JAMB UTME - Physics - 1991

Ajụjụ 1 Ripọtì

A small metal ball thrown vertically upwards from the top of a tower with an initial velocity of 20ms-1. If the ball took a total of 6s to reach ground level, determine the height of the tower [g = 10ms-2]

60m

80m

100m

120m

Akọwa Nkọwa

Time from tower to top, t =

\frac{u - v}{g}

= $\frac{20 - 0}{10}$

= 2s

distance from tower to top

x = ut - $\frac{1}{2}$ gt2

= (20 x 2) - ( $\frac{1}{2}$ x 10 x 2 x 2)

= 20m

Time from the top to the ground, t = 6 - 2

= 4s

distance from top to ground,

x = ut - $\frac{1}{2}$ gt2

= (0 x 4) + ( $\frac{1}{2}$ x 10 x 4 x 4)

= 80cm

therefore, height of tower = 80 - 20

= 60m

Gụọ akwụkwọ nkuzi na Motion (JAMB)

Motion

Lee Azịza

Ajụjụ 2 Ripọtì

Which of the following correctly, describes the image of an object, 4cm from a diverging lens of focal length 12cm?

The image is virtual, 3cm in front of the lens

The image is real, 6cm behind the lens

The image is virtual, 6cm in front of the lens

The image is real, 3cm in front the lens

Akọwa Nkọwa

F =

\frac{u v}{u + v}

F = -ve

-12 = $\frac{4 \times v}{4 + v}$

v = -3cm

Gụọ akwụkwọ nkuzi na Structure Of The Atom (Nigeria Only) (WAEC)

Structure Of The Atom (Nigeria Only)

Lee Azịza

Ajụjụ 3 Ripọtì

In a nuclear fusion process, four protons each of mass Mp were fused to produce a nucleus X of mass Mx. Which of the following equations is CORRECT?

4Mp > Mx

4Mp = Mx

4Mp< Mx

Mp > Mx

Akọwa Nkọwa

In a nuclear fusion process, four protons combine to form a nucleus X. According to the law of conservation of mass, the mass of the products of a chemical reaction must be equal to the mass of the reactants. Since the total mass of the four protons before fusion is 4M_p, the total mass of the resulting nucleus X after fusion must also be 4M_p. Therefore, the correct equation is 4M_p = M_x. Option (b) is correct.

Gụọ akwụkwọ nkuzi na Structure Of The Nucleus (WAEC)

Structure Of The Nucleus

Lee Azịza

Ajụjụ 4 Ripọtì

A man hears his echo from a nearby hill 2s after he shouted. If the frequency of his voice is 260Hz and the wavelength is 1.29m, how far away is the hill?

330.0m

335.4m

660.0m

670.8m

Akọwa Nkọwa

v =

λ

F

260 x 1.29 =335.4ms-1

Echo time = $\frac{2}{2}$ = 1sec.

distance = v x t

= 335.4 x 1 = 335.4m

Gụọ akwụkwọ nkuzi na Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Lee Azịza

Ajụjụ 5 Ripọtì

A body of mass 100g moving with a velocity of 10.0ms-1 collides with a wall. If after collision, it moves with a velocity of 2.0ms-1 in the opposite direction, calculate the change in momentum.

0.8Ns

1.2Ns

12.0Ns

80.0Ns

Akọwa Nkọwa

The change in momentum is calculated by subtracting the initial momentum from the final momentum. The momentum of an object is given by the product of its mass and velocity. Initial momentum = mass × initial velocity Final momentum = mass × final velocity The change in momentum is therefore: change in momentum = final momentum - initial momentum change in momentum = mass × final velocity - mass × initial velocity change in momentum = mass × (final velocity - initial velocity) Substituting the given values: mass = 100g = 0.1kg initial velocity = 10.0ms^-1 final velocity = -2.0ms^-1 (since the body moves in the opposite direction) change in momentum = 0.1kg × (-2.0ms^-1 - 10.0ms^-1) change in momentum = 0.1kg × (-12.0ms^-1) change in momentum = -1.2Ns Therefore, the change in momentum is -1.2Ns. Since momentum is a vector quantity, the negative sign indicates that the direction of momentum has changed, which is in the opposite direction of the initial momentum.

Gụọ akwụkwọ nkuzi na Newton’s Laws Of Motion (WAEC)

Newton’s Laws Of Motion

Lee Azịza

Ajụjụ 6 Ripọtì

Which of the following is TRUE of light and sound waves?

they both transmit energy

they both need a mediumfor propagation

they are both transverse waves

their velocities in air are equal

Akọwa Nkọwa

The correct option is: "they both transmit energy". Both light and sound waves are forms of energy that are transmitted through space. Light waves are electromagnetic waves that can travel through a vacuum, while sound waves are mechanical waves that require a medium, such as air, to propagate. Unlike light waves, sound waves are longitudinal waves, meaning that they vibrate in the direction of their motion, whereas light waves are transverse waves, meaning that they vibrate perpendicular to their direction of motion. The velocities of light and sound waves are not equal in air, with light waves traveling much faster than sound waves.

Gụọ akwụkwọ nkuzi na Sound Waves (WAEC)

Gụọ akwụkwọ nkuzi na Light Energy (JAMB)

Sound Waves

Light Energy

Lee Azịza

Ajụjụ 7 Ripọtì

The electrochemical equivalent of platinum is 5.0 x 10-7kgC-1 to plate-out 1.0kg of platinum, a current of 100A must be passed through an appropriate vessel for

5.6 hours

56 hours

1.4 x 104 hours

2.0 x 104 hours

Akọwa Nkọwa

m = Itz

1 = 100 x 5 x 10-7 x t

t = 20,000sec

= 5.6hrs

Gụọ akwụkwọ nkuzi na Electric Cells (JAMB)

Electric Cells

Lee Azịza

Ajụjụ 8 Ripọtì

A wheel and axle have radii 80cm and 10cm respectively. If the efficiency of the machine is 0.85, an applied force of 1200N to the wheel will raise a load of

8.0N

6.8N

8160N

9600N

Akọwa Nkọwa

Efficiency = m.A x

\frac{1}{V . R}

x 100%

85 = $\frac{l o a d}{1200}$ x $\frac{10}{80}$ x $\frac{100}{1}$

load = 8160N

Gụọ akwụkwọ nkuzi na Simple Machines (JAMB)

Simple Machines

Lee Azịza

Ajụjụ 9 Ripọtì

The absolute temperature of a perfect gas is proportional to the average

potential energy of the molecules

seperation between the molecules

kinetic energy of the molecules

velocity of the molecules

Akọwa Nkọwa

The absolute temperature of a perfect gas is proportional to the average kinetic energy of the molecules. This means that as the temperature of the gas increases, the average kinetic energy of its molecules also increases, and vice versa. The temperature of a gas is a measure of the average kinetic energy of its particles, where the higher the temperature, the more kinetic energy the particles have. Therefore, the option that correctly completes the sentence is "kinetic energy of the molecules".

Gụọ akwụkwọ nkuzi na Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Lee Azịza

Ajụjụ 10 Ripọtì

Which of the following statements is CORRECT about the earth's magnetic field?

the angle of dip is the angle which a freely suspended magnet makes with the vertical

the angle of declination is the angle between the magnetic meridian and the geographic meridian

the angle of declination is the angle which a magnetic compass makes with the magnetic meridian

the angle of inclination is the difference between the angle of dip and the angle of declination

Akọwa Nkọwa

The Earth has a magnetic field that acts like a giant bar magnet with its magnetic field lines directed from the geographic North Pole to the geographic South Pole. The magnetic field of the earth is responsible for the phenomenon of magnetism, and it has three important elements: the angle of dip, the angle of declination, and the angle of inclination. The angle of dip is the angle that a freely suspended magnet makes with the horizontal plane. The angle of dip is also known as the angle of inclination. The angle of declination is the angle between the magnetic meridian and the geographic meridian. It is the angle that a magnetic compass makes with the geographic meridian. The magnetic meridian is the imaginary line that passes through the geographic North Pole and the south magnetic pole. Therefore, the correct statement is "the angle of declination is the angle between the magnetic meridian and the geographic meridian."

Lee Azịza

Ajụjụ 11 Ripọtì

The mass of a stone is 15.0g when completely immersed in water and 10.0g when completely immersed in a liquid of relative density 2.0. The mass of the stone in air is

5.0g

12.0g

20.0g

25.0g

Akọwa Nkọwa

The buoyant force on an object is equal to the weight of the fluid displaced by the object. When the stone is immersed in water, the buoyant force is equal to the weight of the water that is displaced. Similarly, when the stone is immersed in the liquid of relative density 2.0, the buoyant force is equal to the weight of the liquid that is displaced. From the given information, we can calculate the volume of the stone. Let the volume of the stone be V. When completely immersed in water, the buoyant force on the stone is equal to the weight of the water displaced, which is equal to the weight of the stone in air minus the weight of the stone in water. Therefore, we have: Buoyant force = Weight of stone in air - Weight of stone in water V x density of water x g = (10.0g - 15.0g) x g V = 5.0g / (density of water x g) Similarly, when completely immersed in the liquid of relative density 2.0, we have: V x density of liquid x g = (10.0g - weight of liquid displaced) x g V x 2.0 x g = (10.0g - 15.0g) x g V = 5.0g / (2.0 x density of water x g) Since the volume of the stone is the same in both cases, we can set the two expressions for V equal to each other and solve for the density of water: 5.0g / (density of water x g) = 5.0g / (2.0 x density of water x g) density of water = 2.0 Now we can use the buoyant force equation to find the weight of the stone in air: Buoyant force = Weight of stone in air - Weight of stone in water V x density of water x g = (Weight of stone in air - 15.0g) x g Weight of stone in air = V x density of water x g + 15.0g Weight of stone in air = 5.0g + 15.0g Weight of stone in air = 20.0g Therefore, the mass of the stone in air is 20.0g. Answer: 20.0g.

Lee Azịza

Ajụjụ 12 Ripọtì

In which of the following is surface tension important?

the floating of a ship in water

the floating of a dry needle in water

the floating of a baloon in air

the diffusion of sugar solution across a membrane

Temperature And Its Measurement

Pressure

Lee Azịza

Ajụjụ 13 Ripọtì

What is the difference in the amount of heat given out by 4kg of steam and 4kg of water when both are cooled from 100oC to 80oC? [the specific latent heat of steam is 2,260,000Jkg-1, specific heat capacity of water is 4200Jkg-1K-1]

4200J

2,260,OOOJ

9,040,000J

9,380,000J

Akọwa Nkọwa

Heat by water = mc

θ

Heat by steam = mc $θ$ + mL

Difference = mL = 4 x 2,260,000

= 9,040,000J

Gụọ akwụkwọ nkuzi na Heat Energy (WAEC)

Heat Energy

Lee Azịza

Ajụjụ 14 Ripọtì

Which of the following obeys Ohm's Law?

Glass

Diode

All electrolytes

All metals

Akọwa Nkọwa

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points. Therefore, a material that obeys Ohm's Law will have a constant resistance at a given temperature. Metals are good conductors and generally obey Ohm's Law, which means their resistance remains constant over a wide range of voltages and currents. On the other hand, electrolytes, such as solutions of salts and acids, do not obey Ohm's Law because they have a non-linear relationship between current and voltage due to their ionic nature. Glass is an insulator and does not conduct electricity, while a diode is a semiconductor that exhibits nonlinear current-voltage characteristics and does not obey Ohm's Law. Therefore, the material that obeys Ohm's Law is "All metals".

Gụọ akwụkwọ nkuzi na Electrostatics (JAMB)

Electrostatics

Lee Azịza

Ajụjụ 15 Ripọtì

An object moves with uniform speed round a circle. Its acceleration has

constant magnitude and constant direction

constant magnitude and varying direction

varying magnitude but constant direction

varying magnitude and varying direction

Rectilinear Acceleration

Lee Azịza

Ajụjụ 16 Ripọtì

A boy timed 30 oscillations of a certain pendulum thrice and obtained 1min. 10s, 1min. 12s and 1min. 7s respectively.The mean period of oscillation of the pendulum is

0.14s

0.43s

2.32s

6.97s

Akọwa Nkọwa

The mean period of oscillation of the pendulum can be calculated by dividing the total time for the oscillations by the number of oscillations measured. In this case, the boy measured 30 oscillations three times, with times of 1min. 10s, 1min. 12s, and 1min. 7s. First, we need to convert the times to seconds: 1min. 10s = 70s 1min. 12s = 72s 1min. 7s = 67s Next, we can calculate the total time for the 90 oscillations: 70s + 72s + 67s = 209s Finally, we can calculate the mean period by dividing the total time by the number of oscillations: 209s / 90 oscillations = 2.32s Therefore, the mean period of oscillation of the pendulum is 2.32s. Answer option (C) is correct.

Gụọ akwụkwọ nkuzi na Simple Harmonic Motion (WAEC)

Simple Harmonic Motion

Lee Azịza

Ajụjụ 17 Ripọtì

Which of the following are True for $ \alpha $-decay. i. mass number, decrease by four. ii. Atomic number decreases by two. iii. mass number does not change.

i and ii only

ii and iii only

i and iii only

i, ii, and iii only

Akọwa Nkọwa

Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle consisting of two protons and two neutrons, reducing its atomic number by two and its mass number by four. Therefore, statement i is correct. The atomic number is the number of protons in the nucleus, and in alpha decay, two protons are lost. Hence, statement ii is also correct. However, the number of neutrons in the nucleus remains the same in alpha decay, so statement iii is incorrect. Therefore, the correct option is "i and ii only."

Gụọ akwụkwọ nkuzi na Simple A.C. Circuits (WAEC)

Simple A.C. Circuits

Lee Azịza

Ajụjụ 18 Ripọtì

The image in a pin-hole camera is

errect and formed by refraction through a lens

virtual and formed by dispersion

erect and gets sharper as the hole becomes larger

inverted and formed by the light from each point travelling in a straight line

Akọwa Nkọwa

The image in a pin-hole camera is inverted and formed by the light from each point travelling in a straight line. This is because a pin-hole camera works by letting light through a tiny hole (pin-hole), which then creates an inverted image on the opposite side of the box or surface that is receiving the light. The smaller the pin-hole, the sharper the image will be, but it will also be dimmer. The image formed in a pin-hole camera is not refracted by a lens or created through dispersion, but rather through the straight-line propagation of light.

Gụọ akwụkwọ nkuzi na Electrostatics (JAMB)

Electrostatics

Lee Azịza

Ajụjụ 19 Ripọtì

which of the following statements are TRUE of isotopes? i. isotopes of an element have the same chemical properties because they have the same number of electrons. ii. isotopes of elements are normally separated using physical properties. iii. isotopes of an element have the same number of protons in their nuclei.

i and ii only

i and iii only

ii and iii only

i, ii, and iii

Structure Of The Atom (Nigeria Only)

Lee Azịza

Ajụjụ 20 Ripọtì

An insulated rod when rubbed with a material acquires

a negative charge if it is made of glass and rubbed with silk

no charge if it is made of glass and rubbed with fur

no charge if it is made of copper and rubbed with silk

a positive charge if it is made of copper and rubbed with fur

Electrostatics

Lee Azịza

Ajụjụ 21 Ripọtì

4g of a radioactive material of half-life 10 days is spilled on a laboratory floor. How long would it take to disintegrate 3.5g of the material?

1

\frac{1}{4}

days

8

\frac{1}{4}

days

30 days

80 days

Heat Energy

Lee Azịza

Ajụjụ 22 Ripọtì

A pilot records the atmospheric pressure outside its plane as 63cm of Hg while a ground observer records a reading of 75cm of Hg for the atmospheric pressure on the ground. Assuming that the density of the atmosphere is constant, calculate the height of the plane above the ground. (Relative density of Hg = 13.0 and that of air = 0.00013)

1200m

6300m

7500m

13800m

Akọwa Nkọwa

Difference in pressure = 75 - 63

= 12cmHg

pressure = hge

difference in height = $\frac{12}{10 \times \frac{0.00013}{13}}$

= 1200m

Gụọ akwụkwọ nkuzi na Pressure (JAMB)

Pressure

Lee Azịza

Ajụjụ 23 Ripọtì

A moving coil galvanometer of $300\,\Omega$ resistance gives full scale deflection for 1.0mA. The resistance R, of the shunt that is required to convert the galvanometer into a 3.0A ammeter is

899.70

Ω

10.00

Ω

Ω

Ω

0.01

Ω

Electrostatics

Lee Azịza

Ajụjụ 24 Ripọtì

A body is under action of a force F such that the force displacement graph of the body is semi circular as shown above. The work done on the body by the force in moving through 24 meters is

36π J

72π J

144π J

288π J

Work, Energy And Power

Lee Azịza

Ajụjụ 25 Ripọtì

The net capacitance of the circuit shown above is

8.0µF

6.0µF

4.0µF

2.0µF

Capacitors

Current Electricity

Simple A.C Circuits

Measurements And Units

Lee Azịza

Ajụjụ 26 Ripọtì

In a ray diagram for a thin converging lens, a ray that is not parallel to the optic axis but passes through the optic center will

pass through undeviated

pass through the center of curvature after refraction

emerge parallel to the principal axis

pass through the principal focus after refraction

Akọwa Nkọwa

If a ray of light passes through the optic center of a thin converging lens, it will pass through undeviated. This is because the optic center is the point on the lens where the principal axis crosses and it is equidistant from the lens surfaces. Therefore, any ray of light that passes through the optic center will not be refracted or deviated because it is passing through the center of the lens and hitting the surface perpendicularly. A converging lens is thicker in the middle than at the edges and refracts light rays in such a way that they converge at a focal point on the opposite side of the lens. Rays that are not parallel to the principal axis will converge at a point on the opposite side of the lens after refraction. However, if the ray passes through the optic center, it will not bend or deviate from its original path and will pass through undeviated. Therefore, if a ray of light is not parallel to the optic axis but passes through the optic center of a thin converging lens, it will pass through undeviated.

Gụọ akwụkwọ nkuzi na Reflection Of Light At Plane And Curved Surfaces (JAMB)

Gụọ akwụkwọ nkuzi na Optical Instruments (JAMB)

Reflection Of Light At Plane And Curved Surfaces

Optical Instruments

Lee Azịza

Ajụjụ 27 Ripọtì

To convert an alternating current dynamo into a direct current dynamo, the

number of turns in the coil is increased

strength of the field magnet is increased

slip rings are replaced with a split ring commutator

coil is wound on a soft iron armature

Akọwa Nkọwa

To convert an alternating current (AC) dynamo into a direct current (DC) dynamo, you need to replace the slip rings with a split-ring commutator. Slip rings allow the current to change direction every half-turn, resulting in an AC current output. However, a split-ring commutator ensures that the current flows in the same direction at all times, creating a direct current output. The number of turns in the coil, strength of the field magnet, and the material of the armature do not affect the conversion from AC to DC.

Lee Azịza

Ajụjụ 28 Ripọtì

A brass rod is 2m long at a certain temperature. What is the length for a temperature rise of 100K, if the expansivity of brass is 18 x 10-6K-1?

2.0036m

2.0018cm

2.1800

2.0360m

Akọwa Nkọwa

The length of the brass rod changes with a change in temperature due to thermal expansion. The formula for linear expansion is given as: ΔL = αLΔT Where ΔL is the change in length, α is the coefficient of linear expansion, L is the original length of the rod, and ΔT is the change in temperature. Substituting the given values, we get: ΔL = (18 x 10^-6) x 2 x 100 = 0.0036m Therefore, the new length of the brass rod is: L + ΔL = 2m + 0.0036m = 2.0036m Hence, the answer is 2.0036m.

Gụọ akwụkwọ nkuzi na Thermal Expansion (JAMB)

Thermal Expansion

Lee Azịza

Ajụjụ 29 Ripọtì

A thermometer with an arbitrary scale S, of equal division registers -30oS at the ice point and +90oS at the steam point. Calculate the Celsius temperature corresponding to 60oS

25.0oC

50.0oC

66.7oC

75.0oC

Akọwa Nkọwa

To convert from the arbitrary scale S to Celsius, we need to use the following formula: C = (S - 32) x 5/9 First, we need to find the conversion factor between S and Fahrenheit: (Smax - Smin)/(Fmax - Fmin) = (90 - (-30))/(212 - 32) = 120/180 = 2/3 So, for a given temperature T on the arbitrary scale, the corresponding temperature on the Fahrenheit scale is: T_F = (T - Smin) x (Fmax - Fmin)/(Smax - Smin) + Fmin = (T - (-30)) x 180/120 + 32 = (T + 30) x 3/2 + 32 = 1.5T + 77 To convert 60 on the arbitrary scale to Celsius, we first need to find the corresponding temperature on the Fahrenheit scale: 60_S = 1.5 x 60 + 77 = 167 Then, we use the Celsius conversion formula: C = (F - 32) x 5/9 = (167 - 32) x 5/9 = 135 x 5/9 = 75.0°C Therefore, the answer is 75.0oC.

Gụọ akwụkwọ nkuzi na Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Lee Azịza

Ajụjụ 30 Ripọtì

A 20kg mass is to be pulled up a slope inclined at 30o to the horizontal. If the efficiency of the plane is 75%, the force required to pull the load up the plane is [g = 10ms-2]

13.3N

73.5N

133.3N

53.2N

Akọwa Nkọwa

To find the force required to pull the 20kg mass up a slope inclined at 30 degrees to the horizontal, we need to consider the weight of the object and the force needed to overcome friction. The weight of the object is given by the formula W = mg, where m is the mass and g is the acceleration due to gravity (10 m/s^2). So, W = 20 kg x 10 m/s^2 = 200 N. To find the force needed to overcome friction, we need to use the formula F = μN, where μ is the coefficient of friction and N is the normal force. The normal force is the component of the weight that is perpendicular to the slope, which is given by N = W cosθ, where θ is the angle of inclination (30 degrees in this case). So, N = 200 N x cos30 = 173.2 N. Assuming an efficiency of 75%, we know that only 75% of the force applied will go towards moving the object up the slope, while the rest will be lost to friction and other factors. So, the force required to pull the load up the plane is F = (1/0.75) x (200 N x sin30 + μN) = 266.67 N. Therefore, the answer is 266.67 N, which is closest to option C (133.3N).

Gụọ akwụkwọ nkuzi na Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

Lee Azịza

Ajụjụ 31 Ripọtì

Two tuning forks of frequency 256Hz and 260Hz are sounded close to each other. What is the frequency of the beats produced?

2Hz

4Hz

8Hz

258Hz

Akọwa Nkọwa

When two sound waves with slightly different frequencies are played together, they interfere constructively and destructively, resulting in a periodic variation of amplitude, which is called beats. The frequency of beats produced is equal to the difference between the frequencies of the two sound waves. In this case, the two tuning forks have frequencies of 256Hz and 260Hz, respectively. Therefore, the difference between their frequencies is: 260Hz - 256Hz = 4Hz So the beats produced have a frequency of 4Hz. Therefore, the answer is: 4Hz.

Gụọ akwụkwọ nkuzi na Propagation Of Sound Waves (JAMB)

Propagation Of Sound Waves

Lee Azịza

Ajụjụ 32 Ripọtì

The internal resistance of each of the cells E1 and E2 Shown in the figure above is 2 Ohm. Calculate the total current in the circuit

0.80A

0.50A

0.40A

0.004A

Current Electricity

Work, Energy And Power

Lee Azịza

Ajụjụ 33 Ripọtì

An equipment whose power is 1500W and resistance is 375 ohms would draw a current of

0.10A

2.00A

4.00A

77.5A

Akọwa Nkọwa

The current drawn by an equipment can be calculated using Ohm's Law, which states that: V = IR where V is the voltage, I is the current, and R is the resistance. In this case, we know that the power of the equipment is 1500W, and we can also use the formula: P = IV where P is the power, I is the current, and V is the voltage. By combining these two equations, we can eliminate V and get: I = sqrt(P/R) where sqrt represents the square root. Plugging in the values, we get: I = sqrt(1500W / 375 ohms) I = sqrt(4A) I = 2A Therefore, the current drawn by the equipment is 2A. Hence, the answer is 2.00A.

Gụọ akwụkwọ nkuzi na Current Electricity (JAMB)

Current Electricity

Lee Azịza

Ajụjụ 34 Ripọtì

In the above circuit, The fuse wire melts when

K is Opened

K is closed

the 14 ohm resistor is doubled with k closed

the 6 ohm resistor is doubled with k closed

Lee Azịza

Ajụjụ 35 Ripọtì

How long does it take a 750W heater to raise the temperature of 1kg of water from 20oC to 50oC [specific heat capacity of water = 4200Jkg-1K-1]

84s

112s

168s

280s

Akọwa Nkọwa

To calculate the time it takes for a 750W heater to raise the temperature of 1kg of water from 20°C to 50°C, we can use the formula: Q = mcΔT where Q is the heat energy required to heat up the water, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature. Solving for Q, we get: Q = mcΔT Q = 1kg x 4200J/kg°C x (50°C - 20°C) Q = 126000J Now, we can use the formula for power to find out how long it takes for the heater to produce 126000J of heat energy: P = E/t t = E/P where P is the power of the heater, E is the energy produced by the heater, and t is the time taken. Solving for t, we get: t = E/P t = 126000J / 750W t = 168 seconds Therefore, the answer is 168s.

Gụọ akwụkwọ nkuzi na Heat Energy (WAEC)

Heat Energy

Lee Azịza

Ajụjụ 36 Ripọtì

The saturated vapour pressure of a liquid increases as the

volume of the liquid increases

volume of the liquid decreases

temperature of the liquid increases

temperature of the liquid decreases

Akọwa Nkọwa

The saturated vapor pressure of a liquid is the pressure exerted by its vapor when it is in equilibrium with its liquid phase. This means that the rate of evaporation of the liquid is equal to the rate of condensation of its vapor. The saturated vapor pressure of a liquid increases with an increase in the temperature of the liquid. When the temperature increases, the kinetic energy of the liquid molecules increases, and more molecules are able to escape from the liquid surface and enter the vapor phase. This increases the number of vapor molecules in the air above the liquid, which leads to an increase in the vapor pressure. Therefore, the correct option is "temperature of the liquid increases."

Gụọ akwụkwọ nkuzi na Vapours (JAMB)

Vapours

Lee Azịza

Ajụjụ 37 Ripọtì

When an atom loses or gains a charge, it becomes

an electron

an ion

a neutron

a proton

Akọwa Nkọwa

When an atom loses or gains a charge, it becomes an ion. An ion is an atom or molecule that has a net electrical charge because it has either lost or gained one or more electrons. A positively charged ion is called a cation and a negatively charged ion is called an anion.

Gụọ akwụkwọ nkuzi na Structure Of The Atom (Nigeria Only) (WAEC)

Structure Of The Atom (Nigeria Only)

Lee Azịza

Ajụjụ 38 Ripọtì

A trough 12.0cm deep is filled with water of refractive index $ \frac{4}{3} $. By how much would a coin at the bottom of the trough appear to be displaced when viewed vertically from above the water surface?

3.0cm

6.0cm

9.0cm

16.0cm

Akọwa Nkọwa

\frac{4}{3}

=

\frac{12}{Apparent depth}

Apparent depth = 9cm

Height by which coin is raised = 12 - 9

= 3cm

Gụọ akwụkwọ nkuzi na Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Lee Azịza

Ajụjụ 39 Ripọtì

Which of the following is the most suitable for use as an altimeter?

A mecury barometer

A fortin barometer

A mecury manometer

An aneroid barometer

Pressure

Lee Azịza

Ajụjụ 40 Ripọtì

If a current carrying coil is mounted on a metal frame, the back e.m.f. induced in the coil causes

inductance

eddy currents

elecro-magnetism

dipole moment

Electromagnetic Field (Part 2)

Electrostatics

Lee Azịza

Ajụjụ 41 Ripọtì

A room is heated by means of a charcoal fire, An occupant of the room standing away from the fire is warmed mainly by

convection

radiation

conduction

reflection

Akọwa Nkọwa

The occupant of the room standing away from the charcoal fire is mainly warmed by radiation. Radiation is the transfer of heat energy in the form of electromagnetic waves that can travel through empty space. In this case, the heat energy from the charcoal fire is transferred to the surrounding objects and the walls of the room through radiation. The objects and walls then re-radiate the heat energy towards the occupant, warming them up. Convection is the transfer of heat energy through the movement of fluids or gases, but in this scenario, the air in the room is not moving enough to contribute significantly to the warming of the occupant. Conduction is the transfer of heat energy through direct contact between objects, but the occupant is not in direct contact with the fire or the walls of the room. Reflection is the bouncing back of electromagnetic waves from a surface, but it does not play a significant role in the warming of the occupant in this scenario.

Gụọ akwụkwọ nkuzi na Heat Transfer (JAMB)

Gụọ akwụkwọ nkuzi na Structure Of The Nucleus (WAEC)

Gụọ akwụkwọ nkuzi na Electromagnetic Field (Part 1) (WAEC)

Gụọ akwụkwọ nkuzi na Liquids At Rest (JAMB)

Heat Transfer

Structure Of The Nucleus

Electromagnetic Field (Part 1)

Liquids At Rest

Lee Azịza

Ajụjụ 42 Ripọtì

When the bottom tip of a vibrating tuning fork is held in contact with a wooden box, a louder sound is heard. This phenomenon is known as

beats

echoing

resonance

reverberation

Akọwa Nkọwa

The phenomenon you're describing is called resonance. When the tuning fork is brought in contact with the wooden box, the box starts vibrating at the same frequency as the tuning fork, which amplifies the sound waves produced by the tuning fork. This happens because the wooden box is able to resonate, or vibrate, at the same frequency as the tuning fork. Resonance occurs when an object is able to vibrate at its natural frequency in response to another object vibrating at the same frequency. This causes the amplitude (or loudness) of the vibration to increase. In this case, the wooden box is able to resonate at the same frequency as the tuning fork, so the sound produced by the tuning fork is amplified, resulting in a louder sound.

Gụọ akwụkwọ nkuzi na Sound Waves (WAEC)

Sound Waves

Lee Azịza

Ajụjụ 43 Ripọtì

Find the tension T1 in the diagram above if the system is in equilibrium. [G = 10ms-2]

200/√3N

100/√3N

300/√3N

100N

Akọwa Nkọwa

To find the tension T1, we need to analyze the forces acting on the system and use the fact that it is in equilibrium, which means the net force in all directions is zero. First, let's consider the vertical forces. The weight of the hanging mass is acting downwards with a force of 100N. The tension T2 is also acting downwards with a force of T2*cos(30°) because it is at an angle of 30° to the vertical. Therefore, the net vertical force is: Net vertical force = T2*cos(30°) - 100N Since the system is in equilibrium, the net vertical force must be zero. Thus, we can solve for T2: T2*cos(30°) = 100N T2 = 100N/cos(30°) = 115.5N Next, let's consider the horizontal forces. The tension T2 is acting to the left with a force of T2*sin(30°) because it is at an angle of 30° to the horizontal. The tension T1 is acting to the right. Therefore, the net horizontal force is: Net horizontal force = T1 - T2*sin(30°) Since the system is in equilibrium, the net horizontal force must be zero. Thus, we can solve for T1: T1 = T2*sin(30°) = 115.5N*sin(30°) = 57.8N Therefore, the tension T1 is 57.8N. Option (B) 100/√3N
is closest to this value, but it is not the correct answer.

Gụọ akwụkwọ nkuzi na Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

Lee Azịza

Ajụjụ 44 Ripọtì

A spring of force constant 1500Nm-1 is acted upon by a constant force of 75N. Calculate the potential energy stored in the string

1.9J

2.8J

3.45J

0.43J

Akọwa Nkọwa

P.E =

\frac{1}{2}

Fe

= $\frac{1}{2}$ $\frac{F}{K}$

= $\frac{1}{2}$ x $\frac{75 \times 75}{1500}$

= 1.9J

Gụọ akwụkwọ nkuzi na Elasticity (JAMB)

Elasticity

Lee Azịza

Ajụjụ 45 Ripọtì

When a plane mirror at which a ray is incident is rotated through an angle $\theta$, the reflected ray will be rotated through

\frac{1}{2}

θ

θ

2

θ

3

θ

Akọwa Nkọwa

When a plane mirror is rotated through an angle θ, the reflected ray will be rotated through an angle of 2θ. To understand why, imagine a ray of light hitting a plane mirror at an incident angle θ. The angle of incidence is the angle between the incident ray and the perpendicular to the mirror surface at the point of incidence. According to the law of reflection, the angle of reflection is equal to the angle of incidence, and the reflected ray makes the same angle with the perpendicular as the incident ray. Therefore, if we rotate the mirror through an angle θ, the perpendicular to the mirror at the point of incidence will also be rotated through an angle θ. This means that the angle of incidence will also change by θ. Now, since the angle of reflection is equal to the angle of incidence, the angle of reflection will also change by θ. Therefore, the angle between the incident ray and the reflected ray will change by 2θ (θ for the change in angle of incidence and θ for the change in angle of reflection). Hence, the reflected ray will be rotated through an angle of 2θ.

Gụọ akwụkwọ nkuzi na Refraction Of Light Through Plane And Curved Surfaces (JAMB)

Gụọ akwụkwọ nkuzi na Electrostatics (JAMB)

Refraction Of Light Through Plane And Curved Surfaces

Electrostatics

Lee Azịza

Ajụjụ 46 Ripọtì

The spiral spring of a spring balance is 25.0cm long when 5N hangs on it and 30cm long, when the weight is 10N. What is the length of the spring if the weight is 3N assuming Hooke's law is obeyed?

15.0cm

17.0cm

20.0cm

23.0cm

Akọwa Nkọwa

let the length be Lcm

$\frac{5 - 3}{25 - L}$ = $\frac{10 - 5}{30 - 25}$

L = 23cm

Gụọ akwụkwọ nkuzi na Elasticity (JAMB)

Elasticity

Lee Azịza

Ajụjụ 47 Ripọtì

Which of the following phenomena is called photoelectric effect?

high energy electrons impinge on a metalic anode which then emits photons

a high energy photon emits photons as it slowed down

a metal absorbs quanta of light and then emits electrons

Two electrons are created from a quantum of light

Akọwa Nkọwa

The phenomenon called photoelectric effect is when a metal absorbs quanta of light and then emits electrons. This means that when light shines on a metal, the energy from the light is absorbed by the metal atoms, which causes electrons to be emitted from the metal surface. This effect is important in many technologies such as solar cells and photo detectors. The other options are not examples of the photoelectric effect. The first option describes the reverse effect, where high energy electrons cause the emission of photons, which is known as the Bremsstrahlung effect. The second option describes the process of photon emission through the release of energy by a particle as it slows down, which is known as the Cerenkov radiation. The third option describes a similar effect to the photoelectric effect, but instead of emitting electrons, the metal would emit light, which is known as the photoluminescence effect. Finally, the fourth option describes the process of photon absorption, which is not related to the photoelectric effect.

Gụọ akwụkwọ nkuzi na Electromagnetic Field (Part 2) (WAEC)

Electromagnetic Field (Part 2)

Lee Azịza

Ajụjụ 48 Ripọtì

A body of weight W N rest on a smooth plane inclined at an angle $ \theta $^o to the horizontal. What is the resolved part of the weight in Newtons along the plane?

θ

W cos

θ

W sec

θ

W tan

θ

Akọwa Nkọwa

When an object of weight W rests on an inclined plane, the weight force can be resolved into two components: one perpendicular to the plane (W cosθ) and the other parallel to the plane (W sinθ). The component of weight parallel to the plane is what we are interested in, as this is the force that will cause the object to slide down the plane if it is not held in place. Therefore, the resolved part of the weight in Newtons along the plane is W sinθ. This is because the weight force is proportional to the sine of the angle of inclination, meaning that the steeper the slope, the greater the force parallel to the plane. So, option A (W sinθ) is the correct answer.

Gụọ akwụkwọ nkuzi na Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

Lee Azịza

Ajụjụ 49 Ripọtì

The function of the system of granulated carbon mixed with manganese (IV) oxide in a Letanche cell is to

increase the e.m.f. of the cell to 2.0V

prevent local action in the cell

prevent polarisation in the cell

make the cell black and hence a good radiator

Akọwa Nkọwa

The Letanche cell uses a system of granulated carbon mixed with manganese (IV) oxide to prevent polarization in the cell. Polarization happens when the electrodes in a cell become coated with gas bubbles or other substances, hindering the flow of electricity. The mixture of carbon and manganese (IV) oxide acts as a catalyst to speed up the reaction and prevent polarization, ensuring a steady flow of electricity. This allows the cell to maintain a constant voltage and function efficiently. Therefore, option C, "prevent polarisation in the cell," is the correct answer.

Gụọ akwụkwọ nkuzi na Electrostatics (JAMB)

Electrostatics

Lee Azịza






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