JAMB UTME - Physics - 1993 | Green Bridge CBT

Tambaya 1 Rahoto

Two mirrors M1 and M2 are inclined at right angles as shown above calculate the angle of reflection of the ray of light at M2

30∘

45∘

60∘

90∘

Electrostatics

Duba Amsa

Tambaya 2 Rahoto

A rocket burns fuel at the rate of 10kgs-1 and ejects it with velocity of 5 x 103ms-1. The thrust exerted by the gas on the rock is

2.5 x 107N

5.0 x 104N

5.0 x 102N

2.0 x 10-3N

Bayanin Amsa

The thrust exerted by the gas on the rocket can be calculated using the formula: Thrust = Mass flow rate x Exhaust velocity Here, the mass flow rate is given as 10 kg/s and the exhaust velocity is given as 5 x 10^3 m/s. Plugging in the values, we get: Thrust = 10 kg/s x 5 x 10^3 m/s = 50,000 N Therefore, the thrust exerted by the gas on the rocket is 5.0 x 10^4 N, which is.

Karanta bayanin darasi akan Friction (JAMB)

Friction

Duba Amsa

Tambaya 3 Rahoto

Which of the following may be used to explain a mirage? i. layers of air near the road surface have varying refractive indices in hot weather. ii. road surfaces sometimes become good reflectors in hot weathers. iii. light from the sky can be reflected upwards after coming close to the road surface.

ii only

ii and iii only

i and iii only

i, ii, and iii only

Bayanin Amsa

A mirage is a naturally occurring optical phenomenon in which light rays are refracted, or bent, to produce a displaced image of distant objects or the sky. The answer to this question is "i and iii only." A mirage is caused by the refraction of light in layers of air with varying temperatures. In hot weather, the air near the ground is hotter than the air above it, causing light to refract and bend upwards. This creates the appearance of an image or object that is not actually there. This explains option i. Option ii, which states that road surfaces become good reflectors in hot weather, is not a factor in the formation of a mirage. Option iii is also correct because light from the sky can be reflected upwards after coming close to the road surface, which can contribute to the formation of a mirage.

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 4 Rahoto

The property of the eye known as its power of accommodation is controlled by the

pupil

virteous humour

iris

ciliary muscles

Bayanin Amsa

The property of the eye known as its power of accommodation is controlled by the ciliary muscles. These muscles are responsible for changing the shape of the lens in the eye, allowing it to focus on objects at different distances. When we look at something up close, the ciliary muscles contract, causing the lens to thicken and increase its refractive power. When we look at something far away, the ciliary muscles relax, causing the lens to flatten and decrease its refractive power. This process of changing the shape of the lens to adjust focus is known as accommodation, and it is essential for clear vision at different distances.

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

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 5 Rahoto

A 40-W instrument has a resistance of $90\ \Omega$. On what voltage should it be operated normally?

60V

150V

225V

3600V

Bayanin Amsa

To calculate the voltage required to operate the 40-W instrument with a resistance of 90 Ω, we can use Ohm's law, which states that the voltage (V) equals the product of the current (I) and the resistance (R): V = I × R. First, we need to calculate the current that flows through the instrument. We can use the formula for power (P) in terms of current and resistance: P = I^2 × R. Rearranging this equation, we get: I = sqrt(P / R). Substituting the values given in the problem, we get: I = sqrt(40 W / 90 Ω) ≈ 0.632 A Now we can use Ohm's law to find the voltage: V = I × R = 0.632 A × 90 Ω ≈ 56.9 V Therefore, the voltage required to operate the instrument normally is approximately 56.9 volts. (60V) is the closest to this value. Note that this is a simple example of how to apply Ohm's law to a practical problem. In real-life situations, the calculation may be more complicated, and other factors such as the temperature coefficient of resistance may need to be considered.

Karanta bayanin darasi akan Current Electricity (JAMB)

Current Electricity

Duba Amsa

Tambaya 6 Rahoto

The relationship between L and the wavelength I of the stationary wave is

2L =

λ

L =

λ

L =

\frac{λ}{3}

4L = 3

λ

Bayanin Amsa

L = -

λ

\Rightarrow

4L = 3

λ

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

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 7 Rahoto

In the network shown above, determine the potential difference across the 5µF capacitor

3V

6V

12V

18V

Duba Amsa

Tambaya 8 Rahoto

A parallel plate capacitor of area $10\text{cm}^2$ in vacuum has a capacitance of $10^{-2}\,\mu\text{F}$. What is the distance between the plates? $(e = 9 x 10^{-12}\text{fm})$

9 x 10-13m

9 x 10-7m

9 x 10-3m

9 x 107m

Bayanin Amsa

C =

\frac{ε_{o} A}{d}

10-2 x 10-6 = $\frac{9 \times 10^{- 12}}{d}$ x 10-4

= 9 x 10-7m

Karanta bayanin darasi akan Capacitors (JAMB)

Capacitors

Duba Amsa

Tambaya 9 Rahoto

In the figure above,A lever of length 200m is used to lift a load of mass 180kg. The pivot at P is 20m from the load. what minimum force F must be applied at the end of the lever [g = 10ms-2]

9N

18N

20N

200N

Bayanin Amsa

To solve this problem, we can use the principle of moments, which states that the sum of the clockwise moments about a point is equal to the sum of the anticlockwise moments about the same point. In this case, we can take moments about the pivot point P. Let's first calculate the weight of the load: Weight = mass x acceleration due to gravity Weight = 180 kg x 10 m/s^2 Weight = 1800 N Now, let's calculate the anticlockwise moment of the load about point P: Anticlockwise moment = weight x perpendicular distance from pivot to line of action of weight Anticlockwise moment = 1800 N x 20 m Anticlockwise moment = 36000 Nm To balance this moment, we need a clockwise moment produced by the force F applied at the end of the lever. The perpendicular distance from the pivot to the line of action of this force is 200 m - 20 m = 180 m. So, we can calculate the minimum force required as follows: Minimum force = clockwise moment / perpendicular distance from pivot to line of action of force Minimum force = 36000 Nm / 180 m Minimum force = 200 N Therefore, the minimum force required to lift the load is 200 N, which corresponds to option (D).

Karanta bayanin darasi akan Measurements And Units (JAMB)

Measurements And Units

Duba Amsa

Tambaya 10 Rahoto

The thermometric properties of the thermocouple is that is

e.m.f. changes with temperature

resistance changes with temperature

volume changes with temperature

pressure changes with resistance

Bayanin Amsa

The correct thermometric property of a thermocouple is that its "e.m.f. changes with temperature." A thermocouple is a device used to measure temperature, and it works by generating a small electrical voltage in response to changes in temperature. This voltage is called an electromotive force or "e.m.f." The amount of voltage generated by a thermocouple depends on the difference in temperature between its two junctions. As the temperature changes, the e.m.f. also changes, and this change can be measured and used to determine the temperature. The other options given in the question are not correct for a thermocouple. Resistance, volume, and pressure can all change with temperature, but they are not the primary properties that are used to measure temperature with a thermocouple.

Karanta bayanin darasi akan Electrostatics (JAMB)

Electrostatics

Duba Amsa

Tambaya 11 Rahoto

Determine the distance between the consecutive antinodes XX if the wavelength is 66cm

15cm

30cm

60cm

120cm

Bayanin Amsa

Distance XX =

\frac{1}{2}

λ

= $\frac{1}{2}$ x $\frac{60}{1}$

= 30cm

Karanta bayanin darasi akan Propagation Of Sound Waves (JAMB)

Propagation Of Sound Waves

Duba Amsa

Tambaya 12 Rahoto

An element X of an atomic number 88 and mass number 226 decays to form an element Z by emitting two beta particles and an alpha particle. Z is represented by

22282Z

22288Z

22686Z

_{80}^{226} Z

Bayanin Amsa

The atomic number of an element is the number of protons in its nucleus, while the mass number is the sum of protons and neutrons. In this question, the element X has an atomic number of 88 and a mass number of 226, which means it has 88 protons and (226-88) = 138 neutrons. When element X emits two beta particles and an alpha particle, it undergoes radioactive decay. A beta particle is an electron emitted from the nucleus, while an alpha particle is a particle consisting of two protons and two neutrons. The emission of a beta particle increases the atomic number by 1, while the emission of an alpha particle decreases the atomic number by 2 and the mass number by 4. Therefore, after emitting two beta particles and an alpha particle, the new element Z will have an atomic number of (88+2-2) = 88 and a mass number of (226+4-2x2) = 222. The symbol for this element Z is 22288Z, which means it has 88 protons and (222-88) = 134 neutrons. Therefore, the correct answer is 22288Z.

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

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 13 Rahoto

A beam of light is incident from air to water at an angle of $30^\circ\text{C}$. Find the angle of refraction if the refractive index of water is $\frac{4}{3}$

15oC

18oC

22oC

24oC

Bayanin Amsa

\frac{4}{3}

=

\frac{s i n 30}{s i n r}

r = 22oC

Duba Amsa

Tambaya 14 Rahoto

A metal rod 800mm long is heated from 10oC to 95oC. If it expands by 1.36mm, the linear expansivity of the metal is

2.0 x 102K-1

6.0 x 102K-1

5.0 x 10-3K-1

2.0 x 10-5K-1

Bayanin Amsa

The linear expansivity (α) of a material is defined as the change in length per unit length per degree Celsius (or Kelvin) change in temperature. In this problem, we are given the initial length of the metal rod (800 mm), the change in temperature (∆T = 95°C - 10°C = 85°C), and the change in length (∆L = 1.36 mm). We can use the formula: ∆L = αL∆T where L is the initial length of the rod. Solving for α, we get: α = ∆L / (L∆T) Substituting the given values, we have: α = (1.36 mm) / (800 mm x 85°C) Simplifying the denominator: α = (1.36 mm) / 68000 mm°C α = 0.00002°C^-1 Therefore, the linear expansivity of the metal is 2.0 x 10^-5 K^-1. Answer: (D) 2.0 x 10^-5 K^-1.

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Duba Amsa

Tambaya 15 Rahoto

A current carrying conductor experiences a force when placed in a magnetic field because the

conductor is magnetized

magnetic field of the current interacts with external magnetic field

force is due to the motor principle

electric field of the current interacts with the external magnetic field

Bayanin Amsa

The correct option is: "magnetic field of the current interacts with external magnetic field". When a current-carrying conductor is placed in a magnetic field, the magnetic field of the current interacts with the external magnetic field. This interaction creates a force that causes the conductor to move. This phenomenon is known as the Lorentz force. The direction of the force is perpendicular to both the direction of the current and the external magnetic field. The magnitude of the force depends on the strength of the magnetic field, the amount of current flowing through the conductor, and the length of the conductor that is in the magnetic field.

Karanta bayanin darasi akan Force On A Current-Carrying Conductor In A Magnetic Field (JAMB)

Force On A Current-Carrying Conductor In A Magnetic Field

Duba Amsa

Tambaya 16 Rahoto

The external and internal diameter of a tube are measured as (32 ± 2)mm and (21 ± 1)mm respectively. Determine the percentage error in the thickness of the tube.

27%

14%

9%

3%

Bayanin Amsa

Actual thickness = 34 - 22 = 12

error thickness = 30 - 20 = 10

error = 12 - 10 = 2mm that is, (11 ± 1)mm

%error = $\frac{1}{11}$ x $\frac{100}{1}$

= 9%

Karanta bayanin darasi akan Measurements And Units (JAMB)

Measurements And Units

Duba Amsa

Tambaya 17 Rahoto

200g of water at 90oC is mixed with 100g of water at 30oC. What is the final temperature?

50oC

60oC

70oC

80oC

Bayanin Amsa

Heat lost = heat gained

200 x c x (90 - $θ$ )

$θ$ = 70oC

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Duba Amsa

Tambaya 18 Rahoto

Which of the following quantities has the same unit as the watt?

Force x time

Force x distance

Force x acceleration

Force x velocity

Bayanin Amsa

The watt is a unit of power, which represents the rate at which energy is transferred or work is done. It is equal to one joule per second. Out of the given options, only "force x velocity" has the same unit as the watt, which is joules per second. This is because force times velocity gives us the unit of momentum, which is kilogram meters per second (kg m/s), and dividing this by time (in seconds) gives us the unit of power, which is joules per second (J/s) or watts (W). In contrast, "force x time" gives us the unit of impulse (newton seconds), "force x distance" gives us the unit of work or energy (joules), and "force x acceleration" gives us the unit of force times acceleration, which is not a unit of power.

Duba Amsa

Tambaya 19 Rahoto

An object of mass 400g and density 600kgm-3 is suspended with a sting so that half of it is immersed in paraffin of density 900kgm-3. The tension in the string is

1.0N

3.0N

4.0N

5.0N

Bayanin Amsa

volume of object =

\frac{400 \times 10^{- 2}}{600}

= 0.00067m3

volume of paraffin displaced = $\frac{1}{2}$ x 0.00067

= 0.00034m3

mass of paraffin = 900 x 34 x 10-5

= 0.3kg

weight of paraffin = 0.3 x 10

= 3N = upthrust = tension in the string

Karanta bayanin darasi akan Measurements And Units (JAMB)

Measurements And Units

Duba Amsa

Tambaya 20 Rahoto

Radio waves emitted from an antenna are picked up by a radar after reflection from an aircraft in 4 x 10-3S. How far is the aircraft from the antenna? [c = 3 x 108ms-1 ]

6.0 x 102ms-1

1.2 x 103km

3.0 x 103km

6.0 x 10 5km

Bayanin Amsa

Time = 1 x 4 x 10-3sec

distance = velocity x time

= 3 x 108 x 4 x 10-3

= 6 x 105m

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

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Duba Amsa

Tambaya 21 Rahoto

The primary aim in high tension transmission is to

minimize electrical energy losses due to heat production

increase the rate of energy transfer by using high voltage

increase the current in the wire

generate electricity at high current and low voltage

Electric Cells

Duba Amsa

Tambaya 22 Rahoto

A simple cell with mercury-amalgamated zinc electrode prevents

local actions

polarization

buckling

degradation

Electric Cells

Electrostatics

Duba Amsa

Tambaya 23 Rahoto

Which of the following correctly explain(s) why a green leaf appears green in bright day light? i. it absorbs only the green component of sunlight. ii. it absorbs all colour of sunlight except green. iii. it reflects only the green component of sunlight.

i only

ii only

i and iii only

ii and iii only

Properties Of Waves: Reflection, Refraction, Diffraction

Duba Amsa

Tambaya 24 Rahoto

Which of the following waves can propagate through a vacuum?

High velocity sound waves

Ultrasonic waves

Acoustic waves

Infra-red waves

Bayanin Amsa

Infra-red waves can propagate through a vacuum. Unlike sound waves, which require a medium to propagate, and ultrasonic and acoustic waves, which are also sound waves, infra-red waves are electromagnetic waves that can travel through a vacuum. This is because electromagnetic waves consist of oscillating electric and magnetic fields that do not require a medium to propagate.

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

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Duba Amsa

Tambaya 25 Rahoto

The terminal voltage of a battery is 4.0V when supplying a current of 2.0A, and 2.0V when supplying a current of 3.0A. The internal resistance of the battery is

0.5

Ω

1.0

Ω

2.0

Ω

4.0

Ω

Bayanin Amsa

The internal resistance of the battery can be calculated using the formula: R = (V1 - V2) / (I2 - I1) Where: V1 = voltage of the battery when supplying current I1 V2 = voltage of the battery when supplying current I2 I1 = current supplied to the battery when measuring V1 I2 = current supplied to the battery when measuring V2 Substituting the given values in the above formula, we get: R = (4.0V - 2.0V) / (3.0A - 2.0A) = 2.0Ω Therefore, the internal resistance of the battery is 2.0Ω. Explanation: The internal resistance of a battery is the resistance offered by the electrolyte inside the battery to the flow of current. When a battery is connected to an external circuit, the current flows through the external circuit and through the internal resistance of the battery. The voltage drop across the internal resistance reduces the terminal voltage of the battery. By measuring the terminal voltage of the battery at two different currents, we can calculate the internal resistance of the battery using the formula mentioned above.

Karanta bayanin darasi akan Current Electricity (JAMB)

Current Electricity

Duba Amsa

Tambaya 26 Rahoto

Which of the following expressions gives the linear magnification produced by a concave mirror of radius of curvature r, if U and V are the object and image distances respectively?

\frac{V}{r}

- 1

\frac{2 V}{r}

- 1

\frac{U}{r}

- 1

\frac{2 U}{r}

- 1

Bayanin Amsa

The expression that gives the linear magnification produced by a concave mirror of radius of curvature r, if U and V are the object and image distances respectively is "-V/U" or "V/U = -h'/h", where "h'" is the height of the image and "h" is the height of the object. This is a well-known formula in optics, which states that the magnification produced by a mirror is equal to the ratio of the height of the image to the height of the object. Moreover, the negative sign indicates that the image is inverted with respect to the object. None of the options provided match this formula exactly. Option "Vr" and "V/r-1" involve only the image distance and the radius of curvature, but not the object distance, which is necessary to calculate the magnification. Option "Ur" and "U/r-1" involve only the object distance and the radius of curvature, but not the image distance. Finally, option "2Vr" and "2U/r-1" involve a factor of 2 that is not present in the formula for magnification.

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Karanta bayanin darasi akan Gravitational Field (WAEC)

Temperature And Its Measurement

Gravitational Field

Duba Amsa

Tambaya 27 Rahoto

The kinetic theory of photoelectron liberated from a metallic surface depends on the

intensity of the incident radiation

time duration of the incident radiation

temperature of the incident radiation

frequency of the incident radiation

Bayanin Amsa

The kinetic theory of photoelectrons liberated from a metallic surface depends on the frequency of the incident radiation. When a photon with enough energy (frequency) hits a metal surface, it can knock loose an electron, creating a photoelectron. The kinetic energy of the photoelectron will depend on the energy of the incident photon, which is directly related to its frequency. Therefore, higher frequency radiation will result in photoelectrons with greater kinetic energy. The intensity of the incident radiation and the time duration of the incident radiation can affect the total number of photoelectrons released from the metal surface, but not their individual kinetic energy. The temperature of the incident radiation, on the other hand, is not a significant factor in the kinetic theory of photoelectrons.

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

Karanta bayanin darasi akan Thermal Expansion (JAMB)

Electromagnetic Field (Part 2)

Thermal Expansion

Duba Amsa

Tambaya 28 Rahoto

When the volume of a given mass of gas is halved and its temperature doubled, the pressure

remains constant

increases by a factor of 4

increases by a factor of 3

decreases by a factor of 4

Bayanin Amsa

According to the ideal gas law, pressure (P), volume (V), and temperature (T) of a gas are related as follows: P x V = n x R x T, where n is the number of moles of the gas and R is a constant. If the volume of a gas is halved (V/2) and its temperature is doubled (2T), then the new values of P and V can be found by plugging these new values into the ideal gas law: P x (V/2) = n x R x (2T) Simplifying this equation, we get: P/2 = n x R x 2T/V Multiplying both sides by 2, we get: P = n x R x 2T/V Since V is halved, the fraction 2T/V becomes twice as large. Therefore, the pressure (P) of the gas increases by a factor of 2 x 2 = 4. So, the correct option is "increases by a factor of 4".

Karanta bayanin darasi akan Gas Laws (JAMB)

Gas Laws

Duba Amsa

Tambaya 29 Rahoto

Which of the following statements correctly describe(s) cathode rays? i. they consist of tiny particles carrying negative electric charges. ii. they are deflected in a magnetic field but not in an electric field. iii. they consist of fast moving neutrons and deflected in an electric field.

i only

ii only

i and ii only

ii and iii only

Structure Of The Atom (Nigeria Only)

Electrostatics

Duba Amsa

Tambaya 30 Rahoto

A plane inclined at angle $ \theta $ has a velocity ratio of 10 : 1. The inclination of the plane to the horizontal is given by

tan

θ

=

\frac{1}{10}

cot

θ

=

\frac{1}{10}

cos

θ

=

\frac{1}{10}

sin

θ

=

\frac{1}{10}

Bayanin Amsa

V.R =

\frac{1}{s i n θ}

= 10

sin = $\frac{1}{10}$

Karanta bayanin darasi akan Scalars And Vectors (JAMB)

Scalars And Vectors

Duba Amsa

Tambaya 31 Rahoto

A current of 100mA passes through a conductor for 2 minutes. The quantity of electricity transported is

200C

50C

12C

0.002C

Bayanin Amsa

The quantity of electricity transported is given by the formula Q = I x t, where Q is the quantity of electricity, I is the current, and t is the time. In this case, the current is 100mA (which is 0.1A) and the time is 2 minutes (which is 120 seconds). So we can plug these values into the formula: Q = I x t = 0.1A x 120s = 12C Therefore, the quantity of electricity transported is 12C. Option C is the correct answer.

Karanta bayanin darasi akan Current Electricity (JAMB)

Karanta bayanin darasi akan Heat Energy (WAEC)

Karanta bayanin darasi akan Light Waves (WAEC)

Current Electricity

Heat Energy

Light Waves

Duba Amsa

Tambaya 32 Rahoto

The force experienced by an object of mass 60.0kg in the moon's gravitational field is 1.002 x 102N. What is the intensity of the gravitational field?

0.60Nkg-1

1.67Nkg-1

6.12 x 102Nkg-1

9.81ms-2

Bayanin Amsa

The force experienced by an object of mass 60.0kg in the moon's gravitational field is 1.002 x 10^2 N. To find the intensity of the gravitational field, we can use the formula: force = mass x intensity of gravitational field Rearranging this formula gives: intensity of gravitational field = force / mass Plugging in the given values, we get: intensity of gravitational field = 1.002 x 10^2 N / 60.0 kg Simplifying this expression gives: intensity of gravitational field = 1.67 N/kg Therefore, the correct answer is: 1.67N/kg. In simpler terms, the intensity of the gravitational field is the amount of gravitational force per unit mass that an object experiences in that field. The greater the intensity, the stronger the gravitational force, and the more an object will be pulled towards the source of gravity.

Karanta bayanin darasi akan Gravitational Field (JAMB)

Gravitational Field

Duba Amsa

Tambaya 33 Rahoto

The current I in the fig above is

4.00A

1.30A

0.80A

0.75A

Current Electricity

Simple A.C Circuits

Duba Amsa

Tambaya 34 Rahoto

The diagram above shows a velocity time graph representing the motion of a car. Find the total distance covered during the acceleration and retardation of the motion

75m

150m

300m

375m

Motion

Duba Amsa

Tambaya 35 Rahoto

Which of the following are produced after a nuclear fusion process? i. one heavy nucleus. ii. neutrons-1. iii. protons. iv. Energy.

i and ii

i and iv

ii and iii

ii and iv

Magnetic Field (Part 2)

Duba Amsa

Tambaya 36 Rahoto

Which of the following devices may be used to step up the voltage in a d.c. circuit?

a step-up transformer

a d.c. generator

a wattmeter

an induction coil

Electric Field

Electromagnetic Field (Part 2)

Duba Amsa

Tambaya 37 Rahoto

Which of the following correctly describes the energy changes in the generation of light by a hydroelectric power station?

Electrical

\to

mechanical

\to

potential

\to

mechanical

\to

electrical

\to

light

mechanical

\to

sound

\to

electrical

\to

light

kinetic

\to

mechanical

\to

electrical

\to

light

Bayanin Amsa

The correct option that describes the energy changes in the generation of light by a hydroelectric power station is: "Potential energy → mechanical energy → electrical energy → light." This is because a hydroelectric power station converts the potential energy of water stored at a certain height into mechanical energy by allowing the water to flow down through turbines. The turbines are connected to generators that convert the mechanical energy into electrical energy. Finally, the electrical energy is used to power light bulbs and other electrical devices, producing light.

Karanta bayanin darasi akan Current Electricity (JAMB)

Karanta bayanin darasi akan Sound Waves (WAEC)

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

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

Current Electricity

Sound Waves

Magnetic Field (Part 2)

Work, Energy And Power

Duba Amsa

Tambaya 38 Rahoto

If the focal length of a camera lens is 20cm, the distance from the film at which the lens must be set to produce a sharp image of an object 100cm away is

17cm

20cm

25cm

100cm

Bayanin Amsa

To produce a sharp image of an object, the light rays coming from the object must converge at a single point on the camera film. The distance between the lens and the film determines where the image will be in focus. In this case, the focal length of the camera lens is given as 20cm, and the distance from the object to the lens (object distance) is 100cm. To find the distance from the film to the lens (image distance) that will produce a sharp image, we can use the formula: 1/f = 1/o + 1/i where f is the focal length of the lens, o is the object distance, and i is the image distance. Substituting the given values, we get: 1/20 = 1/100 + 1/i Simplifying this equation gives: 1/i = 1/20 - 1/100 = 0.04 i = 1/0.04 = 25cm Therefore, the distance from the film at which the lens must be set to produce a sharp image of an object 100cm away is 25cm, which corresponds to option (C) in the given choices.

Karanta bayanin darasi akan Structure Of The Nucleus (WAEC)

Structure Of The Nucleus

Duba Amsa

Tambaya 39 Rahoto

Two forces of magnitudes 7N and 3N act at right angles to each other. The angle $ \theta $ between the resultant and the 7N force is given by

cos

θ

=

\frac{3}{7}

sin

θ

=

\frac{3}{7}

tan

θ

=

\frac{3}{7}

cot

θ

=

\frac{3}{7}

Bayanin Amsa

θ

= tan-1

\frac{3}{7}

Duba Amsa

Tambaya 40 Rahoto

Natural radioactivity is the random spontaneous disintegration of the nuclei of heavy isotopes of certain elements with the emission of

α

, X, and

β

α

,

β

and

γ

-rays

X,

α

,

γ

-ray

α

,

β

and X-rays

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Electric Field

Refraction Of Light Through Plane And Curved Surfaces

Duba Amsa

Tambaya 41 Rahoto

Determine the inductive reactance when a 30.0mH inductor with a negligible resistance is connected to a 1.3 x 103Hz oscillator.

39.0

Ω

122.5

Ω

245.0

Ω

39000.0

Ω

Bayanin Amsa

The inductive reactance (XL) of a 30.0 mH inductor with a negligible resistance when connected to a 1.3 x 10^3 Hz oscillator can be calculated using the formula: XL = 2πfL where f is the frequency of the oscillator in Hz, and L is the inductance in Henries. Substituting the given values, we get: XL = 2π(1.3 x 10^3 Hz)(30.0 x 10^-3 H) XL = 245.0 Ω Therefore, the inductive reactance is 245.0 Ω.

Duba Amsa

Tambaya 42 Rahoto

Which of the following correctly explains why soft iron is proffered to steel in electromagnets? i. soft iron is more readily magnetized than steel. ii. soft iron is more readily demagnetized than steel. iii. soft iron retains magnetism more than steel.

i only

ii and iii only

i and ii only

i, ii, and iii.

Bayanin Amsa

Soft iron is preferred to steel in electromagnets because it is more readily magnetized than steel, which means it can be easily magnetized by a small amount of electric current. This property of soft iron allows electromagnets to be more responsive and efficient. On the other hand, soft iron is also more readily demagnetized than steel, which means it loses its magnetism quickly once the electric current is turned off. This property is important because it allows the electromagnet to be turned on and off quickly, which is necessary for many applications. Therefore, the correct answer is option "i and ii only". Soft iron is not preferred to steel because it retains magnetism more than steel.

Karanta bayanin darasi akan Electric Field (WAEC)

Electric Field

Duba Amsa

Tambaya 43 Rahoto

A weightless vessel of dimensions 4m x 3m x 2m is filled with a liquid of density 1000kgm-3 and sealed. What is the maximum pressure this container can exert on a flat surface? [g = 10ms-2]

9 x 104NM-2

4 x 104NM-2

3 x 104NM-2

2 x 104NM-2

Bayanin Amsa

P = 2 x 1000 x 10

= 2 x 104NM-2

Karanta bayanin darasi akan Pressure (JAMB)

Pressure

Duba Amsa






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