2017 WAEC GCE Physics Theory/Essay Questions and Answers | Verified Expo Site
GCE WAEC 2017 physics theory, specimen and waec physics practical specimen - In this article, I settled down and drafted some very important Physics past questions and answers for students who will need it to prepare for their exams. Read carefully to see the physics answers on the this page. Do not use it as an expo. ts for study purposes only.
Question 11
Define: work; energy.
A pump is used to raise water from a depth of 20 m to fill a reservoir of volume 1800 m3 in 5 hours. Calculate the power of the pump. ( Density of water = 1000 kg m-3; g = 10 ms-2)
(i) What is terminal velocity?
(ii) Explain the action of a parachute.
(iii) Why is it important that a defender in football should have a lot of mass?
Comments
Most candidates gave the correct definition of energy but could not define work correctly. The phrase “ in the direction of the applied force “ was missing in their definition of work. Some candidates mistook the definition of energy for work.
The calculation of the power of the pump was well tackled by most respondent candidates. However, very few candidates had problems deriving the mass of water m from volume v and density ρ of water. Some candidates did not convert the time from hours to seconds before substitution while few other candidates could not relate Pt = mgh.
Many candidates were not able to explain the concept of terminal velocity nor the action of a parachute as many of them failed to establish the fact that the amount of air resistance experienced is increased by the umbrella of the parachute. Performance was low.
The reason why a defender in football should have a lot of mass was poorly tackled by many candidates as they could not relate mass to inertia.
The expected answers are:
a)(i) Work is the product of the force and displacement/ distance in the direction of the force. Accept W = Fs provided the letters are correctly defined.
(ii) Energy is the capacity / ability to do work.
(b) m = D x V
= 1000 x 1800
= 1,800,000
Pt = mgh
P x 5 x 60 x 60 = 1,800,000 x 10 x 20
P = 20,000W
(c) (i) The steady/constant speed with which an object/body moves freely downward through a fluid when the resultant force on it is zero
OR
The steady/constant speed attained by an object/body falling freely through a fluid when the viscous force and upthrust of the fluid on it are (jointly) equal to the weight of the object/body
(ii) The umbrella of the parachute increases the amount of air resistance experienced by the sky driver and this in turn slows down the (rate of )fall until terminal velocity is attained.
(d) An attacker runs towards a defender’s goal with high velocity that generates high momentum in order to stop/block the attacker the defender must have great inertia which is a measure of mass so the defender must have a lot of mass.
Question 12
(a) State the principle of conservation of energy.
(b) State three effects of heat on a substance.
(c) Explain why when the bulb of thermometer is dipped into ice-cold water the mercury level first rises before falling.
(d) Water of volume 200 cm3 in an aluminum container at 20o C is cooled to – 10oC when put in a freezer.
Calculate the quantity of heat energy extracted from the water. If heat is extracted at a rate of 120 Js-1 by the freezer, calculate the time taken to produce this ice. State a reason why the calculated time will be less than the actual time. (Density of water = 1000 kg m-3; specific heat capacity of water = 4200 J kg-1; specific latent heat of fusion of ice = 336000\j kg-1; specific heat capacity of ice = 2100J kg-1 K-1)
Question 13
(a) Define:
(i) principal focus for a diverging lens;
(ii) principal axis.
(b) State a situation under which the speed of light can change.
(c) The separation between the lenses of an astronomical telescope in normal adjustment is 35.0 cm and the angular magnification is 8
Calculate the focal lengths of the lenses.
The lenses separation are adjusted to produce the final image 30 cm from the eye piece lens. Calculate the new separation of the lenses
Sketch a ray diagram to illustrate the image formed in 13(c)(ii)above.
Comments
(a) Most respondent candidates were not able to define correctly principal focus for a diverging lens and principal axis.
(b) Relatively few candidates succeeded in stating the required situation under which the speed of light can change. Many of them failed to realize that the speed changes as light travels from one medium to another of different optical densities.
(c) Most candidates messed up with their responses on this numerical on astronomical telescope in normal adjustment. This state of affairs could readily be ascribed to the effect of rote-learning as candidates reactions showed lack of proper understanding of the concept. The sketching of the ray diagram was a mirage.
The expected answers are:
(a)(i) This is a point on the principal axis to which all rays parallel and close to the principal axis appear to diverge after refraction / passing through the lens.
(ii) This is a line joining the centre of curvature (principal focus) of the lens to the optical centre.
(b) When the light travels from one medium to another of different refractive index/optical density.
(c) Angular magnification = fo
fe
fo + fe = 35
8fe + fe = 35
9fe = 35
fe = 3.89 cm
fo = 35 – 3.89
fo = 31.11 cm
1 = 1 + 1
fe = u v
1 = 1 + 1
3.89 u -30
1 = 1 + 1 = 30 + 3.89
u 3.89 30 116.7
u = 116.7 33.89 = 3.44cm
- New separation of lenses = 3.44 + 31.11 = 34.55cm
- Parallel rays incident on objective lens
- Correct position of real image formed by objective lens
- Virtual image formed close to the objective lens
Question 14
(a) Define magnetic flux.
(b) Sketch a diagram to illustrate magnetic lines of force for a bar magnet placed in a uniform earth’s magnetic field with its north pole pointing north. Indicate the neutral points.
(c) State two ways in which the earth’s magnetic field is important to life on the earth surface.
(d) List two applications of electromagnets
(e) In a certain house, three ceiling fans each of 80 W, an air-conditioner rated 1500 W are operated for 6 hours each day. Seven lamps each rated 40 W are switched on for 10 hours each day. The home theatre 100 W and television set 80 W are also operated each day for 5 hours after which they are kept in the standby mode. In the standby mode, the power consumption is 5% of the power rating. Calculate the:
- Total energy consumed in the house for 30 days in kWh;
- Cost of operating all the appliances for 30 days at N10.00 per kWh.
Comments
- The description of magnetic flux posed some problem to candidates who failed to see it as basically a collection of (imaginary) lines of force in a magnetic field. This part called for a sketch of a given magnetic field. It was poorly reacted to by candidates.
- Some sketches were rough while some did not indicate the correct direction of the arrows on the magnetic lines.
- Most candidates were able to state ways in which magnetic field is important to life on earth. Performance was fairly alright.
- Listing of applications of electromagnets was well answered by virtually all the respondent candidates. Performance was good.
- The numerical part was poorly tacked most respondent candidates mishandled the standby mode part of the question. Performance was poor.
The expected responses are:
(a) Magnetic flux consists of lines which represent the direction and strength of a magnetic field at any point in the field.
(b) correct shape of lines of force correct arrows / directions two neutral points P correctly indicated
(c) Importance of earth’s magnetic field to life on earth e.g
- Navigation by sailors and pilots
- Plotting of contour lines by surveyors
- shielding of dangerous radiations / cosmic rays from the sun
(d) Application of electromagnetic e.g
- loud speakers
- door bells
- automatic switches / relays
- separation of magnetic materials from non magnetic materials
- lifting of heavy magnetic materials
NOTE: Standby mode was for 19 hours
- Energy consumed in a day = 14.311 kWh
- Energy consumed in 30 days = 14.311 x 30 = 429.33 kWh
(ii) Cost of operating for 30 days
1 KWh N10.00
429.33KWh 429.33 x 10 = N 4293.30
Question 15
Explain radioactivity under the following headings: meaning; types.
State two: advantages of nuclear fusion over nuclear fission; methods of detection of radioactivity.
A nuclear fusion equation for generation of power is shown below
+ +
[ H = 1.00728u; H = 2.00141u He = 3.00160u He = 4.00150u; c = 3.0 x 108 ms-1; Iu = 1.7 x 10-27 kg]
An electron of mass 9.1 x 10-31 kg moves with a speed of 4.8 x 108 ms-1 in a vacuum. Calculate the wavelength of the associated wave. [h = 6.6 x 10-34 Js]
Many candidates gave a fair attempt at this question as they found it very tractable. However, few candidates were still using the word “spontaneous” in defining radioactivity.
Advantages of nuclear fusion over nuclear fission was poorly attempted. However, they performed well in their attempt to state two methods of detecting radioactivity.
Comments
In the calculation of the energy released in the reaction, some candidates made themselves creditably equal to the required task. However, few candidates found the conversion of the unified atomic mass to kilogram difficult. Very few candidates did the correct thing while most of other respondent candidates derailed by starting their responses on a wrong note; i.e with wrong formula. Generally, performance was fair.
The expected responses are:
(a)(i) Radioactivity is the decay/disintegration of the nucleus of an unstable nucleus/atom (of an element) with the release of any one, two or all of alpha, beta and gamma radiations.
(ii) Types of radioactivity: Natural and Artificial
(b)(i) Advantages of nuclear fusion over nuclear fission e.g
- greater energy is released
- less dangerous radiations are released
- raw materials required are cheaper
- raw materials are readily available.
(ii) Methods of detecting radioactivity. Any correct 2. e.g
- photographic films
- Geiger muller (G-M) tube
- (Electron) – e loud chamber
- semiconductor detectors
(c) m = mass defect = (mass of products ) – (mass of reactants)
= 4.00150 2.00141
+ 1.00728 - + 3.00160
5.00878 5.00301
Mass defect = 0.00577u
= 5.77 x 10-3u
= 5.77 x 10-3 x 1.7 x 10-27 kg
= 9.81 x 10-30 kg
E = mc2
= 9.81 x 10-3 x ( 3 x 108 )2
= 8.83 x 10-13 J
(d) λ = h
mV = 6.6 x 10-34
9.1 x 10-31 x 4.8 x 108
= 1.51 x 10-12 m
GENERAL COMMENTS ON THE PAPER AS A WHOLE
- The standard of the paper compared favourably with those of previous years.
- The questions and rubrics were clear and the time allocated is very adequate.
- The marking scheme was detailed and generous.
- The performance of the candidates were slightly better than that of last year.
- The candidature of 167986 had a raw mean score of 10 and a standard deviation of a 05.74 as against a raw mean score of 10 and a standard deviation of 05.92 in November/December 2011 WASSCE with a population of 159123.
CANDIDATES WEAKNESSES AND SUGGESTED REMEDIES
Weaknesses observed are as follows:
- Many candidates could not differentiate between the demands of questions such as define, what is and explain.
- poor grammatical expressions
- poor computational skills
- lack of understanding of some concepts such as; work, terminal velocity, principal focus for a diverging lens, principal axis, capillary action, astronomical telescope in normal adjustment, magnetic flux and application of energy consumed in kilowatt-hour.
The following remedies were suggested:
- WAEC should reverse back to conducting qualifying examinations before the conduct of WASSCE.
- Society should discourage harbouring “miracle” examination centres to thrive.
- Parents to provide candidates with text books.
- Government to employ qualified teachers
- Teachers to employ basic concepts and principles and application of such to everyday life.
- Government to provide well-equipped laboratories and libraries
- Teachers to drill candidates appropriately while preparing them for the examination
- Candidates should be more studious
- Government to provide adequate teaching facilities in schools
- Teachers to discourage rote-learning by candidates and improve on teaching and learning processes.
CANDIDATES’ STRENGTHS
Majority of the candidates were able to:
- list two uses of Polaroid
- calculate the work done by a loaded wire
- list two properties of cathode rays;
- state one phenomenon that can only be explained in terms of the wave nature of light.
- define energy
- state three effects of heat on a substance
- explain the meaning of radioactivity
- state two methods of detection of radioactivity
- made conscious effort to attempt all the questions.
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