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Ajụjụ 1 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 2 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 3 Ripọtì
Which of the following statement about the electromagnet shown above is correct?
Akọwa Nkọwa
A - B = S - N.
Also, starting end of the current is south while terminating end is North.
Ajụjụ 4 Ripọtì
In a slide wire bridge, the balance is obtained at a point 25cm from one end of wire 1m long. The resistance to be tested is connected to that end and a standard resistance of 3.6Ω is connected to the other end of the wire. Determine the value of the unknown resistance
Akọwa Nkọwa
R3.6=7525=13
3R = 3.6
R = 1.2Ω
Ajụjụ 5 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 8 Ripọtì
The diagram above represents the stress-strain graph of a loaded wire. Which of these statements is correct?
Akọwa Nkọwa
- I is the elastic limit
- the end of the constant part J is the yield point
- L is the break point.
Ajụjụ 9 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 10 Ripọtì
A microscope is focused on a mark on a table, when the mark is covered by a plate of glass 2m thick, the microscope has to be raised 0.67cm for the mark to be once more in focus. Calculate the refractive index.
Akọwa Nkọwa
R = th = 2cm, d = 0.67cm
| n | = | RA | = | RR.d | = | 22-0.67 | = | 1.52 |
Ajụjụ 11 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 12 Ripọtì
Workdone on an object to bring it to a certain point in space is called
Akọwa Nkọwa
The work done on an object to bring it to a certain point in space is called "Potential Energy". Potential energy is a form of energy that an object possesses due to its position relative to other objects. When an object is lifted or moved to a higher point against gravity, work is done on it, and this work is stored as potential energy. The potential energy of an object is directly proportional to its height and mass. It can be converted into other forms of energy, such as kinetic energy, when the object is released or allowed to move freely. Therefore, potential energy is a type of stored energy that an object has due to its position, and it can be released to do work.
Ajụjụ 13 Ripọtì
One newton × One meter equals?
Akọwa Nkọwa
One newton times one meter is equal to one Joule. A newton is the unit of measurement for force, and a meter is the unit of measurement for distance. When force is applied over a distance, work is done, which is measured in Joules. Therefore, one newton multiplied by one meter results in one Joule of work done. The other options listed (one water, one ampere, one kilogram) are not correct units of measurement for this calculation.
Ajụjụ 14 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 15 Ripọtì
Heat may be transferred by conduction, convention and radiation. By which of these methods does heat travel through vacuum?
Akọwa Nkọwa
Heat can be transferred by conduction, convection, and radiation. Conduction is the transfer of heat through a material by the movement of heat-carrying particles, such as atoms or molecules, from one part of the material to another. This method of heat transfer is not possible in a vacuum, as there are no particles present to carry heat. Convection is the transfer of heat by the movement of a fluid, such as air or water. This method of heat transfer is also not possible in a vacuum, as there are no fluids present to carry heat. Radiation is the transfer of heat through electromagnetic waves, such as light or infrared radiation. This method of heat transfer does not require any material or fluid medium, and can therefore occur in a vacuum. Therefore, the answer is "Radiation only".
Ajụjụ 16 Ripọtì
Which of the following is/are the limitations to the Rutherford's atomic models?
I. It is applicable when energy is radiated as electrons are revolving
II. It is applicable when energy is radiated in a continuous mode
III. It is applicable to an atom with only one electron in the other shell
Akọwa Nkọwa
Rutherford assumed that (I) energy is radiated when electrons are revolving (II) energy is radiated in a continuous mode. These are limitations of Rutherford's model
Ajụjụ 17 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 18 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 19 Ripọtì
Ripple in a power supply unit is caused by
Akọwa Nkọwa
The correct option is "Using a zener diode" as fluctuation of d.c signal results from the rectification of a.c to d.c.
Ajụjụ 20 Ripọtì
The equilibrium position of objects in any field corresponds to situation of
Akọwa Nkọwa
The equilibrium position of an object in any field corresponds to the situation of minimum potential energy. This means that at the equilibrium position, the object has the lowest possible potential energy within the field. In other words, the forces acting on the object are balanced, and the object is not being pushed or pulled in any direction. Therefore, the object will remain at rest at the equilibrium position unless it is acted upon by an external force. Of the options given, the correct answer is "minimum potential energy".
Ajụjụ 21 Ripọtì
- angle of dip is zero at the magnetic equator
- angle of variation is the same as angle of declination.
Akọwa Nkọwa
- angle of dip is zero at the magnetic equator
- angle of variation is the same as angle of declination.
Ajụjụ 22 Ripọtì
The Earth's magnetic equator passes through Jos in Nigeria. At Jos, the
Akọwa Nkọwa
The Earth has a magnetic field that is generated by the movement of molten iron in its core. The magnetic field has different properties at different locations on the Earth's surface. The magnetic equator is an imaginary line on the Earth's surface where the inclination or tilt of the Earth's magnetic field is zero, meaning that the magnetic field lines are parallel to the Earth's surface. At Jos, Nigeria, the Earth's magnetic equator passes through, which means that the angle of inclination or dip of the Earth's magnetic field is zero. Therefore, the correct answer is that the angle of dip is zero. This means that a magnetic needle suspended by a thread or placed on a horizontal surface would remain horizontal and not point downwards or upwards, as it would at other locations on the Earth's surface. This makes Jos an important location for studying the Earth's magnetic field and for conducting experiments related to magnetism.
Ajụjụ 23 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 24 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 25 Ripọtì
A copper rod, 5m long when heated through 20c, expands by 1mm. If a second copper rod, 2.5m long is heated through 5c, by how much will it expand?
Akọwa Nkọwa
l1
= 5m, ΔT = 10c, l2
- l1
= 1mm
l1
= 2.5m, ΔT = 5c, l2
- l1
= ?
| using | α | = | l2 - l1 l1 ΔT |
| 15(10) | = | l2 - l1 2.5(5) |
| l2 | - | l2 | = | 2.5(5)5(10) | = | 14 | = | 0.25mm |
Ajụjụ 26 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 27 Ripọtì
Which of the following statements is/are correct for a freely falling body?
I. the total is entirely kinetic
II. the ratio of potential energy to kinetic energy is constant
III. the sum of potential and kinetic energy is constant
Akọwa Nkọwa
The correct answer is "III only". A freely falling body is one that is falling under the influence of gravity and experiences no other force or constraint. In this situation, the total energy of the body is conserved, meaning that the sum of its potential and kinetic energy remains constant. The potential energy of a body is directly proportional to its height above the ground, and its kinetic energy is directly proportional to its velocity. As the body falls, its potential energy decreases and its kinetic energy increases, but the total energy remains constant. Statement III is correct because the sum of potential and kinetic energy is indeed constant for a freely falling body. Statement I is incorrect because the body has both potential and kinetic energy, so the total energy is not entirely kinetic. Statement II is incorrect because the ratio of potential energy to kinetic energy is not constant for a freely falling body, as both are changing as the body falls.
Ajụjụ 28 Ripọtì
A vibrator causes water ripples to travel across the surface of a tank. The wave travels 50cm in 2s and the distance between successive crests is 5cm. Calculate the frequency of the vibrator
Akọwa Nkọwa
The frequency of the vibrator can be calculated using the formula: frequency = speed / wavelength where speed is the speed of the wave, and wavelength is the distance between successive crests. In this case, we are given that the wave travels 50cm in 2s, which means the speed of the wave is: speed = distance / time = 50cm / 2s = 25cm/s We are also given that the distance between successive crests is 5cm, which is the wavelength. Therefore, the frequency of the vibrator is: frequency = speed / wavelength = 25cm/s / 5cm = 5Hz So the correct answer is 5Hz.
Ajụjụ 29 Ripọtì
In the molecular explanation of conduction, heat is transferred by the
Akọwa Nkọwa
In the molecular explanation of conduction, heat is transferred by the Free electrons. In metals, free electrons move randomly and collide with other particles as they gain kinetic energy. These free electrons transfer the energy to the adjacent particles, which in turn gain kinetic energy and transmit it to other adjacent particles, thus transferring heat energy from one part of the material to another. This process of heat transfer by free electrons is called conduction. Therefore, the correct option is "Free electrons."
Ajụjụ 30 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 31 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 32 Ripọtì
The lower fixed part of a faulty thermometer reads 2°C while the upper fixed point is 100°C.
What is the true temperature when the thermometer reads 51°C?
Akọwa Nkọwa
Since the thermometer is faulty, it is not measuring the temperature accurately. To find the true temperature, we need to determine the extent of the error in the thermometer. We can do this by comparing the difference between the lower fixed point and the reading with the difference between the upper fixed point and the true temperature. Since the lower fixed point reads 2°C and the upper fixed point reads 100°C, and the thermometer reading is 51°C, we can calculate the error as follows: True temperature = (51°C - 2°C) / (51°C - 2°C) * (100°C - 51°C) + 51°C = 50°C So, the true temperature when the thermometer reads 51°C is 50°C, which is option B.
Ajụjụ 33 Ripọtì
If the attraction of the sun is suddenly ceased, the earth would continue to move in a straight line making a tangent with the original orbit. This statement is derived from Neutron's
Akọwa Nkọwa
The correct answer is the First law of motion. The First law of motion, also known as the law of inertia, states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. In this case, the earth is moving in its orbit around the sun because of the force of gravity between the two objects. If the force of gravity suddenly ceased, the earth would no longer be acted upon by an external force and would continue to move in a straight line, making a tangent with its original orbit. This idea is attributed to Sir Isaac Newton, who developed the laws of motion and the law of universal gravitation. However, the specific statement mentioned in the question is derived from the First law of motion.
Ajụjụ 34 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 35 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 36 Ripọtì
Which of the following characteristics of a wave is used in the measurement of the depth of the Sea?
Akọwa Nkọwa
Depth of sea can be measured by echo, a reflected sound waves.
Ajụjụ 37 Ripọtì
When two objects A and B are supplied with the same quantity of heat, the temperature change in A is obtained to be twice that of B. The mass of P is half that of Q. The ratio of the specific heat capacity of A to B is
Akọwa Nkọwa
θA = 2θB ,
| mA | = | 12 | mB |
H = MCθ
mA
cA
θA
= mB
cB
θB
( 1/2 mB
)CA
(2θB
) = mB
cB
θB
| CA CB | = | 11 |
⇒ 1 : 1
Ajụjụ 38 Ripọtì
The distance between an object and its real image in a convex lens is 40cm. If the magnification of the image is 3, calculate the focal length of the lens
Akọwa Nkọwa
u + v = 40
vu = 3
v = 3u
u + 3u = 40
4u = 40
u = 10cm
v = 3u = 30cm
f = uvu+v=10(30)10+30=30040
= 7.5 cm
Ajụjụ 39 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 40 Ripọtì
A well 1km deep is filled with a liquid of density 950kg/m3 and g = 10m/s2 , the pressure at the bottom of the well is
Akọwa Nkọwa
P = Pa + ρgh = (1.00 × 105
) + (950 × 10 × 1000)
P = 105
+ (95 × 105
) = 105
(1 + 95) = 96 × 105
P = 9.6 × 106
N/m2
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