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Physics
UTME / JAMB 2023
Heat and Temperature
Solution
Which of the following thermometers measures temperature from the thermal radiation emitted by objects?
A
Pyrometer thermometer
B
Platinum resistance thermometer
C
Thermocouple thermometer
D
Constant pressure gas thermometer
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
A pyrometer is a type of remote-sensing thermometer used to measure the temperature of a surface. It does this by measuring the thermal radiation or infrared energy being emitted from the object. Therefore, a pyrometer thermometer measures temperature from the thermal radiation emitted by objects.
Source: WAEC / JAMB / NECO Classroom Archive (Record #67883)
Physics
UTME / JAMB 2023
Heat and Temperature
Solution
An open-tube mercury manometer is used to measure the pressure in a gas tank. When the atmospheric pressure is 101,325 \(P_a\), what is the absolute pressure in \(P_a\) in the tank if the height of the mercury in the open tube is 25 cm higher. density of mercury = \(13600kg/m^3, g = 9.8m/s^2\)
A
108,986 Pa
B
165,238 Pa
C
122,364 Pa
D
134,645 Pa
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
density (ρ) = 13600 \(kgm^{-3}\); g =\(9.8 ms^{-2}\)
\(P_{atm}\) =101,325 Pa; ρ=13600 \(kgm^{-3}\)
Pabs=\(P_{atm}\)+ρgh
⇒Pabs=101,325+13600 x 9.8 x 0.25
⇒Pabs=101,325+33320
⇒ Pabs=134,645Pa
Source: WAEC / JAMB / NECO Classroom Archive (Record #68118)
Physics
UTME / JAMB 2023
Heat and Temperature
Solution
The sensitivity of a thermometer is
A
All of the above
B
how quickly a temperature change can be detected
C
the difference between the maximum and the minimum temperature
D
the smallest temperature change that can be detected or measured
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The sensitivity of a thermometer is defined as the smallest temperature change that can be detected or measured, NOT how quickly that temperature change can be detected.
In contrast, the difference between the maximum and the minimum temperature is the range of a thermometer
Source: WAEC / JAMB / NECO Classroom Archive (Record #68119)
Physics
UTME / JAMB 2023
Heat Transfer
Solution
The surface temperature of a swimming pool on a warm day is 25ºC and the temperature at the bottom is 15ºC. If the swimming pool has a surface area of 620 \(m^2\) and a depth of 1.5m. Find the rate at which energy is transferred by conduction from the surface to the bottom of the swimming pool.
[Thermal conductivity of water (k) = 0.6071 Wm-1K-1]
A
2.5kw
B
250kw
C
300kw
D
3.0kw
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
L=1.5m, A=\(620m^2\), ø=25-15=10^\circc
K=0.6071 , \(Wm^{-1}k^{-1}\) , \(\frac{q}{t}\) =?
\(\frac{q}{t}\) =KA, \(\frac{►t}{L}\)
where k is the thermal conductivity constant and \(\frac{q}{t}\) is the rate of heat transfer
=\(\frac{q}{t}\)=0.6071\times620\times\(\frac{10}{1.5}\)
\(\frac{q}{t}\) = 2509.35w ≈ 2.5kw
Source: WAEC / JAMB / NECO Classroom Archive (Record #67441)
Physics
UTME / JAMB 2023
Measurement of Heat Energy
Solution
What is the amount of heat required to raise the temperature of a 0.02 kg of ice cube from \(-10^oC\) to \(10^oC\) ?
[specific latent heat of fusion of ice = 3.34 x \(10^5\) \(Jkg^-1\), Specific heat capacity of water = 4200 \(Jkg^-1\) \(k^-1\)
Specific heat capacity of ice = 2100 \(Jkg^-1\) \(k^-1\)
A
6680 J
B
1680 J
C
7520 J
D
7940 J
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Quantity of heat required to raise the temperature of the ice cube from \(-10^oC\) to \(0^oC\)
⇒H=mc►ø = 0.02 \times 2100 \times 0 -(-10)
⇒ H = 0.02 \times 2100 \times (0 + 10) = 0.02 \times 2100 \times 10
⇒ H = 420 J
quantity of heat required to melt ice at \(0^oC\)
⇒ H = mL = 0.02 \times 3.34 \times \(10^5\)
⇒H=6680 j
Quantity of heat required to raise the temperature of the melted ice cube (water) from \(0^oC\) to \(-10^oC\)
⇒H=mc►ø = 0.02 \times 4200 \times (10-0)
⇒H = 0.02 \times 4200 \times 10
⇒H = 840j
;420 + 6680 + 840 = 7940 j
Source: WAEC / JAMB / NECO Classroom Archive (Record #67877)
Physics
UTME / JAMB 2023
Measurement of Heat Energy
Solution
Which of the following is/are not true about the heat capacity of a substance?
(i) It is an intensive property
(ii) Its S.I unit is \(jK-1\)
(iii) It is an extensive property
(iv) Its S.I unit is \(jkg-1\)
A
(i) and (iii) only
B
(i), (ii) and (iv) only
C
(i) and (iv) only
D
(ii) and (iii) only
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The S.I unit of heat capacity is JK-1 and it is an extensive property as it depends on the amount of substance present.
Therefore, (i) and (iv) only are not true
Source: WAEC / JAMB / NECO Classroom Archive (Record #67984)
Physics
UTME / JAMB 2023
Electric Field
Solution
A positively charged particle is placed near a negatively charged particle. What is the direction of the electric force between the two particles?
A
The electric force is directed from the negative particle to the positive particle.
B
The electric force is zero.
C
The direction of the electric force cannot be determined.
D
The electric force is directed from the positive particle to the negative particle.
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The electric force between two charged particles is always directed from the positive particle to the negative particle. This is because opposite charges attract each other, while like charges repel each other.
Source: WAEC / JAMB / NECO Classroom Archive (Record #68580)
Physics
UTME / JAMB 2023
Electric Field
Solution
A metal sphere is placed on an insulating stand. A negatively charged rod is brought close to it. If the sphere is earthed and the rod is taken away, what will be the charge on the sphere?
A
The sphere will remain neutral.
B
The sphere will have a net negative charge.
C
The sphere will have a net positive charge.
D
The sphere will become polarized
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
When a negatively charged rod is brought close to a metal sphere, the free electrons in the sphere are repelled from the rod and move to the other end of the sphere. This creates a region of positive charge on the side of the sphere closest to the rod, and a region of negative charge on the opposite side. The process of charge distribution stops when the net force on the free electrons inside the metal is equal to zero.
If the sphere is then earthed, the free electrons will flow from the sphere to the ground, leaving the sphere with a net positive charge.
Source: WAEC / JAMB / NECO Classroom Archive (Record #69768)
Physics
UTME / JAMB 2023
Magnetic Field
Solution
Which of the following statements is correct about the angle of dip at various points on Earth?
A
The angle of dip is zero at the equator and 90 degrees at the magnetic poles.
B
The angle of dip is greater at higher altitudes than at lower altitudes
C
The angle of dip is positive in the northern hemisphere and negative in the southern hemisphere.
D
The angle of dip is constant at all points on Earth
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
The angle of dip is the angle between the direction of the intensity of the total magnetic field of Earth and a horizontal line in the magnetic meridian. It is also known as the magnetic inclination. The angle of dip ranges from 0 degrees at the magnetic equator to 90 degrees at the magnetic poles.
Source: WAEC / JAMB / NECO Classroom Archive (Record #68582)
Physics
UTME / JAMB 2023
Magnetic Field
Diagram
Solution
In the diagram above, if the south poles of two magnets stroke a steel bar, the polarities at X and Y will respectively be
Attached Figure / Diagram
A
south and north
B
north and south
C
north and north
D
south and south
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The polarities at X and Y would be north and north.
Source: WAEC / JAMB / NECO Classroom Archive (Record #68583)
Physics
UTME / JAMB 2023
Electromagnetic Field
Solution
A step-down transformer is used on a 2.2 kV line to deliver 110 V. How many turns are on the primary windings if the secondary has 25 turns?
A
20
B
250
C
200
D
500
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Ep = 2.2kV = 2200V; Es=110V; Ns=25; Np=?
=\(\frac{E_p}{E_s}\) = \(\frac{N_p}{N_s}\)
= \(\frac{2200}{110}\) = \(\frac{N_p}{25}\)
=110 \times Np = 2200 \times 25
=\(N_p\) = \(\frac{55,000}{110}\)
∴\(N_p\) = 500 turns
Source: WAEC / JAMB / NECO Classroom Archive (Record #68000)
Physics
UTME / JAMB 2023
Electromagnetic Field
Solution
A generator manufacturing company accidentally made an AC generator instead of a DC generator. To fix this error,
A
the magnetic field needs to be made stronger
B
the split rings should be replaced with slip rings
C
the number of turns of the armature coil needs to be decreased
D
the slip rings should be replaced with split rings
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
An AC generator uses slip rings to transfer the induced current smoothly to the circuit. A DC generator uses split rings to transfer the induced current to the circuit and also convert the induced AC into pulsating DC. So, to convert an AC generator into a DC generator, the slip rings needs to be replaced with split rings.
Source: WAEC / JAMB / NECO Classroom Archive (Record #68577)
Physics
UTME / JAMB 2023
Electric Field
Solution
The terminals of a battery of emf 24.0 V and internal resistance of 1.0 Ω is connected to an external resistor 5.0 Ω. Find the terminal p.d.
A
18.0V
B
12.0V
C
16.0V
D
20.0V
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
ε = 24.0 V, r=1\Omega, R=5.0\Omega
terminal p·d = ?
ε = IR+Ir = I(R+r)
⇒ IR+Ir = I(R+r)
⇒ I = \(\frac{24}{5+1}\)
⇒ I =\(\frac{24}{6}\)
⇒ I = 4 A
terminal p·d = 4 \times 5
terminal p·d = 20.0 V
Source: WAEC / JAMB / NECO Classroom Archive (Record #67374)
Physics
UTME / JAMB 2023
Electric Field
Solution
A parallel plate capacitor separated by an air gap is made of \(0.8m^2\) tin plates and 20 mm apart. It is connected to 120 V battery. What is the charge on each plate?
Take \(ε_o\) = \(8.85×10^-12 Fm^-1\)
A
3.54nC
B
42.5nC
C
35.4nC
D
4.25nC
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
A= \(0.8m^2\) d= 20mm =\(\frac{20}{1000}\) = 0.02m
v =120v; \(ε_oA\)= \(8.85 \times 10^-12 fm^-1\)
C = \(\frac{ε_oA}{d}\)
C =\(\frac{8.85 \times 10^-12\times0.8}{0.02}\)
C= \(3.54 \times 10^-10 F\)
Q= CV
⇒ \(3.54 \times 10^-10\) \times 120 = \(4.25\times10^-8c\)
Q=\(42.5 \times 10^-9c\) = 42.5nC
Source: WAEC / JAMB / NECO Classroom Archive (Record #67863)
Physics
UTME / JAMB 2023
Electric Field
Solution
A charge of \(4.6 × 10^{-5}\)C is placed in an electric field of intensity \(3.2 × 10^4\) \(Vm^{-1}\). What is the force acting on the electron?
A
1.5 N
B
3.7 N
C
4.2 N
D
2.5 N
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
q=\(4.6 \times 10^{-5}\), C= \(3.2 \times 10^4\) \(Vm^{-1}\) F= ?
F=qE
⇒ \(4.6 \times 10^{-5}\)C \times \(3.2 \times 10^4\)
; F= 1.5 N
Source: WAEC / JAMB / NECO Classroom Archive (Record #68565)