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Physics
UTME / JAMB 2020
Waves, Sound & Vibrations
Solution
A ship travelling towards a cliff receives the echo of its whistle after 3.5 seconds. A short while later, it receives the echo after 2.5 seconds. If the speed of sound in air under the prevalling condition is 250ms -1, how much closer is the ship to the cliff?
A
10m
B
350m
C
175m
D
125m
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Difference in time = 1 sec
echo time = 1 \(\div\) 2 = 0.5 sec
Difference of cliff from ship = 0.5 x 250
= 125m
Source: Myschool Past Questions Archive (Record #56915)
Physics
UTME / JAMB 2020
Lenses & Optical Instruments
Solution
Which of the following statements about defects of vision is/are CORRECT?
I. For a long sighted person, close objects appear blurred
II. For a short sighted person, distant objects appear blurred
III. Short sight is corrected by using a pair of converging lenses
A
I only
B
II only
C
I and II only
D
II and III only
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Statements I & II are correct
Statement III is incorrect because short sightedness is corrected by using suitable concave or diverging lens
Hence; option C is the correct option
Source: Myschool Past Questions Archive (Record #56916)
Physics
UTME / JAMB 2020
Reflection, Refraction & Dispersion of Light
Solution
Which of the following conditions are necessary and sufficient for total internal reflection to take place at the boundary between two optical media?
I. Light is passing from optical denser medium to optically less dense medium
II. Light is passing from optically less dense medium to optically denser medium
III. Angle of incidence is greater than a certain critical value
IV. Angle of incidence is lesser than a certain critical value
A
I and II only
B
II only III only
C
III and IV only
D
I and III only
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The conditions for total internal reflection include light traveling from an optically denser medium (higher refractive index) to an optically less dense medium (lower refractive index) and that the angle of incidence is greater than the critical angle.
Source: Myschool Past Questions Archive (Record #56917)
Physics
UTME / JAMB 2020
Temperature, Thermometry & Thermal Expansion
Solution
The linear expansivity of brass is 2 x 10\(^{-5}\) \(^oC^{-1}\). If the volume of piece of brass is 10cm\(^3\) at 0\(^o\)C. What will be its volume at 100\(^o\)C?
A
10.06cm\(^3\)
B
10.04 cm\(^3\)
C
10.02 cm\(^3\)
D
10.20 cm\(^3\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
3 x \(\alpha\) = \(\frac{V_2 - V_1}{V_1 \times \Delta \theta}\)
3 x 2 x 10\(^{-5}\) = \(\frac{V_2 - 10}{100 \times 10}\)
V\(_2\) = (100 x 10 x 3 x 2 10\(^{-2}\)) + 10
= 10.06 cm\(^3\)
Source: Myschool Past Questions Archive (Record #56918)
Physics
UTME / JAMB 2020
Temperature, Thermometry & Thermal Expansion
Solution
The lower and upper fixed points marked on a Mercury-in-glass thermometer are 210mm apart. The end of the mercury column in the tube is 49mm above the lower fixed point in a room. What is the temperature of the room in degrees Celsius?
A
55.3\(^o\)C
B
23.3\(^o\)C
C
49.0\(^o\)C
D
16.43\(^o\)C
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
\(\frac{49 - 0}{210 - 0}\) = \(\frac{\theta - 0}{100 - \theta}\)
\(\frac{49}{210} = \frac{\theta}{100}\)
49 x 100 = 210 x \(\theta\)
\(\theta = \frac{4900}{210}\) = 23.3\(^o\)
Source: Myschool Past Questions Archive (Record #56919)
Physics
UTME / JAMB 2020
Current Electricity & Electric Circuits
Diagram
Solution
Two mirrors of the same length are arranges as show in the diagram. A ray of light NO strikes the system at O and emerges along PQ. The emergent ray has been deviated through
Attached Figure / Diagram
A
220\(^o\)
B
200\(^o\)
C
210\(^o\)
D
180\(^o\)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Deviation angle
= \(\theta_1 + \theta_2\) = 50 + 50 + 40 + 40
= 180\(^o\)
Source: Myschool Past Questions Archive (Record #56910)
Physics
UTME / JAMB 2020
Current Electricity & Electric Circuits
A magnetic needle is suspended first at the earth's north magnetic pole and then at a point on the magnetic equator. The respective angles between the needle and the horizontal are
A
0\(^o\) and 0\(^o\)
B
60\(^o\) and 60\(^o\)
C
90\(^o\) and 90\(^o\)
D
90\(^o\) and 0\(^o\)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #56911)
Physics
UTME / JAMB 2020
Equilibrium of Forces, Moments & Flotation
A hydrometer is an instrument for measuring the
A
a depth of water of water in a vessel
B
relative density of a liquid by the method of flotation
C
relative density of a liquid by finding the apparent loss in weight
D
relative humidity of the air
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 #56912)
Physics
UTME / JAMB 2020
Current Electricity & Electric Circuits
Diagram
Solution
Consider the three forces acting at O and in equilibrium as shown in the figure. Which of the following equations is/are CORRECT?
I. P\(_1\) cos \(\theta_1\) = p\(_1\) cos \(\theta_2\)
II. P\(_3\) = P\(_1\) cos \(\theta_4\) + P\(_2 cos_2\)
III. P\(_1 \sin \theta_1 = P_2 \sin \theta_2\)
Attached Figure / Diagram
A
I only
B
II only
C
III only
D
I, II and III only
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
resolving \(P_1\), \(P_2\) and \(P_3\) into x and y component
\(P_x\) = \(P_1 Cos\theta_1\)
\(P_y\) = \(P_1Sin\theta_1\)
\(P_x\) = \(P_2Cos\theta_2\)
\(P_y\) = \(-P_2Sin \theta_2\)
\(P_x\) = \(- P_3Cos \theta0\)
therefore
total \(P_x\) = \(P_1 Cos\theta_1\) + \(P_2Cos \theta_2\) + \(- P_3Cos \theta0\) = 0
total \(P_y\) = \(P_1Sin\theta_1\) + \(-P_2Sin \theta_2\) = 0
so that \(P_1Sin\theta_1\) = \(P_2Sin \theta_2\)
\( P_3Cos \theta0\) = \(P_1 Cos\theta_1\) + \(P_2Cos \theta_2\)
Source: Myschool Past Questions Archive (Record #56913)
Physics
UTME / JAMB 2020
Reflection, Refraction & Dispersion of Light
Solution
The refractive index of a liquid is 1.5. If the velocity of light in vacuum is 3.0 x 10\(^8ms^{-1}\), the velocity of light in the liquid is
A
1.5 x 10\(^8ms^{-1}\)
B
2.0 x 10\(^8ms^{-1}\)
C
3.0 x 10\(^8ms^{-1}\)
D
4.5 x 10\(^8ms^{-1}\)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
n = \(\frac{\text{speed in air}}{\text{speed in glass}}\)
1.5 = \(\frac{3 \times 10^4}{\text{speed in glass}}\)
Speed in glass = 3 x 10\(^8\)m/s / 1.5
= 2 x 10\(^8\)m/s
Source: Myschool Past Questions Archive (Record #56914)
Physics
UTME / JAMB 2020
Calorimetry, Latent Heat & Change of State
Solution
Heat is supplied uniformly at the rate of 100W to 1.0 x 10\(^{-2}\)kg of a liquid for 20 seconds. If the temperature of the liquid rises by 5\(^o\)C, then the specific heat capacity of the liquid is
A
2.0 x 10\(^2\)Jkg\(^-1\)K\(^-1\)
B
2.0 x 10\(^2\)Jkg\(^-1\)
C
4.0 x 10\(^4\)Jkg\(^-1\)K\(^-1\)
D
4.0 x 10\(^4\)Jkg\(^-1\)
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Heat energy = Electrical Energy
MC\(\theta\) = power x time
5 x 1 x 10\(^{-2}\) x C = 100 x 20
C = \(\frac{200}{5 \times 10^2}\)
= 4 x 10\(^4\) J/kg/\(^oC\)
Source: Myschool Past Questions Archive (Record #56920)
Physics
UTME / JAMB 2020
Work, Energy, Power & Simple Machines
Diagram
Solution
The fiqure above represents a block-and-tackle pulley system on which an effort of W Newtons supports a load of 120.0N. If the efficiency of the machine is 40\(^o\), then the value of W is
Attached Figure / Diagram
A
28.0N
B
48.0N
C
288.0N
D
50.0N
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
V.E = 6, load = 120N
Efficiency = M.A x \(\frac{1}{V.R}\) x 100%
40 = \(\frac{120}{effort}\) x \(\frac{1}{6}\) x \(\frac{100}{1}\)
Effort = 50N
Source: Myschool Past Questions Archive (Record #56922)
Physics
UTME / JAMB 2020
Calorimetry, Latent Heat & Change of State
Diagram
In an experiment in which molten naphthalene is allowed to cool, the cooling curve i n the figure was obtained. The temperature 80\(^o\)C is known as the
Attached Figure / Diagram
A
cooling temperature
B
boiling point
C
melting point
D
vaporization point
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 #56923)
Physics
UTME / JAMB 2020
Current Electricity & Electric Circuits
Diagram
Solution
In the figure, the value of R is
Attached Figure / Diagram
A
3\(\Omega\)
B
6\(\Omega\)
C
4\(\Omega\)
D
5\(\Omega\)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
R\(_T\) = \(\frac{V}{1} = \frac{12}{2}\) = 6\(\Omega\)
R\(_T\) = 3 + \(\frac{R \times 6}{R + 6}\)
R = 6\(\Omega\)
Source: Myschool Past Questions Archive (Record #56924)
Physics
UTME / JAMB 2020
Current Electricity & Electric Circuits
Diagram
Solution
Which of the following instruments consumes the highest current?
I. Electric iron
II. Television set
III. Torchlight
IV. Immersion heater
250v
220v
6v
110v
1kW
110W
30W
500W
Attached Figure / Diagram
A
I
B
II
C
III
D
IV
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
(I) I = \(\frac{P}{V} = \frac{1000}{25}\) = 4A
(II) I = \(\frac{110}{220}\) = 0.5A
(III) I = \(\frac{30}{6}\) = 5A
(IV) I = \(\frac{500}{110}\) = 4.55A
Source: Myschool Past Questions Archive (Record #56925)