WAEC SSCE - Physics - 1990 (Objective)

A particle of mass 2.5 x 10^-6kg revolving around the earth has a radial acceleration of 4 X 10⁷ ms^-2. What is the centripetal force of the particles?

6.25 x 10^-14N

1.60 x10^-13N

5.00 x 10N

1.00x10²N Amsar da ta dace

2.00x1O²N

Bayanin Amsa

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

Newton’s Laws Of Motion

Duba Amsa

Tambaya 5 Rahoto

Heat transfer by convection in a liquid is due to the

translatory motion of the molecules of the liquid Amsar da ta dace

increased vibration of the molecules of the liquid about their mean positions

expansion of the liquid as it is heated

latent heat of vaporization of the liquid

ability of the liquid to evaporate at all temperatures

Bayanin Amsa

Heat transfer by convection in a liquid is due to the translatory motion of the molecules of the liquid. When a liquid is heated, the molecules gain kinetic energy and move faster, which results in the formation of convection currents. These currents carry the heat energy from one point to another in the liquid. As the heated liquid rises, it transfers the heat to the surrounding cooler liquid molecules by collisions, causing the cooler liquid to become less dense and rise as well. This creates a continuous cycle of rising and sinking of the liquid, which transfers heat throughout the liquid. This process is known as convection, and it is the main mechanism of heat transfer in fluids like liquids and gases.

Karanta bayanin darasi akan Heat Transfer (JAMB)

Heat Transfer

Duba Amsa

Tambaya 6 Rahoto

The diagram above represents a portion of a micrometer screw 20 gauge. What is its reading?

3.70mm

3.67mm

3.50mm

3.33mm

3.17mm Amsar da ta dace

Bayanin Amsa

To determine the reading of the micrometer screw gauge, we need to add the main scale reading to the circular scale reading. From the image, we can see that the main scale reading is 3.0 mm, which is the distance shown between the left end of the sleeve and the central zero mark on the thimble. To find the circular scale reading, we count the number of divisions on the circular scale that line up with the main scale. We see that the 20th line on the circular scale is aligned with the main scale. Since each division on the circular scale represents 0.01 mm, we can calculate the circular scale reading as 20 x 0.01 mm = 0.20 mm. Therefore, the total reading is 3.0 mm + 0.20 mm = 3.20 mm. None of the answer options match our calculated value. The closest option is 3.17 mm.

Duba Amsa

Tambaya 7 Rahoto

An object weighs 10.0 N in air and 7.0 N in water. What is its weight when immersed in a liquid of relative density 1.5?

4.50N

4.67N

5.50N Amsar da ta dace

6.67N

8.50N

Bayanin Amsa

Karanta bayanin darasi akan Gravitational Field (JAMB)

Gravitational Field

Duba Amsa

Tambaya 8 Rahoto

What is the difference between a crystalline and an amorphous solid? A crystalline solid

has cubic structure while amorphous solid has hexagonal structure

has regularly-repeating pattern while an amorphous solid has not Amsar da ta dace

is white while amorphous solid is yellow

is always hard while amorphous solid is always soft

melts on heating while an amorphous solid sublimates on heating.

Bayanin Amsa

The main difference between a crystalline and an amorphous solid is in their structure. A crystalline solid has a highly ordered, repeating pattern of molecules or atoms, which creates a well-defined shape and structure. This means that a crystalline solid has a regular geometric shape, with flat faces and sharp edges. On the other hand, an amorphous solid lacks this repeating pattern and does not have a well-defined shape. Instead, it has a more random arrangement of molecules or atoms, which creates a less ordered, less defined structure. This means that an amorphous solid has a more irregular shape, with rounded or curved surfaces. Examples of crystalline solids include diamonds, salt, and sugar, while examples of amorphous solids include glass, rubber, and plastic. In summary, the main difference between crystalline and amorphous solids is in their molecular or atomic arrangement, with crystalline solids having a highly ordered, repeating pattern and amorphous solids having a more random, less defined structure.

Karanta bayanin darasi akan Elasticity (JAMB)

Elasticity

Duba Amsa

Tambaya 9 Rahoto

When the pressure of the gas in a discharge tube is gradually lowered. I. there is a violet glow in the tube II. the glow in the tube splits into bands which get further apart. III, the glow and the bands eventually disappear and there is a green fluorescence in the tube. IV. the tube breaks into pieces. Which of the following are correct?

I and II only

I and III only Amsar da ta dace

I and IV only

I, II and III only

I, II III and IV

Bayanin Amsa

Karanta bayanin darasi akan Pressure (JAMB)

Karanta bayanin darasi akan Liquids At Rest (JAMB)

Pressure

Liquids At Rest

Duba Amsa

Tambaya 10 Rahoto

The diagram above represents part of a wave motion in air. If the wave travels with a speed of 300ms^-1calculate the frequency of the wave.

360Hz

300Hz

250Hz Amsar da ta dace

150Hz

100Hz

Bayanin Amsa

Karanta bayanin darasi akan Production And Propagation Of Waves (WAEC)

Karanta bayanin darasi akan Sound Waves (WAEC)

Production And Propagation Of Waves

Sound Waves

Duba Amsa

Tambaya 11 Rahoto

A parallel beam of light is to be obtained from the headlamp ofa car. At which of the following positions should the source of light be placed from the pole of its spherical minor?

At the focal point Amsar da ta dace

At the centre of curvature

Beyond the centre of curvature

Between the focal point and the pole

Between the focal point and the center of curvature

Bayanin Amsa

In order to obtain a parallel beam of light from the headlamp of a car using a spherical mirror, the light source should be placed at the focal point of the mirror. This is because a spherical mirror can converge or diverge the light rays depending on where the light source is placed relative to the mirror. If the light source is placed at the focal point of the mirror, the light rays will be reflected in parallel to each other, creating a parallel beam of light. Placing the light source at any other position, such as the centre of curvature, beyond the centre of curvature, or between the focal point and the pole, will cause the light rays to converge or diverge, creating a beam of light that is not parallel. Therefore, the correct option is "At the focal point".

Duba Amsa

Tambaya 12 Rahoto

Dry hydrogen is trapped by pellet of mercury in a uniform capillary tube closed at one end. If the length of the column of hydrogen at 27°C is 1.0m, at what temperature will the length be 1.20m?

22.5^oC

32.4^oC

77.0^oC

87.0^oC Amsar da ta dace

360.0^oC

Bayanin Amsa

The problem involves a trapped gas at a fixed volume in a closed-end tube, and we are asked to find the temperature at which the length of the gas column will increase to a specified value. This can be solved using the relationship between the volume of a gas and its temperature, known as Charles's Law. According to Charles's Law, the volume of a gas is directly proportional to its temperature at constant pressure. Mathematically, this can be expressed as: V1/T1 = V2/T2 where V1 and T1 are the initial volume and temperature, respectively, and V2 and T2 are the final volume and temperature, respectively. In this problem, the volume of the gas is constant since it is trapped by the mercury pellet, so we can simplify the equation to: T1/T2 = L1/L2 where L1 and L2 are the initial and final lengths of the gas column, respectively. We are given that the initial length of the gas column is 1.0 m at a temperature of 27°C (which can be converted to 300.15 K). We are asked to find the temperature at which the length of the gas column will increase to 1.20 m. Using the above equation, we can solve for T2: T2 = (L2/L1) x T1 T2 = (1.20 m / 1.0 m) x 300.15 K T2 = 360.18 K We can convert this temperature to degrees Celsius by subtracting 273.15: T2 = 87.03°C Therefore, the correct option is (d) 87.0°C.

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Duba Amsa

Tambaya 13 Rahoto

A girl whose mass is 55kg stands on a spring - weighing machine inside a lift. When the lift starts to ascend, its acceleration is 2ms1. What will be the reading on the machine? (Take g = ms^-2)

66 kg Amsar da ta dace

55 kg

44 kg

22 kg

11 kg

Bayanin Amsa

The reading on the spring - weighing machine will be greater than the actual weight of the girl. This is because when the lift starts to ascend, the girl and the weighing machine experience an upward force due to the acceleration of the lift. This force adds to the force of gravity acting on the girl and thus, the reading on the weighing machine increases. To find the reading on the weighing machine, we need to calculate the total force acting on the girl. Using Newton's second law, F=ma, the total force acting on the girl is the sum of the force of gravity and the force due to the lift's acceleration: F_total = (55 kg)(9.81 m/s^2) + (55 kg)(2 m/s^2) F_total = 676.05 N Therefore, the reading on the weighing machine will be: Reading = F_total = 676.05 N So the answer is 66 kg when converted to mass.

Duba Amsa

Tambaya 14 Rahoto

A work of 30 joules is done is transferring 5 millicoulombs of charge from a point B to a point A in an electric Held. The potential difference between B and A is

1.7 x 10^-4V

3.4x10^-4V

1.5x10^-1V

6.0 x10³V Amsar da ta dace

1.2 x10⁴V

Bayanin Amsa

The work done (W) in transferring charge (q) between two points in an electric field is given by the formula: W = qV Where V is the potential difference between the two points. In this question, the work done is 30 J and the charge transferred is 5 mC (or 5 x 10^-3 C). Therefore, we can rearrange the formula to find the potential difference: V = W/q = 30 J / 5 x 10^-3 C = 6,000 V So the potential difference between points B and A is 6,000 volts. Therefore, the correct option is: - 6.0 x 10^3 V

Karanta bayanin darasi akan Electrostatics (JAMB)

Electrostatics

Duba Amsa

Tambaya 15 Rahoto

The absolute zero temperature is defined as the temperature at which

thermal motion ceases Amsar da ta dace

the temperature of a gas is 273^oC

ice melts

the volume of a real gas is maximum

the pressure of a real gas is maximum

Bayanin Amsa

The absolute zero temperature is defined as the temperature at which thermal motion ceases. This is the lowest possible temperature that can be reached, and it is equivalent to zero Kelvin (0 K) or approximately -273.15 degrees Celsius (-459.67 degrees Fahrenheit). At this temperature, the particles that make up matter have the lowest possible energy and therefore stop moving completely. This is because temperature is a measure of the average kinetic energy of particles in a substance, and at absolute zero, the kinetic energy is zero. This concept is important in many areas of physics, such as thermodynamics and quantum mechanics, and it has practical applications in areas such as cryogenics, where temperatures approaching absolute zero are used to achieve superconductivity and other unique properties of matter.

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Duba Amsa

Tambaya 16 Rahoto

Which of the following reduce(s) the effect of the back e.m.f. generated in the primary coil of an induction coil? The capacitor in the circuit II. The make and break contact in the circuit III. The ratio of turns in the secondary to that in the primary

l only

II only

I and II only Amsar da ta dace

II and III only

I, II and III.

Bayanin Amsa

An induction coil is a device that converts high voltage, low current power to low voltage, high current power through electromagnetic induction. When an alternating current passes through the primary coil, a changing magnetic field is produced which induces an alternating current in the secondary coil. The back e.m.f. generated in the primary coil of an induction coil opposes the flow of current in the coil, causing a decrease in the current. This effect can be reduced by using a capacitor in the circuit, which stores the charge and releases it when the current drops. This maintains the flow of current and reduces the effect of the back e.m.f. Therefore, option I is correct. The make and break contact in the circuit also helps to reduce the effect of the back e.m.f. The contact repeatedly breaks the circuit, causing the current to stop flowing momentarily, which interrupts the magnetic field and reduces the back e.m.f. Therefore, option II is also correct. The ratio of turns in the secondary to that in the primary does not affect the back e.m.f. generated in the primary coil. It only affects the voltage induced in the secondary coil. Therefore, option III is incorrect. Hence, the correct answer is option C, I and II only.

Karanta bayanin darasi akan Capacitors (JAMB)

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

Capacitors

Simple A.C Circuits

Duba Amsa

Tambaya 17 Rahoto

A force F is applied to a body P as shown in the diagram above. If the body P moves through a distance r, which of the following represents the work done?

F^cos?/_x

Fx tan?

Fx sin?

Fx cos? Amsar da ta dace

Bayanin Amsa

Work done (W) is defined as the product of force (F) and displacement (s) in the direction of force. In the given diagram, the force F is acting at an angle ? to the direction of displacement, and the displacement is given by r. Therefore, the work done can be calculated as follows: W = F x r cos? Here, F represents the magnitude of the force, r represents the displacement of the body, and cos? represents the cosine of the angle between the force and displacement vectors. Therefore, the correct option for the work done is Fx cos?.

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

Work, Energy And Power

Duba Amsa

Tambaya 18 Rahoto

When the direction of vibration of the particles of a medium is perpendicular to the direction of travel of a wave the wave transmitted is known as

sound wave

transverse wave Amsar da ta dace

longitudinal wave

stationary wave

mechanical wave

Bayanin Amsa

The wave transmitted in which the direction of vibration of the particles of a medium is perpendicular to the direction of travel of the wave is called a transverse wave. In a transverse wave, the oscillations of the particles of the medium occur at right angles to the direction of propagation of the wave. Examples of transverse waves include waves on a string, electromagnetic waves, and seismic S-waves.

Karanta bayanin darasi akan Sound Waves (WAEC)

Karanta bayanin darasi akan Production And Propagation Of Waves (WAEC)

Karanta bayanin darasi akan Reflection Of Light At Plane And Curved Surfaces (JAMB)

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

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

Sound Waves

Production And Propagation Of Waves

Reflection Of Light At Plane And Curved Surfaces

Properties Of Waves: Reflection, Refraction, Diffraction

Simple A.C Circuits

Duba Amsa

Tambaya 19 Rahoto

A cell of e.m.f. 1.5 V and internal resistance of 2.59 is connected in series with an ammeter of resistance 0.59 and a resistor of resistance 7.09. Calculate the current in the circuit.

0.15 A Amsar da ta dace

0.20 A

0.60A

3.00A

6.67 A.

Bayanin Amsa

To calculate the current in the circuit, we need to use Ohm's law and Kirchhoff's circuit laws. First, we can calculate the total resistance of the circuit by adding the resistance of the ammeter and the resistor in series: R = 0.59 Ω + 7.09 Ω = 7.68 Ω Using Kirchhoff's circuit laws, we know that the current I through the circuit is equal to the e.m.f. E of the cell divided by the total resistance R of the circuit plus the internal resistance r of the cell: I = E / (R + r) Substituting the given values, we get: I = 1.5 V / (7.68 Ω + 2.59 Ω) = 0.15 A Therefore, the current in the circuit is 0.15 A. So, the answer is 0.15 A.

Karanta bayanin darasi akan Current Electricity (JAMB)

Karanta bayanin darasi akan Electric Cells (JAMB)

Karanta bayanin darasi akan Simple Harmonic Motion (WAEC)

Current Electricity

Electric Cells

Simple Harmonic Motion

Duba Amsa

Tambaya 20 Rahoto

Water does not drip through an open umbrella of silk material unless the inside of the umbrella is touched. Which of the following phenomena is responsible for this?

Surface tension Amsar da ta dace

Hydrostatic upthrust

Viscosity

Diffusion

Osmosis

Bayanin Amsa

The phenomenon responsible for water not dripping through an open silk umbrella is surface tension. Surface tension is the property of a liquid that causes its surface to behave like a stretched elastic membrane. The water droplets adhere to the silk fabric of the umbrella and are held in place by the cohesive forces between water molecules. The umbrella's fabric has a network of fibers, which creates capillary action that results in an upward force that balances the weight of the water, making it difficult for the water to fall through the umbrella. Only when the inside of the umbrella is touched and the tension is disrupted, the water droplets fall through the fabric.

Karanta bayanin darasi akan Measurements And Units (JAMB)

Measurements And Units

Duba Amsa

Tambaya 21 Rahoto

The nucleon number and the proton number of a neutral atom of an element are 238 and 92 respectively. What is the number of neutrons in the atom?

330

165

146 Amsar da ta dace

119

Bayanin Amsa

The nucleon number of an atom is the sum of its protons and neutrons. Therefore, the number of neutrons in an atom can be calculated by subtracting the number of protons (also called the atomic number) from the nucleon number. In this case, the nucleon number is given as 238 and the proton number (atomic number) is given as 92. Thus, the number of neutrons can be calculated as: Number of neutrons = nucleon number - proton number Number of neutrons = 238 - 92 Number of neutrons = 146 Therefore, the answer is 146.

Karanta bayanin darasi akan Structure Of The Nucleus (WAEC)

Structure Of The Nucleus

Duba Amsa

Tambaya 22 Rahoto

What is the electric potential energy between two protons each of charge q and at a distance of r apart? (Permittivity of free space = ε_o)

4πε_oq²r

qr/4πε_o

q/4πε_or Amsar da ta dace

q²/4πε_or

q 2 /4πε o r 2

Bayanin Amsa

Karanta bayanin darasi akan Electrostatics (JAMB)

Electrostatics

Duba Amsa

Tambaya 23 Rahoto

An oscillating pendulum has a velocity of 2ms^-1 at the equilibrium position O and velocity at same point P. Using the diagram above, calculate the height of P above O. (Take g = 10ms²)

5.0m

2.0m

0.4m

0.2m Amsar da ta dace

0.1m

Bayanin Amsa

To solve this problem, we need to use the conservation of mechanical energy of the pendulum. At the highest point P, the pendulum is momentarily at rest, and all its kinetic energy has been converted to potential energy. Therefore, the sum of kinetic and potential energies at P is equal to the sum of kinetic and potential energies at O. At position O, the pendulum has only kinetic energy, which is given by the formula 1/2 * m * v^2, where m is the mass of the pendulum and v is its velocity. We are not given the mass of the pendulum, but we can assume that it is constant throughout the motion. Therefore, the kinetic energy at O is 1/2 * m * (2)^2 = 2m. At position P, the pendulum has only potential energy, which is given by the formula m * g * h, where h is the height of P above O. Therefore, the potential energy at P is m * g * h. Since the total mechanical energy of the pendulum is conserved, we can equate the kinetic energy at O to the potential energy at P: 2m = m * g * h Simplifying this equation, we get: h = 2 / g = 2 / 10 = 0.2m Therefore, the height of P above O is 0.2m. Answer option (D) is correct.

Karanta bayanin darasi akan Scalars And Vectors (WAEC)

Karanta bayanin darasi akan Simple Harmonic Motion (WAEC)

Karanta bayanin darasi akan Gravitational Field (WAEC)

Scalars And Vectors

Simple Harmonic Motion

Gravitational Field

Duba Amsa

Tambaya 24 Rahoto

A rectangular coil of wire can rotate in a magnetic field. The ends of the coil are soldered to the two halves of a split ring. Two carbon brushes are made to press lightly against the split ring and when these are connected in circuit with a battery and rheostat, the coil rotates. This is a description of

a suspended- coil vanometer

a moving-cell ammeter

a d.c generator Amsar da ta dace

an electric motor

an induction coil,

Bayanin Amsa

Karanta bayanin darasi akan Magnets And Magnetic Fields (JAMB)

Karanta bayanin darasi akan Gravitational Field (JAMB)

Magnets And Magnetic Fields

Gravitational Field

Duba Amsa

Tambaya 25 Rahoto

Which of the following pairs of light rays shows the widest separation in the spectrum of white light?

Yellow and red

Orange and green

Violet and blue

Blue and green

Red and indigo Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Light Energy (JAMB)

Light Energy

Duba Amsa

Tambaya 26 Rahoto

Given that the gravitational constant is 7 X 10^-11 Nm2kg^-2, what is the force of attraction between 10⁶kg mass of lead hanging one metre away from a 10³kg mass of iron?

7 x10^-20N

7 x 10^-5N

7x10^-2N Amsar da ta dace

7x10³N

7x10⁰N

Bayanin Amsa

The force of gravitational attraction between two masses can be calculated using the formula F = G(m1m2)/r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers of mass. Using this formula, and plugging in the given values, we have: F = (7 x 10^-11 Nm^2/kg^2) x (10^6 kg) x (10^3 kg) / (1 m)^2 F = 7 x 10^-2 N Therefore, the force of attraction between the lead mass and the iron mass is 7 x 10^-2 N.

Karanta bayanin darasi akan Gravitational Field (JAMB)

Karanta bayanin darasi akan Gravitational Field (WAEC)

Gravitational Field

Duba Amsa

Tambaya 27 Rahoto

How much heat is given out when a piece of iron of mass 50g and specific heat capacity 460 Jkg^-1k^-1 cools from 85^oC to 25^oC?

1.38 x 10⁶ J

2.53x10⁴J

7.66x10³J

1.38x10³J Amsar da ta dace

1.27 X10³J

Bayanin Amsa

The amount of heat given out by the iron can be calculated using the formula: Q = mcΔT Where Q is the heat energy transferred, m is the mass of the iron, c is the specific heat capacity of iron, and ΔT is the change in temperature of the iron. Substituting the given values, we have: Q = (0.05 kg) × (460 J/kg·K) × (85°C - 25°C) Q = 0.05 kg × 460 J/kg·K × 60°C Q = 1380 J Therefore, the amount of heat given out by the iron when it cools from 85°C to 25°C is 1380 J. Hence the correct option is (d) 1.38x10³J.

Karanta bayanin darasi akan Heat Energy (WAEC)

Heat Energy

Duba Amsa

Tambaya 28 Rahoto

A plane inclined at an angle at 30^o to the horizontal has an efficiency of 50%. The force parallel to the plane required to push a load of 120 N uniformly up the plane is

40√3N

60N

120N Amsar da ta dace

200N

240N

Bayanin Amsa

Karanta bayanin darasi akan Motion (JAMB)

Karanta bayanin darasi akan Measurements And Units (JAMB)

Motion

Measurements And Units

Duba Amsa

Tambaya 29 Rahoto

What is the value of R when G shows no deflection in the circuit illustrated above?

80?

75?

45?

20? Amsar da ta dace

10?

Bayanin Amsa

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

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

Electromagnetic Field (Part 2)

Electromagnetic Field (Part 1)

Duba Amsa

Tambaya 30 Rahoto

Which of the following is not an evidence of the particle nature of matter?

Diffusion

Brownian motion

Diffraction Amsar da ta dace

Crystal structure

Photoelectricity

Bayanin Amsa

Diffraction is not an evidence of the particle nature of matter. Diffraction is the bending of waves around an obstacle or through an opening. It is a property of waves, and it is not specific to the particle nature of matter. In fact, diffraction is often used to demonstrate the wave-like behavior of particles, such as electrons. On the other hand, the other options listed are evidences of the particle nature of matter: - Diffusion is the process of particles spreading out from an area of high concentration to an area of low concentration. It is a result of the movement of individual particles, and it is evidence of their particle-like nature. - Brownian motion is the random movement of particles in a fluid due to collisions with other particles. This motion is caused by the individual particles that make up the fluid, and it is evidence of their particle-like nature. - Crystal structure refers to the orderly arrangement of atoms or molecules in a solid. This structure is a result of the individual particles that make up the solid, and it is evidence of their particle-like nature. - Photoelectricity is the emission of electrons from a material when light shines on it. This phenomenon is evidence of the particle-like nature of light, which is made up of particles called photons. Therefore, the correct answer is "Diffraction".

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

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 31 Rahoto

Calculate the resistance of the filament of a lamp rates 240 V, 40 W

240Ω

360Ω

720Ω

1440Ω Amsar da ta dace

2880Ω

Bayanin Amsa

The power (P) of the lamp is given as 40 W and the voltage (V) is given as 240 V. Using Ohm's Law, we can calculate the resistance (R) of the filament of the lamp. Ohm's Law states that V = IR, where V is the voltage, I is the current, and R is the resistance. Using the formula for power, we know that P = IV. Substituting Ohm's Law into this equation gives P = (V²/R). Rearranging this formula gives R = V²/P. Substituting the given values gives R = (240²/40) = 1440 Ω. Therefore, the resistance of the filament of the lamp is 1440 Ω. Answer: 1440Ω

Karanta bayanin darasi akan Current Electricity (JAMB)

Current Electricity

Duba Amsa

Tambaya 32 Rahoto

Two solids P and Q of equal mass are heated to the same temperature and dropped simultaneously into two identical containers X and Y containing equal volumes of water at room temperature. The temperature of X will rise higher than that of Y only if the

volume of P is greater than that of Q

specific heat capacity of P is higher than that of Q Amsar da ta dace

density of P is greater than that of Q

density of P is lower than that of Q

specitic heat capacity of P is lower than that of Q.

Bayanin Amsa

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Duba Amsa

Tambaya 33 Rahoto

Birds can perch safely on overhead electric power cables without being electrocuted because

birds can withstand high p.d. of the electric power cables

any accumulated charge on the birds is quickly dissipated

the feet of the birds are well insulated Amsar da ta dace

the current is a.c. and has particular direction

there is no p.d between the birds and the cables

Bayanin Amsa

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

Karanta bayanin darasi akan Magnets And Magnetic Fields (JAMB)

Karanta bayanin darasi akan Gravitational Field (JAMB)

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

Karanta bayanin darasi akan Motion (WAEC)

Simple A.C. Circuits

Magnets And Magnetic Fields

Gravitational Field

Simple A.C Circuits

Motion

Duba Amsa

Tambaya 34 Rahoto

A radioactive substance has a half-life of 20 hours. What fraction of the original radioactive nuclei will remain after 80 hours?

1/32

1/16 Amsar da ta dace

1/8

1/4

1/2

Bayanin Amsa

The half-life of a radioactive substance is the time taken for half of the radioactive nuclei to decay. After one half-life, half of the radioactive nuclei have decayed and only half remain. Similarly, after two half-lives, only a quarter of the original radioactive nuclei remain, and after three half-lives, only an eighth remain. In this question, the half-life is given as 20 hours, and the time elapsed is 80 hours, which is 4 times the half-life. Therefore, the number of half-lives that have passed is 4. So, after 4 half-lives, the fraction of the original radioactive nuclei remaining is (1/2) x (1/2) x (1/2) x (1/2) = 1/16. Therefore, the answer is: 1/16 of the original radioactive nuclei will remain after 80 hours.

Karanta bayanin darasi akan Wave-particle Paradox (WAEC)

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

Wave-particle Paradox

Properties Of Waves: Reflection, Refraction, Diffraction

Duba Amsa

Tambaya 35 Rahoto

Calculate the current in the 3? resistor shown in
the diagram above.

3.0A

4.0A Amsar da ta dace

4.3A

12.0A

39.0A

Bayanin Amsa

Karanta bayanin darasi akan Current Electricity (JAMB)

Karanta bayanin darasi akan Scalars And Vectors (JAMB)

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

Current Electricity

Scalars And Vectors

Simple A.C Circuits

Duba Amsa

Tambaya 36 Rahoto

A cube made of a metal of linear expansivity is warmed through a temperature of t. lf the initial volume of the cube is V what is the increase in volume of the cube?

1/3α Vt

1/2α Vt

α Vt

2α Vt

3α Vt Amsar da ta dace

Bayanin Amsa

Karanta bayanin darasi akan Thermal Expansion (JAMB)

Thermal Expansion

Duba Amsa

Tambaya 37 Rahoto

A charge of 1.0 x 10^-5 Coulombs experience force of 40N at a certain point in space. What is the electric field intensity?

8.00 x 10⁶NC^-1

4.00 x 10⁶NC^-1 Amsar da ta dace

4.00 x 10^-4NC^-1

2.00 x 10^-4NC^-1

2.50 x 10^-1NC^-1

Bayanin Amsa

The electric field intensity (E) at a point in space is defined as the electric force (F) per unit charge (q) at that point. Therefore, we can use the formula: E = F/q Substituting the given values, we get: E = 40 N / 1.0 x 10^-5 C = 4.00 x 10⁶ N/C Thus, the electric field intensity at the given point in space is 4.00 x 10⁶ N/C. Therefore, the correct option is: 4.00 x 10⁶NC^-1.

Karanta bayanin darasi akan Electric Field (WAEC)

Electric Field

Duba Amsa

Tambaya 38 Rahoto

A concave mirror of radius of curvature 20cm has a pin placed at 15cm from its pole. What will be the magnification of the image formed?

4.00

2.00 Amsar da ta dace

1.33

1.50

0.25

Bayanin Amsa

When an object is placed in front of a concave mirror, the mirror forms an image of the object. The image can either be real or virtual, and it can be magnified or diminished depending on the location of the object and the focal length of the mirror. In this case, the concave mirror has a radius of curvature of 20cm, which means its focal length is half of that or 10cm. The pin is placed at a distance of 15cm from the pole of the mirror, which means it is located beyond the focal point. Since the object is placed beyond the focal point, the image formed will be real and inverted. The magnification of the image can be calculated using the formula: magnification = height of image / height of object = -v/u where v is the distance of the image from the mirror, u is the distance of the object from the mirror, and the negative sign indicates that the image is inverted. Using the mirror formula (1/f = 1/v + 1/u), we can find the distance of the image from the mirror: 1/f = 1/v + 1/u 1/10 = 1/v + 1/15 v = 30cm Now, we can use the magnification formula to find the magnification of the image: magnification = height of image / height of object = -v/u magnification = -30/15 magnification = -2.00 The negative sign indicates that the image is inverted, and the magnification is 2.00. Therefore, the answer is 2.00.

Duba Amsa

Tambaya 39 Rahoto

Which of the following is/are necessary for the production of interference with two wave trains? I. The sources must be close to each other. II The waves must have the same frequency. III. The waves must have the same amplitude. IV. The waves must travel with the speed of light.

I, II and IV only

I, II and III only Amsar da ta dace

I and II only

II, III and IV only

l, III and IV only

Bayanin Amsa

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

Karanta bayanin darasi akan Types Of Waves (WAEC)

Properties Of Waves: Reflection, Refraction, Diffraction

Types Of Waves

Duba Amsa

Tambaya 40 Rahoto

The diagram above shows an incident ray A, inclined at angle of 50^o to the interface CB. The ray OB is found to lie along the surface. What is the refractive index of the medium X with respect to air?

sin50^o/sin40^o

sin40^osin50^o

sin90^o/sin50^o

sin40^o/sin90^o

sin90^o/sin40^o Amsar da ta dace

Bayanin Amsa

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

Refraction Of Light Through Plane And Curved Surfaces

Duba Amsa

Tambaya 41 Rahoto

Which of the following is not crystal?

Sodium chloride

Lead Amsar da ta dace

Copper (II) tetraoxosulphate (VI)

Glass

Germanium

Bayanin Amsa

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

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 42 Rahoto

Which of the following is/are characteristics of sound? I Pitch II. Loudness III. Quality IV Noise

I only

II only

I and II only

l, II and III only Amsar da ta dace

I, II, III and IV

Bayanin Amsa

Karanta bayanin darasi akan Current Electricity (JAMB)

Current Electricity

Duba Amsa

Tambaya 43 Rahoto

lf the temperature of a small quantity of water in a closed container is gradually increased from 0^o to 40^oC then the density of the water within this range

increases for a while and then decreases Amsar da ta dace

decreases for a while and then increases

increases gradually

decreases gradually

remains the same

Bayanin Amsa

The density of water decreases as its temperature increases from 0^oC to 4^oC, after which it increases. Therefore, as the temperature of the small quantity of water in a closed container is gradually increased from 0^oC to 40^oC, its density will first decrease until it reaches a minimum value at 4^oC, and then it will increase. Therefore, the correct answer is (A) "increases for a while and then decreases".

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Duba Amsa

Tambaya 44 Rahoto

How long will it take to heat 3 kg of water from 28^oC to 88^oC in an electric taking a current of 6 A from an e.m.f. source of 220V?

96s

1.20s

570s Amsar da ta dace

600s

1200s

Bayanin Amsa

To solve this problem, we need to use the formula: Q = m c ΔT where Q is the heat energy required to heat the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. The specific heat capacity of water is 4.18 J/g°C, and since we have 3 kg of water, the mass of water is 3000 g. So, Q = 3000 g × 4.18 J/g°C × (88°C - 28°C) = 376440 J Now, we can use the formula for electrical energy: E = V I t where E is the electrical energy, V is the voltage, I is the current, and t is the time. Rearranging the formula to solve for t, we get: t = E / (V I) Substituting the given values, we get: t = 376440 J / (220 V × 6 A) = 570 s Therefore, the time it will take to heat 3 kg of water from 28°C to 88°C in an electric taking a current of 6 A from an e.m.f. source of 220V is 570 seconds or 9.5 minutes. So, the correct answer is 570s.

Karanta bayanin darasi akan Heat Energy (WAEC)

Karanta bayanin darasi akan Gravitational Field (JAMB)

Heat Energy

Gravitational Field

Duba Amsa

Tambaya 45 Rahoto

Viscosity in a liquid does not depend on the

nature of the liquid

relative velocity between the liquid layers

area of the surfaces in contact

temperature of the liquid

normal reaction between the liquid layers Amsar da ta dace

Bayanin Amsa

Viscosity is a measure of a liquid's resistance to flow, or how "thick" or "sticky" it is. The viscosity of a liquid depends on the nature of the liquid, the temperature of the liquid, and the relative velocity between the liquid layers. However, it does not depend on the area of the surfaces in contact or the normal reaction between the liquid layers. To put it simply, imagine you have two identical cups filled with the same liquid. If you pour the liquid from one cup to the other, you will notice that the liquid flows more slowly than water. This is because the liquid has a higher viscosity. The viscosity depends on the type of liquid (e.g. honey is more viscous than water), and how fast or slow the liquid is moving (e.g. syrup flows more slowly than water). The viscosity also changes with temperature, as higher temperatures tend to decrease the viscosity of a liquid. However, the area of the surfaces in contact and the normal reaction between the liquid layers do not affect the viscosity of a liquid. These factors might affect how the liquid flows, but they don't change the liquid's intrinsic property of viscosity.

Karanta bayanin darasi akan Pressure (JAMB)

Karanta bayanin darasi akan Liquids At Rest (JAMB)

Pressure

Liquids At Rest

Duba Amsa

Tambaya 46 Rahoto

Which of the following particles conduct electricity through salty water?

Atom

molecules

Electrons

Neutrons

Ions Amsar da ta dace

Bayanin Amsa

Particles that conduct electricity through salty water are called ions. When salt is dissolved in water, it dissociates into positively charged ions (such as sodium ions) and negatively charged ions (such as chloride ions). These ions are able to move freely in the water, carrying an electric charge and allowing an electric current to flow through the solution. In contrast, atoms and molecules do not conduct electricity in water because they do not have a net electric charge and are not able to move freely. Electrons and neutrons are subatomic particles that can be found in atoms, but they are not able to move freely in water and therefore cannot conduct electricity in this medium. Therefore, the correct option is "Ions".

Karanta bayanin darasi akan Electrostatics (JAMB)

Electrostatics

Duba Amsa

Tambaya 47 Rahoto

A bat emits a sound wave at a speed of 1650.0ms^-1 and receives the echoes 0.15s later. Calculate the distance of the bat from the reflector

8.75m

16.50m

87.75m

123.75m Amsar da ta dace

330.00m

Bayanin Amsa

The distance of the bat from the reflector can be calculated using the speed of sound and the time taken for the echo to return. The speed of sound is given as 1650.0ms^-1. When the bat emits a sound wave, it travels to the reflector and back, covering a total distance equal to twice the distance from the bat to the reflector. The time taken for this round trip is 0.15s (as given in the question). Using the equation: Distance = Speed x Time We can calculate the distance of the bat from the reflector as: Distance = (Speed of sound x Time taken)/2 Distance = (1650.0ms^-1 x 0.15s)/2 Distance = 123.75m Therefore, the correct answer is (d) 123.75m.

Karanta bayanin darasi akan Current Electricity (JAMB)

Karanta bayanin darasi akan Sound Waves (WAEC)

Current Electricity

Sound Waves

Duba Amsa

Tambaya 48 Rahoto

A metal is illuminated with a radiation of energy 6.88eV If the kinetic energy of the emitted electrons is 1.5OeV, calculate the work function of the metal.

0.22eV

4.59eV

5.38eV Amsar da ta dace

8.38eV

10.32eV

Bayanin Amsa

Karanta bayanin darasi akan Elementary Modern Physics (JAMB)

Elementary Modern Physics

Duba Amsa

Tambaya 49 Rahoto

A simple microscope forms an image twice the size of the object. If the focal length of the lens of the microscope is 20m, how far is the object from the lens?

10m

20m

30m Amsar da ta dace

40m

60m

Bayanin Amsa

Karanta bayanin darasi akan Electrostatics (JAMB)

Electrostatics

Duba Amsa

Tambaya 50 Rahoto

Which of the following are contact forces? I. Force of tension II. Force of fiction, III. Magnetic force III Force of reaction.

I,II and III only

I, II and IV only Amsar da ta dace

I, III and IV only

II, III and IV only

l and IV only.

Bayanin Amsa

Contact forces are forces that require physical contact between two objects to occur. Based on this definition, only forces of tension and friction are considered contact forces among the options given. Force of tension is a force that is transmitted through a string, rope, or cable when it is pulled tight by forces acting from opposite ends. It is a contact force because it requires physical contact between the string or cable and the objects pulling on it. Force of friction is a force that opposes motion between two surfaces in contact. It is a contact force because it requires physical contact between the two surfaces. Magnetic force and force of reaction are not contact forces because they do not require physical contact between two objects to occur. Magnetic force occurs between two magnetic objects, and force of reaction is the force exerted by a surface in response to an object pressing against it. Therefore, the correct answer is (b) I, II, and IV only.

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Karanta bayanin darasi akan Gravitational Field (WAEC)

Karanta bayanin darasi akan Equilibrium Of Forces (WAEC)

Temperature And Its Measurement

Gravitational Field

Equilibrium Of Forces

Duba Amsa

Tambaya 51 Rahoto

A simple pendulum makes 50 oscillations in one minute. What is its period of oscillation?

0.02s

0.20s

0.83s

1.20s Amsar da ta dace

50.00s

Bayanin Amsa

The period of a pendulum is the time it takes for one complete oscillation, which is the time it takes for the pendulum to swing back and forth once. In this question, we are given that the simple pendulum makes 50 oscillations in one minute. To find the period of oscillation, we need to divide the total time taken by the number of oscillations. We can convert one minute to seconds by multiplying it by 60, since there are 60 seconds in one minute. So, the time taken for 50 oscillations is: Time = 1 minute/50 oscillations = 60 seconds/50 oscillations = 1.2 seconds/oscillation Therefore, the period of oscillation of the pendulum is 1.2 seconds. Therefore, the correct option is (d) 1.20s.

Karanta bayanin darasi akan Simple Harmonic Motion (WAEC)

Simple Harmonic Motion

Duba Amsa

Tambaya 52 Rahoto

The eye defect illustrated by the diagram above is

astigmatism

long sightedness Amsar da ta dace

loss accommodation

short sightedness

colour blindness

Bayanin Amsa

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

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 53 Rahoto

The body P shown in the diagram above is in equilibrium. If the mass Q of the body is 10kg, calculate the tension T in the string. (Take g = 10ms²)

20/?3N

20N

50N

200?3N

200N Amsar da ta dace

Bayanin Amsa

The body P is in equilibrium, which means that the net force acting on it is zero. The weight of P acts downwards and is balanced by the tension in the string acting upwards. The weight of P is given by: Weight of P = mP * g = 20 kg * 10 ms^-2 = 200 N Since the net force acting on P is zero, the tension in the string is equal to the weight of P. Therefore: Tension in string = Weight of P = 200 N So, option (E) is the correct answer.

Karanta bayanin darasi akan Equilibrium Of Forces (JAMB)

Karanta bayanin darasi akan Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

Duba Amsa

Tambaya 54 Rahoto

The eye defect illustrated by the diagram above is

astigmatism

long sightedness Amsar da ta dace

loss accommodation

short sightedness

colour blindness

Bayanin Amsa

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

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 55 Rahoto

If the frequency of the ac. circuit illustrated above is 500/? Hz what would be the reactance in the circuit?

0.0009?/?

100?

1030?

1400? Amsar da ta dace

2500?

Bayanin Amsa

Duba Amsa

Tambaya 56 Rahoto

Two forces each of 4 N act on the opposite sides of a rectangular plate as shown in the diagram above. Calculate the magnitude of g the couple acting on the plate.

6.4 Nm

3.2 Nm

1.6 Nm Amsar da ta dace

0.8 Nm

0.0 Nm

Bayanin Amsa

Karanta bayanin darasi akan Equilibrium Of Forces (JAMB)

Karanta bayanin darasi akan Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

Duba Amsa

Tambaya 57 Rahoto

A stone is dropped into the middle of a pool of water. Which of the following statements is/are correct? I, Spherical waves are set up in the water. II. The water moves outwards to the sides of the pool. III. Energy is transmitted outwards from the centre of disturbance

I only

II only

III only

I and III only Amsar da ta dace

II and III only.

Bayanin Amsa

Karanta bayanin darasi akan Rectilinear Acceleration (WAEC)

Rectilinear Acceleration

Duba Amsa

Tambaya 58 Rahoto

A tuning fork sounds louder when its stem is pressed against a table top than when held in air because

a larger mass of air is set vibrating by the table top

the whole table vibrates in resonance

the whole table has acquired a larger frequency Amsar da ta dace

the fork and table have the same frequency

the amplitude of the vibrating fork is increased.

Bayanin Amsa

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

Karanta bayanin darasi akan Rectilinear Acceleration (WAEC)

Karanta bayanin darasi akan Gravitational Field (WAEC)

Newton’s Laws Of Motion

Rectilinear Acceleration

Gravitational Field

Duba Amsa

Tambaya 59 Rahoto

Which of the following statements about the atom is/are correct? I. Energy is needed to remove electrons from an atom II. Particles in the nucleus of an atom are bound together by strong forces. III. A large amount of energy is required to separate the particles in the nucleus of an atom. IV. When the nucleus of a uranium atom is split, the energy released accounts for the difference in the masses of the products and the parent nucleus

I only

I and II only

I and III only

II and IV only Amsar da ta dace

I, II, III and IV

Bayanin Amsa

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

Structure Of The Atom (Nigeria Only)

Duba Amsa

Tambaya 60 Rahoto

sin50^o/sin40^o

sin40^osin50^o

sin90^o/sin50^o

sin40^o/sin90^o

sin90^o/sin40^o Amsar da ta dace

Bayanin Amsa

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

Refraction Of Light Through Plane And Curved Surfaces

Duba Amsa

Tambaya 61 Rahoto

An object R leaves a platform XY with a horizontal velocity of 7m^-1 and lands at Q. lf it takes the same object 0.3s to fall freely from Y to P, calculate the distance PQ.(Take g = 1Oms²)

7.30m

2.55m

2.10m Amsar da ta dace

1.65m

0.45m.

Bayanin Amsa

The time of flight of an object in projectile motion can be calculated using the formula: t = 2u sin?/g where u is the initial velocity of the object, ? is the angle of projection, and g is the acceleration due to gravity. In this question, the object R leaves the platform with a horizontal velocity of 7m/s, which means its initial vertical velocity is 0. We can therefore use the formula: t = 2h/g where h is the height of the platform XY. We can rearrange this equation to solve for h: h = 1/2gt^2 We are given that it takes 0.3s for the object to fall freely from Y to P, so: h = 1/2 × 10 × (0.3)² = 0.45m Now, we can use the fact that the horizontal distance traveled by the object is given by: d = ut where d is the horizontal distance, u is the initial horizontal velocity, and t is the time of flight. Since there is no horizontal acceleration, the time of flight is the same as the time it takes for the object to fall from Y to P. Therefore: d = 7 × 0.3 = 2.1m Finally, we can use Pythagoras' theorem to find the distance PQ: PQ = ?(d² + h²) = ?(2.1² + 0.45²) ? 2.16m Therefore, the answer is 2.10m.

Karanta bayanin darasi akan Gravitational Field (JAMB)

Gravitational Field

Duba Amsa






Latsa & Riƙe don Ja Shi Gabaɗaya
Danna nan don rufewa