Physics 2015 JAMB Past Questions
Physics 2015 JAMB Past Questions
1.If the volume of a gas is recoded as 76cm3 at a temperature of 27°C with a pressure of 800mmHg. What will be the volume of the gas at S.T.P?
- A. 36.2cm3
- B. 25.7cm3
- C. 72.8cm3
- D. 24.3cm3
Explanation
General formular of Gas laws(P1V1) ÷ T1= (P2V2) ÷ T2
P1 = 800mmHg, V, = 76cm3, Tr = 273 + 27 = 300k
P2 = 760mmHg, V2 = ? T2= 273 + 0 = 273 + 6 = 277k
(800 × 76) ÷ 300 = (760 × V2) ÷ 273
V2 = (300 × 76 × 273) ÷ (300 × 760) = 72.8 cm3
The volume of gas as S.T.P = 72.8 cm3
2. Boyle’s law may be Witten as
- A. P1V1= P2V2
- B. V1 = V2 ÷ T1T2
- C. P1T1 = P2T2
- D. P1V1 ÷ T1 = P1V1 ÷ T2
Correct Answer: Option A
Explanation
Boyle‘s law stats that the volume of a given mass of a gas, at a constant temperature is inversely proportional to us pressure.Mathematically
V ∝ l/p, v = k/p (where k is constant)
K = VP
Let k = k1
Then, if k = k, V and P = V1P1
V1P1 = V2P2
3. The direction of the path taken by light is called
- A. Locus
- B. Lines
- C. Ray
- D. Beam
Correct Answer: Option C
Explanation
Ray – a beam of light or radiation
4. What is the inductance (L) of an inductor whose reactance is 1Ω at 50/ π hertz (H)?
- A. 100 H
- B. 0.01H
- C. 10H
- D. 1H
Correct Answer: Option B
Explanation
XL = WL = 2πFL (w = 2πf)XL = reactance inductor (S.I unit Ω)
L = inductance (L) in Henry
XL = 1Ω, f = 50/π, L = ?
1 = 2× π × 50/π × L
1 = 100L
i.e. 100L = 1
L = 1/100
0.01H
∴ Inductance (L) of a inductor = 0.01H
5. A stroboscope can be used to make the ware appear …………..
- A. Stationary
- B. In motion
- C. Dipper
- D. Ripple
Correct Answer: Option A
Explanation
Stroboscope is a scientific instrument used to make a wave appears stationary when used
6
In an elastic collision
I. Energy is conserved
II. Energy is decreased
III. Energy is increased
IV. Linear momentum conversed
I. Energy is conserved
II. Energy is decreased
III. Energy is increased
IV. Linear momentum conversed
- A. I only
- B. I and IV only
- C. II only
- D. III only
Correct Answer: Option B
Explanation
In an inelastic collision, linear momentum is conserved and energy is decreasedMomentum and kinetic energy are conserved in elastic collision
7. Body floats in fluid when its …………………
- A. Displaces its own volume of the liquid (fluid)
- B. Displaces its own mass of the fluid
- C. Displace its own density of the fluid
- D. Is no longer subjected to gravity
Correct Answer: Option B
Explanation
Note:The principle of floatation states that a floating body displaces its own weight of the fluid or which its float.For object to sink weight of the object (W) > upthrust of the fluid (v) i.e. w > u
(i.e. weight of the body (or object) mass of the fluid)
8. The general gas law can be written as
- A. PV ÷ T = a constant
- B. PV = a constant
- C. v/t = a constant
-
Correct Answer: Option A
Explanation
Formula for general gas law
V × 1/T × P
N [(PV) ÷ T] = K is constant D. p/T = a constant
9. A motor boy of a
driver saw what seem to appear to be a pool of water a head of him,
while his driver’s drives a new truck on a funny day. Which of the
following explains his observation?
- A. He saw a flying objects
- B. He must be thinking of water and food
- C. He was seeing the shadow of the car
- D. The sun ray from the sky were passing from a hot air (layer) to cold layer
Correct Answer: Option D
Explanation
The appearance of pool of water on road surfaces is called MIRAGEHis observation show that light from the sun ray refracted through a layer of hot air(layer) which is closed to the ground, passes from a hot air layer to the cold layer
10. Which of the following is not luminous object?
- A. The sun
- B. The moon
- C. light candle
- D. Star
Correct Answer: Option B
Explanation
Sun are self luminous sourceMoon is a non – luminous
11. The resultant force of a couple is ………?
- A. Infinity
- B. Zero
- C. One
- D. Half
Correct Answer: Option B
Explanation
Note: moment of a couple = force × length (fL)The tangential force is actingin an opposite direction alongside its length
i.e f1 L1 = f2L2
f2L2 − f1 L1 = 0(i.e Zero)
12. A vector quantity can only be completely described when ……… and …………are mentioned.
- A. Magnitude and motion
- B. Magnitude and direction
- C. Distance and specified displacement
- D. Moment and momentum
Correct Answer: Option B
Explanation
A vector quantity has both magnitude and direction
13. In which of the following device in the expansion of solid is a disadvantages?
I. Simple pendulum
II. Fire alarms
III. Thermostat
IV. Bimetallic thermometer
- A. III only
- B. II & III only
- C. I only
- D. II, III & IV only
Correct Answer: Option C
Explanation
Simple pendulum does not depend on thermal expansionFive alarms, Thermostat and Bimetallic thermometer are application of thermal expansion of solid
14. If an object is placed at a height of tem above the ground at a stationary point. It posses what types of energy?
- A. Mechanical energy
- B. Stationary energy
- C. Potential energy
- D. Vibrational energy
Correct Answer: Option C
Explanation
Energy due to position T is potential energy (P.E)Energy due to motion is kinetic energy (K.E)
15. Calculate the gravitation force of attraction between two planets of 1024kg and 1027kg separated by a distance of 1020meters.
- A. 6.67N
- B. 6.56N
- C. 7.5N
- D. 8.6N
Correct Answer: Option A
Explanation
F = (Gm1m2) ÷ r2M1 = 1024Kg, M2 = 1027Kg , r = 102
(Note: the gravitational constant of G =
F = \( (6.67 × 10^{– 11 + 24 + 27)} 10^40 = 6.67 × (10^40 ÷ 10^40)
= 6.67N
f = 6.67N
16. Mr. F. Abioye
observed his image though a plane mirror, kept a distance way from him
at 4.7m in his room at what distance is the image found from Abioye eye?
- A. 9.4m
- B. 7.5m
- C. 5.7m
- D. 6.9m
Correct Answer: Option A
Explanation
As object and image are usually at equal distance apart from the mirror /p>Distance between Abioye (a) and hip image (I) = AI
i.e. AI = 4.7 + 47
= 9.4m
17. Primary colours of light are
- A. Red, blue and green
- B. Cyan, magneta and yellow
- C. Black, blue and green
- D. None of these colour
Correct Answer: Option A
Explanation
Primary colours of light are red, blue and greenSecondary colours of light are cyan, magneta and yellow
18. Electrons were discovered by
- A. James Chadwick B.
- B. J.J Thompson
- C. Sir Isaac Newton
- D. Charles Newton
Correct Answer: Option B
Explanation
J.J. Thompson discovered ELECTRONS
19. What is the relationship mechanical advantage (MA) velocity Ratio (V.R) and efficiency with percentage?
- A. e = M.A/V.R × 100 %
- B. M.A = E.V.R
- C. V.R = 1 ÷ sin θ M.A
- D. M.A = 1/E × 100%
Correct Answer: Option A
Explanation
he relationship between M.A, V.R, e and 100% can be written ase = M.A/V.R × 100 %
i.e E/100 = M.A/V.R
= L ÷ E/x ÷ y = Ly/Ex
E/100 = Ly/Ex
20. The external parts of an object can be measure accurately through the use of which scientific instrument?
- A. Micrometer screw gauge
- B. Vernier Caliper
- C. Meter rule
- D. None of the instrument
Correct Answer: Option A
Explanation
Micrometer screw gauge measure an objects accurately (mainly on the external parts of the object). It measures the external parts of objects only
21. Convert 45°C to kelvin
- A. 318K
- B. 250K
- C. 184K
- D. 120K
Correct Answer: Option A
Explanation
Note: T = 273 + θθ = 45°
T = 273 + 45
= 318K
22. A member of the crew of a SPACESHIP experiences weightlessness in a space when they
- A. Fall freely from the earth’s gravitational field
- B. Is walking on the planet
- C. In between the sun and the earth
- D. Holding anything in space is negative
Correct Answer: Option A
Explanation
Weightlessness:– The state of being free from the effects of gravity. (zero gravity)
23. Radio waves belong to the class of wave whose velocity is about
- A. 34m/s
- B. 3 ×106m/s
- C. 3 ×108m/s
- D. 3 × 10-8m/s
Correct Answer: Option C
Explanation
A radio waves is an electromagnetic waves which travels with a constant velocity of 3 × 108m/s
24. At 35°C and 5
atmospheres pressure the density of a gas is 0. 600g per liter. What is
the density at 7 atmospheric pressure and 350°C.
- A. 0.27g / liters
- B. 0.25g / liters
- C. 1.7g / liters
- D. 0.15g / liters
Correct Answer: Option C
Explanation
atmd1 = 0.600g/ litre
θ2 = 350°C, T2 = 2 + 3 + θ
= 273 + 350 = 623k, d2 = ?
P2 = 7atm
T a P a 1/d = TPK/d (where k is a constant)
Tp1 ÷ d1 = T1P2 ÷ d2, [(308 × 5) ÷ 0.600]
= (623 × 7) ÷ d2
d2 = (623 × 7 ×0.600) ÷ (308 × 5)
= 2616.6 ÷ 1540
= 1.69g / litresd2 = 1.69g / litres
∴ d2 = 1.70g / litres (Approx.)
25. The compressed regions and the space-out regions in longitude wave are refers to as
- A. Compression and Rarefaction
- B. Comparison and Explosion
- C. Reflection only
- D. Refraction only
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