Question Library
Browse, filter, and study verified past examination questions with step-by-step mathematical reasoning.
Searching 18,000+ Questions...
Sifting topics, KaTeX formulas & verified solutions
Showing 841–855 of 1709 verified questions
Sort by:
Physics
WAEC / WASSCE 2004
Mechanics and Properties of Matter
A body is said to be moving with uniform acceleration if it experiences equal
A
increase in velocity at equal time intervals
B
decreases in velocity at equal time intervals
C
increases in speed at equal time intervals
D
decreases in speed at equal time intervals
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 #34279)
Physics
WAEC / WASSCE 2004
Mechanics and Properties of Matter
Solution
A body is dropped from a certain height above the ground level, falls with uniform
A
speed
B
velocity
C
acceleration
D
retardation
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
For an object falling from a height, the acceleration is constant and is equal to g. This uniform acceleration is produced due to the gravitational force and is known as acceleration due to gravity.
Source: Myschool Past Questions Archive (Record #34280)
Physics
WAEC / WASSCE 2004
Mechanics and Properties of Matter
Solution
A body of mass 8kg moving with a speed of 30ms-1 is brought to rest in 2.5s by a constant retarding force. Calculate the magnitude of the force
A
240N
B
96N
C
20N
D
9N
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
F = ma = \(\frac{mv}{t} = \frac{8 \times 30}{2.5} = 96\)
Source: Myschool Past Questions Archive (Record #34281)
Physics
WAEC / WASSCE 2004
Mechanics and Properties of Matter
Solution
A force acting on a body causes a change in the momentum of the body from 12kg ms-1 to 16kg ms-1 in 0.2s. calculate the magnitude of the force
A
140.0N
B
20.0N
C
4.0N
D
0.8N
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
F = \(\frac{mv - mu}{t} = \frac{16 - 12}{0.2} = 20N\)
Source: Myschool Past Questions Archive (Record #34282)
Physics
WAEC / WASSCE 2004
Mechanics and Properties of Matter
Solution
A ball bearing is projected vertically upwards from the ground with a velocity of 15ms-1. Calculate the time taken by the ball to return to the ground [g = 10ms-1]
A
1.5s
B
3.0s
C
5.0s
D
7.5s
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
t = 2\(\frac{(v - u)}{g} = 2\frac{(0 - 15)}{-10}\)
\(\frac{2 \times 15}{10}\) = 3.0s
Source: Myschool Past Questions Archive (Record #34284)
Physics
WAEC / WASSCE 2004
Mechanics and Properties of Matter
Solution
A faulty barometer reads 72.6cm Hg when the atmospheric pressure is 75.0cm Hg. Calculate the atmospheric pressure when the barometer reads 72.0cn Hg
A
79.4cm Hg
B
74.38cm Hg
C
74.2cm Hg
D
69.4cm Hg
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
At 77cm Hg, A.P = \(\frac{75.0 \times 72.0}{72.6} = 74.38\)
Source: Myschool Past Questions Archive (Record #34325)
Physics
WAEC / WASSCE 2004
Mechanics and Properties of Matter
Solution
The temperature of 900cm3 of an ideal gas at a pressure of 114cm Hg is 27oC. Calculate its volume at 76cm Hg and 0oC
A
50.0cm3
B
546.0cm3
C
659.3cm3
D
1228.5cm3
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
\(V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{114 \times 900 \times 273}{300 \times 76}\)
= 1228.5cm3
Source: Myschool Past Questions Archive (Record #34332)
Physics
WAEC / WASSCE 2004
Mechanics and Properties of Matter
Solution
Calculate the energy stored in a 20\(\mu F\) capacitor. If the p.d between the plates is 40V
A
3.2 x 10-2J
B
1.6 x 10-2J
C
8.0 x 10-4J
D
4.0 x 10-4J
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
W = \(\frac{cv^2}{2} = \frac{20 \times 10^{-6} \times 40^{2}}{2}\)
= 1.6 x 10-2J
Source: Myschool Past Questions Archive (Record #34400)
Physics
WAEC / WASSCE 2004
Mechanics and Properties of Matter
Given that the gravitational constant is G. Newton's universal law of gravitation states that the force of attraction between two masses M1 and M2 separation by a distance r is
A
\(Gr^2 M_1M_2\)
B
\(Gr M_1M_2\)
C
\(\frac{GM_1 M_2}{r^2}\)
D
\(\frac{GM_1 M_2}{r}\)
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 #34401)
Physics
WAEC / WASSCE 2004
Mechanics and Properties of Matter
Solution
A charge of 1.6 x 10-10C is placed in a uniform electric field of intensity 2.0 x 105 NC-1. calculate the magnitude of the electric force exerted on the change
A
3.2 x 105N
B
1.8 x 105N
C
3.2 x 10-5N
D
1.8 x 10-5N
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
F = Exq = 2.0 x 105 x 1.6 x 10-10
= 3.2 x 10-5N
Source: Myschool Past Questions Archive (Record #34402)
Physics
WAEC / WASSCE 2004
Mechanics and Properties of Matter
The main function of the mouth piece of a telephone is that it converts sound energy to
A
mechanical energy
B
electrical energy
C
chemical energy
D
thermal energy
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 #34404)
Physics
UTME / JAMB 2004
Mechanics and Properties of Matter
Solution
If a force of 50N stretches a wire from 20m to 20.01m, what is the amount of force required to stretch the same material from 20m to 20.05m?
A
250N
B
200N
C
100N
D
50N
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Original length = 20m
first new length = 20.01m
first increase in length = 20.01m - 20m = 0.01m
Second new length = 20.05m
second increase in length = 20.05 - 20m = 0.05m
From hooke's law = \( \frac{F_1}{e_1} = \frac{F_2}{e_2}\)
\(\implies \frac{50}{0.01} = \frac{F_2}{0.05}\)
\(\implies F_2 = \frac{50 \times 0.05}{0.01} = 250N \)
Source: Myschool Past Questions Archive (Record #41447)
Physics
NECO SSCE 2004
Mechanics and Properties of Matter
Which of the following conclusion(s) can be drawn from Newton’s second law of motion? Force is proportional to
I. the product of mass and acceleration
II. the product of mass and velocity
III. Impulse
A
I only
B
II only
C
III only
D
II and III only
E
I, II and III
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option E.
Source: Myschool Past Questions Archive (Record #42954)
Physics
NECO SSCE 2004
Mechanics and Properties of Matter
A body of mass 40g loses 80J of heat energy. If the specific heat capacity of the body is 400 Jkg-1, calculate the change in temperature of the body.
A
0.2K
B
0.4K
C
2.0K
D
5.0K
E
10.0K
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 #42957)
Physics
NECO SSCE 2004
Mechanics and Properties of Matter
In which of these fields are repulsive forces NOT experienced?
I. Magnetic field
II. Gravitational field
III. Electric field
A
I only
B
II only
C
III only
D
I and III only
E
II and III only
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 #42959)