WAEC SSCE - Physics - 1998 (Objective)

Tambaya 1 Rahoto

The intensity of light falling on the film in a camera depend on the i. brightness of the object. ii. diameter of the aperture stop. iii. speed of the shutter. Which of the statements is/are correct/

i, ii and iii

i and ii only

i and iii only

ii and iii only

iii only Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Light Energy (JAMB)

Karanta bayanin darasi akan Structure Of The Nucleus (WAEC)

Light Energy

Structure Of The Nucleus

Duba Amsa

Tambaya 2 Rahoto

An electron is accelerated from rest through a potential difference of 70kV in a vacuum. Calculate the maximum speed acquired by the electron (electronic charge = -1.6 x 10^-19; mass of an electron = 9.1 x 10^-31kg)

3.00 x 10⁸ms^-1

2.46 x 10⁸ms^-1

1.57 x 10⁸ms^-1 Amsar da ta dace

1.32 x 10⁸ms^-1

1.11 x 10⁸ms^-1

Bayanin Amsa

Duba Amsa

Tambaya 3 Rahoto

The direction of the magnetic field at a point in the vicinity of a bar magnetic is

along the line joining the point to the neutral point

always away from the south pole of the magnet

opposite the direction of theresultant field at that point

always towards the north pole of the magnet

the direction towards which the north pole of a compass needle would point Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Magnetic Field (Part 1) (WAEC)

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

Magnetic Field (Part 1)

Electromagnetic Field (Part 2)

Duba Amsa

Tambaya 4 Rahoto

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

1.43

1.33

1.00

0.83

0.75 Amsar da ta dace

Bayanin Amsa

When an object is immersed in a liquid, it displaces some volume of the liquid. According to Archimedes' principle, the weight of the liquid displaced is equal to the weight of the object. Since the object floats, the weight of the liquid displaced must be equal to the weight of the object. Let V be the volume of the rectangular block of wood and W its weight. Also, let the density of water be ρ1 and the density of the other liquid be ρ2. When the block is immersed in water, two-thirds of its volume is immersed. Therefore, the volume of water displaced is two-thirds of the volume of the block, which is 2V/3. According to Archimedes' principle, the weight of the water displaced is equal to the weight of the block, so: ρ1 x (2V/3) x g = W where g is the acceleration due to gravity. Rearranging, we get: ρ1 = W / [(2V/3) x g] Similarly, when the block is immersed in the other liquid, half of its volume is immersed. Therefore, the volume of the other liquid displaced is half of the volume of the block, which is V/2. According to Archimedes' principle, the weight of the other liquid displaced is equal to the weight of the block, so: ρ2 x (V/2) x g = W where g is the acceleration due to gravity. Rearranging, we get: ρ2 = W / [(V/2) x g] Dividing the second equation by the first equation gives: ρ2/ρ1 = [(V/2) x g] / [(2V/3) x g] Simplifying, we get: ρ2/ρ1 = 3/4 Therefore, the relative density of the liquid is 0.75. So the correct option is (E) 0.75.

Karanta bayanin darasi akan Friction (JAMB)

Karanta bayanin darasi akan Vapours (JAMB)

Friction

Vapours

Duba Amsa

Tambaya 5 Rahoto

A piece of brass of mass 170kg has its temperature raised from 0^oC to 30^oC. Calculate its increase in volume, given the density of brass at 0^oC as 8.5 x 10³kgm³ and its cubic expansivity as 5.7 x 10^-1k¹

5.7 x 10 -5 m 3

4.3 x 10^-5m³

3.4 x 10^-5m³ Amsar da ta dace

3.4 x 10^-3m³

3.4 x 10^-2m³

Bayanin Amsa

When a material is heated, it expands. The increase in volume can be calculated using the following formula: ΔV = VαΔT Where ΔV is the change in volume, V is the original volume, α is the coefficient of linear or cubic expansivity depending on the type of expansion, and ΔT is the change in temperature. For a cubic expansivity, the formula can be written as: ΔV = VβΔT Where Vβ is the coefficient of cubic expansivity. Using the given values, the increase in volume of the brass can be calculated as: ΔV = VβΔT = Vβ(Tf - Ti) Where Tf is the final temperature and Ti is the initial temperature. The initial volume of the brass can be calculated using its mass and density: V = m/ρ = 170/8.5 x 10^3 = 0.02 m^3 Substituting the values, we get: ΔV = Vβ(Tf - Ti) = 0.02 x 5.7 x 10^-3 x (30 - 0) = 3.42 x 10^-5 m^3 Therefore, the increase in volume of the brass is 3.42 x 10^-5 m^3. The correct option is (a) 3.4 x 10^-5 m^3.

Karanta bayanin darasi akan Elasticity (JAMB)

Elasticity

Duba Amsa

Tambaya 6 Rahoto

which of the following pairs of physical quantities comprises vectors?

capacitance and inductance

force ratio and velocity ratio Amsar da ta dace

friction and momentum

energy and power

electric field potential and electric field intensity

Bayanin Amsa

The physical quantities that have both magnitude and direction are called vectors. Among the given options, the pair of physical quantities that comprise vectors are "force and velocity". Both force and velocity have a direction and magnitude associated with them. The direction of force is along the line of action of the force and the direction of velocity is the direction of motion. Therefore, the correct answer is "force and velocity".

Karanta bayanin darasi akan Scalars And Vectors (WAEC)

Karanta bayanin darasi akan Scalars And Vectors (JAMB)

Karanta bayanin darasi akan Gravitational Field (WAEC)

Scalars And Vectors

Gravitational Field

Duba Amsa

Tambaya 7 Rahoto

A piece of stone attached to one end pf string is whirled round in a horizontal circle. When the string is suddenly cut, the stone will

move towards te centre of the circle

stop moving immediately

move along a circular path of smaller radius

fly off in a direction tangential to the circular path Amsar da ta dace

acquire greater centripetal force

Bayanin Amsa

When an object is whirled round in a horizontal circle, it moves with a constant speed and acceleration towards the centre of the circle. This acceleration is known as the centripetal acceleration and is provided by the tension in the string. When the string is suddenly cut, the centripetal force acting on the stone is removed, and it will fly off in a direction tangential to the circular path. Therefore, the correct option is "fly off in a direction tangential to the circular path."

Karanta bayanin darasi akan Simple Harmonic Motion (WAEC)

Karanta bayanin darasi akan Motion (JAMB)

Karanta bayanin darasi akan Equilibrium Of Forces (WAEC)

Simple Harmonic Motion

Motion

Equilibrium Of Forces

Duba Amsa

Tambaya 8 Rahoto

The diagram below illustrates an a.c.source of 50V(r.m.s.), \(\frac{100}{\pi}\)Hz connected in series with an inductor of inductor of inductance L and a resistor of resistance R. The current in the circuit is 2A and the p.d across L and R are 30V and 40V respectively. Calculate the average power dissipated in the circuit

100W

80W Amsar da ta dace

60W

20W

10W

Bayanin Amsa

Duba Amsa

Tambaya 9 Rahoto

As the end Q of the plank is raised, the component of W, normal to plank, will

increase

decrease Amsar da ta dace

remain the same

be slightly greater than W

be independent of w

Bayanin Amsa

As the end Q of the plank is raised, the component of W, normal to the plank will decrease. This is because the weight, W, acting vertically downwards on the plank can be resolved into two components: one perpendicular to the plank and the other parallel to the plank. The component perpendicular to the plank is Wcosθ, where θ is the angle between the weight vector and the normal to the plank. As Q is raised, the angle θ increases, causing the value of cosθ to decrease, hence decreasing the component of W perpendicular to the plank.

Karanta bayanin darasi akan Scalars And Vectors (WAEC)

Karanta bayanin darasi akan Equilibrium Of Forces (WAEC)

Scalars And Vectors

Equilibrium Of Forces

Duba Amsa

Tambaya 10 Rahoto

An object is heated from 30^oC to 57^oC. The increases in its temperature on the kelvin scale is

17k

27k

81k

246k

300k Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Duba Amsa

Tambaya 11 Rahoto

The process of increasing the energy of an atom via inelastic collision with an electron is known as

ionization

excitation Amsar da ta dace

field emission

photoemission

thermonic emission

Bayanin Amsa

The process of increasing the energy of an atom via inelastic collision with an electron is known as excitation. When an electron collides with an atom, it can transfer some of its energy to the atom. This results in the atom's electrons jumping to higher energy levels, or excited states. Excitation can occur through various processes, such as collisions with other particles or the absorption of light. In this case, the excitation occurs through an inelastic collision with an electron. The energy transferred to the atom can be used for various purposes, such as to emit light or to initiate chemical reactions.

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

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 12 Rahoto

I the box is pulled from P to Q, which of the following expresses the velocity ratio of the inclined plane?

PR/PQ

OR/PQ

OR/PR

PQ/PR

PQ/QR Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Motion (JAMB)

Motion

Duba Amsa

Tambaya 13 Rahoto

A proton of charge 1.6 x 10¹⁹C is projected into a uniform magnetic field of flux density with a constant speed of 1.6 x 10⁶ms^-1, calculate the magnitude of the force exerted on it by the field

0.0N

2.0 x 10^-21N

1.3 x 10^-17N Amsar da ta dace

5.1 x 10^-14N

2.3 x 10¹³N

Bayanin Amsa

Karanta bayanin darasi akan Simple A.C. Circuits (WAEC)

Karanta bayanin darasi akan Electric Field (WAEC)

Simple A.C. Circuits

Electric Field

Duba Amsa

Tambaya 14 Rahoto

The diagram above illustrates a freely suspended bar magnet NS hanging from a point in a horizontal ceiling. The thread used for suspending the magnet and the axis of the magnet are as indicated on the diagram. The angle marked \(\theta\) is called the angle os

depression

declination

variation

elevation

dip Amsar da ta dace

Bayanin Amsa

The angle marked theta in the diagram is called the "dip" of the magnet. The dip is the angle between the magnetic field lines of the Earth and a horizontal plane at the location of the magnet. The dip angle is important in navigation, especially for compasses, because it helps determine the latitude of the location.

Karanta bayanin darasi akan Electric Field (WAEC)

Electric Field

Duba Amsa

Tambaya 15 Rahoto

An astronomical telescope having an objective of focal length 100cm and an eyepiece of focal length 10cm is used in normal adjustment, calculate the separation of the lenses

0.10m

0.90cm

1.10m Amsar da ta dace

1.80m

2.20m

Bayanin Amsa

The magnifying power of a telescope in normal adjustment is given by the ratio of the focal length of the objective to that of the eyepiece: Magnifying power, M = fo/fe where fo is the focal length of the objective and fe is the focal length of the eyepiece. The separation of the lenses in a telescope is the sum of the focal lengths of the lenses: separation of lenses = fo + fe Given that fo = 100cm and fe = 10cm, we can find the magnifying power and the separation of the lenses: M = fo/fe = 100cm/10cm = 10 separation of lenses = fo + fe = 100cm + 10cm = 110cm = 1.10m Therefore, the separation of the lenses is 1.10m. Answer:.

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

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 16 Rahoto

Dry oxygen is trapped by a pellet of mercury in a uniform capillary tube which is sealed at one end. The length of the column of oxygen at 27^oC is 50cm. if the pressure of the oxygen is constant, at what temperature will the length be 60cm?

360.0^oC

240.6^oC

237.0^oC

87.0^oC Amsar da ta dace

36^oC

Bayanin Amsa

This is a question about the relationship between the temperature and volume of a gas when pressure is held constant, known as Charles' Law. According to Charles' Law, the volume of a gas is directly proportional to its temperature when the pressure is constant. In this question, the length of the column of oxygen in the capillary tube is directly proportional to its volume. This means that if we increase the temperature of the oxygen while keeping the pressure constant, the length of the column of oxygen will increase proportionally. We can use the formula for Charles' Law, V1/T1 = V2/T2, to solve for the final temperature T2. Let V1 be the initial volume of the oxygen when the length of the column is 50cm, T1 be the initial temperature of the oxygen, V2 be the final volume of the oxygen when the length of the column is 60cm, and T2 be the final temperature of the oxygen that we want to find. We know that V1 = V2 since the pressure is constant, and we can substitute the given values into the formula to get: V1/T1 = V2/T2 V1 = 50 cm V2 = 60 cm T1 = 27°C + 273.15 = 300.15 K Solving for T2, we get: T2 = (V2/V1) x T1 = (60/50) x 300.15 = 360.18 K Converting back to Celsius, we get: T2 = 360.18 K - 273.15 = 87.03°C Therefore, the answer is option (D) 87.0^oC.

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Duba Amsa

Tambaya 17 Rahoto

The phenomenon which occurs when light changes direction as it passes from one medium to another is called

reflection

polarisation

interference

diffraction

refraction Amsar da ta dace

Bayanin Amsa

The phenomenon which occurs when light changes direction as it passes from one medium to another is called refraction. Refraction is the bending of light as it passes from one medium to another, such as from air to water. The amount of bending depends on the angle at which the light strikes the surface, and the difference in the refractive indices of the two media. This phenomenon is the reason why objects can appear distorted when viewed through a curved surface like a lens, and it is also why a straw in a glass of water appears to bend at the water's surface.

Karanta bayanin darasi akan Refraction Of Light Through Plane And Curved Surfaces (JAMB)

Karanta bayanin darasi akan Properties Of Waves: Reflection, Refraction, Diffraction (WAEC)

Refraction Of Light Through Plane And Curved Surfaces

Properties Of Waves: Reflection, Refraction, Diffraction

Duba Amsa

Tambaya 18 Rahoto

The diagram above represents a section of a pair of vernier calipers. The reading on the instrument is

5.22cm

5.24cm

5.26 Amsar da ta dace

5.35

Bayanin Amsa

Karanta bayanin darasi akan Measurements And Units (JAMB)

Measurements And Units

Duba Amsa

Tambaya 19 Rahoto

A lantern gives an image of 3m square of a slide 7.62cm square on a screen. If the screen is 10m from the projection lens of the lantern, calculate the focal length of the lens

40.3cm

26.1cm

24.8cm Amsar da ta dace

0.7cm

0.3cm

Bayanin Amsa

Karanta bayanin darasi akan Thermal Expansion (JAMB)

Karanta bayanin darasi akan Gravitational Field (WAEC)

Thermal Expansion

Gravitational Field

Duba Amsa

Tambaya 20 Rahoto

An eclipse of the sun by the moon occurs when the sun, the moon and the earth are in a straight line and the

earth casts a shadow on the moon

moon casts a shadow on the sun

moon is between the sun and the earth Amsar da ta dace

earth is between the sun and the moon

sun is between the moon and the earth

Bayanin Amsa

An eclipse of the sun by the moon occurs when the sun, the moon and the earth are in a straight line and the moon is between the sun and the earth. The moon casts its shadow on the earth, blocking the light from the sun, and this causes a temporary darkness on the earth. This phenomenon is known as a solar eclipse, and it occurs when the new moon comes between the sun and the earth in its monthly orbit.

Karanta bayanin darasi akan Description Property Of Fields (WAEC)

Karanta bayanin darasi akan Gravitational Field (WAEC)

Description Property Of Fields

Gravitational Field

Duba Amsa

Tambaya 21 Rahoto

Use the following data to determine the length L of a wire when a fore of 30N is applied, assuming Hooke's law is obeyed
\(\begin{array}{c|c}\text{Force applied/N} & 0 & 5 & 10 & 30\\ \hline \text{length of wire/mm} & 500.0 & 500.0 & 501.0 & L \end{array}\)

3.0mm

3.5mm

503.00mm Amsar da ta dace

503.5mm

506.0mm

Bayanin Amsa

Hooke's Law states that the force applied to a spring is directly proportional to the extension or compression of the spring, provided the elastic limit is not exceeded. This can be expressed mathematically as F=kx, where F is the force, k is the spring constant, and x is the extension or compression. To determine the length L of the wire when a force of 30N is applied, we need to first find the spring constant k. We can do this by calculating the extension x for each force applied, using the formula: x = (L - L₀) where L₀ is the original length of the wire (500.0 mm). From the data given, we can calculate the extensions as follows: For 0 N: x = (500.0 - 500.0) = 0 mm For 5 N: x = (500.0 - 500.0) = 0 mm For 10 N: x = (501.0 - 500.0) = 1.0 mm Using Hooke's Law, we can find the spring constant k for the wire: k = F/x = 10 N / 1.0 mm = 10 N/mm Now we can use Hooke's Law again to find the extension x for a force of 30 N: 30 N = 10 N/mm * x x = 3.0 mm Finally, we can find the length L of the wire when a force of 30 N is applied: L = L₀ + x = 500.0 mm + 3.0 mm = 503.0 mm Therefore, the length of the wire when a force of 30N is applied is 503.0 mm. Answer is the correct answer.

Karanta bayanin darasi akan Elasticity (JAMB)

Elasticity

Duba Amsa

Tambaya 22 Rahoto

Air at temperature 527^oC 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 127^oC as it leaves the engine

1.5 atmospheres

2.5 atmospheres

3.0 atmospheres Amsar da ta dace

12.8 atmospheres

15.0 atmospheres

Bayanin Amsa

Karanta bayanin darasi akan Gas Laws (JAMB)

Karanta bayanin darasi akan Quantity Of Heat (JAMB)

Gas Laws

Quantity Of Heat

Duba Amsa

Tambaya 23 Rahoto

It is always not possible to determine exactly and simultaneously the position and momentum of a particle. This statement is known as the

De broglie's law

Heisenberg's uncertainty principle Amsar da ta dace

Compton effect

Frank-Hertz experimental law

Wave particle paradox

Bayanin Amsa

Heisenberg's uncertainty principle states that it is impossible to simultaneously determine the exact position and momentum of a particle. This principle is a fundamental concept in quantum mechanics, and it means that the more precisely the position of a particle is known, the less precisely its momentum can be known, and vice versa. The principle applies to all particles, including subatomic particles like electrons and photons. The uncertainty principle has important implications for the behavior of quantum systems and the interpretation of quantum mechanics. It is named after Werner Heisenberg, a German physicist who first described the principle in 1927.

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

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 24 Rahoto

the shortest length of the air column in a resonance tube at resonance is 0.12m and the next resonance length is 0.37m. calculate the frequency of vibration given that the speed of sound in air is 340m^-1

1360Hz

694Hz

680Hz Amsar da ta dace

34Hz

340Hz

Bayanin Amsa

The difference between the first and second resonance lengths is equal to half the wavelength of the sound wave. Hence, the wavelength is: λ = 2 × (0.37m – 0.12m) = 0.5m The speed of sound is given as 340m/s. We can use the formula: v = fλ where v is the velocity of sound, f is the frequency of vibration, and λ is the wavelength of the sound wave. Substituting the values we get: 340 = f × 0.5 Solving for f, we get: f = 680 Hz Therefore, the frequency of vibration is 680Hz. So, the answer is 680Hz.

Karanta bayanin darasi akan Change Of State (JAMB)

Change Of State

Duba Amsa

Tambaya 25 Rahoto

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?

half way between the equator and the north pole

near the equator

half way between the equator and the south pole Amsar da ta dace

at either pole

at the equator

Bayanin Amsa

Duba Amsa

Tambaya 26 Rahoto

A converging lens of focal length 5cm is used as a magnifying glass by a man whose nearpoint is 35cm. Calculate the magnification given by the lens

6 Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Duba Amsa

Tambaya 27 Rahoto

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

20.0kgm^-1 Amsar da ta dace

5.0kgm^-1

1.3kgm^-1

0.0kgm^-1

10.0kgm^-1

Bayanin Amsa

The impulse experienced by an object is equal to the change in momentum of the object. Momentum is defined as the product of mass and velocity, i.e., p=mv. When the ball hits the wall, it experiences an impulse that causes a change in its momentum. Initially, the ball is moving towards the wall with a velocity of 2m/s, and the velocity changes direction after the collision, so the final velocity is -2m/s. The change in velocity, Δv = (-2) - (2) = -4m/s. Using the formula for impulse, J = Δp = mΔv, we can calculate the impulse experienced by the ball: J = 5.0kg × (-4m/s) = -20.0kgm/s. Therefore, the impulse experienced by the ball is 20.0 kgm/s. Note that the negative sign indicates that the impulse is in the opposite direction to the initial velocity of the ball.

Duba Amsa

Tambaya 28 Rahoto

An object is protected with a velocity of 100m^-1 at an angle of 60^o to the vertical. Calculate the time taken by the object to reach the highest point (Take g as 10ms^-2

5.0s

8.7s Amsar da ta dace

10.0s

17.3s

20.0s

Bayanin Amsa

The motion of the object can be divided into two components, horizontal and vertical. The horizontal component of the velocity will remain constant, while the vertical component will change due to the acceleration due to gravity acting in the downward direction. We can use the vertical component of the velocity to determine the time taken by the object to reach the highest point. The initial vertical velocity of the object can be calculated using the formula: v₀sinθ = (100m/s)sin60^o = 86.6m/s where v₀ is the initial velocity, and θ is the angle of the velocity vector with the horizontal. At the highest point, the vertical component of the velocity will be zero. Using the formula: v = v₀ + at where v is the final velocity, a is the acceleration due to gravity (-10m/s²), and t is the time taken to reach the highest point, we can calculate the time taken as follows: 0 = 86.6 - 10t t = 8.66s Therefore, the time taken by the object to reach the highest point is approximately 8.7 seconds. The closest option to this answer is (b) 8.7s.

Karanta bayanin darasi akan Motion (JAMB)

Motion

Duba Amsa

Tambaya 29 Rahoto

A waterfall is 1.260m high. Calculate the change in temperature of a quantity of water that falls from the top to the bottom of the waterfall. (Neglect heat losses to the surroundings, take g as 10m^-2 and specific heat capacity of water as 4200Jkg^-1K^-1)

0.3^oC

3.0^oC Amsar da ta dace

33.3^oC

42.0^oC

100.0^oC

Bayanin Amsa

Karanta bayanin darasi akan Heat Energy (WAEC)

Heat Energy

Duba Amsa

Tambaya 30 Rahoto

The frictional effect between the layers of a moving fluid is called

capillarity

turbulence

diffusion

osmosis

viscosity Amsar da ta dace

Bayanin Amsa

The frictional effect between the layers of a moving fluid is called viscosity. Viscosity is a measure of a fluid's resistance to flow, which arises due to the friction between the layers of the fluid as they move past each other. A highly viscous fluid has a thick and sticky consistency and resists flowing, while a less viscous fluid flows more easily. Viscosity is an important property of fluids, and it has many practical applications in areas such as engineering, chemistry, and biology.

Karanta bayanin darasi akan Friction (JAMB)

Friction

Duba Amsa

Tambaya 31 Rahoto

Lenz's law of electromagnet states that

the induced charge is constant for a fixed change of force

a force is exerted on a current-carrying conductor in a magnetic field

the magnitude of the induced e.m.f in a circuit is proportional to the rate of change of the number of lines of force linking the circuit

the induced current in a conductor is in such a direction as to oppose the change producing it Amsar da ta dace

e.m.f. is induced in a circuit whenever there is a change in the magnetic flux linked with the circuit

Bayanin Amsa

Lenz's law of electromagnetism states that the direction of an induced current in a conductor is always such as to oppose the change that produced it. In other words, when there is a change in magnetic flux through a circuit, it produces an induced electromotive force (emf) which in turn generates a current. This induced current flows in such a direction that it opposes the change that produced it. This law is a consequence of the conservation of energy principle, as the induced current creates a magnetic field that opposes the change in the magnetic field that caused the current to be induced.

Karanta bayanin darasi akan Electric Field (WAEC)

Karanta bayanin darasi akan Properties Of Waves: Reflection, Refraction, Diffraction (WAEC)

Karanta bayanin darasi akan Electrostatics (JAMB)

Electric Field

Properties Of Waves: Reflection, Refraction, Diffraction

Electrostatics

Duba Amsa

Tambaya 32 Rahoto

Calculate he wavelength of a note which is one octave lower than a note of 256Hz in a medium in which the speed of sound is 352ms^-1

0.69m

1.38m Amsar da ta dace

2.75m

5.50m

9.30m

Bayanin Amsa

Karanta bayanin darasi akan Sound Waves (WAEC)

Karanta bayanin darasi akan Propagation Of Sound Waves (JAMB)

Sound Waves

Propagation Of Sound Waves

Duba Amsa

Tambaya 33 Rahoto

In the electrical method of magnetisation, the polarity of the magnet depends on the

amount of current passed

direction of current Amsar da ta dace

magnetic material used

size of the magnetic material

orientation of the magnetic material in space

Bayanin Amsa

The electrical method of magnetisation involves passing an electric current through a magnetic material to create a magnetic field. The polarity of the magnet that is produced depends on the direction of the current. The direction of the current can be determined by the direction of the flow of electrons in the wire, which is usually indicated by an arrow in a circuit diagram. Therefore, the correct answer is "direction of current."

Karanta bayanin darasi akan Electric Field (WAEC)

Electric Field

Duba Amsa

Tambaya 34 Rahoto

a pilot records the atmospheric pressure outside his plane as 63cm of Hg while a ground observer records a reading of 75cm of Hg with his barometer. Assuming that the density of air is constant, calculate the height of the plane above the ground. (take the relative densities of air and mecury as 0.00136 and 13.6 respectively)

120m

138m

274m

1200m Amsar da ta dace

120,000m

Bayanin Amsa

When the pilot recorded the atmospheric pressure outside the plane as 63cm of Hg, it means that the pressure due to the atmosphere is balancing the pressure exerted by a column of mercury 63cm high in a barometer tube. Likewise, when the ground observer recorded a reading of 75cm of Hg, it means that the pressure due to the atmosphere is balancing the pressure exerted by a column of mercury 75cm high in a barometer tube. The pressure due to the atmosphere at any point is proportional to the height of the column of air above that point. This implies that the difference in the atmospheric pressure between the pilot's location and the ground observer's location is directly proportional to the height difference between the two points. Using the given data, the pressure difference between the two points is 75 - 63 = 12 cm of Hg. Let h be the height of the plane above the ground, then by the principle of hydrostatics, the pressure difference is also given by: h = (pressure difference) x (density of mercury/density of air) x (gravitational acceleration) Substituting the given values and solving for h, we have: h = 12 x (13.6/0.00136) x (9.81) = 1200m Therefore, the height of the plane above the ground is 1200m. Answer: 1200m

Karanta bayanin darasi akan Pressure (JAMB)

Pressure

Duba Amsa

Tambaya 35 Rahoto

A 90-W immersion heater is used to supply energy for five minutes. The energy supplied is used to completely melt 180g of a solid at is melting point. Neglecting energy losses to the surroundings, calculate the specific latent heat of fusion of the solid

0.5Jg^-1

2.5Jg^-1 Amsar da ta dace

15.0Jg^-1

450.0Jg^-1

150Jg^-1

Bayanin Amsa

Karanta bayanin darasi akan Quantity Of Heat (JAMB)

Quantity Of Heat

Duba Amsa

Tambaya 36 Rahoto

The term torque means

the moment Amsar da ta dace

the resultant of several forces acting on a body in equilibrum

two equal and oposite forces whose lines of action do not coincide

two coplaner forces at right angles to each other

the equilibrant of two concurrent forces

Bayanin Amsa

Karanta bayanin darasi akan Scalars And Vectors (JAMB)

Karanta bayanin darasi akan Equilibrium Of Forces (WAEC)

Scalars And Vectors

Equilibrium Of Forces

Duba Amsa

Tambaya 37 Rahoto

If \(\theta\) is equal to \(\alpha\) at the time the box is just about to slide down the plank, the coefficient of static friction between the plank and box is

sin\(\alpha\)

cos\(\alpha\)

tan\(\alpha\) Amsar da ta dace

sec\(\alpha\)

cot\(\alpha\)

Bayanin Amsa

When the box is just about to slide down the plank, the force of static friction acting on the box is equal and opposite to the component of the weight of the box parallel to the surface of the plank. This component is given by mg sin\(\alpha\), where m is the mass of the box, g is the acceleration due to gravity and \(\alpha\) is the angle of inclination of the plank. Since \(\theta\) is equal to \(\alpha\), the angle of inclination of the plank is also \(\theta\). Therefore, the coefficient of static friction \(\mu\) is given by: \(\mu = \frac{\text{Force of friction}}{\text{Normal force}} = \frac{mg \sin \theta}{mg \cos \theta} = \tan \theta = \tan \alpha\) Hence, the correct option is tan\(\alpha\).

Karanta bayanin darasi akan Friction (JAMB)

Friction

Duba Amsa

Tambaya 38 Rahoto

Which of the following explains the concave meniscus of water in a clean glass tube? the

adhension between water and glass molecules is greater than the sdhension between water molecules Amsar da ta dace

cohension between water molecules is greater then the adhension between glass and water molecule

molecules of water near the glass move faster than the molecules at the centre of the tube

molecules of water at the water-air boundary are often attracted to the centre of the tube

weight of the water pulls the centre part of the surface down

Bayanin Amsa

The concave meniscus of water in a clean glass tube is due to the adhesive forces between the water molecules and the glass tube being greater than the cohesive forces between the water molecules themselves. This causes the water to be attracted to the glass, resulting in the formation of a concave meniscus. Therefore, the correct option is "adhesion between water and glass molecules is greater than the cohesion between water molecules."

Karanta bayanin darasi akan Liquids At Rest (JAMB)

Liquids At Rest

Duba Amsa

Tambaya 39 Rahoto

The images formed by diverging lenses are always

diminished, virtual and inverted

diminished, virtual and errect Amsar da ta dace

diminished inverted and real

magnidied, virtual and errect

magnified, real and inverted

Bayanin Amsa

Diverging lenses are also known as concave lenses. These lenses are thinner at the centre and thicker at the edges. The images formed by diverging lenses are always diminished, virtual and erect. Diminished means that the image is smaller than the object. Virtual means that the image appears to be behind the lens and cannot be projected on a screen. Erect means that the image appears to be in the same orientation as the object. The reason for these characteristics is that the diverging lens spreads out the light rays that enter it, causing them to diverge or move apart. This creates an image that is smaller and appears to be located behind the lens.

Karanta bayanin darasi akan Properties Of Waves: Reflection, Refraction, Diffraction (WAEC)

Karanta bayanin darasi akan Electrostatics (JAMB)

Properties Of Waves: Reflection, Refraction, Diffraction

Electrostatics

Duba Amsa

Tambaya 40 Rahoto

Calculate the resistance of the filament of lamp rated at 240V 60W

240\(\Omega\)

360\(\Omega\)

960\(\Omega\) Amsar da ta dace

144\(\Omega\)

2880\(\Omega\)

Bayanin Amsa

The power rating of the lamp is given as 60W, and the voltage rating is given as 240V. We can use the formula for power, which is P = V^2/R, to calculate the resistance of the filament. First, we rearrange the formula to solve for resistance R: R = V^2 / P Plugging in the given values, we get: R = (240V)^2 / 60W Simplifying, we get: R = 960\(\Omega\) Therefore, the resistance of the filament of the lamp is 960\(\Omega\).

Karanta bayanin darasi akan Current Electricity (JAMB)

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

Current Electricity

Work, Energy And Power

Duba Amsa

Tambaya 41 Rahoto

The average kinetic energy of the molecules of a perfect gas is directly proportional to the

pressure exerted by the gas

volume of the gas molecule

kelvin temperature of the gas

volume of the gas container

celcius temperature of the gas Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Karanta bayanin darasi akan Measurements And Units (JAMB)

Temperature And Its Measurement

Measurements And Units

Duba Amsa

Tambaya 42 Rahoto

How much heat is emitted when a body of mass 200g cools from 37^oC to 31^oC? (specific heat capacity of the body = 0.4Jg^-1K^-1,)

4800J

1200J

480J Amsar da ta dace

202J

80J

Bayanin Amsa

Karanta bayanin darasi akan Heat Energy (WAEC)

Heat Energy

Duba Amsa

Tambaya 43 Rahoto

If an ebonite rod is rubbed with fur

both the ebonite rod and the fur will be negatively charged

the ebonite rod will be positively charged while the fur will have no charge

there will be electron transfer only if the ebonite rod is earthed

the ebonite rod will be negatively charged while the fur will be positively charged Amsar da ta dace

the ebonite rod will be positively charged while the fur will be negatively charged

Bayanin Amsa

When an ebonite rod is rubbed with fur, electrons are transferred from the fur to the ebonite rod. Electrons are negatively charged, so when electrons move from one object to another, it leaves the object with a positive charge, and the other with a negative charge. Therefore, the ebonite rod will be negatively charged while the fur will be positively charged. Thus, the correct answer is: "the ebonite rod will be negatively charged while the fur will be positively charged".

Karanta bayanin darasi akan Electrostatics (JAMB)

Electrostatics

Duba Amsa

Tambaya 44 Rahoto

1.33

1.25

0.80

0.75

0.60 Amsar da ta dace

Bayanin Amsa

Duba Amsa

Tambaya 45 Rahoto

A piece of metal is heated until it becomes red-hot, it is then quickly transferred into a beaker containing boiling water. What effect will this have on the boiling water?

the boiling water rises slightly

the boiling point rises sharply

the rate of vaporisation of the boiling water increases Amsar da ta dace

the boiling point first rises and then falls

neither the boiling point nor the rate of vaporisation increases

Bayanin Amsa

Karanta bayanin darasi akan Heat Energy (WAEC)

Heat Energy

Duba Amsa

Tambaya 46 Rahoto

Which of the following are both mechanical and transverse?

infra-red rays

gamma rays

sound waves

water waves Amsar da ta dace

micro waves

Bayanin Amsa

Karanta bayanin darasi akan Sound Waves (WAEC)

Karanta bayanin darasi akan Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Sound Waves

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Duba Amsa

Tambaya 47 Rahoto

If the cubic expansivity of brass between 27^oC and 100^oC is 5.7 x 10^-5K^-1, What is its linear expansivity?

2.85 x 10^-5K^-1

1.90 x 10^-5K^-1 Amsar da ta dace

1.86 x 10^-5K^-1

1.70 x 10^-5K^-1

1.63 x 10^-5K^-1

Bayanin Amsa

Karanta bayanin darasi akan Elasticity (JAMB)

Elasticity

Duba Amsa

Tambaya 48 Rahoto

A sonometer wire under a tension of N10N, produces a frequency of 250Hz when plucked. Keeping the length of the wire constant, the tension is adjusted to produce a new frequency of 350Hz. Calculate the new tension

39.2N

19.6N Amsar da ta dace

14.2N

7.4N

5.1N

Bayanin Amsa

The relationship between tension (T) and frequency (f) for a string fixed at both ends, such as a sonometer wire, is given by the equation: f = (1/2L)√(T/μ) where L is the length of the string, μ is the linear mass density of the string. Since the length of the wire is kept constant, we can equate the two expressions for frequency to find the relationship between the two tensions: (1/2L)√(T1/μ) = f1 = 250 Hz (1/2L)√(T2/μ) = f2 = 350 Hz Dividing the two equations gives: (f2/f1) = √(T2/T1) (350/250) = √(T2/N10N) Squaring both sides of the equation gives: (350/250)^2 = T2/N10N Solving for T2, we get: T2 = (350/250)^2 x N10N T2 = 19.6 N (to 2 significant figures) Therefore, the new tension required to produce a frequency of 350 Hz is 19.6 N. Answer option (B) is correct.

Duba Amsa

Tambaya 49 Rahoto

A cell of e.m.f. 1.5v is connected in series with a resistor of 3\(\Omega\). A high resistance voltmeter connected across the cell registers only 0.9v. Calculate the internal resistance of the cell

5.0\(\Omega\) Amsar da ta dace

4.5\(\Omega\)

2.4\(\Omega\)

2.0\(\Omega\)

1.8\(\Omega\)

Bayanin Amsa

Karanta bayanin darasi akan Electric Cells (JAMB)

Electric Cells

Duba Amsa

Tambaya 50 Rahoto

Two radioactive elements X and Y have half-lives of 100 and 50 years respectively. Samples of X and Y initially contain equal number of atoms. What is the ratio of the number of the remaining atoms of X to that of Y after 200 years?

4:1

3:1

1:1

1:2 Amsar da ta dace

1:4

Bayanin Amsa

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

Karanta bayanin darasi akan Wave-particle Paradox (WAEC)

Magnetic Field (Part 2)

Wave-particle Paradox

Duba Amsa

Tambaya 51 Rahoto

An electromagnetic wave of frequency 5.0 x 10¹⁴Hz, is incident on the surface of water of refractive index \(\frac{4}{3}\). Taking the speed of the wave in air as 3.0 x 10⁸ms^-1, calculate the wavelength of the wave in water.

2.2 x 10^-6m

1.7 x 10^-7m

8.0 x 10^-7m

6.0 x 10^-7m

4.5 x 10^-7m Amsar da ta dace

Bayanin Amsa

The formula for calculating the wavelength of a wave is given by: \begin{equation*} \lambda = \frac{v}{f} \end{equation*} where - \(\lambda\) is the wavelength of the wave - v is the velocity of the wave in the medium - f is the frequency of the wave. The velocity of the electromagnetic wave in air is given as 3.0 x 10⁸ms^-1. The refractive index of water is given as \(\frac{4}{3}\). The refractive index is the ratio of the speed of light in a vacuum to the speed of light in a medium. Hence the speed of the wave in water is given as: \begin{equation*} v_{water} = \frac{v_{air}}{n} = \frac{3.0\times10^8}{4/3} = 2.25\times10^8ms^{-1} \end{equation*} Substituting the values of v and f in the formula above, we have: \begin{equation*} \lambda = \frac{v}{f} = \frac{2.25\times10^8}{5.0\times10^{14}} = 4.5\times10^{-7}m \end{equation*} Therefore, the wavelength of the electromagnetic wave in water is 4.5 x 10^-7m. Hence the correct option is: 4.5 x 10^-7m.

Karanta bayanin darasi akan Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Duba Amsa

Tambaya 52 Rahoto

What is the image distance of an object placed at a distance of 2f from a converging lens of focal length f?

f/4

f/2

2f Amsar da ta dace

Bayanin Amsa

The image distance of an object placed at a distance of 2f from a converging lens of focal length f can be found using the lens formula: 1/f = 1/v - 1/u where f is the focal length of the lens, v is the image distance, and u is the object distance. Since the object is placed at a distance of 2f, u = 2f. Substituting these values in the lens formula, we get: 1/f = 1/v - 1/2f Multiplying both sides by 2f, we get: 2 - f/f = v Simplifying this equation, we get: v = 2f Therefore, the image distance of an object placed at a distance of 2f from a converging lens of focal length f is 2f.

Duba Amsa

Tambaya 53 Rahoto

The inside of a vacuum flak is usually coated with silver to reduce heat loss by

evaporation

radiation Amsar da ta dace

conduction

convention

condensation

Bayanin Amsa

The correct answer is **radiation**. Heat can be transferred through vacuum by radiation, which is the emission of electromagnetic waves from a body due to its temperature. A vacuum flask consists of two walls of glass with a vacuum in between them, and this vacuum is a poor conductor of heat, which reduces heat loss by conduction and convection. However, the inner wall of a vacuum flask can still lose heat by radiation. Hence, to reduce heat loss by radiation, the inside of a vacuum flask is usually coated with a reflective surface, typically silver, which reflects heat back into the flask, keeping the contents warmer for longer.

Karanta bayanin darasi akan Heat Transfer (JAMB)

Karanta bayanin darasi akan Heat Energy (WAEC)

Heat Transfer

Heat Energy

Duba Amsa

Tambaya 54 Rahoto

The lower and upper fixed points of a mecury-in-glass thermometer are marked Y and 180mm respectively. On a particular the mecury meniscus in the thermometer rises to 45mm. If the corresponding reading on a scale is 10^o, calculate the value of Y.

4mm

18mm

30mm Amsar da ta dace

135mm

450mm

Bayanin Amsa

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Duba Amsa

Tambaya 55 Rahoto

A radioactive nuclide of proton number X emits a \(\beta\)-particle to form a new nuclide to proton number Y. The correct equation relating X and Y is

X = Y - 1

X = Y + 1 Amsar da ta dace

X = 1 - Y

X = \(\frac{1}{y}\)

X = Y

Bayanin Amsa

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

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 56 Rahoto

When light from a source is sent through a gas. i. certain wavelengths are absorbed. ii. a dark line is left. Which of the following is/are correct?

i only

iii only

i and ii only Amsar da ta dace

i and iii only

ii and iii only

Bayanin Amsa

Karanta bayanin darasi akan Structure Of The Nucleus (WAEC)

Structure Of The Nucleus

Duba Amsa

Tambaya 57 Rahoto

The gravitational force of the moon is one-sixth that of the earth. If a body weighs 6.0N on the moon. Calculate its weight on the earth

36.0N Amsar da ta dace

12.0N

6.0N

2.0N

1.0N

Bayanin Amsa

Weight is the force experienced by a body due to gravity. It is given by the product of mass and gravitational acceleration, which is 9.8m/s² on earth and 1.6m/s² on the moon. Let's assume the mass of the body to be m kg. On the moon, weight, W_m = mg_m = 6.0N, where g_m = 1.6m/s² is the gravitational acceleration on the moon. On the earth, weight, W_e = mg_e, where g_e = 9.8m/s² is the gravitational acceleration on the earth. We are required to find W_e, given that the gravitational force of the moon is one-sixth that of the earth, which means the gravitational acceleration on the moon is one-sixth that of the earth. So, g_m = (1/6)g_e Substituting this value of g_m in the equation for weight on the moon, we get: W_m = mg_m = m(1/6)g_e = (1/6)(mg_e) Hence, weight of the body on earth, W_e = 6W_m = 6*(1/6)(mg_e) = mg_e Therefore, W_e = 6.0N * 9.8m/s² = 58.8N Hence, the weight of the body on earth is 58.8N. Therefore, the answer is 36.0N.

Karanta bayanin darasi akan Gravitational Field (JAMB)

Gravitational Field

Duba Amsa






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