JAMB UTME - Physics - 2019
Ajụjụ 1 Ripọtì
The lead-acid accumulator consists of
Akọwa Nkọwa
- the positive pole is lead peroxide (PbO2
)
- the negative pole is head
- the electrolyte is H2
SO4
Ajụjụ 2 Ripọtì
A body was slightly displaced from its equilibrium position. Which one of the following is a condition for its stable equilibrium
Akọwa Nkọwa
The condition for stable equilibrium of a body that has been slightly displaced from its equilibrium position is "an increase in the potential energy of the body." When an object is at its equilibrium position, it has a minimum potential energy. When the object is displaced from its equilibrium position, it has a higher potential energy. For the object to be in stable equilibrium, it must be able to return to its equilibrium position after it has been displaced. If the potential energy of the object increases as it is displaced, it means that the equilibrium position is a point of stable equilibrium. This is because the object will experience a restoring force that will push it back towards its equilibrium position, as the potential energy decreases. Therefore, an increase in potential energy is a condition for a body to be in stable equilibrium after it has been slightly displaced from its equilibrium position. An increase in kinetic energy or height does not necessarily indicate stability, as it depends on the specific situation and other factors at play.
Ajụjụ 3 Ripọtì
A thermocouple thermometer is connected to a millivoltmeter which can read up to 10mV. When one junction is in ice at 0°C and the other is steam at 100°C, the millivoltmeter reads 4mV. What is the maximum temperature which this arrangement can measure
Akọwa Nkọwa
The maximum temperature which this arrangement can measure is 250°C. A thermocouple thermometer works by using the thermoelectric effect, which is the phenomenon that occurs when two dissimilar metals are joined together to form a loop and a temperature difference is established between the two junctions. This temperature difference generates a small electrical voltage, which can be measured using a millivoltmeter. The voltage generated is proportional to the temperature difference between the two junctions. In the case of the thermocouple thermometer described, one junction is in ice at 0°C and the other is steam at 100°C, and the millivoltmeter reads 4mV. This means that the voltage generated by the thermocouple is 4 millivolts, which corresponds to a temperature difference of 100°C. However, the millivoltmeter can only read up to 10mV, so the maximum temperature difference it can measure is 10mV / 4mV/°C = 250°C. This means that the maximum temperature which this arrangement can measure is 250°C.
Ajụjụ 4 Ripọtì
Which of the following equations is the correct definition of the reactance of an indicator L?
Akọwa Nkọwa
The correct definition of the reactance of an inductor L is: Reactance = (Amplitude of voltage) ÷ (Amplitude of current) The reactance of an inductor is a measure of the opposition offered by the inductor to the flow of alternating current (AC). It is denoted by the symbol Xl and is measured in ohms. When AC flows through an inductor, a magnetic field is generated around the inductor, which opposes any changes in the current flowing through it. This opposition to the flow of current is called reactance. The reactance of an inductor depends on its inductance, frequency of the AC signal, and the amplitude of the AC signal. However, the reactance of an inductor is directly proportional to the frequency of the AC signal and the inductance of the inductor. The reactance of an inductor is also affected by the amplitude of the AC signal, but this effect is not as significant as the other two factors. is the correct definition of the reactance of an inductor, as it expresses the ratio of the amplitude of voltage to the amplitude of current, which is a common way to define reactance. is incorrect, as it represents the power delivered by the AC signal, not the reactance. and are also incorrect, as they involve squaring either the amplitude of current or the amplitude of voltage, which is not a valid method of calculating reactance. Therefore, the correct option is.
Ajụjụ 5 Ripọtì
A man on a bench will exert the greatest pressure on the bench when he
Akọwa Nkọwa
The man on the bench will exert the greatest pressure when he stands on the toes of one foot. This is because when he stands on one foot, all his weight is concentrated on a smaller surface area of the bench, resulting in more pressure. The pressure he exerts is calculated by dividing his weight by the surface area in contact with the bench. When he stands on one foot, the surface area is smaller, which means the pressure exerted is greater. In comparison, when he lies flat on his back or belly, or when he stands on both feet, his weight is distributed over a larger surface area, resulting in less pressure.
Ajụjụ 6 Ripọtì
The height at which the atmosphere cases to exist is about 80km. If the atmospheric pressure on the ground level is 760mmHg, the pressure at a height of 20km above the ground level is
(ρm = 13.6g/cm3 ρ = 0.00013g/cm3 )
Akọwa Nkọwa
ρm
hm
= ρa
ha
13.68(760 - p) × 10−3
= 13 × 10−5
(20 × 103
)
| 760 | - | p | = | 13 × 10−5 × 20 × 103 13.68 × 10−3 | = | 19.00 | × | 101 |
760 - p = 190
p = 760 - 190 = 570mmHg
Ajụjụ 7 Ripọtì
Calculate the velocity ratio of a screw jack of pitch 0.2cm if the length of the tommy bar is 23cm
Akọwa Nkọwa
P = 0.2cm, L = r = 23cm
| VR | = | 2?rP | = | 2?LP | = | 2?×230.2 | = | 230? |
Ajụjụ 8 Ripọtì
The volume of a stone having an irregular shape can be determined using?
Akọwa Nkọwa
The volume of a stone with an irregular shape can be determined using a measuring cylinder. A measuring cylinder is a glass or plastic container with a narrow cylindrical shape and markings on the side to indicate the volume it contains. To determine the volume of an irregularly shaped stone, you would fill the measuring cylinder with water, carefully lower the stone into the water, and note the increase in the volume of the water. The difference in the volume of the water before and after the stone was added is equal to the volume of the stone. The meter rule, vernier calliper, and micrometer screw gauge are all measuring instruments, but they are not designed to measure the volume of irregularly shaped objects. The meter rule is a measuring tool used for measuring length. The vernier calliper is used for measuring the diameter of objects, and the micrometer screw gauge is used for precise measurements of small distances.
Ajụjụ 9 Ripọtì
A supply of 400V is connected across capacitors of 3μf and 6μf in series. Calculate the charge
Akọwa Nkọwa
| CT | = | C1 × C2 C1 + C2 |
| = | 3 × 63 + 6 |
= 189
= 2μf
Q = CV
⇒ 2 × 10−6
× 400
⇒ 800 × 10−6
C = 8 × 10−4
C
Ajụjụ 10 Ripọtì
Which of the following statements are correct of the production and propagation of waves?
I. vibration produces waves
II. waves transmit energy along the medium
III. the medium through which the wave travels does not travel with the wave
IV. waves do not require any medium for transmission
Akọwa Nkọwa
The correct statement is: I and II and III only. Explanation: - Statement I is correct because the production of waves involves some kind of disturbance that creates a vibration in the medium, which then propagates as a wave. - Statement II is correct because waves carry energy along the medium as they propagate. This is why waves can be used to transmit information or power over long distances. - Statement III is correct because the medium through which a wave travels does not move with the wave. Instead, the wave passes through the medium, causing it to oscillate or vibrate, but not to move along with the wave. - Statement IV is incorrect because most waves require a medium through which to propagate. For example, sound waves require air, water waves require water, and seismic waves require the Earth's crust. There are some types of waves, such as electromagnetic waves, that can propagate through a vacuum, but this is not true for all waves.
Ajụjụ 11 Ripọtì
Which of the following is consistent with Charles' law?
I
II 
III
IV. 
Akọwa Nkọwa
This is the correct graph. The graph is volume against 1/ temperature where temperature is in Celsius.
Ajụjụ 12 Ripọtì
A rectangular solid black has length 10cm, breadth 5cm and height 2cm. If it lies on a horizontal surface, and has density 100kg/m3 , calculate the pressure it exerts on the surface.
Akọwa Nkọwa
To calculate the pressure that the rectangular solid exerts on the surface, we need to use the formula for pressure: Pressure = Force / Area In this case, the force is the weight of the rectangular solid, which we can calculate using the formula: Weight = Mass x Gravity The mass of the rectangular solid can be calculated using its density and volume: Mass = Density x Volume The volume of the rectangular solid is simply its length x breadth x height: Volume = Length x Breadth x Height = 10 cm x 5 cm x 2 cm = 100 cm3 We need to convert this volume to cubic meters to use the density given in kg/m3: Volume = 100 cm3 = 0.0001 m3 Now we can calculate the mass: Mass = Density x Volume = 100 kg/m3 x 0.0001 m3 = 0.01 kg The gravity is the acceleration due to gravity, which we can assume to be 9.81 m/s2. Therefore, the weight is: Weight = Mass x Gravity = 0.01 kg x 9.81 m/s2 = 0.0981 N Now we can use this weight to calculate the pressure on the surface. The surface area in contact with the rectangular solid is simply its length x breadth: Area = Length x Breadth = 10 cm x 5 cm = 50 cm2 We need to convert this area to square meters: Area = 50 cm2 = 0.005 m2 Therefore, the pressure is: Pressure = Force / Area = 0.0981 N / 0.005 m2 = 19.62 N/m2 We can convert this to units of N/cm2 or N/mm2 if desired. This is equivalent to: Pressure = 0.1962 N/cm2 = 0.0001962 N/mm2 So the pressure that the rectangular solid exerts on the surface is 19.62 N/m2, which is approximately 20 N/m2. Therefore, the answer is 200 N/m2.
Ajụjụ 13 Ripọtì
A metal rod has a length of 100cm at 200oC . At what temperature will its length be 99.4cm. If the linear expansivity of the material of the rod is 2 × 10−5C−1
Akọwa Nkọwa
The linear expansivity of a material describes how its length changes with temperature. If the linear expansivity is given as 2 × 10^-5/°C, this means that for every 1°C change in temperature, the length of the material will change by 2 × 10^-5 times its original length. Given that the rod has a length of 100 cm at 200°C, we can use this information to find its length at a different temperature. If we let L be the length of the rod at temperature T, we can write the relationship as follows: L = 100 cm * (1 + 2 × 10^-5 * (T - 200°C)) To find the temperature at which the rod will have a length of 99.4 cm, we can set L equal to 99.4 cm and solve for T: 99.4 cm = 100 cm * (1 + 2 × 10^-5 * (T - 200°C)) 99.4 cm / 100 cm = 1 + 2 × 10^-5 * (T - 200°C) 0.994 = 1 + 2 × 10^-5 * (T - 200°C) -0.006 = 2 × 10^-5 * (T - 200°C) -0.006 / 2 × 10^-5 = T - 200°C -0.006 / (2 × 10^-5) = T - 200°C -0.006 / (2 × 10^-5) + 200°C = T So the temperature at which the rod will have a length of 99.4 cm is approximately equal to -0.006 / (2 × 10^-5) + 200°C, or -100°C. Therefore, the answer is -100°C.
Ajụjụ 14 Ripọtì
A ray of light passes through the centre of curvature of a concave mirror and strikes the mirror. At what angle is the ray reflected?
Akọwa Nkọwa
When a light ray passes through the center of curvature of a concave mirror and strikes the mirror, the reflected ray will be reflected back on itself, creating an angle of 0 degrees. Therefore, the correct answer is 0o.
Ajụjụ 15 Ripọtì
Water and Kerosine are drawn respectively into the two limbs of a Hare's apparatus. The destiny of water is 1.0gcm−3 and the density of kerosine is 0.80gcm−3 . If the height of the water column is 20.0cm, calculate the height of the kerosine column.
Akọwa Nkọwa
Devices with different liquids
d1
h1
= d2
h2
1 × 20 = 0.8 × h
| h | = | 200.8 | = | 25cm |
Ajụjụ 16 Ripọtì
A body moves in SHM between two point 20m on the straight line Joining the points. If the angular speed of the body is 5 rad/s. Calculate its speed when it is 6m from the center of the motion.
Akọwa Nkọwa
From two parts 20m apart
a = 10m, x = 6m, A = 5
V = ω√A2−X2
= 5√102−62
= 40m/s
Ajụjụ 17 Ripọtì
A car moving at 20m/s with its horn blowing (f = 1200Hz) is chasing another car going at 15m/s. What is the apparent frequency of the horn as heard by the driver being chased?
Akọwa Nkọwa
| f1 | = | f(v - vo )v - vs | = | 1200(340 - 15)340 - 20 | = | 1.22KHz |
Ajụjụ 18 Ripọtì
Neutrons were discovered by
Akọwa Nkọwa
Neutrons were discovered by James Chadwick. In 1932, he conducted an experiment in which he bombarded a thin sheet of beryllium with alpha particles. He observed that a new type of radiation was emitted that was not affected by electric or magnetic fields. He concluded that this radiation was composed of particles that were neutral and had a mass similar to that of a proton. He called these particles "neutrons," and his discovery revolutionized our understanding of atomic structure and led to the development of nuclear energy.
Ajụjụ 19 Ripọtì
The pin-hole camera produces a less sharply defined image when the
Akọwa Nkọwa
The pin-hole camera produces a less sharply defined image when the pin-hole is larger. A pin-hole camera works by allowing light to pass through a small hole (the pin-hole) and project an inverted image of the outside world onto a screen or surface located behind the hole. The smaller the pin-hole, the sharper the resulting image, as light passing through a smaller hole produces less diffraction or spreading out of the light. When the pin-hole is larger, more light enters the camera, but the light rays also become more scattered, resulting in a less well-defined image. This is because the larger opening allows more light rays to enter at different angles, creating a wider range of paths that the light can take as it travels through the camera and onto the screen. As a result, the image is less clear and less defined, with less sharp edges and more blurring. is the correct answer because it correctly identifies the effect of a larger pin-hole on the image produced by the pin-hole camera. less illumination, would actually produce a dimmer image, but it would not affect the sharpness or definition of the image. the distance of the screen from the pin-hole, and the distance of the object from the pin-hole, would affect the size of the image and the scale of the objects, but they would not affect the sharpness or definition of the image.
Ajụjụ 20 Ripọtì
A cone is in unstable equilibrium has its potential energy
Akọwa Nkọwa
In unstable equilibrium, potential energy decreases as the height decreases.
Ajụjụ 22 Ripọtì
During the transformation of matter from the solid to the liquid state, the heat supplied does not produce temperature increase because
Akọwa Nkọwa
When a solid is heated to its melting point, the heat supplied is used to overcome the intermolecular forces holding the molecules in a fixed position, resulting in the breaking of these bonds. As a result, the solid transforms into a liquid without any change in temperature. This is because the heat energy supplied is used in breaking the bonds between molecules rather than increasing the kinetic energy of the molecules, which is what causes an increase in temperature. Therefore, the correct option is: "all the heat is used to break the bonds holding the molecules of the solid together."
Ajụjụ 23 Ripọtì
Which of the following bodies, each with centre of gravity G, lying on a horizontal table, is/are in unstable equilibrium?
Akọwa Nkọwa
- I and II are in neutral equilibrium. They will roll continuously on the table
- III is a body with high centre of gravity (unstable)
- IV is a body with high centre of gravity (stable)
Ajụjụ 24 Ripọtì
According to kinetic molecular model, in gases
Akọwa Nkọwa
In kinetic molecular model, gases are energised and thus moves freely, fast as they occupy specific space
Ajụjụ 26 Ripọtì
The pitch of a screw jack is 0.45cm and the arm is 60cm long. If the efficiency of the Jack is 75/π %, calculate the mechanical advantage.
Akọwa Nkọwa
P = 0.45cm, L = 60cm, Eff = 75/π%
| VR | (Screw | system) | = | 2πrP | = | 2πLP |
| M.A | = | Eff% × VR100 | = | 75π | × | 1100 | × | 2π × 600.45 | = | 75 × 800300 | = | 200 |
Ajụjụ 28 Ripọtì
The volume of 0.354g of helium at 273°C and 114cm of mercury pressure is 2667cm3 . Calculate the volume
Akọwa Nkọwa
m = 0.354g, T1
= 273°C = 273 + 273 = 576K
P1
= 114cmHg, V1
= 2667cm3
at STP
T2
= 273K, P2
= 76cmHg, V2
= ?
| P1 V1 T1 | = | P2 V2 T1 |
| V2 | = | 114 × 2667 × 27376 × 576 | = | 2000.25cm3 |
Ajụjụ 29 Ripọtì
A boy pushes a 500kg box along a floor with a force of 2000N. If the velocity of the box is uniform, the co-efficient of friction between the box and the floor is
Akọwa Nkọwa
The coefficient of friction is a measure of the amount of friction between two surfaces. It is represented by the symbol "μ" and is a dimensionless quantity. The coefficient of friction between two surfaces depends on the nature of the surfaces in contact and the force pressing them together. In this problem, the boy is pushing the box with a force of 2000N. If the box is moving with a uniform velocity, then the force of friction acting on the box is equal and opposite to the pushing force applied by the boy. We can calculate the force of friction using the formula: frictional force = coefficient of friction x normal force where the normal force is the force exerted by the floor on the box in a direction perpendicular to the floor. Since the box is not moving up or down, the normal force is equal to the weight of the box. The weight of the box can be calculated using the formula: weight = mass x gravity where mass is the mass of the box and gravity is the acceleration due to gravity (9.8 m/s^2). So, the weight of the box is: weight = 500 kg x 9.8 m/s^2 = 4900 N The force of friction is equal to the pushing force of 2000N, so we can set these two equal to each other and solve for the coefficient of friction: frictional force = 2000N coefficient of friction x normal force = 2000N coefficient of friction x 4900N = 2000N coefficient of friction = 2000N / 4900N = 0.408 So, the coefficient of friction between the box and the floor is approximately 0.4. Therefore, the correct answer is 0.4.
Ajụjụ 30 Ripọtì
The earth's gravitational field intensity at its surface is about
(G = 6.7 × 10−11 Nm2 /kg2 , mass of the earth is 6 × 1024 kg, radius of the earth is 6.4 × 106 m, g on the earth = 9.8m/s2 )
Akọwa Nkọwa
The earth's gravitational field intensity at its surface can be calculated using the formula: g = G * M / r^2 where G is the gravitational constant, M is the mass of the earth, r is the radius of the earth, and g is the gravitational field intensity at the surface of the earth. Substituting the given values, we get: g = (6.7 × 10^-11 Nm^2/kg^2) * (6 × 10^24 kg) / (6.4 × 10^6 m)^2 g = 9.8 N/kg (approx.) Therefore, the answer is 9.8N/kg.
Ajụjụ 31 Ripọtì
A mass of 0.5kg is whirled in a vertical circle of radius 2m at a steady rate of 2 rev/s. Calculate the centripetal force
Akọwa Nkọwa
The centripetal force is the force that acts towards the center and keeps an object moving in a circular path. To calculate the centripetal force, we can use the following formula: f = m * v^2 / r where: - f = centripetal force - m = mass of the object (0.5 kg) - v = velocity of the object (2 rev/s * 2 * pi m/rev = 12.57 m/s) - r = radius of the circle (2 m) Plugging in the values, we get: f = 0.5 kg * 12.57 m/s^2 / 2 m f = 31.43 N Rounding to the nearest whole number, the centripetal force is 31 N. So, the closest answer from the options is 160N.
Ajụjụ 32 Ripọtì
If a body moves with a constant speed and at the same time undergoes an acceleration, its motion is said to be
Akọwa Nkọwa
If a body moves with a constant speed but at the same time undergoes an acceleration, its motion is called rectilinear motion. This means that the body moves in a straight line and its speed changes at a constant rate, causing an acceleration. It is different from oscillation, circular and rotational motions which involve changes in direction, as well as changes in speed.
Ajụjụ 33 Ripọtì
The conductivity of gases at low pressure can be termed as
I. hot cathode emission
II. thermo ionic emission
III. cold cathode emission
IV. Field emission
Akọwa Nkọwa
As conduction of gases is at low pressure and high voltage, called field or cold cathode emission.
Ajụjụ 34 Ripọtì
When blue and green colours of light are mixed, the resultant colour is
Ajụjụ 35 Ripọtì
A train has an initial velocity of 44m/s and an acceleration of -4m/s2 . Calculate its velocity after 10 seconds
Akọwa Nkọwa
The velocity of the train after 10 seconds can be calculated using the formula: v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Substituting the given values, we get: v = 44 m/s + (-4 m/s^2) x 10 s v = 44 m/s - 40 m/s v = 4 m/s Therefore, the velocity of the train after 10 seconds is 4m/s. Answer option D is correct. Explanation: The train has an initial velocity of 44 m/s and an acceleration of -4 m/s^2. The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which means that the train is slowing down. After 10 seconds, the train's velocity decreases by 40 m/s (4 m/s^2 x 10 s) to reach a final velocity of 4 m/s.
Ajụjụ 36 Ripọtì
The value of T in the figure above is
Akọwa Nkọwa
Tsin30 + Tsin30 =40
2Tsin30 = 40
Tsin30 = 40/2 = 20
T(12 ) = 20
T = 20 x 2 = 40N
Ajụjụ 37 Ripọtì
According to kinetic molecular model, in gases
Akọwa Nkọwa
According to the kinetic molecular model, in gases, the molecules are very fast apart and occupy all the space made available. This means that gas molecules are in constant random motion and they move freely in all directions without any regular arrangement. They collide with each other and with the walls of the container, exerting pressure. The temperature of the gas is related to the average kinetic energy of the gas molecules. The higher the temperature, the faster the gas molecules move, and the higher the kinetic energy.
Ajụjụ 38 Ripọtì
The mass of water vapour in a given volume of air is 0.05g at 20°C, while the mass of water vapour required to saturate it at the same temperature is 0.15g. Calculate the relative humidity of the air.
Akọwa Nkọwa
Relative humidity is a measure of how much water vapor the air is holding compared to the maximum amount it could hold at a given temperature. It is expressed as a percentage. To calculate the relative humidity of the air in this problem, we need to use the formula: Relative humidity = (mass of water vapor in air / mass of water vapor required for saturation) x 100% We are given that the mass of water vapor in the air is 0.05g and the mass of water vapor required for saturation at the same temperature is 0.15g. Plugging these values into the formula, we get: Relative humidity = (0.05 / 0.15) x 100% = 33.33% Therefore, the relative humidity of the air is 33.33%. So the answer is 33.33%.
Ajụjụ 39 Ripọtì
The momentum of a car moving at a constant speed in a circular track
Akọwa Nkọwa
Movement of an object in a circle with an acceleration towards its center is provided by change in velocity and centripetal force a α V α Fc
Ajụjụ 40 Ripọtì
Gases conduct electricity under
Akọwa Nkọwa
Gases conduct electricity under low pressure and high voltage
