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Physics
WAEC / WASSCE 2024
Measurement of Heat Energy
Solution
A non-conducting vessel contains a liquid of mass 70 g at 25º C. Another 70.0g of the same liquid at 35ºC is quickly poured into the vessel. Determine the final temperature of the mixture
A
15ºC
B
20ºC
C
30ºC
D
40ºC
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Heat gained by 25º C body = Heat lost by 35º C body
M\(_c\) c \(\delta\)\(\theta\) = M\(_h\) c\(\delta\)\(\theta\)
70 x c x (f - 25) = 70 x c x (35 - f)
where f = final temperature
f - 25 = 35 - f (f and 70 cancel out)
2f = 35 + 25 = 60
f = \(\frac{60}{2}\) = 30ºC
Source: WAEC / JAMB / NECO Classroom Archive (Record #72075)
Physics
WAEC / WASSCE 2024
Measurement of Heat Energy
Solution
The amount of heat energy that will be extracted when 0.02kg of water vapor condenses to water at 97ºC is [specific heat capacity of water = 4200Jkg\(^{-1}\)k\(^{-1}\), Specific heat of vaporization of water = 2.26 x 10\(^6\)Jkg\(^{-1}\)]
A
45,452J
B
45,200J
C
25,200J
D
252J
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
specific heat capacity of water = 4200Jkg\(^{-1}\)k\(^{-1}\), Specific heat of vaporization of water = 2.26 x 10\(^6\)Jkg\(^{-1}\), mass of water = 0.02kg, water vapour temp. = 100ºC, water at 97ºC
The amount of heat required = ML + Mc\Delta\(\theta\)
= 0.02 x 2.26 x 10\(^6\) + 0.02 x 4200 x (100 - 97)
= 45,200 + 252 = 45,452J.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72080)
Physics
UTME / JAMB 2024
Evaporation and Boiling
Solution
If the S.V.P of water vapour was 13.5mmHg at 33ºC and 7.3mmHg at 7ºC. Find the percentage relative of the air on a day when average air temperature was 33ºC and dew point was 7ºC.
A
55%
B
60%
C
54%
D
51%
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Relative humidity = \(\frac{\text{S.V.P at dew point}}{\text{S.V.P at air temperature}}\)
S.V.P at dew point = 7.3mmHg
S.V.P at air temperature = 13.5mmHg
R.H = \(\frac{7.3}{13.5}\) x 100 \(\approxeq\) 54%
Source: WAEC / JAMB / NECO Classroom Archive (Record #70366)
Physics
UTME / JAMB 2024
Evaporation and Boiling
Solution
If the rate of evaporation is y, and the rate of condensation is x, then, a vapour is saturated when
A
x > y
B
x
C
x = y
D
x \(\leq\) y
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
In a saturated vapour, the rate of evaporation (y) equals the rate of condensation (x). This means that the amount of vapor entering the gaseous phase is equal to the amount of vapour returning to the liquid phase, resulting in a stable concentration of vapour in the air.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70437)
Physics
UTME / JAMB 2024
Evaporation and Boiling
Solution
Newton's law of cooling is valid only for a
A
small temperature range
B
large temperature range
C
free flow range
D
presence of heat source
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Newton's Law of Cooling states that the rate of heat loss of an object is directly proportional to the difference in temperature between the object and its surroundings, provided that this temperature difference is small.
Therefore, this law is only valid within a small temperature range.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70539)
Physics
UTME / JAMB 2024
Evaporation and Boiling
Solution
When thermal energy in a solid is increased, the change in state is called
A
melting
B
evaporation
C
freezing
D
condensation
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
When thermal energy is increased, the change in state is called melting (solid to liquid) or vaporization (liquid to gas), depending on the initial and final states of the substance.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70541)
Physics
WAEC / WASSCE 2024
Evaporation and Boiling
Solution
The constant temperature at which a liquid change into gaseous state is known as
A
evaporation point
B
boiling point
C
melting point
D
dew point
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The constant temperature at which a liquid changes into gaseous state is known as its boiling point.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72082)
Physics
UTME / JAMB 2024
Gravitational Field
Solution
The gravitational force between two objects is 10N, what is the new value of the force if the distance between them is halved?
A
10N
B
20N
C
40N
D
80N
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Gravitational force Fg = \(\frac{Gm_1m_2}{r^2}\)
Fg = 10N
distance = \(\frac{r}{2}\)
F'g = \(\frac{Gm_1m_2}{(\frac{r}{2})^2}\) = 4 (\(\frac{Gm_1m_2}{r^2}\)) = 4Fg = 4 x 10 = 40N
Source: WAEC / JAMB / NECO Classroom Archive (Record #70482)
Physics
UTME / JAMB 2024
Gravitational Field
Solution
The gravitational force between two objects masses 10\(^{24}\)kg and 10\(^{27}\)kg is 6.67N. Calculate the distance between them [ G = 6.6 x 10\(^{-11}\)Nm\(^2\)kg\(^{-2}\) ]
A
1.0 x 10\(^{20}\)m
B
2.0 x 10\(^{-20}\)m
C
1.0 x 10\(^{-20}\)m
D
2.0 x 10\(^{20}\)m
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Gravitational force Fg = \(\frac{Gm_1m_2}{r^2}\) = 6.67N
\(m_1\) = 10\(^{24}\)kg, \(m_2\) = 10\(^{27}\)kg, G = 6.6 x 10\(^{-11}\)Nm\(^2\)kg\(^{-2}\) and r = ?
6.67 = \(\frac{ 6.6\times 10^{-11}\times10^{24}\times10^{27}}{r^2}\)
6.67r\(^2\) = 6.6 x 10\(^{40}\)
r\(^2\) = \(\frac{6.6\times10^{40}}{6.67}\) \(\approx\) 9.89 \times 10\(^{39}\)
r = \(\sqrt{ 9.89\times10^{39}}\) \(\approxeq\) 9.94 \times 10\(^{19}\)m → 1.0 x 10\(^{20}\)m
Source: WAEC / JAMB / NECO Classroom Archive (Record #70486)
Physics
UTME / JAMB 2024
Electric Field
Solution
When a charged ebonite rod is brought near a charged glass rod, there will be
A
attraction
B
repulsion
C
spark
D
no force experienced
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Ebonite rods are negatively charged, while glass rods are positively charged when rubbed. Opposite charges attract each other, leading to an attractive force between the rods.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70360)
Physics
UTME / JAMB 2024
Magnetic Field
Solution
Calculate the magnetic force on an electron in a magnetic field of flux density 10T, with a velocity of 3 x 10\(^7\)m/s at 60º to the magnetic field (e = 1.6 x 10\(^{-19}\)C)
A
4.8 x 10\(^{-11}\)N
B
4.6 x 10\(^{-12}\)N
C
4.16 x 10\(^{-11}\)N
D
4.3 x 10\(^{-11}\)N
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The magnetic force on an electron in a magnetic field (F) = q v Bsin\(\theta\)
B = 10T, q = 3 x 10\(^7\)m/, \(\theta\) = 60º and q = 1.6 x 10\(^{-19}\)C
F = 1.6 x 10\(^{-19}\) x 3 x 10\(^7\) x 10 x sin 60º \(\approxeq\) 4.162 \times 10\(^{-11}\)N
Source: WAEC / JAMB / NECO Classroom Archive (Record #70527)
Physics
UTME / JAMB 2024
Magnetic Field
Solution
The device for measuring the angle of dip is
A
Dip circle
B
Dip inclination
C
Dip variation
D
Dip meter
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
The device for measuring the angle of dip is called a dip circle or declinometer.
A dip circle contains a magnetic needle that can move freely, enabling it to align with the Earth's magnetic field. The angle formed between the needle and the horizontal plane represents the angle of dip.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70687)
Physics
WAEC / WASSCE 2024
Magnetic Field
Solution
A galvanometer of full-scale deflection 10 mA has a resistance of 5\(\Omega\). Calculate the shunt needed to convert it to measure currents of 2 A.
A
0.025\(\Omega\)
B
0.050\(\Omega\)
C
0.250\(\Omega\)
D
0.500\(\Omega\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Voltage across shunt = Voltage across galvanometer
I\(_s\)R\(_s\) = I\(_g\)R\(_g\)
R\(_g\) = 5\(\Omega\), I\(_g\) = 10mA = 0.01A, I\(_s\) = 2 - 0.01 = 1.99A
R\(_s\) = \(\frac{I_g \times R_g}{I_s}\) = \(\frac{0 .01 \times 5}{1.99}\) = 0.025\(\Omega\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #72076)
Physics
WAEC / WASSCE 2024
Magnetic Field
Solution
In storing magnets, the keeper is used to
A
cancel the effect of the earth's magnetism
B
reduce self demagnetization
C
protect the magnet from stray electric field
D
increase the strength of the magnet.
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Keepers reduce self-demagnetization of magnets by providing a low reluctance path for magnetic flux, thereby containing the flux within the keeper.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72077)
Physics
UTME / JAMB 2024
Electromagnetic Field
Solution
Which of the following operates based on magnetic effect of electric current?
A
electric heater
B
photocell
C
Dynamo
D
Cold cathode
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
A dynamo converts mechanical energy into electrical energy using the principle of electromagnetic induction, which is a magnetic effect of electric current.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70359)