) A ball of mass 200 g released from a height of 2.0 m hits a horizontal floor and rebounds to a height of 1.8 in. Calculate the impulse received by the floor. (g = 10 ms\(^{-2}\)).
(d) A body of mass 20 g 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:
(i) maximum displacement from the mean position;
(ii) maximum velocity;
(iii) maximum potential energy. (g = 10 ms\(^{-2}\) \(\pi\) = 3.14)
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Net force is the effective or resultant force resulting from the actions of a system of forces on a body.
(b) The principle ofconservation of linear momentum explains that in an isolated or closed system of colliding bodies the total linear momentum in a fixed direction remains constant, e.g. the recoil of gun, colliding trolleys, rocket propulsion, etc.
(c) V\(_1^2\) = \(U_1^2 + 2gh_1\) = 0 + 2 x 10 x 2
= 40
\(V_1 = \sqrt{40}\) = 6.325m/s
\(v_2^2 = u_2^2 - 2gh_2\)
O = \(U^2_2\) - 2 x 10 x 1.8
u = \(\sqrt{36}\)
= 6ms\(^{-1}\)
Impulse Change in momentun
mv\(_1\) - (-mu\(_2\)) = m (v\(_1\) + u\(_2\))
= 0.2(6.325 + 6.000)
= 2.46 Ns
(d) v = \(\omega \sqrt{r^2 - y^2}\)
2\(\pi\)f\(\sqrt{r^2 - y^2}\)
2 = 2 x 3.14 x 5 x \(\sqrt{r^2 - 0.1^2}\)
r = 0.12m
(ii) Vmax = \(\omega\)r = 2\(\pi\)fr
= 2 x 3.142 x 5 x 0.12
= 3,77 m/s
(iii) P.E = \(\frac{1}{2}m \omega^2e^2 = \times \frac{1}{2}m(2 \pi f)^2r^2\)
= \(\frac{1}{2} \times 0.02(2 \times 3.142 \times 5)^2 \times (0.12)^2\)
= 1.42 x 10\(^{-1}J\)
Source: Myschool Past Questions Archive (Record #52952)
Physics
WAEC / WASSCE 2016
Atomic, Quantum, Nuclear Physics & Electronics
Solution
(a) Explain the terms:
(i) thermal equilibrium;
(ii) fundamental interval.
(b) List two uses of the hydraulic press.
(c) Name the material used to reset the steel index in the Six's maximum and minimum thermometer.
(d)(i) A nursing mother prepared her baby's milk mixture at 85^\circC, in a feeding bottle. In order to cool it to 40^\circC, she immersed the bottle in an aluminium bowl of heat capacity 90 JK\(^{-1}\) containing 500 g of water at 26^\circC. If the mass of the mixture is 300g, calculate the specific heat capacity of the mixture. [Neglect heat losses and heat capacity of the bottle; specific heat capacity of water = 4200 J kg\(^{-1}\) K\(^{-1}\)]
(ii) (\(\alpha\)) Name two ways through which the bottle losses heat.
(\(\beta\)) Name two industrial processes in which heat exchanger is used.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(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 between the lower fixed point, i.e. temperature of melting ice at standard atmospheric pressure and that upper fixed point, i.e. temperature of steam at standard atmospheric pressure.
(b) Uses of hydraulic press are:-
-Compressing metal sheets;
-Compressing soft materials into bales;
-Compressing paper in printing industries;
-Lifting of objects.
(c) The material used to reset the steel index in the six's maximum and minimum thermometer is magnet.
(d)(i) Heat lost by mixture = Heat gained by water + Heat gained by aluminium.
M\(_mC_w\Delta \theta_m\) + C\(_a\Delta\theta_a\)
0.3 x C\(_m\) x (85 - 40) = 0.5 x 4200 (40 - 26) + 90 (40 - 26)
13.5 C\(_m\) = 29400 + 1260
C\(_m\) = \(\frac{30660}{13.5}\)
= 2271.1 Jkg\(^{-1}\)K\(^{-1}\)
(ii) (\(\alpha\)) The bottle losses heat through conduction and convection. (\(\beta\)) Heat exchanger in: Incubation, refriaeration, air-conditioning, sewage treatment, cooling byradiators, generation of electricity, etc.
Source: Myschool Past Questions Archive (Record #52953)
Physics
WAEC / WASSCE 2016
Reflection, Refraction & Dispersion of Light
Solution
(a) Define critical angle.
(b) How are anti-nodes created in a stationary wave?
(c) The angle of minimum deviation of an equilateral triangular glass prism is 46.2^\circ. Calculate the refractive index of the glass.
(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.
(e) An illuminated object is placed at a distance of 75 cm from a converging lens of focal length 30 cm.
(i) Determine the image distance.
(ii) 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.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(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 90^\circ.
(b) Antinodes are created in a stationary wave when an incident and its reflected waves are superposed and 180^\circ out of phase.
(c) n = \(\frac{Sin\frac{(A + Dm)}{2}}{\sin (\frac{A}{2})}\)
= \(\frac{Sin\frac{(60 + 46.2)}{2}}{\sin (\frac{60}{2})}\)
= \(\frac{sin 53.1}{sin 30}\)
= 1.6
(d) No image is obtained on the screen because
- 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.
(e) (i) \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{30} - \frac{1}{7}\)
= \(\frac{5 - 2}{150} = \frac{3}{150}\)
U = 75 + 25 = 100cm
\(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\)
\(\frac{1}{100} + \frac{1}{50} = \frac{1}{50} = \frac{1}{50}\)
f = 50cm
Source: Myschool Past Questions Archive (Record #52954)
Physics
WAEC / WASSCE 2016
Electrostatics & Capacitors
Solution
(a) Explain briefly dielectric strength.
(b) An electromagnetic wave has its wavelength shorter than those of radiowave and microwave but longer than that of visible light.
(i) Identify the wave.
(ii) Name one suitable detector for the wave.
(iii) Name one source of the wave.
(c) An oil drop carrying a charge of 1.0 x 10\(^{-19 }\)C is found to remain at rest in a uniform electric field of intensity 1200 NC\(^{-1}\). Calculate the weight of the oil drop.
(d) An RLC series circuit consists of a 100\(\Omega\) resistor, 0.05 H inductor and a 25 \(\mu\) capacitor. A 220 V, 50 Hz mains voltage is applied across the circuit. Calculate the:
(i) impedance;
(ii) current. (\(\pi\) = 3.14)
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
a) Dielectric strength is the maximum potential gradient or electric field strength an insulator can withstand without breaking down. It is measured in Vmm\(^{-1}\) or Vm\(^{-1}\).
(b)(i) The wave is infra-red wave.
(ii) Suitable detector can be -Infra-red camera;
-Thermopile;
-Bolometer or
-Photographic film.
(iii) Source of the wave can be - Hot bodies or sun
(c) E = \(\frac{F}{q} = \frac{W}{q}\)
W =mF = Eq
= 1200 x 1.0 x 10\(^{-19}\)
= 1200 x 1.0 x 10\(^{-19}\)N
(d) X\(_L\) = 1.2\(\pi fL\)
= 2 x 3.142 x 50 x 0.50 = 15.7\(\Omega\)
X\(_c\) \(\frac{1}{2 \pi fL}\)
= \(\frac{1}{2 \times 3.14 \times 50 \times 25 \times 10^{-16}}\)
= 127.4 \(\Omega\)
(i) Z = \(\sqrt{R^2 + (X_1 - X_c)^2}\)
= \(\sqrt{100^2 + (127.4 - 15.7)^2}\)
= 149.9\(\Omega\)
(ii) I \(\frac{V}{Z} = \frac{220}{149.9}\)
= 1.47A
Source: Myschool Past Questions Archive (Record #52955)
Physics
WAEC / WASSCE 2016
Atomic, Quantum, Nuclear Physics & Electronics
Solution
(a) Explain the following terms:
(i) mass defect;
(ii) binding energy of a nucleus.
(b)(i) Assuming the wave nature of an electron, what is the effect of decreasing the speed of a photoelectron on its; (\(\alpha\)) wavelength? (\(\beta\)) energy?
(ii) A particle of friasS 4.4 x 10\(^{-23}\) kg moves with a velocity of 10\(^5\)ms\(^{-1}\). Calculate its wavelength. (h = 6.6 x 10\(^{-34}\) Js)
The diagram above shows part of a radioactive decay series. Use it to answer the following questions.
(i) Name a pair of isotopes.
(ii) Name the isotopes with which the series starts.
(iii) Write down a nuclear equation for two ekgmples of each of: (\(\alpha\)) alpha decay; (\(\beta\)) beta decay.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) Mass defect is the difference between the mass of a nucleus and the total mass of its nucleons. It expressed in atomic mass unit (a.m.u.). It is also a measure of binding energy.
Mass defect = total mass of nucleus
– mass of nucleons.
(ii) Binding energy is the work done or energy needed to separate the nucleons in a nucleus. It is measured in eV or Joules (J).
(b)(i) The effect of decreasing the speed of a photoelectron on: (\(\alpha\)) wavelength --> wavelength increases. (\(\beta\)) energy ---> energy decreases.
(ii) \(\lambda = \frac{h}{mv}\)
= \(\frac{6.6 \times 10^{-34}}{4.4 \times 10^{-2} \times 10^{-16}}\)m
(c) Pairs of isotopes in the figure are;
\(^{208}_{82}Pb\) and \(^{212}_{82}Pb\) or \(^{212}_{84}Po\) and \(^{216}_{84}Po\) or
\(^{224}_{88}Ra\) and \(^{288}_{88}Ra\) or \(^{288}_{90}Th\) and \(^{232}_{90}Th\)
(ii) The series started with isotopes \(^{232}_{90}Th\)
(iii) Nuclear equation for;
\(\alpha\)Alpha decay
\(^{232}_{90}Th\) \(\to\) \(^{228}_{88}Ra\) + \(^{4}_{2}He\) + Energy;
\(^{228}_{90}Th\) \(\to\) \(^{22}_{88}Ra\) + \(^{4}_{2}He\) + Energy;
\(^{224}_{88}Ra\) \(\to\) \(^{220}_{86}Rn\) + \(^{4}_{2}He\) + Energy;
\(^{212}_{84}Po\) \(\to\) \(^{208}_{82}Pb\) + \(^{4}_{2}He\) + Energy;
\(^{216}_{84}Po\) \(\to\) \(^{212}_{82}Pb\) + \(^{4}_{2}He\) + Energy;
\(^{212}_{84}Rn\) \(\to\) \(^{208}_{82}Po\) + \(^{4}_{2}He\) + Energy;
\(^{220}_{84}Rn\) \(\to\) \(^{216}_{84}Po\) + \(^{4}_{2}He\) + Energy;
\(^{216}_{84}Rn\) \(\to\) \(^{212}_{82}Pb\) + \(^{4}_{2}He\) + Energy;
(\(\beta\)) Beta decay
\(^{224}_{88}Ra\) \(\to\) \(^{224}_{89}Ac\) + \(^{o}_{-1}e\) + Energy;
\(^{228}_{88}Ac\) \(\to\) \(^{228}_{90}Ac\) + \(^{o}_{-1}e\) + Energy;
\(^{212}_{82}Pb\) \(\to\) \(^{212}_{83}Bi\) + \(^{o}_{-1}e\) + Energy;
\(^{212}_{83}Ra\) \(\to\) \(^{212}_{84}Ac\) + \(^{o}_{-1}e\) + Energy;
Source: Myschool Past Questions Archive (Record #52956)