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Physics
UTME / JAMB 1998
Mechanics and Properties of Matter
Solution
The atmospheric pressure due to water is 1.3 x 106Nm-2. What is the total pressure at the bottom of an ocean 10m deep? [Density of water = 1000kgm-3, g = 10ms-2]
A
1.3 x 107Nm-2
B
1.4 x 106Nm-2
C
1.4 x 105Nm-2
D
1.0 x 10 5Nm-2
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Pressure = heg = 1000 x 10 x 10 = 105Nm-2
Total pressure = (1.3 x 106) + (1 x 105)
= 1.4 x 106Nm-2
Source: Myschool Past Questions Archive (Record #27615)
Physics
UTME / JAMB 1998
Mechanics and Properties of Matter
Solution
A solid of weight 0.600N is totally immersed in oil and water respectively. If the upthrust of oil is 0.210N and the relative density of oil is 0.875. Find the upthrust in water.
A
0.600N
B
0.360N
C
0.240N
D
0.180N
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Rd = \(\frac{\text{upthrust in oil}}{\text{upthrust in }H_2O}\)
thus; 0.875 = \(\frac{0.21}{\text{upthrust in }H_2O}\)
upthrust in H2O = 0.24N
Source: Myschool Past Questions Archive (Record #27617)
Physics
UTME / JAMB 1998
Mechanics and Properties of Matter
Solution
A 500W heater is used to heat 0.6kg of water from 25ºC to 100ºC in t\(_1\) seconds. If another 1000W heater is used to heat 0.2kg of water from 10ºC to 100ºC in t\(_2\), seconds, find \(\frac{t_1}{t_2}\)
A
50
B
5
C
5/4
D
1/5
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Heat energy = Electrical energy
mc\(\theta\) = power x time
0.6 x c x (100 - 25) = 500 x t\(_1\)
and, 0.2 x c x (100 - 10) = 1000 x t\(_2\)
\(\frac {t_1}{t_2}\) = \(\frac{0.09}{0.018}\) = 5
Source: Myschool Past Questions Archive (Record #27630)
Physics
UTME / JAMB 1998
Mechanics and Properties of Matter
Solution
The force of repulsion between two point positive charges 5\(\mu\)C and 8\(\mu\)C separated at a distance of 0.02m apart is. [\(\frac{1}{4{\pi}{\varepsilon}_o}\) = 9 x 10\(^9\)Nm\(^2\)C\(^{-2}\)]
A
1.8 x 10-10N
B
9.0 x 10-8N
C
9.0 x 102N
D
4.5 x 103N
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
F =\(\frac{1}{4{\pi}{\varepsilon}_o}\) x \(\frac {q_1q_2}{r^2}\)
F =\(\frac {5 {\times} 10^{-6} {\times} 8 {\times} 10^{-6}}{0.002 {\times} 0.002}\) x 9 x 10\(^9\)
F = 900N
Source: Myschool Past Questions Archive (Record #27660)
Physics
UTME / JAMB 1998
Mechanics and Properties of Matter
Solution
Find the work done in moving a 2 C charge between two points X and Y in an electric field if the potential difference is 100 volts
A
50J
B
100J
C
200J
D
400J
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
V = \(\frac{W}{Q}\)
W = VQ
W = 100 x 2
W = 200J
Source: Myschool Past Questions Archive (Record #27663)
Physics
UTME / JAMB 1998
Mechanics and Properties of Matter
Solution
Electricity is supplied to a school along a cable of total resistance 0.5\(\Omega\) with the maximum current drawn from the mains as 100A. The maximum energy dissipated as heat for 1 hr is
A
3.6 x 103J
B
5.0 x 103J
C
3.0 x 105J
D
1.8 x 107J
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Elect. energy = I2Rt
= 100 x 100 x 0.5 x 3600
= 1.8 x 107J
Source: Myschool Past Questions Archive (Record #27666)
Physics
UTME / JAMB 1998
Mechanics and Properties of Matter
Solution
In a nuclear fusion experiment, the loss of mass amounts to 1.0 x 10-6kg. The amount of energy obtained from the fusion is
A
3.0 x 10-4J
B
3.0 x 10-1J
C
9.0 x 104J
D
9.0 x 1010J
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Energy = mc2
= 1 x (3 x 108)2 x 106
= 9 x 1010J
Source: Myschool Past Questions Archive (Record #27675)
Physics
WAEC / WASSCE 1998
Mechanics and Properties of Matter
Solution
which of the following pairs of physical quantities comprises vectors?
A
capacitance and inductance
B
force ratio and velocity ratio
C
friction and momentum
D
energy and power
E
electric field potential and electric field intensity
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Friction and Momentum both have magnitudes and directions.
Source: Myschool Past Questions Archive (Record #33484)
Physics
WAEC / WASSCE 1998
Mechanics and Properties of Matter
Solution
A rectangular block of wood floats in water with two-thirds of its volume immersed. When placed in another liquid, it floats with half of its volume immersed. Calculate the relative density of the liquid
A
1.43
B
1.33
C
1.00
D
0.83
E
0.75
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
from the first statement, mg = \(\frac{2}{3}Vp_wg\) ........................1
from second statement, mg = \(\frac{1}{2}Vp_lg \)...................2
equating equations 1 and 2 we have that
\(\frac{2}{3}Vp_wg\) = \(\frac{1}{2}Vp_lg\)
\(\frac{4}{3} = \frac{p_l}{p_w}\)
= 1.33
Source: Myschool Past Questions Archive (Record #33507)
Physics
WAEC / WASSCE 1998
Mechanics and Properties of Matter
Solution
The earth is not a perfect sphere because its equatorial axis is longer than the polar axis. where on the earth's surface would an object have its greatest weight?
A
half way between the equator and the north pole
B
near the equator
C
half way between the equator and the south pole
D
at either pole
E
at the equator
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The acceleration due to gravity is maximum at the poles and minimum at the equator, therefore the weight of the object will be maximum at the poles.
Source: Myschool Past Questions Archive (Record #33542)
Physics
WAEC / WASSCE 1998
Mechanics and Properties of Matter
Solution
A ball of mass 5.0kg hits a smooth vertical wall normally with a speed of 2m-1 and rebounds with the same speed. Determine the impulse experienced by the ball
A
20.0kgm-1
B
5.0kgm-1
C
1.3kgm-1
D
0.0kgm-1
E
10.0kgm-1
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Using: \(\text{Impulse} = \Delta \text{Momentum} = m(v - u)\)
Initial Momentum: \(5\text{ kg} \times 2\text{ m/s} = +10\text{ kg m/s}\)
Final Momentum: \(5\text{ kg} \times (-2\text{ m/s}) = -10\text{ kg m/s}\)
Change: \(-10 - 10 = -20\text{ kg m/s}\)
Source: Myschool Past Questions Archive (Record #33543)
Physics
WAEC / WASSCE 1998
Mechanics and Properties of Matter
Solution
A piece of brass of mass 170kg has its temperature raised from 0ºC to 30ºC. Calculate its increase in volume, given the density of brass at 0 ^\circC as 8.5 x 103kgm\(^3\) and its cubic expansivity as 5.7 x 10\(^{-5}\)k\(^{-1}\)
A
3.4 x 10-4m3
B
4.3 x 10-5m3
C
3.4 x 10-5m3
D
3.4 x 10-3m3
E
3.4 x 10-2m3
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
V = \(\frac{m}{p}= \frac{170}{8.5 x 10^{-3}}\) = 0.02
= \(\Delta\)\(\frac{\delta V}{V \delta T}\)
\(\delta V\) = \(\Delta\) x V x \(\delta\)T
= 5.7 x \(10^{-5}\) x 0.02 x 30
= 3.42 x \(10^{-5}\)
Source: Myschool Past Questions Archive (Record #33555)
Physics
WAEC / WASSCE 1998
Mechanics and Properties of Matter
Solution
The average kinetic energy of the molecules of a perfect gas is directly proportional to the
A
pressure exerted by the gas
B
volume of the gas molecule
C
kelvin temperature of the gas
D
volume of the gas container
E
celcius temperature of the gas
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The average kinetic energy of the molecules of a perfect gas is directly proportional to the Absolute temperature that is the temperature measured using the kelvin scale.
Source: Myschool Past Questions Archive (Record #33559)
Physics
WAEC / WASSCE 1998
Mechanics and Properties of Matter
Solution
Air at temperature 527oC and pressure 30 atmospheres is admitted into the cylinder of an engine. Calculate the pressure of the gas when it has expanded to 5 times its volume and cooled to 127oC as it leaves the engine
A
1.5 atmospheres
B
2.5 atmospheres
C
3.0 atmospheres
D
12.8 atmospheres
E
15.0 atmospheres
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
\(\frac{V_1P_1}{T_1} = \frac{V_2P_2}{T_2}\)
\(\frac{1 \times 30}{527 + 2733} = \frac{5P_2}{127 + 273}\)
\(\frac{30}{800} = \frac{5p_2}{400}\)
P2 = \(\frac{400 \times 30}{800 \times 5}\)
= 3
Source: Myschool Past Questions Archive (Record #33580)
Physics
WAEC / WASSCE 1998
Mechanics and Properties of Matter
Diagram
I the box is pulled from P to Q, which of the following expresses the velocity ratio of the inclined plane?
Attached Figure / Diagram
A
PR/PQ
B
OR/PQ
C
OR/PR
D
PQ/PR
E
PQ/QR
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 #33517)