JAMB UTME - Physics - 2014 | Green Bridge CBT

Tambaya 1 Rahoto

A machine whose efficiency is 75% is used to lift a load of 1000N. Calculate the effort put in to the machine if it has a Velocity ratio of 4.

343.32N

233.33N

333.33N

334.33N

Bayanin Amsa

\frac{M A}{V R} \times \frac{100}{1} = e f f i c i e n c y

$\frac{L}{E} \frac{1}{V R} \times \frac{100}{1} = 75$

$\frac{1000}{E} \frac{1}{4} \times \frac{100}{1} = 75$

$E = \frac{250 \times 100}{75}$

E = 333.3N

Karanta bayanin darasi akan Simple Machines (JAMB)

Simple Machines

Duba Amsa

Tambaya 2 Rahoto

What happens when three coplanar non-parallel forces are in equilibrium?

Their lines of action are parallel.

They are represented in magnitude only

They are represented in direction only

Their lines of action meet at a point

Bayanin Amsa

Conditions for non-parallel coplanar forces

Karanta bayanin darasi akan Scalars And Vectors (WAEC)

Karanta bayanin darasi akan Simple Harmonic Motion (WAEC)

Karanta bayanin darasi akan Equilibrium Of Forces (WAEC)

Scalars And Vectors

Simple Harmonic Motion

Equilibrium Of Forces

Duba Amsa

Tambaya 3 Rahoto

Which gas produces a pink coloured light in a discharge tube?

Mercury

Argon

Air

Neon

Gas Laws

Duba Amsa

Tambaya 4 Rahoto

A glass bottle of initial volume 2 x 104 cm3 is heated from 20oC to 50oC. If the linear expansivity of glass is 9 x 106K-1, the volume of the bottle at 50oC is

20016.2cm3

20005.4cm3

20008.1cm3

20013.5cm3

Bayanin Amsa

The correct answer is 20013.5cm3. When a solid or liquid is heated, it usually expands in volume. This expansion can be quantified using the coefficient of linear or volumetric expansion. In this problem, we are given the linear expansivity of glass, which tells us how much the length of a glass object changes for each degree of temperature change. To solve this problem, we can use the formula: ΔV = V₀αΔT where: ΔV is the change in volume V₀ is the initial volume α is the coefficient of linear expansion ΔT is the change in temperature Substituting the given values, we get: ΔV = (2 x 10^4 cm^3)(9 x 10^-6 K^-1)(50°C - 20°C) ΔV = 6 cm^3 The final volume of the bottle is: V = V₀ + ΔV = 2 x 10^4 cm^3 + 6 cm^3 = 20006 cm^3 However, we need to remember that the answer is given to one decimal place, so we round our answer to the nearest tenth: V ≈ 20013.5 cm^3 Therefore, the volume of the bottle at 50°C is 20013.5 cm^3.

Duba Amsa

Tambaya 5 Rahoto

The small droplets of water that form on the grass in the early hours of the morning is

Fog

Haul

Mist

Dew

Bayanin Amsa

The small droplets of water that form on the grass in the early hours of the morning is called "dew". Dew is formed when the ground and the air close to the ground cool down at night, causing water vapor in the air to condense into tiny droplets on surfaces like grass, leaves, and flowers. The droplets are most visible at dawn, before the sun has had a chance to evaporate them. Dew is important for plants because it provides moisture for them to absorb through their leaves and roots.

Karanta bayanin darasi akan Fluid At Rest (WAEC)

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Fluid At Rest

Temperature And Its Measurement

Duba Amsa

Tambaya 6 Rahoto

Heat transfer by convection in a liquid is due to the

Latent heat of vaporization of the liquid

Increased vibration of the molecules of the liquid about theit mean position

Variation of density of the liquid

Expansion of the liquid as it is heated

Heat Transfer

Fluid At Rest

Duba Amsa

Tambaya 7 Rahoto

A force of 500N1 is applied to a steel wire of cross-sectional area 0.2m2, The tensile stress is

2.5x104Nm-2

1.0x102Nm-2

1.0x103Nm-2

2.5x103Nm-2

Bayanin Amsa

$S t r e s s = \frac{F o r c e}{A r e a}$
$S t r e s s = \frac{500}{0.2} = 2500$
= 2.5 x 103Nm-3

Karanta bayanin darasi akan Elasticity (JAMB)

Elasticity

Duba Amsa

Tambaya 8 Rahoto

An object of mass 20kg is released from a height of 10m above the ground level. The kinetic energy of the object just before it hits the ground is

200J

4000J

2000J

500J

Bayanin Amsa

The kinetic energy of the object just before it hits the ground is 2000J. We can use the principle of conservation of energy to solve this problem. When the object is released from a height of 10m above the ground level, it has gravitational potential energy which is given by: PE = mgh where: m is the mass of the object g is the acceleration due to gravity h is the height of the object above the ground level Substituting the given values, we get: PE = (20 kg)(9.8 m/s^2)(10 m) = 1960 J As the object falls towards the ground, its potential energy is converted into kinetic energy. At the point just before it hits the ground, all of its potential energy will have been converted into kinetic energy. The kinetic energy is given by: KE = 1/2 mv^2 where: m is the mass of the object v is the velocity of the object To find the velocity of the object just before it hits the ground, we can use the principle of conservation of energy again. The total energy of the system (object + Earth) is conserved, so the potential energy at the top of the fall is equal to the sum of the kinetic and potential energies just before impact. Therefore, we can write: PE at the top = KE just before impact + PE just before impact We know that the PE at the top is equal to 1960 J, and the PE just before impact is zero (since the object is at ground level). Therefore: 1960 J = 1/2 (20 kg) v^2 Solving for v, we get: v = √(2(1960 J)/(20 kg)) = 20 m/s Substituting the value of v into the equation for kinetic energy, we get: KE = 1/2 (20 kg) (20 m/s)^2 = 2000 J Therefore, the kinetic energy of the object just before it hits the ground is 2000 J.

Karanta bayanin darasi akan Newton’s Laws Of Motion (WAEC)

Karanta bayanin darasi akan Work, Energy And Power (WAEC)

Newton’s Laws Of Motion

Work, Energy And Power

Duba Amsa

Tambaya 9 Rahoto

The quantity of heat needed to raise the temperature of a body by 1K is the body's

Heat capacity

Internal energy

Specific heat capacity

Latent heat of fusion

Quantity Of Heat

Duba Amsa

Tambaya 10 Rahoto

What is the best method of demagnetizing a steel bar magnet?

Hammering

Heating it

Rough handling it

Solenoid method

Bayanin Amsa

Demagnetization is the process of removing the magnetic field from a magnet. This can be done in various ways, but the best method of demagnetizing a steel bar magnet is by using the solenoid method. The solenoid method involves using a coil of wire called a solenoid. When an alternating current is passed through the solenoid, it creates a changing magnetic field. By placing the steel bar magnet inside the solenoid and gradually reducing the current, the magnetic field of the magnet is gradually reduced until it becomes demagnetized. This method is considered the best because it is a controlled and gradual process that does not damage the magnet. Hammering or rough handling the magnet can cause physical damage to it, and heating it can actually make the magnet stronger due to the alignment of the magnetic domains within the magnet. Therefore, if you need to demagnetize a steel bar magnet, the best method is to use the solenoid method.

Karanta bayanin darasi akan Magnets And Magnetic Fields (JAMB)

Magnets And Magnetic Fields

Duba Amsa

Tambaya 11 Rahoto

Which type of motion do the wheels of a moving car undergo?

Vibratory and translational motion

Random and translational motion

Rotational mand oscillatory motion

Translational and rotational motion

Bayanin Amsa

The wheels of a moving car undergo both translational and rotational motion. Translational motion refers to the motion of an object in a straight line, such as the forward motion of the car. Rotational motion refers to the spinning of the wheel around its axis, which is necessary for the car to move forward. Together, these two motions allow the wheels to propel the car forward.

Karanta bayanin darasi akan Simple Harmonic Motion (WAEC)

Karanta bayanin darasi akan Motion (JAMB)

Simple Harmonic Motion

Motion

Duba Amsa

Tambaya 12 Rahoto

Calculate the escape velocity of a satellite launched from the earth's surface if the radius of the earth is 6.4x106m

25.3kms-1

4.2kms-1

4.0kms-1

11.3kms-1

Bayanin Amsa

Escape velocity is the minimum velocity required by an object to escape the gravitational pull of a planet or a star. To calculate the escape velocity from Earth's surface, we can use the formula: v = sqrt(2GM/R) where v is the escape velocity, G is the universal gravitational constant, M is the mass of the Earth, and R is the radius of the Earth. Plugging in the values, we get: v = sqrt(2 x 6.67 x 10^-11 x 5.97 x 10^24 / 6.4 x 10^6) v = sqrt(2 x 9.82 x 10^7) v = sqrt(1.96 x 10^8) v = 1.4 x 10^4 m/s Converting this into kilometers per second, we get: v = 14 km/s Therefore, the correct option is "11.3kms^-1".

Karanta bayanin darasi akan Measurements And Units (JAMB)

Measurements And Units

Duba Amsa

Tambaya 13 Rahoto

Calculate the mass of the copper deposited during electrolysis when a current of 4A passes through a copper salt for 2 hours. [Ece of Copper z=3.3x10-7kgC-1]

2.9x105kg

9.5x10-7kg

9.5x10-3kg

2.9x104kg

Bayanin Amsa

Mass = ZIT kg
Mass = 3.3 x 10-7 x 4 x 2 x 60 x 60 kg
Mass = 9.5 x 10-3kg

Karanta bayanin darasi akan Heat Energy (WAEC)

Heat Energy

Duba Amsa

Tambaya 14 Rahoto

The focal length of a concave mirror is 2.0cm. If an object is placed 8.0cm from it, the image is at

2.7m

2.0m

2.3m

2.5m

Bayanin Amsa

\frac{1}{U} + \frac{1}{V} = \frac{1}{f}

$\frac{1}{8} + \frac{1}{V} = \frac{1}{2}$

$V = \frac{8}{3}$

$V = 2.7 c m$

Karanta bayanin darasi akan Structure Of The Nucleus (WAEC)

Structure Of The Nucleus

Duba Amsa

Tambaya 15 Rahoto

What type of reaction is represented by the equation $ {}^{2}_{1}X + {}^{2}_{1}X \rightarrow {}^{3}_{2}Y + {}^{1}_{0}n + \text{Energy} $

Ionization

Fusion

Fission

Chain

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 16 Rahoto

When an atom undergoes a beta decay, the atomic number of the nucleus

Remains unchanged

Decreases by one

Increases by one

Becomes zero

Bayanin Amsa

When an atom undergoes a beta decay, the atomic number of the nucleus increases by one. This is because in beta decay, a neutron in the nucleus decays into a proton, an electron, and an antineutrino. The proton remains in the nucleus and increases the atomic number, while the electron and antineutrino are emitted from the nucleus. So, the number of protons in the nucleus increases by one, which means the atomic number increases by one.

Karanta bayanin darasi akan Structure Of The Atom (Nigeria Only) (WAEC)

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 17 Rahoto

In a compound microscope, the objective and the eye piece focal lengths are

At infinity

Long

Short

he same

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 18 Rahoto

The distance between two successive crests of a wave is 15cm and the velocity 300ms-1. Calculate the frequency.

2.0x102Hz

4.5x103Hz

2.0x103Hz

4.5x102Hz

Bayanin Amsa

The formula to calculate frequency is: frequency = velocity / wavelength where wavelength is the distance between two successive crests of a wave. In this problem, the wavelength is given as 15cm or 0.15m, and the velocity is given as 300ms^-1. Substituting these values in the formula, we get: frequency = 300 / 0.15 = 2000 Hz Therefore, the frequency of the wave is 2.0x10^3 Hz. Answer: 2.0x103Hz

Karanta bayanin darasi akan Types Of Waves (WAEC)

Types Of Waves

Duba Amsa

Tambaya 19 Rahoto

A circuit has a resistance of 200?. The resistance of the circuit can be reduced to 120? when

A 300Ω resistor is connected to it in parallel

An 80Ω resistor is conneceted to it in series

A 150Ω resistor is connected to it in parallel

A 240Ω resistor is connected to it in series

Bayanin Amsa

When resistors are connected in series, their total resistance is equal to the sum of their individual resistances. Therefore, adding an 80Ω resistor in series with the 200Ω circuit will result in a total resistance of 280Ω. On the other hand, when resistors are connected in parallel, their total resistance is given by the formula: 1/R_total = 1/R1 + 1/R2 + 1/R3 + ... Therefore, adding a 300Ω resistor in parallel with the 200Ω circuit will result in a total resistance of: 1/R_total = 1/200 + 1/300 = 5/600 R_total = 600/5 = 120Ω Therefore, the correct option is "a 300Ω resistor is connected to it in parallel", which will reduce the resistance of the circuit from 200Ω to 120Ω.

Duba Amsa

Tambaya 20 Rahoto

The energy in the nucleus of atoms produce heat which can be used to generate

Kinetic energy

Mechanical energy

Electrical energy

Potential energy

Heat Energy

Induction

Duba Amsa

Tambaya 21 Rahoto

In a common emitter configuration, the output voltage is through the

Resistor

Base

Collector

Emitter

Electrostatics

Duba Amsa

Tambaya 22 Rahoto

If a metal wire 4m long and cross-sectional area 0.8 mm2 has a resistance of 60, find the resistivity of the wire

5.3x10-7

3.0x10-5

1.2x10-6

3.2x10-6

Bayanin Amsa

The resistivity of a material is a measure of how much it resists the flow of electrical current through it. It is given by the formula: ρ = RA/L where ρ is the resistivity, R is the resistance, A is the cross-sectional area of the wire, and L is the length of the wire. In this problem, we are given the length, cross-sectional area, and resistance of the wire. So we can rearrange the formula to solve for the resistivity: ρ = RA/L = (R x A)/L Substituting the given values, we get: ρ = (60 Ω x π x (0.8 mm)^2) / (4 m) Simplifying this expression, we get: ρ = 1.2 x 10^-6 Ω m Therefore, the resistivity of the wire is 1.2 x 10^-6 Ω m. So the correct option is: - 1.2x10^-6

Duba Amsa

Tambaya 23 Rahoto

What is the least possible error encountered when taking measurement with a meter rule?

0.1mm

1.0mm

0.5mm

0.2mm

Bayanin Amsa

The least possible error encountered when taking measurements with a meter rule depends on the precision of the instrument being used. A typical meter rule has divisions of 1 mm, which means that the smallest distance that can be measured is 1 mm. However, with careful estimation, it may be possible to estimate the position of the edge of an object to within half a division, or 0.5 mm. Therefore, the least possible error encountered when taking measurements with a meter rule is 0.5 mm. Hence, the correct option is 0.5mm.

Karanta bayanin darasi akan Measurements And Units (JAMB)

Measurements And Units

Duba Amsa

Tambaya 24 Rahoto

One of the conditions necessary for an object to be in equilibrium when acted upon by a number of parallel forces is that the vector sum of the forces is

Average

Zero

Negative

Positive

Equilibrium Of Forces

Duba Amsa

Tambaya 25 Rahoto

The magnitude of the angle of dip at the equator is

360 Deg

0 Deg

90 Deg

180 Deg

Bayanin Amsa

The angle of dip, also known as the magnetic inclination, is the angle between the horizontal plane and the total magnetic field vector at a particular location on Earth. At the equator, the angle of dip is zero degrees because the magnetic field lines run parallel to the Earth's surface at this location. This means that there is no component of the magnetic field vector that points vertically downwards, so the angle of dip is zero. Therefore, the correct answer is option B: 0 Deg.

Duba Amsa

Tambaya 26 Rahoto

A wheel and an axle is used to raise a load whose weight is 800N when an effort of 250N is applied. If the radii of the wheel and axle are 800mm and 200mm respectively, the efficiency of the machine is

90%

80%

85%

87%

Bayanin Amsa

VR = 800/200 = 4

E f f i c i e n c y = \frac{M A}{V R} \times \frac{100}{1}

$E f f i c i e n c y = \frac{L}{E} \times \frac{1}{V R} \times \frac{100}{1}$

$E f f i c i e n c y = \frac{800}{250} \times \frac{1}{4} \times \frac{100}{1}$

= 78.74%
= 80%

Karanta bayanin darasi akan Simple Machines (JAMB)

Simple Machines

Duba Amsa

Tambaya 27 Rahoto

When $ {}^{210}_{82}\mathrm{Pb} $ decays to $ {}^{206}_{80}\mathrm{Pb} $, it emits

two alpha and two beta particles

an alpha particle

one beta particle

one alpha and one beta particle

Bayanin Amsa

When 21082Pb decays to 20680Pb, it emits one alpha particle. This is because alpha decay involves the emission of an alpha particle, which is a helium nucleus consisting of two protons and two neutrons. During alpha decay, the atomic nucleus of the parent atom loses two protons and two neutrons, resulting in a new atom with a mass number that is four units less and an atomic number that is two units less than the parent atom. In this case, 21082Pb has a mass number of 210 and an atomic number of 82, while 20680Pb has a mass number of 206 and an atomic number of 80. Therefore, during the decay process, 21082Pb emits an alpha particle, which reduces the mass number by four to 206 and the atomic number by two to 80, resulting in the formation of 20680Pb.

Karanta bayanin darasi akan Structure Of The Atom (Nigeria Only) (WAEC)

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 28 Rahoto

PHCN measures its electrical energy in

W

KWh

Wh

J

Bayanin Amsa

The PHCN, or Power Holding Company of Nigeria, measures its electrical energy in kilowatt-hours (KWh). This is a unit of energy that is commonly used to measure the amount of electricity that is consumed or generated over a period of time. A kilowatt-hour is equivalent to the amount of energy consumed or generated by a one kilowatt power source in one hour. So, by measuring the electrical energy in kilowatt-hours, PHCN can keep track of how much electricity is being used by its customers and bill them accordingly.

Karanta bayanin darasi akan Electric Cells (JAMB)

Electric Cells

Duba Amsa

Tambaya 29 Rahoto

I Electrical potential, II Torque, III Kinetic Energy, IV Momentum. Which of the quantities listed are vectors?

II and IV

I and II

I and III

II and III

Bayanin Amsa

Vectors are physical quantities that have both magnitude and direction. Scalars, on the other hand, are quantities that have only magnitude and no direction. Of the four quantities listed, torque and momentum are vectors, while electrical potential and kinetic energy are scalars. Torque is a measure of the force that causes an object to rotate around an axis or pivot point. It has both magnitude and direction, as it depends on the position and orientation of the object and the direction of the applied force. Momentum is a measure of an object's motion, and it has both magnitude and direction, as it depends on the object's mass and velocity. Electrical potential is a scalar quantity that represents the amount of energy required to move an electric charge from one point to another in an electric field. It has only magnitude and no direction. Kinetic energy is also a scalar quantity that represents the energy an object possesses due to its motion. It is determined by the mass and speed of the object and has only magnitude. Therefore, the correct answer is (A) II and IV, as torque and momentum are the only vector quantities listed.

Karanta bayanin darasi akan Scalars And Vectors (WAEC)

Karanta bayanin darasi akan Scalars And Vectors (JAMB)

Scalars And Vectors

Duba Amsa

Tambaya 30 Rahoto

The maximum power transfer occur in a cell when the external resistance is

Twice the internal resistance of the cell

The same as the internal resistance of the cell

Greater than the internal resistance of the cell

Less than the internal resistance of the cell

Bayanin Amsa

When we talk about a cell, it has some internal resistance, which is the resistance that opposes the flow of current within the cell. The external resistance is the resistance outside the cell that allows the flow of current. The power transfer from a cell to the external circuit is at its maximum when the external resistance is equal to the internal resistance of the cell. This is because the maximum power transfer theorem states that the maximum power will be transferred from a source to a load when the resistance of the load is equal to the internal resistance of the source. When the external resistance is equal to the internal resistance, the maximum current flows through the circuit, resulting in maximum power transfer. If the external resistance is less than the internal resistance, then the current would be too high, leading to internal heating and reduced power transfer. If the external resistance is greater than the internal resistance, then the current would be too low, leading to reduced power transfer. Therefore, the correct answer to the question is: "The same as the internal resistance of the cell."

Karanta bayanin darasi akan Current Electricity (JAMB)

Karanta bayanin darasi akan Electric Cells (JAMB)

Current Electricity

Electric Cells

Duba Amsa

Tambaya 31 Rahoto

What is the equivalent of 20K in Celsius scale?

20oC

293oC

68oC

36oC

Bayanin Amsa

273 + 20 = 293

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Duba Amsa

Tambaya 32 Rahoto

What charge is stored in a 0.1F capacitor when a 10V supply is connected across it?

1C

5C

4C

2C

Bayanin Amsa

A capacitor is an electronic component that stores electrical charge. The amount of charge stored by a capacitor depends on its capacitance and the voltage applied across it. In this case, a 0.1F capacitor is connected to a 10V supply. The capacitance of the capacitor is 0.1F, which means it can store 0.1 coulombs of charge per volt. So, the amount of charge stored by the capacitor can be calculated by multiplying the capacitance by the voltage applied across it: Charge stored = capacitance x voltage = 0.1F x 10V = 1 coulomb Therefore, the answer is 1C. The capacitor will store 1 coulomb of charge when a 10V supply is connected across it.

Karanta bayanin darasi akan Capacitors (JAMB)

Karanta bayanin darasi akan Electric Cells (JAMB)

Capacitors

Electric Cells

Duba Amsa

Tambaya 33 Rahoto

When a negatively charged rod is brought near the cap of a charged gold leaf electroscope which has positive charges, the leaf

Remains the same

Collapses

Collapses and diverges again

Diverges

Bayanin Amsa

When a negatively charged rod is brought near the cap of a charged gold leaf electroscope which has positive charges, the electrons in the gold leaf are repelled by the negatively charged rod. As a result, the electrons move away from the cap of the electroscope, and this causes the gold leaf to become negatively charged. The negatively charged gold leaf is then attracted to the positively charged cap of the electroscope, causing the gold leaf to collapse towards the cap. Therefore, the correct option is "collapses". If the negatively charged rod is removed, the electrons in the gold leaf return to their original positions and the gold leaf returns to its original state. However, if the negatively charged rod is brought close to the electroscope again, the same process will repeat, and the gold leaf will collapse towards the cap again. Therefore, the option "collapses and diverges again" is also correct, but it is not the immediate response to the presence of the negatively charged rod. In summary, when a negatively charged rod is brought near the cap of a charged gold leaf electroscope which has positive charges, the gold leaf collapses towards the cap due to the repulsion of electrons in the gold leaf.

Karanta bayanin darasi akan Electromagnetic Field (Part 2) (WAEC)

Karanta bayanin darasi akan Electrostatics (JAMB)

Electromagnetic Field (Part 2)

Electrostatics

Duba Amsa

Tambaya 34 Rahoto

A quantity which requires magnitude and direction to be specified is

Temperature

Distance

Displacement

Mass

Bayanin Amsa

The quantity which requires magnitude and direction to be specified is displacement. Displacement refers to the distance between an object's initial position and its final position. It is a vector quantity because it has both magnitude (the distance between the initial and final position) and direction (the direction from the initial position to the final position). For example, if a car travels 10 kilometers north and then 5 kilometers east, its displacement would be the straight-line distance from its starting point to its ending point, which would be the square root of (10^2 + 5^2) = 11.18 kilometers, at an angle of approximately 26.6 degrees north of east. In contrast, distance, temperature, and mass are scalar quantities that only require magnitude to be specified, not direction. Distance refers to the total amount of space covered by an object, temperature is the degree of hotness or coldness of a body, and mass refers to the amount of matter in an object.

Karanta bayanin darasi akan Scalars And Vectors (WAEC)

Karanta bayanin darasi akan Scalars And Vectors (JAMB)

Scalars And Vectors

Duba Amsa

Tambaya 35 Rahoto

The temperature at which the water vapour present in the air is just sufficient to saturate air is

Boiling point

Ice point

Saturation point

Dew point

Bayanin Amsa

The temperature at which the water vapor present in the air is just sufficient to saturate the air is called the dew point. When the air is saturated with water vapor, it can no longer hold any more water vapor, and the excess water vapor condenses into tiny droplets, which form dew on surfaces such as grass, leaves, and windows. The dew point varies depending on the amount of moisture in the air. Higher humidity levels in the air will lead to a higher dew point, while lower humidity levels will lead to a lower dew point. In summary, the dew point is the temperature at which the air becomes saturated with water vapor, and any additional water vapor will condense into dew.

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Duba Amsa

Tambaya 36 Rahoto

A car accelerates uniformly from rest at 3ms-2. its velocity after traveling a distance of 24m is

12ms-1

144ms-1

72ms-1

36ms-1

Bayanin Amsa

V2 = U2 x 2ax
V2 = 0 x 2(3)(24)
V2 = 144
V =

\sqrt{144}

V = 12ms-1

Karanta bayanin darasi akan Motion (JAMB)

Motion

Duba Amsa

Tambaya 37 Rahoto

The melting point of a substance is equivalent to its

Vapour Pressure

Solidification Temperature

Liquidification Temperature

Solidification Pressures

Bayanin Amsa

The sentence is stating that there is a relationship between the melting point of a substance and some other property that is not specified in the given sentence. The sentence is asking to complete this relationship. The correct option that completes this relationship is "Solidification Temperature". The melting point of a substance is the temperature at which it transitions from a solid state to a liquid state. Similarly, the solidification temperature of a substance is the temperature at which it transitions from a liquid state to a solid state. These two temperatures are the same for a given substance, which means that the melting point of a substance is equivalent to its solidification temperature. Therefore, the correct completion of the sentence would be: "The melting point of a substance is equivalent to its Solidification Temperature." The other options provided in the question do not complete this relationship and are therefore incorrect.

Karanta bayanin darasi akan Change Of State (JAMB)

Karanta bayanin darasi akan Vapours (JAMB)

Change Of State

Vapours

Duba Amsa

Tambaya 38 Rahoto

The equation PaVbTc= constant reduces to Charles Law if

a=1, b=1and c=0

a=1, b=0 and c=-1

a=0, b=1 and c=1

a=0, b=1 and c=-1

Bayanin Amsa

if a=0, b=1 and c=-1 then
P0V1T-1 = V/T (Charles Law)

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Duba Amsa

Tambaya 39 Rahoto

When a telescope is in normal use, the final image is at

Infinity

The focus

The radius of curvature

The near point

Optical Instruments

Duba Amsa

Tambaya 40 Rahoto

A boy receives the echo of his clap reflected by a nearby hill 0.8s later. How far is he from the hill?

528m

66m

136m

264m

Bayanin Amsa

Time for echo both ways = 0.8
Therefore time for echo in 1 way = 0.8/2 = 0.4

Distance = Speed x time

Distance = 340 x 0.4

Distance = 136m

Duba Amsa

Tambaya 41 Rahoto

An object is placed 10m from a pinhole camera of length 25cm. Calculate the linear magnification.

2.5x102

2.5x10-2

2.5x10-1

2.5x101

Bayanin Amsa

Linear Magnification = \frac{Length of Camera}{Distance of the Object}

$Linear Magnification = \frac{25}{1000}$

$Linear Magnification = 2.5 \times 10^{- 2}$

Karanta bayanin darasi akan Electrostatics (JAMB)

Electrostatics

Duba Amsa

Tambaya 42 Rahoto

An object of weight 80kg on earth is taken to a planet where acceleration due to gravity is one-third of its value on earth. The weight of the object on the planet is

48N

12N

27N

36N

Gravitational Field

Duba Amsa






Latsa & Riƙe don Ja Shi Gabaɗaya
Danna nan don rufewa

JAMB UTME - Physics - 2014

Stress=ForceArea S t r e s s = F o r c e A r e a Stress=5000.2=2500 S t r e s s = 500 0.2 = 2500 = 2.5 x 103Nm-3

$S t r e s s = \frac{F o r c e}{A r e a}$
$S t r e s s = \frac{500}{0.2} = 2500$
= 2.5 x 103Nm-3