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Physics
WAEC / WASSCE 1994
Mechanics and Properties of Matter
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
Mechanics and Properties of Matter
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
Mechanics and Properties of Matter
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
Mechanics and Properties of Matter
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
Mechanics and Properties of Matter
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)
Physics
UTME / JAMB 1994
Mechanics and Properties of Matter
Solution
Which of the following is TRUE of a particle moving in a horizontal circle with constant angular velocity?
A
The energy is costant but the linear momentum varies
B
The linear momentum is conatant but the energy varies
C
both energy and linear momentum are constant
D
The speed and the linear velocity are both constant
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Energy is constant: Since the angular velocity (\(\omega\)) and radius (r) are constant, the speed remains constant. Kinetic energy depends only on mass and speed, so it remains constant.
Linear momentum varies: Linear momentum is a vector quantity. Although its magnitude is constant, its direction changes continuously in circular motion, so the momentum varies.
Source: Myschool Past Questions Archive (Record #27865)
Physics
UTME / JAMB 1994
Mechanics and Properties of Matter
Solution
An object of mass 50kg is released from a height of 2m.Find the kinetic energy just before it strikes the ground
A
250J
B
1000J
C
10 000J
D
100 000J
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
P.E = K.E = mgh
= 50 x 10 x 2 = 1000J
Source: Myschool Past Questions Archive (Record #27866)
Physics
UTME / JAMB 1994
Mechanics and Properties of Matter
A cone in an unstable equilibrium has its potential energy
A
decreased
B
increased
C
unchanged
D
oscillating
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 #27867)
Physics
UTME / JAMB 1994
Mechanics and Properties of Matter
Solution
Calculate the magnitude of the force required to just move a 20kg object along a horizontal surface if the coefficient of friction is 0.2. [g = 10ms-2]
A
400.0N
B
40.0N
C
4.0N
D
0.4N
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
F = \(\mu\)mg
= 0.2 x 20 x 10
= 40N
Source: Myschool Past Questions Archive (Record #27869)
Physics
UTME / JAMB 1994
Mechanics and Properties of Matter
Solution
Calculate the velocity ratio of a screw jack of pitch 0.3cm if the length of the tommy bar is 21cm
A
\(\frac{1}{140}\)\(\pi\)
B
14\(\pi\)
C
70\(\pi\)
D
140\(\pi\)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
V.R = \(\frac{2\pi L}{p}\)
= \(\frac{2 \times \pi \times 2}{0.3}\)
V.R = 140\(\pi\)
Source: Myschool Past Questions Archive (Record #27880)
Physics
UTME / JAMB 1994
Mechanics and Properties of Matter
Solution
A force of 15N stretches a spring to a total length of 30cm. An additional force of 10N stretches the spring 5cm further. Find the natural length of the spring
A
25.0cm
B
22.5cm
C
20.0cm
D
15.0cm
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
\(\frac{15}{30 - L}\) = \(\frac{10}{5}\)
L = 22.5cm
Source: Myschool Past Questions Archive (Record #27882)
Physics
UTME / JAMB 1994
Mechanics and Properties of Matter
Which of these statements are TRUE of pressure in liquids? i. pressure acts equally in all direction. ii. pressure decreases with depth. iii. pressure at the same level of a liquid is the same. iv pressure is dependent on the cross-sectional area of the barometer tube
A
i, iii only
B
i, ii, iii only
C
i, iii, and iv only
D
i, ii, iii and iv
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 #27883)
Physics
UTME / JAMB 1994
Mechanics and Properties of Matter
Solution
The mass of a specific gravity bottle is 15.2g when it is empty. it is 24.8g when filled with kerosene and 27.2g when filled with distilled water. Calculate the relative density of kerosene
A
1.25
B
1.10
C
0.90
D
0.80
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Masses of kerosene = 24.8 - 15.2
= 9.6
mass of H2O = 27.2 - 15.2
= 12
R.d = \(\frac{9.6}{12}\) = 0.8
Source: Myschool Past Questions Archive (Record #27887)
Physics
UTME / JAMB 1994
Mechanics and Properties of Matter
Solution
If a solid X floats in liquids P of relative density 2.0 and in liquid Q of relative density 1.5, it can be inferred that the
A
weight of P displaced is greater than that of Q
B
weight of P displaced is less than that of Q
C
volume of P displaced is greater than that of Q
D
volume of P displaced is less than that of Q
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
R.d \(\alpha\) weight of liquid displaced
\(\frac{\text {R.d of P}}{\text {R.d of Q}}\) = \(\frac{\text{weight of p displaced}}{\text{weight of Q displaced}}\)
\(\frac{2}{1.5}\) = \(\frac{\text {weight of p displaced}}{\text {weight of Q displaced}}\)
A solid floats when the weight of the liquid it displaces is equal to the weight of the solid.
The relative density indicates that liquid P is denser than liquid Q.
Conclusion: Since liquid P is denser, for the same volume of the solid X displaced, the weight of the liquid displaced will be greater in liquid P than in liquid Q.
Source: Myschool Past Questions Archive (Record #27889)
Physics
UTME / JAMB 1994
Mechanics and Properties of Matter
Solution
A Motor Tyre is inflated to pressure of 2.0 x 105Nm-2 when the temperature of air is 27oC. What will be the pressure in it at 87oC assuming that the volume of the Tyre does not change?
A
2.6 x 105Nm-2
B
2.4 x 105Nm-2
C
2.2 x 105Nm-2
D
1.3 x 105Nm-2
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
\(\frac{p_1}{T_1}\) = \(\frac{p_2}{T_2}\)
\(\frac{2 \times 10^5}{27 + 273}\) = \(\frac{p_2}{87 + 273}\)
P2 = 2.4 x 105Nm-2
Source: Myschool Past Questions Archive (Record #27903)