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Physics
WAEC / WASSCE 1994
Current Electricity & Electric Circuits
Solution
A cell of e.m.f. 1.5V and internal resistance of 2.5Ω is connected in series with an ammeter of resistance 0.5W and a resistor of resistance of 7.0Ω. Calculate the current in the circuit.
A
0.15A
B
0.20A
C
2.10A
D
3A
E
6.6A
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
I =\(\frac{E}{R+r} = \frac{1.5}{2.5+0.5+7}\) = 0.15A
Source: Myschool Past Questions Archive (Record #26276)
Physics
WAEC / WASSCE 1994
Linear Motion, Projectiles & Kinematics
Solution
A ball is dropped from a height of 45m above the ground. Calculate the velocity of the ball just before it strikes the ground. (Neglect air resistance and take g as \(10ms^{-2}\)
A
21.2\(ms^{-1}\)
B
30\(ms^{-1}\)
C
300\(ms^{-1}\)
D
450\(ms^{-1}\)
E
900\(ms^{-1}\)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
\(v^2 = u^2+2as=0^2 + 2\times 10\times 45 \Rightarrow
v^2 = 900m^2s^{-2} \Rightarrow \\
v = \sqrt{900m^2s^{-2}} \Rightarrow
v = 30ms^{-1}\)
Source: Myschool Past Questions Archive (Record #26277)
Physics
WAEC / WASSCE 1994
Linear Motion, Projectiles & Kinematics
Solution
A body accelerates uniformly from rest at \(2ms^{-2}\). Calculate the velocity after traveling 9m
A
36\(ms^{-1}\)
B
18\(ms^{-1}\)
C
6\(ms^{-1}\)
D
4.5\(ms^{-1}\)
E
4.24\(ms^{-1}\)
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
\(V^2= U^2+2as\) =\(0+ 2\times 2ms^{-2}\times 9 =36ms^{-1}\) thus
V=\(\sqrt{36m^2s^{-2}}\) = \(6ms^{-1}\)
Source: Myschool Past Questions Archive (Record #26278)
Physics
WAEC / WASSCE 1994
Atomic, Quantum, Nuclear Physics & Electronics
Solution
A metal rod of length 40cm at 20°C is heated to a temperature of 45°C. If the new length of the rod is 40.05cm. Calculate its linear expansitivity
A
1.2 x \(10^{-5}K{-1}\)
B
2.5x \(10^{-5}K{-1}\)
C
3.5x \(10^{-5}K{-1}\)
D
4.5x \(10^{-5}K{-1}\)
E
5x \(10^{-5}K{-1}\)
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
\(\alpha = \frac{I_2-I_1}{I_1(\theta_2-\theta_1)} \\ \Rightarrow \frac{40.05 - 40}{40 \times (45-20)} = 5.0 \times10^{-5}\)
Source: Myschool Past Questions Archive (Record #26279)
Physics
WAEC / WASSCE 1994
Calorimetry, Latent Heat & Change of State
Solution
500g of water is heated so that its temperature rises from 30°C to 72°C in 7 minutes. calculate the supplied per minute. (Specific heat capacity of water is 4200J\(kg^{-1}K^{-1}\)
A
9000J
B
12600J
C
21000J
D
25200J
E
88200J
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
\(\frac{Q}{t} = \frac{0.5\times4200\times(72-30)}{7}=12600J\)
Source: Myschool Past Questions Archive (Record #26280)
Physics
WAEC / WASSCE 1994
Calorimetry, Latent Heat & Change of State
Solution
An electric kettle rated 1500W boils away 0.3kg of a liquid at its boiling point every 300s. Calculate the specific latent heat of the vaporisation of the liquid
A
1.5 x \(10^6Jkg^-1\)
B
1.35 x \(10^6Jkg^-1\)
C
1.5 x \(10^5Jkg^-1\)
D
4.5 x \(10^2Jkg^-1\)
E
3 x \(10^2Jkg^-1\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Pt=mI; I=\(\frac{1500\times300}{0.3}=1.5\times10^6Jkg^-1\)
Source: Myschool Past Questions Archive (Record #26281)
Physics
WAEC / WASSCE 1994
Current Electricity & Electric Circuits
Solution
\(500cm^{-3}\) of a gas is collected at 0°C and at a pressure of 72cm of mercury. What is the volume of the the gas at the same temperature and at a pressure of 76cm of the same mercury?
A
\(\frac{76\times500}{72}cm^3\)
B
\(\frac{72\times500}{76}cm^3\)
C
\(\frac{72\times76}{500}cm^3\)
D
\(\frac{76}{72\times500}cm^3\)
E
\(\frac{72}{76\times500}cm^3\)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
\(P_1V_1=P_2V_2, V_2= \frac{72\times500}{76}cm^3\)
Source: Myschool Past Questions Archive (Record #26282)
Physics
WAEC / WASSCE 1994
Current Electricity & Electric Circuits
Solution
The pressure of air in a tire is \(22.5Nm^{-2}\) at 27°C. If the air in the tyre heats up to 47°C, calculate the new pressure of the air, assuming that no air leaks out and that the change in volume of the air can be neglected.
A
\(21.10Nm^{-2}\)
B
\(24Nm^{-2}\)
C
\(39.07Nm^{-2}\)
D
\(45Nm^{-2}\)
E
\(97.25Nm^{-2}\)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
\(\frac{P_1}{T_1}=\frac{P_2}{T_2};P_2= \frac{22.5\times320K}{300K}=24Nm^{-2}\)
Source: Myschool Past Questions Archive (Record #26283)
Physics
WAEC / WASSCE 1994
Calorimetry, Latent Heat & Change of State
Solution
Calculate the time taken to heat 2kg of water from 50^\circC to 100^\circC in an electric kettle taking 5A, from a 210V supply. (Specific heat capacity of water = 4200\(kg^{-1}K{-1}\)
A
2.5 minutes
B
5.7 minutes
C
6.7 minutes
D
7.5 minutes
E
9.5 minutes
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
t = \(\frac{mc\theta}{p}\)
= \(\frac{2\times4200\times50}{210\times5}\) = 400s = 6.7minutes
Source: Myschool Past Questions Archive (Record #26284)
Physics
WAEC / WASSCE 1994
Waves, Sound & Vibrations
Solution
A light wave travels from air into a medium of refractive index 1.54. If the wavelength of the light in air is \(5.9\times10^{-7}\)m, calculate its wavelength in the medium
A
\(2.5\times10^{-7}\)m
B
\(3.8\times10^{-7}\)m
C
\(4.6\times10^{-7}\)m
D
\(5.2\times10^{-7}\)m
E
\(6.1\times10^{-7}\)m
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
n =\(\frac{wavelength\, in\, air}{wavelength\, in \,medium}= \frac{5.9\times10^{-7}}{1.54}= 3.8\times10^{-7}\)m
Source: Myschool Past Questions Archive (Record #26285)
Physics
WAEC / WASSCE 1994
Equilibrium of Forces, Moments & Flotation
Solution
The density of water is 1g\(cm^3\) while that of ice is 0.9\(cm^3\), calculate the volume change when 90g of ice is completely melted
A
0\(cm^3\)
B
9\(cm^3\)
C
10\(cm^3\)
D
90\(cm^3\)
E
100\(cm^3\)
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
V = \(\frac{m}{p} \, ice volume= 90 cm^3;water volume =100 cm^3\\
Volume change = (100-90) cm^3= 10 cm^3\)
Source: Myschool Past Questions Archive (Record #26287)
Physics
WAEC / WASSCE 1994
Equilibrium of Forces, Moments & Flotation
Solution
The density of a body is \(5\times10^3kgm{^-3}\) and it weighs 1N in air. Calculate the apparent weight of the body when totally immersed in water. (Density of water =\(10^3kgm{-3}, g= 10ms^{-2}
A
0.02N
B
0.08N
C
0.20N
D
0.80N
E
1N
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Upthrust = weight of the water displaced =pvg= 1000 x 2 x \(10^{-5}\)=0.20N \\ Apparent weight = weight in air – upthrust = 1-0.2N=0.8N
Source: Myschool Past Questions Archive (Record #26286)
Physics
UTME / JAMB 1994
Equilibrium of Forces, Moments & Flotation
Solution
If it takes 5.0 hrs to drain a container of 540.0m\(^3\) of water, what is the flow rate of water from the container in kgs\(^{-1}\)? [Density of water = 1000kgm\(^{-3}\)]
A
32.5kg-1 s
B
31.5kg-1
C
30.8kg-1 s
D
30.0kg-1 s
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Flow rate = \(\frac{mass}{time}\)
= \(\frac{1000 \times 540}{5 \times 3600}\)
= 30.0kgs\(^{-1}\)
Source: Myschool Past Questions Archive (Record #27858)
Physics
UTME / JAMB 1994
Current Electricity & Electric Circuits
Solution
A boat travels due east at a speed 40ms\(^{-1}\)s across a river flow due south at 30ms\(^{-1}\)s. What is the resultant speed of the boat?
A
1.3ms-1
B
10.0ms-1
C
50.0ms-1
D
70.0ms-1
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
R\(^2\) = 40\(^2\) + 30\(^2\)
= 1600 + 900 = 2500
R = \(\sqrt{2500}\) = 50m/s
Source: Myschool Past Questions Archive (Record #27859)
Physics
UTME / JAMB 1994
Linear Motion, Projectiles & Kinematics
Solution
An object is projected with a velocity of 80ms-1 at an angle of 30o to the horizontal.The maximum height reached is
A
20m
B
80m
C
160m
D
320m
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
max. height = \(\frac{U^2\sin^2\theta}{2g}\)
= \(\frac{80 \times 80 \times \sin^2(30)}{2 \times 10}\)
= 80m
Source: Myschool Past Questions Archive (Record #27860)