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Physics
UTME / JAMB 2018
Current Electricity & Electric Circuits
A particle of mass M initially at rest splits into two. If one of the particles of mass M1 moves with velocity V1 , the second particle moves with velocity
A
\(\frac{M1V1}{M-M1}\)
B
\(\frac{M1V1}{M1-M}\)
C
\(\frac{M1V1}{M}\)
D
\(\frac{MV}{M-M1}\)
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 #47736)
Physics
UTME / JAMB 2018
Current Electricity & Electric Circuits
Solution
To keep a vehicle moving at a constant speed V requires power P from the engine. The force provided by the engine is
A
\(\frac{P}{v}\)
B
(\(\frac{1}{2}\))V
C
PV
D
\(\frac{P}{V^2}\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
P = F \times \(\frac{S}{1}\)
P = F x V
F = \(\frac{P}{V}\)
Source: Myschool Past Questions Archive (Record #47737)
Physics
UTME / JAMB 2018
Units, Measurements & Dimensional Analysis
Solution
Which of the following is the dimension of pressure
A
ML\(^{−1}\)T\(^{−2}\)
B
MLT\(^{−2}\)
C
ML\(^{2}\)T\(^{3}\)
D
ML\(^{−3}\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Pressure = \(\frac{Force}{Area}\)
Pressure = m \times \(\frac{a}{t}\)
= \(\frac{MLT^{−2}}{L \times L}\)
= ML\(^{−1}\)T\(^{−2}\)
Source: Myschool Past Questions Archive (Record #47738)
Physics
UTME / JAMB 2018
Elasticity, Density, Pressure & Fluids
Solution
Palm oil from a bottle flows out more easily after it has been heated because the
A
Molecules do not increase speed during the heating
B
Viscosity of oil is reduced
C
Oil molecules force each other out
D
Adhesion between the oil molecules and those of the bottle is increased
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
If a liquid is heated the particles are given more energy and move faster and faster expanding the liquid.The result is that liquids show a reduction in viscosity with increasing temperature. With high temperatures, viscosity increases in gases and decreases in liquids, the drag force will do the same.
Source: Myschool Past Questions Archive (Record #47739)
Physics
UTME / JAMB 2018
Newton's Laws, Momentum & Collisions
Solution
A ball of mass 5.0kg hits a smooth vertical wall normally with a speed of 2ms\(^{-1}\). Determine the magnitude of the resulting impulse
A
20.0kgms\(^{-1}\)
B
10.0kgms\(^{-1}\)
C
5.0kgms\(^{-1}\)
D
2.5kgms\(^{-1}\)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Impulse = Change in momentum
m (v - u)
5 (2.0)
= 5 x 2 = 10.0kgm\(^{-1}\)
= 10.0kgms\(^{-1}\)
Source: Myschool Past Questions Archive (Record #47740)
Physics
UTME / JAMB 2018
Equilibrium of Forces, Moments & Flotation
Solution
An object weighs 30N in air and 21N in water. The weight of the object when completely immersed in a liquid of relative density 1.4 is
A
25.2N
B
17.4N
C
12.6N
D
9.0N
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Weight of water displaced = upthrust = 30 - 21 = 9N
Mass of water displaced = \(\frac{9}{10}\) = 0.9kg
Volume of object = 9 \times 10\(^{-4}\)m\(^3\)
= (9 \times 10\(^{-4}\)) (1.4 \times103)
= 1.26kg = 12N
30 - 12.6 = 17.4N
Source: Myschool Past Questions Archive (Record #47743)
Physics
UTME / JAMB 2018
Units, Measurements & Dimensional Analysis
Which of the following instruments is most suitable for measuring the outside diameter of a narrow pipe in a few millimeters in diameter?
A
Pair of calipers
B
Meter rule
C
Micrometer screw gauge
D
Tape rule
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #47745)
Physics
UTME / JAMB 2018
Current Electricity & Electric Circuits
Solution
Electrical power is transmitted at a high voltage rather than a low voltage because the amount of energy loss is due to
A
Heat dissipation
B
Production of eddy currents
C
Excessive current discharge
D
Excessive voltage discharge
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
The primary reason that power is transmitted at high voltages is to increase efficiency. As electricity is transmitted over long distances, there are inherent energy losses along the way. High voltage transmission minimizes the amount of power lost as electricity flows from one location to the next. How? The higher the voltage, the lower the current. The lower the current, the lower the resistance losses in the conductors. And when resistance losses are low, energy losses are low also. Electrical engineers consider factors such as the power being transmitted and the distance required for transmission when determining the optimal transmission voltage
Source: Myschool Past Questions Archive (Record #47746)
Physics
UTME / JAMB 2018
Lenses & Optical Instruments
Solution
If the focal length of a camera is 20cm, the distance from the film at which the lens must be set to produce a sharp image of 100cm away is
A
17cm
B
20cm
C
25cm
D
100cm
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Focal length: \( f = 20 \, \text{cm} \)
Object distance: \( u = 100 \, \text{cm} \) (positive since it's in front of the lens)
\end{itemize}
Using the lens formula:
\(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
Substituting the known values:
\(\frac{1}{20} = \frac{1}{v} + \frac{1}{100}\)
Rearranging to solve for \( v \):
\(\frac{1}{v} = \frac{1}{20} - \frac{1}{100}\)
Finding a common denominator (100):
\(\frac{1}{20} = \frac{5}{100}\)
Thus, \(\frac{1}{v} = \frac{5}{100} - \frac{1}{100} = \frac{4}{100}\)
Solving for \( v \):
\(v = \frac{100}{4}\) = 25 cm
Source: Myschool Past Questions Archive (Record #47747)
Physics
UTME / JAMB 2018
Lenses & Optical Instruments
Solution
A lens of focal length 15cm forms an erect image which is three times the size of the object. The distance between the object and the image is ___.
A
10cm
B
20cm
C
30cm
D
40cm
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
F= 15cm
M = 3
M = \(\frac{v}{u}\)
V =3u
\(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)
\(\frac{1}{15}\) = \(\frac{1}{u}\) - \(\frac{1}{3u}\)
\(\frac{1}{15}\) = \(\frac{2}{3u}\)
U =10cm
D = u + v
= 10 +30 = 40cm
Source: Myschool Past Questions Archive (Record #47754)
Physics
UTME / JAMB 2018
Gas Laws & Heat Transfer
The photo cell works on the principle of the
A
Voltaic cell
B
Emission of electron by incident radiation
C
Emission of protons by incident radiation
D
Photographic plate
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 #47756)
Physics
UTME / JAMB 2018
Current Electricity & Electric Circuits
Solution
When a known standard resistor of 2.0 is connected to the 0.0cm end of a metre bridge, the balance point is found to be at 55.0cm
A
1.10Ω
B
1.64Ω
C
2.44Ω
D
27.50Ω
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
\(\frac{R_1}{R_2}\) = \(\frac{(100- L_1)}{( L_1)}\)
\(\frac{R_1}{2}\) = \(\frac{45}{55}\)
R\(_1\) = 2 \times \(\frac{45}{55}\)
= 1.64\Omega
this is because the known resistor is connected to 0.00cm end and balances at 55cm
Source: Myschool Past Questions Archive (Record #47761)
Physics
UTME / JAMB 2018
Current Electricity & Electric Circuits
Solution
A resistor connected to a 12V battery draws a current of 2A. The energy dispatched in the resistor in 5 minutes is ___.
A
3,600J
B
6,000J
C
7,200J
D
14,400J
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Power drawn by the resistor = IV = (2)(3) = 24W = 24J/s
Energy dissipated = power \times time = (24)(5)(60) = 7200J
Source: Myschool Past Questions Archive (Record #47763)
Physics
UTME / JAMB 2018
Equilibrium of Forces, Moments & Flotation
Solution
A solid weighs 45N and 15N respectively in air and water. Determine the relative density of the solid
A
0.33
B
0.50
C
1.50
D
3.00
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Weight in Air = 45N
Weight in Water = 15N
R.D = \(\frac{45}{45-15}\) = 1.50
Source: Myschool Past Questions Archive (Record #47766)
Physics
UTME / JAMB 2018
Waves, Sound & Vibrations
Solution
A boy receives the echo of his clap reflected by a nearby hill 0.8s later. If the speed of sound in air is 340ms\(^-1\), how far is he from the hill?
A
66m
B
136m
C
264m
D
528m
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
2 \times Distance = speed \times time
2 \times Distance = (340) \times (0.8)
= 272/2
Distance = 136m
Source: Myschool Past Questions Archive (Record #47767)