JAMB UTME - Physics - 2007 | Green Bridge CBT

Ibeere 1 Ìròyìn

A conductor has a diameter of 1.00 mm and length 2.00m. if the resistance of the material is 0.1Ω, its resistivity is?

2.55 x 105 Ωm

2.55 x 102 Ωm

3.93 x 10-6 Ωm

3.93 x 10-8 Ωm

Awọn alaye Idahun

R = (1)/a, P = (R.a)/1 = (0.1 x 3.14)/2 (10-3/2)3
∴ P = (0.1 x 3.14 x 10-6)/2 x 4
= 0.3925 x 10-7 OR -3.930 X 10-8 Ωm

Wo Idahun

Ibeere 2 Ìròyìn

A particle of weight 120N is placed on a plane inclined at 300 to the horizontal. If the plane has an efficiency of 60% to the floor what is the force required to push the weight uniformly up the plane?

50N

100N

175N

210N

Awọn alaye Idahun

Efficiency =

\frac{W (120 N) I N P U T}{W O R K O U T P U T} \times \frac{100}{1}

And for incline plane

V.R = $\frac{1}{S i n θ} = V . R = \frac{1}{S i n 30} = \frac{1}{0.5} = 2 therefore \frac{60}{120} = \frac{M . A}{2} = M . A = \frac{120}{100} = 1.2 but M.A = \frac{l o a d}{e f f o r t} = \frac{120}{E} therefore \frac{120}{E} = 1.2 = E = \frac{120}{1.2} = 100 N$

Therefore Effort up the plane = 100N

Ka ìwé ẹ̀kọ́ nípa Scalars And Vectors (JAMB)

Ka ìwé ẹ̀kọ́ nípa Equilibrium Of Forces (WAEC)

Ka ìwé ẹ̀kọ́ nípa Measurements And Units (JAMB)

Scalars And Vectors

Equilibrium Of Forces

Measurements And Units

Wo Idahun

Ibeere 3 Ìròyìn

The difference observed in solids, liquids and gases can be accounted for by

The different molecules in each of them

The spaces and forces acting between the molecules

Thier melting points

Their relative massess

Awọn alaye Idahun

The question is asking what accounts for the observed differences in solids, liquids, and gases. The options provide four possible answers to choose from. The correct answer is: "The spaces and forces acting between the molecules." Solids, liquids, and gases have different physical properties because of the way their molecules are arranged and how they interact with each other. The molecules in solids are tightly packed together and vibrate in place, while the molecules in liquids and gases have more space between them and can move around more freely. The spaces and forces between molecules determine how they behave under different conditions, such as temperature and pressure. This is why solids, liquids, and gases have different melting points and boiling points, and why they can have different densities and viscosities.

Ka ìwé ẹ̀kọ́ nípa Structure Of Matter And Kinetic Theory (JAMB)

Structure Of Matter And Kinetic Theory

Wo Idahun

Ibeere 4 Ìròyìn

A car of mass 1500 kg goes round a circular curve of radius 50m at a speed of 40 ms-1. The magnitude of the centripetal force on the car is ?

4.8 x 104 N

4.8 x 103 N

1.2 x 103 N

1.2 x 102 N

Awọn alaye Idahun

Centipetal force of the car = ma = (MV)2/r
= (1500 x 40 x40)/50
= 48000 = 4.8 x 104 N

Ka ìwé ẹ̀kọ́ nípa Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

Wo Idahun

Ibeere 5 Ìròyìn

.

Three cells each of e.m.f 1.5 V and internal resistance 2.5Ω are connected as shown in the diagram above. Find the net e.m.f and the internal resistance

1.5V, 7.50Ω

1,5V, 0.83Ω

4.5V, 7.50Ω

4.5V, 0.83Ω

Awọn alaye Idahun

Since the cells are connected in parallel the net e.m.f is that of one cell, E.
However the effective internal resistance should be added as parallel resistance.
=>1/v + 1/25 + 1/25 + 1/25 = 1.2
∴ v = 0.83Ω
∴ Effective e.m.f and internal resistance are respectively = 1.5V and 0.83Ω

Wo Idahun

Ibeere 6 Ìròyìn

A 120V, 60W lamp is to be operated on 220V ac supply mains. calculate the value of non inductive resistance that would be required to ensure that the lamp is run on correct value

100Ω

200Ω

300Ω

500Ω

Awọn alaye Idahun

When the voltage across a resistive load is increased, the power consumed by the load increases. In this problem, the voltage across the lamp is increased from 120V to 220V. This would cause an increase in power consumption, which may lead to the lamp being damaged. To prevent this from happening, a non-inductive resistor is connected in series with the lamp. To calculate the resistance required, we can use the formula: R = (V2 - V1)2 / P Where: V1 = initial voltage (120V) V2 = final voltage (220V) P = power (60W) Substituting the values, we get: R = (220V - 120V)2 / 60W R = 10000 / 60 R = 166.67 Ω Thus, a non-inductive resistance of approximately 167 Ω would be required to ensure that the lamp is run on the correct value. However, since none of the options match this value exactly, we would choose the option closest to the calculated value, which is 200 Ω.

Wo Idahun

Ibeere 7 Ìròyìn

A string of length 4m is extended by 0.02m when a load of 0.4kg is suspended at its end. What will be the length of the string when the applied force is 15N?

4.05m

4.08m

5.05m

6.08m

Awọn alaye Idahun

The problem involves finding the final length of a string when a certain force is applied to it, given its initial length and how much it extends under a certain load. To solve the problem, we can use Hooke's law, which states that the extension of an elastic object is directly proportional to the applied force, provided the elastic limit is not exceeded. In this case, we know that the string of length 4m extends by 0.02m when a load of 0.4kg is suspended at its end. This means that the spring constant k is: k = F/x = (0.4 kg * 9.81 m/s^2)/0.02 m = 196.2 N/m where F is the force, x is the extension, and 9.81 m/s^2 is the acceleration due to gravity. To find the length of the string when a force of 15N is applied, we can use Hooke's law again: F = kΔL where ΔL is the change in length of the string. Rearranging the equation to solve for ΔL, we get: ΔL = F/k = 15 N/196.2 N/m = 0.0764 m Therefore, the final length of the string is: L = 4m + ΔL = 4m + 0.0764m = 4.0764m So the answer is 4.08m.

Ka ìwé ẹ̀kọ́ nípa Scalars And Vectors (JAMB)

Scalars And Vectors

Wo Idahun

Ibeere 8 Ìròyìn

The fundamental frequency of a plucked wire under a tension of 400N is 250Hz. When the frequency is changed to 500 Hz at constant length, the tension is

1600N

400N

160N

40N

Awọn alaye Idahun

The tension in a wire affects its fundamental frequency. As tension increases, so does the frequency. The relationship between tension and frequency is given by the following formula: f = (1/2L)√(T/μ) where f is the frequency, L is the length of the wire, T is the tension, and μ is the linear mass density of the wire. We are given that the wire has a tension of 400N and a frequency of 250Hz. If we keep the length of the wire constant and increase the frequency to 500Hz, we can use the formula to find the new tension: f = (1/2L)√(T/μ) Rearranging the formula, we get: T = (4L^2μ)(f^2) Since we are keeping the length of the wire constant, we can simplify the formula to: T ∝ f^2 This means that the tension is proportional to the square of the frequency. If we increase the frequency by a factor of 2 (from 250Hz to 500Hz), the tension will increase by a factor of 2^2 = 4. Therefore, the new tension is: T = 4(400N) = 1600N So the answer is 1600N.

Wo Idahun

Ibeere 9 Ìròyìn

The energy E of a photon and it's wavelength are related by Eλ = X. the numerical value of X is?
[h = 6.63 x 10-34js, c = 3 x 108 ms-1]

6.6 X 10-28

1.99 X 10-27

6.60 x 10-26

1.99 x 10-25

Awọn alaye Idahun

The equation Eλ = X relates the energy E of a photon to its wavelength λ. The value of X can be found by substituting the given values for Planck's constant (h) and the speed of light (c) in the equation. Using the given values, X = hc = (6.63 x 10^-34 J s) x (3 x 10^8 m/s) = 1.99 x 10^-25 J m. Therefore, the correct option is 1.99 x 10-25.

Wo Idahun

Ibeere 10 Ìròyìn

A radioactive substance has a half-life of 20 days what fraction the original radiation nuclei will remain after 80 days?

(1)/32

(1)/16

(1)/ 8

(1)/4

Awọn alaye Idahun

The half-life of a radioactive substance is the time it takes for half of the original radioactive nuclei to decay. If the half-life of a substance is 20 days, this means that after 20 days, half of the original nuclei would have decayed, and half would remain. After another 20 days (a total of 40 days), half of the remaining nuclei would have decayed, leaving a quarter of the original nuclei. Similarly, after 60 days, half of the remaining quarter of nuclei would have decayed, leaving one-eighth of the original nuclei. Finally, after 80 days, half of the remaining one-eighth of nuclei would have decayed, leaving one-sixteenth of the original nuclei. Therefore, the fraction of the original radiation nuclei that will remain after 80 days is 1/16. So, the correct option is: (1)/16.

Ka ìwé ẹ̀kọ́ nípa Wave-particle Paradox (WAEC)

Wave-particle Paradox

Wo Idahun

Ibeere 11 Ìròyìn

118.8cm2 surface of the copper cathode of a voltameter is to be coated with 10-6m thick copper of density 9 x 103kgm-3. How long will the process run with 10A constant current?

[3.3 x 10-7kgC-1]

5.4 min

10.8 min

15.0 min

20.0 min

Awọn alaye Idahun

The problem is asking for the time it will take to deposit a certain thickness of copper on a copper cathode in a voltameter given the surface area of the cathode, the desired thickness of the deposit, the density of copper and a constant current. The formula to use is: time = (mass of copper deposited) / (current x electrochemical equivalent of copper) The mass of copper deposited can be calculated as: mass = volume x density = (surface area x thickness) x density The electrochemical equivalent of copper is the amount of copper deposited by one coulomb of charge passing through a copper ion in the electrolyte, and its value is 0.000329 gC^-1. Substituting the given values in the above formulas, we get: mass = (118.8 cm^2 x 10^-6 m) x (9 x 10^3 kgm^-3) = 1.0692 x 10^-3 kg electrochemical equivalent of copper = 0.000329 kgC^-1 time = (1.0692 x 10^-3 kg) / (10 A x 0.000329 kgC^-1) = 3.2548 s Converting the time to minutes, we get: time = 3.2548 s x (1 min / 60 s) = 0.0542 min ≈ 5.4 min Therefore, the answer is 5.4 min.

Wo Idahun

Ibeere 12 Ìròyìn

The instrument that measures both a.c and d.c is?

a current balance

a moving current ammeter

a moving iron ammeter

an inverter

Awọn alaye Idahun

The instrument that measures both AC and DC is the "moving iron ammeter." This type of ammeter measures the average value of the current, and its design allows it to work for both AC and DC. The other options listed are not suitable for measuring both AC and DC.

Ka ìwé ẹ̀kọ́ nípa Current Electricity (JAMB)

Current Electricity

Wo Idahun

Ibeere 13 Ìròyìn

A blacksmith heated a metal whose cubic expansivity is 6.3 x 10-6K-1. The areal expansivity is

2.0 x 10-6K-1

2.1 x 10-6K-1

4.2 x 10-6K-1

6.3 x 10-6K-1

Awọn alaye Idahun

If the lineal expansivity = ∝
=> area expansivity β= 2∝.
And cubic expansivity δ = 3∝.
=> 6.3 x 10-6 = 3∝
∴ ∝ =

∴ ∝ =	6.3 x 10-6	= 2.1x 10-6
	__3

∴ Area = 2∝ = 2 x 2.1x10-6
= 4.2 x 10-6K-1

Ka ìwé ẹ̀kọ́ nípa Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Wo Idahun

Ibeere 14 Ìròyìn

A gramophone record takes 5s to reach its constant angular velocity of $4\pi\ \text{rads}^{-1}$ from rest. Find its constant angular acceleration.

0.4

π

rads-2

0.8

π

rads-2

1.3

π

rads-2

20.0

π

rads-2

Awọn alaye Idahun

The gramophone record has an initial angular velocity of 0 and reaches a final angular velocity of 4π π rads-1 in 5 seconds. The constant angular acceleration can be found using the following formula: angular acceleration = (final angular velocity - initial angular velocity) / time Substituting the given values, we get: angular acceleration = (4π π rads-1 - 0) / 5s = 4π π rads-1 / 5s = 0.8π π rads-2 Therefore, the constant angular acceleration is 0.8π π rads-2.

Ka ìwé ẹ̀kọ́ nípa Newton’s Laws Of Motion (WAEC)

Newton’s Laws Of Motion

Wo Idahun

Ibeere 15 Ìròyìn

The ratio of the coefficient of linear expansion of two metals ∝1/∝2 is 3:4. If, when heated through the same temperature change, the ratio of the increase in length of the two metals, e1/e2 is 1:2, the ratio of the original lengths l1/l2 is

8/3

3/2

2/3

3/8

Awọn alaye Idahun

Ratio of their linear expansion = ∝1/∝2= 3:4.
When heated to the same temperature range, the ratio of their increase in length e1/e2 = 1:2
But the increase in length of
1 = e1 = ∝1l1Δθ and increase i length of
2 = e2 = ∝2l2Δθ

=>	e1	=	∝1l1Δθ	=	∝1l1	=	1
	e2		∝2l2Δθ		∝2l2		2

But	∝1	=	3	, ∴	3	x	l1	=	1
	∝2		4		4		l2		2

=> 6l1 = 4l2

∴	l1	=	4	=	2	OR 2 : 3
	l2		6		3

Ka ìwé ẹ̀kọ́ nípa Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Wo Idahun

Ibeere 16 Ìròyìn

I. Mass

II. Density

III. Temperature

IV. Nature of substance

Which of the above affect diffusion?

I and II only

I,II and IV only

II, III and IV only

I, II, III and IV

Awọn alaye Idahun

Rates of diffusion of a substance generally depend on density temp. and nature of substance II,III and IV

Ka ìwé ẹ̀kọ́ nípa Fundamental And Derived Quantities And Units (WAEC)

Fundamental And Derived Quantities And Units

Wo Idahun

Ibeere 17 Ìròyìn

A clinical thermometer is different from other mercury in glass thermometers owing to

the constriction on its stem

the grade of mercury used in it

its long stem

its wide range of temperatures

Awọn alaye Idahun

A clinical thermometer is different from other mercury in glass thermometers because of the constriction on its stem. The constriction prevents the mercury from falling back when the thermometer is removed from the patient's body. This allows the temperature reading to remain constant until it is reset. The other options are not correct because the grade of mercury used in a thermometer is usually the same, the length of the stem can vary depending on the thermometer's purpose, and the wide range of temperatures is not necessarily unique to clinical thermometers.

Ka ìwé ẹ̀kọ́ nípa Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Wo Idahun

Ibeere 18 Ìròyìn

Two long parallel wires X and Y carry currents 3A and 5A each. if the force experienced by unit length is 5 x 10-5N. the force per unit length experienced by wire Y is

3 x 10-5

3 x 10-6

5 x 105

5 x 10-5

Awọn alaye Idahun

since the wires are parallel to each other, the forces which they exert on each other are equal and opposite, since they must agree with newton's law of action an reaction. Hence the force exerted by the wire is equally
5 x 10-5

Ka ìwé ẹ̀kọ́ nípa Electrostatics (JAMB)

Electrostatics

Wo Idahun

Ibeere 19 Ìròyìn

A boy drags a bag of rice along a smooth horizontal floor with a force of 2N applied at the angle of 60o to the floor. The work done after a distance of 3m is

3 J

4 J

5 J

6 J

Awọn alaye Idahun

Effective component of the force along the horizontal = F cos 60o
Work done by this force : 60
F cos 60 x X
= 2 x 0.5 x 3
= 3J

Ka ìwé ẹ̀kọ́ nípa Work, Energy And Power (WAEC)

Ka ìwé ẹ̀kọ́ nípa Temperature And Its Measurement (JAMB)

Work, Energy And Power

Temperature And Its Measurement

Wo Idahun

Ibeere 20 Ìròyìn

The blade of a hoe feel colder to touch in the morning than the wooden handle because the

handle is a better conductor of heat than blade

handle contains stored energy in form of heat

blade is a better conductor of heat than the handle

blade is placed at a lower temperature than the heat

Awọn alaye Idahun

The blade of a hoe feels colder to touch in the morning than the wooden handle because the blade is a better conductor of heat than the handle. In the morning, the temperature of the environment is usually lower than the temperature of the objects in it. The blade being a better conductor of heat than the handle conducts the heat from our hand more efficiently, making it feel colder. The handle, on the other hand, being a poor conductor of heat, does not conduct the heat from our hand very well, so it feels warmer.

Ka ìwé ẹ̀kọ́ nípa Heat Transfer (JAMB)

Ka ìwé ẹ̀kọ́ nípa Heat Energy (WAEC)

Ka ìwé ẹ̀kọ́ nípa Conduction Of Electricity Through (JAMB)

Heat Transfer

Heat Energy

Conduction Of Electricity Through

Wo Idahun

Ibeere 21 Ìròyìn

A machine has a velocity ratio of 4. if it requires 800 N to overcome a load of 1600 N,what is the efficiency of the machine?

60%

50%

40%

2%

Awọn alaye Idahun

The efficiency of a machine is the ratio of output work to input work. In this case, the load is 1600 N and it requires 800 N to overcome it. So, the input work is 800 N × distance moved by effort. The output work is 1600 N × distance moved by load. Velocity ratio is the ratio of the distance moved by effort to the distance moved by load. Velocity Ratio = distance moved by effort / distance moved by load Velocity Ratio = 4 distance moved by effort = 4 × distance moved by load Input work = force × distance moved by effort Input work = 800 N × 4 × distance moved by load Output work = force × distance moved by load Output work = 1600 N × distance moved by load Efficiency = Output work / Input work Efficiency = (1600 N × distance moved by load) / (800 N × 4 × distance moved by load) Efficiency = 0.5 = 50% Therefore, the efficiency of the machine is 50%. Option (B) is the correct answer.

Ka ìwé ẹ̀kọ́ nípa Work, Energy And Power (WAEC)

Ka ìwé ẹ̀kọ́ nípa Simple Machines (JAMB)

Work, Energy And Power

Simple Machines

Wo Idahun

Ibeere 22 Ìròyìn

A 40 kW electric cable is used to transmit electricity through a resistor of resistance 2.0Ω at 800 V. The power loss as internal energy is?

4.0 x 102W

5.0 x 102W

4.0 x 103W

5.0 x 103W

Awọn alaye Idahun

In general, power = IV; 40kw = IV.
∴ 40,000 = 1 x 800
1 = (40.000)/800 = 50A. tune the current through resistor = 50A.
∴ power loss = 12R = 502 X 2 = 2500 X 2 = 5.0 X 103W

Ka ìwé ẹ̀kọ́ nípa Heat Energy (WAEC)

Heat Energy

Wo Idahun

Ibeere 23 Ìròyìn

A piece of iron weights 250 N in air and 200 N in a liquid of density 1000 kgm-3. the volume of the iron is?

5.0 x 10-2 ms-2

4.5 x 10--2 ms-3

2.5 x 10-2 ms-8

2.0 x 10-2 ms-3

Awọn alaye Idahun

With in the air = 250N
wt. in the liquid = 200N
loss in wt. in the liquid = 250 - 200 = 50N
∴ Wt. the liquid of equal vol. of the iron = 50N
∴If the density of the liquid is 1000 lgm-3
then the volume of the liquid = (N)/D = V = (25kg)/1000 =0.025m3 = 2.5 x 10-2 m8.

Ka ìwé ẹ̀kọ́ nípa Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Wo Idahun

Ibeere 24 Ìròyìn

I. They should be identical

II. They should originate from the same source.

III. They should be coherent.

IV. They should be monochromatic.

From the statements above, the conditions for two waves to interfere are?

I,II and III only

I, III and IV only

II, III and IV only

I, II, III and IV

Awọn alaye Idahun

The statements above describe the conditions for two waves to interfere. Interference occurs when two waves meet and overlap. For interference to occur, the waves should be identical (statement I), which means they have the same amplitude, frequency, and wavelength. The waves should also originate from the same source (statement II), which means they should have the same phase. In addition, the waves should be coherent (statement III), which means they have a constant phase difference. Finally, the waves should be monochromatic (statement IV), which means they have a single frequency. Therefore, the correct answer is option II, which includes statements I, III, and IV.

Ka ìwé ẹ̀kọ́ nípa Properties Of Waves: Reflection, Refraction, Diffraction (WAEC)

Ka ìwé ẹ̀kọ́ nípa Types Of Waves (WAEC)

Properties Of Waves: Reflection, Refraction, Diffraction

Types Of Waves

Wo Idahun

Ibeere 25 Ìròyìn

Silicon doped with aluminum and germanium doped with arsenic become?

p- and n- types respectively

n- en p- type respectively

p- type semiconductor

n- type semiconductors

Awọn alaye Idahun

When silicon, a four valency element, is doped with aluminum, a three valency element, a p-type semiconductor is produced where the major carriers are the electrons.
Thus, the correct option is p-type and n-type respectively

Ka ìwé ẹ̀kọ́ nípa Structure Of The Atom (Nigeria Only) (WAEC)

Structure Of The Atom (Nigeria Only)

Wo Idahun

Ibeere 26 Ìròyìn

The production of pure spectrum could easily be achieved using a?

glass prism with a pin

triangular prism with two convex lens

traingular prism only

triangular prism with two concave lenses

Light Energy

Reflection Of Light At Plane And Curved Surfaces

Wo Idahun

Ibeere 27 Ìròyìn

From the diagram above, the bob of the pendulum has the fastest speed at

X

Y

Z

W

Awọn alaye Idahun

The ball has its fastest speed at its equilibrium position when its displacement X = O; at X

Ka ìwé ẹ̀kọ́ nípa Motion (JAMB)

Motion

Wo Idahun

Ibeere 28 Ìròyìn

If two charged plates are maintained at a potential difference of 3 kv, the work done in taking a charge of 600 µC across the field is?

0.8 J

1.8 J

9.0 J

18.0 J

Awọn alaye Idahun

The work done in taking a charge across a potential difference is given by the formula: work = charge x potential difference In this case, the charge taken is 600 µC (which is 600 x 10^-6 C) and the potential difference between the plates is 3 kV (which is 3 x 10^3 V). Therefore, substituting these values into the above formula, we get: work = 600 x 10^-6 C x 3 x 10^3 V = 1.8 J Hence, the work done in taking a charge of 600 µC across the field is 1.8 J. Therefore, the answer is 1.8 J.

Ka ìwé ẹ̀kọ́ nípa Electrostatics (JAMB)

Electrostatics

Wo Idahun

Ibeere 29 Ìròyìn

A 120 V, 60 W lamp is to be operated on 220 V a.c. the supply mains. calculate the values of non-inductive resistance that would be required to ensure that the lamp is run on current value.

100 Ω

200 Ω

300 Ω

500 Ω

Awọn alaye Idahun

Power of the lamp = 60W = (V2)/R and the voltage rate = 120v.
R = (V2)/60 = (120 X 120)/60 = 240 Ω, the fuse wire that should ensure current does not exceed this value should be rated 200 Ω

Wo Idahun

Ibeere 30 Ìròyìn

When an object is placed between the principal focus and the optical center of a convex lens, it could be used as a?

Simple microscope

refecting lens

compound microscope

projector

Awọn alaye Idahun

When an object is placed between the principal focus and the optical center of a convex lens, it forms a virtual and magnified image. This image can be seen through the lens and can be used as a simple microscope to observe small objects or details. Therefore, the correct answer is a "Simple microscope".

Ka ìwé ẹ̀kọ́ nípa Structure Of The Atom (Nigeria Only) (WAEC)

Structure Of The Atom (Nigeria Only)

Wo Idahun

Ibeere 31 Ìròyìn

I. Use a liquid with a high melting point.

II. use a liquid of high volume expansivity.

III. use a capillary tube of large diameter.

Which of the above best describes how the sensitivity of a liquid-in-glass thermometer can be enhanced?

I only

II. only

I and II only

II and III only

Awọn alaye Idahun

The question is asking how the sensitivity of a liquid-in-glass thermometer can be enhanced. The three options provided are: (I) use a liquid with a high melting point, (II) use a liquid of high volume expansivity, and (III) use a capillary tube of large diameter. Option I alone cannot enhance the sensitivity of the thermometer. The melting point of a liquid only affects the upper range of temperatures that the thermometer can measure, but it has no effect on the sensitivity or the precision of the measurements. Therefore, option I is not the correct answer. Option II can enhance the sensitivity of the thermometer. The volume expansivity of a liquid refers to how much the volume of the liquid changes with temperature. A liquid with high volume expansivity will change its volume more for a given change in temperature, resulting in a larger change in the length of the column in the thermometer. This means that the thermometer will be more sensitive to temperature changes. Therefore, option II is a correct answer. Option III alone cannot enhance the sensitivity of the thermometer. The diameter of the capillary tube only affects the rate at which the thermometer responds to temperature changes, but it has no effect on the sensitivity or the precision of the measurements. Therefore, option III is not the correct answer. Therefore, the correct answer is option II only, which is "II. use a liquid of high volume expansivity."

Ka ìwé ẹ̀kọ́ nípa Temperature And Its Measurement (JAMB)

Ka ìwé ẹ̀kọ́ nípa Pressure (JAMB)

Temperature And Its Measurement

Pressure

Wo Idahun

Ibeere 32 Ìròyìn

The instrument used by designers to obtain different color pattern is called?

periscop

sextant

episcope

kaleidoscope

Awọn alaye Idahun

The instrument used by designers to obtain different color patterns is called a kaleidoscope. A kaleidoscope is a tube-shaped optical instrument that reflects multiple mirrors to produce a symmetrical pattern of colors and shapes. Designers use a kaleidoscope to get inspiration for creating new patterns and designs in various fields, including textile design, graphic design, and art.

Ka ìwé ẹ̀kọ́ nípa Optical Instruments (JAMB)

Optical Instruments

Wo Idahun

Ibeere 33 Ìròyìn

In the diagram above , if the magnetic field points into the paper and the force on a current - carrying conductor points upwards, what is the direction of the current?

Upwards

Downwards

Right

Left

Awọn alaye Idahun

According to Fleming's left-hand rule, which relates the direction of the magnetic field, current, and force in a conductor, the direction of the current in the conductor will be perpendicular to both the direction of the magnetic field and the direction of the force acting on it. In this case, the magnetic field points into the paper (i.e., downwards, from our perspective), and the force on the conductor points upwards. Therefore, the direction of the current in the conductor must be perpendicular to both, which is to the left. Hence, the answer is "Left".

Ka ìwé ẹ̀kọ́ nípa Magnets And Magnetic Fields (JAMB)

Magnets And Magnetic Fields

Wo Idahun

Ibeere 34 Ìròyìn

During the nuclear reaction described by

235	W →	235	X→	231	Y
92		93		91

The particles emitted are respectively

α and α

α and β

β and β

β and α

Awọn alaye Idahun

The decay process illustrated above shows that the lost β -particle to decay to the X by increase its atomic number ( from 92 to 93) and leaving the mass no. unaffected. There after, x decayed by -α emission to give Y, by decreasing the mass no. by 2 ( from 93 to 91).
Thus , the emission is β and α particles respectively

Ka ìwé ẹ̀kọ́ nípa Position, Distance And Displacement (WAEC)

Position, Distance And Displacement

Wo Idahun

Ibeere 35 Ìròyìn

The instantaneous value of the induced e.m.f as a function of time is ε = εo sin ωt where εo is the peak value of the e.m.f. The instantaneous value of the e.m.f., one quarter of the period is

0.00

εo/4

εo/2

εo

Awọn alaye Idahun

The question is asking for the instantaneous value of the induced e.m.f at one quarter of the period. The given equation for the e.m.f as a function of time is ε = εo sin ωt, where εo is the peak value of the e.m.f. One quarter of the period corresponds to a time t = T/4, where T is the period of the waveform. At this time, the argument of the sine function is ωt = π/2. Therefore, the instantaneous value of the e.m.f. at one quarter of the period is ε = εo sin(π/2) = εo * 1 = εo. The answer is therefore (c) εo/2.

Ka ìwé ẹ̀kọ́ nípa Current Electricity (JAMB)

Current Electricity

Wo Idahun

Ibeere 36 Ìròyìn

Two spheres of masses 5.0 kg and 10.0 kg are 0.3 m apart. Calculate the force of attraction between them?
[G = 6.67 X 10-11 Nm2 Kg2]

3.50 x 10-10N

3.71 x 10-8 N

3.57 x 10-2N

4.00 x 102N

Awọn alaye Idahun

The force of attraction between the two spheres is given by F (GM1 M2)/r2 = (6.67X 10-11 X5X10)/(0.3)2
= 3.71 X 10-8N.

Ka ìwé ẹ̀kọ́ nípa Gravitational Field (JAMB)

Gravitational Field

Wo Idahun

Ibeere 37 Ìròyìn

2 kg of water is heated with a heating coil which draws 3.5 A from a 200 V mains for 2 minutes. what is the increase in temperature of the water?
[specific heat capacity of water = 4200 jkg -1k-1]

30°

25°

15°

10°

Awọn alaye Idahun

The increase in temperature of the water can be calculated using the formula: Q = mcΔT Where Q is the heat energy transferred to the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. First, let's calculate the heat energy transferred to the water: Power = VI = 200V × 3.5A = 700W Energy = Power × time = 700W × 2 minutes = 84000J Now, let's calculate the mass of the water: 1 kg of water has a volume of 1 liter, and the density of water is 1000 kg/m³. Therefore, 2 kg of water has a volume of 2 liters or 0.002 m³. Next, we can use the formula to calculate the change in temperature: ΔT = Q / (mc) ΔT = 84000J / (2kg × 4200J/kg.K) ΔT = 10°C Therefore, the increase in temperature of the water is 10°C. Option (d) is the correct answer.

Ka ìwé ẹ̀kọ́ nípa Heat Energy (WAEC)

Heat Energy

Wo Idahun

Ibeere 38 Ìròyìn

In a large telecommunications auditorium, perforated absorbent materials are used to line the ceiling so as to

Increase the reverberation of sound in the hall

increase the amount of echo in the hall

reduce the hight of the ceiling from the floor

reduce the reverberation of sound in the hall

Awọn alaye Idahun

The purpose of perforated absorbent materials in a telecommunications auditorium is to reduce the reverberation of sound in the hall. Reverberation is the persistence of sound in an enclosed space after the sound source has stopped. The use of absorbent materials helps to reduce this effect, creating a clearer and more intelligible sound. Therefore, option D is the correct answer.

Ka ìwé ẹ̀kọ́ nípa Thermal Expansion (JAMB)

Ka ìwé ẹ̀kọ́ nípa Simple A.C Circuits (JAMB)

Ka ìwé ẹ̀kọ́ nípa Propagation Of Sound Waves (JAMB)

Thermal Expansion

Simple A.C Circuits

Propagation Of Sound Waves

Wo Idahun

Ibeere 39 Ìròyìn

The figure above shows 4 forces 3N, 10N, 3√3N and 6N acting on a particle P. The resultant of he four forces is

5N

5√ 3N

10N

10√ 3N

Awọn alaye Idahun

Resolution of force:
Total vertical upward force:
= 3√3 + 10 cos 60o
= 3√3 + 5.
Total downward force
= 6 cos 30o
= 6 x √(3/2) = 3√3
Total horizontal forces towards right;
= 3N + cos 30o
= 3N + 10 x √(3/2)
= 3N + 5√3
Total horizontal forces towards left:
= 6 cos 60o
= 6 x 0.5
= 3.0N
Summary:
Vertical resultant forces upward:
= 3√3 + 5 - 3√3
= 5N
Horizontal resultant forces towards right:
= 3N + 5√3 - 3√3
= 5N
Horizontal resultant force towards right
= 3N + 5√3 - 3N
= 5√3N

∴ R2 = 52 + (5√3)2
= 25 + 25 x 3
= 100
∴ R = √100
= 10N

Ka ìwé ẹ̀kọ́ nípa Scalars And Vectors (JAMB)

Ka ìwé ẹ̀kọ́ nípa Equilibrium Of Forces (WAEC)

Scalars And Vectors

Equilibrium Of Forces

Wo Idahun

Ibeere 40 Ìròyìn

The bond that forms a semiconductor is?

covalent

electrovalent

ionic

metallic

Awọn alaye Idahun

The bond that forms a semiconductor is a covalent bond. In a covalent bond, two or more atoms share electrons in order to achieve a stable configuration. In semiconductors, such as silicon and germanium, each atom has four valence electrons which form covalent bonds with neighboring atoms, resulting in a crystal lattice structure. This lattice structure creates a small energy gap between the valence and conduction bands, which allows some electrons to move from the valence band to the conduction band when energy is added to the system, resulting in electrical conductivity.

Ka ìwé ẹ̀kọ́ nípa Structure Of The Atom (Nigeria Only) (WAEC)

Structure Of The Atom (Nigeria Only)

Wo Idahun

Ibeere 41 Ìròyìn

The unit of moment of a couple can be expressed in?

N m-1

N m-2

N m

N m2

Awọn alaye Idahun

The unit of moment of a couple can be expressed in "N m", which stands for Newton-meters. This is because moment is defined as the product of the magnitude of the force and the perpendicular distance between the line of action of the force and the point about which the moment is being calculated. In the case of a couple, there are two equal and opposite forces acting at a perpendicular distance from each other, and the moment of the couple is the product of one of the forces and the distance between the forces. The unit of force is Newtons (N) and the unit of distance is meters (m), so the unit of moment is Newton-meters (N m).

Ka ìwé ẹ̀kọ́ nípa Measurements And Units (JAMB)

Measurements And Units

Wo Idahun

Ibeere 42 Ìròyìn

Lenz's law is a law of conservation of

electric charge

electric current

Energy

Momentum

Awọn alaye Idahun

Lenz's law is a fundamental law in physics that states that the direction of an induced current in a conductor will be such that it opposes the change that produced it. In simpler terms, Lenz's law is a law of conservation of energy that describes the direction of induced current in relation to a changing magnetic field. When a magnetic field changes, an electric current is induced in a conductor, and the direction of the induced current is such that it opposes the change in the magnetic field. Therefore, the correct answer is Energy.

Ka ìwé ẹ̀kọ́ nípa Electrostatics (JAMB)

Electrostatics

Wo Idahun

Ibeere 43 Ìròyìn

A glass plate 0.9cm thick has a reflective index of 1.50. how long does it take for a pulse of light to pass through the plate?
[c = 3.0 x 8 ms-1]

4.5 x 10-11s

4.5 x 10-10s

4 .5 x 10-12 s

3.0 x 10-10s

Awọn alaye Idahun

N = 1.5 = (speed of light in glass)/speed of light in air
(1.5)/1 = (3.0 x 108)/V
V = (3.0 x 108)/1.5
V = 2.0 x 108 ms-1
But x = Vt
t = (x)/Vt
t = (x)/v = (9 x 10)4/2.0x 108 = 4.5 x 10-2S
∴ Time = 4.5 x 10-12S

Ka ìwé ẹ̀kọ́ nípa Electromagnetic Field (Part 1) (WAEC)

Ka ìwé ẹ̀kọ́ nípa Gravitational Field (WAEC)

Electromagnetic Field (Part 1)

Gravitational Field

Wo Idahun

Ibeere 44 Ìròyìn

I. Increase the melting point of the liquid.

II. Increase the boiling point of the liquid.

III. Decrease the melting point of the liquid.

IV. Decrease the boiling point of the liquid.

Which of the statement above about the effect on increase in pressure in a liquid are correct?

I and II only

I and III only

II and III only

III and IV only

Awọn alaye Idahun

Increase in pressure generally raises the boiling point but but lowers the melting point. II and III

Ka ìwé ẹ̀kọ́ nípa Change Of State (JAMB)

Change Of State

Wo Idahun

Ibeere 45 Ìròyìn

The phenomenon of light bending round an obstacle is?

diffraction

refraction

reflection

polarization

Awọn alaye Idahun

The phenomenon of light bending around an obstacle is called diffraction. When light waves encounter an obstacle or an opening in an object that is comparable in size to the wavelength of the light, the waves will bend and spread out. This bending and spreading out of waves is known as diffraction. It is a characteristic of waves that causes them to bend around obstacles or to spread out after passing through small openings. Refraction is the bending of light as it passes through a medium, such as a lens or a prism. Reflection is the bouncing of light off a surface. Polarization refers to the orientation of the electric field in a light wave, which can be either parallel or perpendicular to the direction of the wave.

Ka ìwé ẹ̀kọ́ nípa Reflection Of Light At Plane And Curved Surfaces (JAMB)

Reflection Of Light At Plane And Curved Surfaces

Wo Idahun

Ibeere 46 Ìròyìn

A reservoir 500 m deep is filled with a fluid of density 850 kgm-3. If the atmospheric pressure is 1.05 x 105 Nm-2, the pressure at the bottom of the reservoir is.
[g = 10ms-2]

4.72 x 106 Nm-2

4.36 x 106 Nm-2

4.28 x 106 Nm-2

4.25 x 106 Nm-2

Awọn alaye Idahun

In general, pressure varies with depth in the relation p = lgh
= 850 x 10 x 500
= 4250000
= 4.25 x 106 NM-2

Ka ìwé ẹ̀kọ́ nípa Pressure (JAMB)

Pressure

Wo Idahun

Ibeere 47 Ìròyìn

From the diagram above, the inductive reactance and the resistance R are respectively

50Ω and 45Ω

25Ω and 51Ω

20Ω and 50Ω

10Ω and 50Ω

Awọn alaye Idahun

The inductive reactance is given as
Xl = 2πfl
= 2 x 50/π x 0.1
= 10Ω
Again from ohms law,
R = V/I = 75/1.5 = 50Ω
∴ Thus the inductive reactance and the resistance are respectively 10Ω, and 50Ω

Ka ìwé ẹ̀kọ́ nípa Capacitors (JAMB)

Ka ìwé ẹ̀kọ́ nípa Simple A.C Circuits (JAMB)

Capacitors

Simple A.C Circuits

Wo Idahun

Ibeere 48 Ìròyìn

A 40KW electric cable was uses to transmit electricity through a resistor of resistance 2.00Ω at 800V. The power loss as internal energy is

4.0 x 102W

5.0 x 102W

4.0 x 103W

5.0 x 103W

Awọn alaye Idahun

In general, Power = IV;

⟹ 40 K W = I V Therefore 40000 = 1 \times 800 ⟹ I = \frac{40000}{800} = 50 A, tune the current through

Resistor = 50A
power loss= I2R = 502 x 2
= 2500 x 2 = 5.0 x 103W

Ka ìwé ẹ̀kọ́ nípa Electrical Energy And Power (JAMB)

Electrical Energy And Power

Wo Idahun

Ibeere 49 Ìròyìn

Which of the following gas laws is equivalent to the work done?

Boyle' law

Charles law

Pressure law

van der Waal's law

Awọn alaye Idahun

Mathematically Boyle's law = PV = K
Where P = Pressure
and V = Volume

By definition Pressure = $\frac{F}{A} = \frac{F}{M \times M}$

Again Workdone = force x displacement

therefore $P V = \frac{F}{A} \times V = \frac{F}{M^{2}} \times \frac{M^{3}}{1} P V = \frac{F}{x \times x} \times \frac{x \times x \times x}{1} = F \times x$

= force x displacement

Ka ìwé ẹ̀kọ́ nípa Gas Laws (JAMB)

Gas Laws

Wo Idahun

Ibeere 50 Ìròyìn

A charge 50 µc has electric field strength of 360 NC-1 at a certain point. The electric field strength due to another charge 120 µc kept at the same distance apart and in the same medium is?

864 NC-1

150 NC-1

144 NC-1

18 NC-1

Awọn alaye Idahun

E =