JAMB UTME - Physics - 2001
Frage 1 Bericht
If a spherical metal bob of radius 3cm is fully immersed in a cylinder containing water and the water level rises by 1cm, what is the radius of the cylinder?
Antwortdetails
If the bob is fully immersed in the water, then the volume of water displaced will be equal to the volume of the bob = 4/3πr3, where r is the radius of the bob.
This volume of water in the cylinder = πr2l, where r = radius of the cylinder, and l = level of water rose in the cylinder.
Therefore, π x r2 x l = 4/3 x π x 33
Then r2 = 36
r = √36
r = 6cm
Frage 2 Bericht
What is the angle of dip at the magnetic equator
Antwortdetails
The angle of dip at the magnetic equator is 0 degrees. The magnetic dip angle or dip angle is the angle between the magnetic field lines and the horizontal plane. At the magnetic equator, the magnetic field lines are horizontal, so the angle of dip is zero. The magnetic equator is the location on the Earth's surface where the magnetic field is horizontal. At this location, the magnetic field lines are parallel to the Earth's surface, and there is no north-south component of the magnetic field.
Frage 3 Bericht
In the diagram above, which of the simple pendula will resonate with P when set into oscillation
Antwortdetails
The pendulum that will resonate with P is T. being of the same length and therefore of the same frequency.
Frage 4 Bericht
when a piece of rectangular glass block is inserted between two parallel capacitor,at constant plate area and distance of separation, the capacitance of the capacitor will
Antwortdetails
Frage 5 Bericht
The cost of running five 60 W lamps and four 100 W lamps for 20 hours if electrical energy cost #10.00 per kwh is
Antwortdetails
To calculate the cost of running the lamps, we need to know the total energy consumed by the lamps. The energy consumed is given by the product of the power rating of each lamp, the number of hours it is used, and the total number of lamps. For the 60 W lamps, the energy consumed is: 60 W x 5 lamps x 20 hours = 6000 Wh = 6 kWh For the 100 W lamps, the energy consumed is: 100 W x 4 lamps x 20 hours = 8000 Wh = 8 kWh Therefore, the total energy consumed by all the lamps is: 6 kWh + 8 kWh = 14 kWh Now we can calculate the cost of running the lamps: Cost = Energy consumed x Cost per kWh Cost = 14 kWh x $10.00/kWh Cost = $140.00 So the answer is $140.00.
Frage 6 Bericht
The resultant of two forces acting on an object is maximum if the angle between them is
Antwortdetails
When two forces are acting on an object, their resultant is the total force that produces the same effect as the individual forces combined. The magnitude and direction of the resultant depend on the magnitudes and directions of the individual forces. The angle between the forces affects the magnitude of the resultant force. The maximum resultant force occurs when the forces are acting in the same direction, which means the angle between them is 0 degrees. As the angle between the forces increases, the magnitude of the resultant force decreases. Therefore, the answer is 0 degrees.
Frage 7 Bericht
The process of energy production in the sun is
Antwortdetails
The process of energy production in the sun is: - nuclear fusion The sun produces energy through a process called nuclear fusion, which involves the fusion of atomic nuclei to form heavier elements. Specifically, hydrogen atoms combine to form helium, which releases a tremendous amount of energy in the form of light and heat. This process is the source of the sun's energy, and it is what allows it to emit light and heat for billions of years.
Frage 8 Bericht
The thermometric substance of an absolute thermometer is
Frage 9 Bericht
If ray travelling in air is incident on a transparent medium as shown in the diagram, the refractive index of the medium is given as
Antwortdetails
| From the diagram above, refractive index of the medium = | sin(90- α ) |
| _sin β |
| Reflective index of the medium = | cos α |
| sin β |
Frage 10 Bericht
A stone of mass 1kg is dropped from a height of 10m above the ground and falls freely under gravity. Its kinetic energy 5m above the ground is then equal to
Antwortdetails
At the height of 10m, its total energy is potential, given by P.E. = mgh = 1 x 10 x 10 = 100J.
But as the stone descends, its potential energy decreases, while its kinetic energy increases.
Half way during the fall, (h = 5m), its P.E. = 50J, and consequently, its K.E. should be 50J, such that from the principle of conservation of energy, the total energy of the stone at any instant should be conserved.
Frage 11 Bericht
The height at which the atmosphere ceases to exist is about 80km. If the atmospheric pressure on the ground level is 760mmHg, the pressure at a height of 20km above the ground level is
Antwortdetails
(p - 760)/(0 - 760) = (h - 0)/(80 - 0)
∴ (p - 760)/-760 = (20 - 0)/80
80(p-760) = -760(20-0)
∴80p - 60800 = -15200
80p = 60800 - 15200
80p = 45600
∴ p = 45600/80
p = 570mmHg
Frage 12 Bericht
A bread toaster uses a current of 4A when plugged in a 24oV line. it takes 1 minute to toast slices of bread. what is the energy consumed by the toaster
Antwortdetails
The energy consumed by the toaster can be calculated using the formula: Energy (E) = Power (P) x Time (t) We can find the power used by the toaster using the formula: Power (P) = Current (I) x Voltage (V) So, P = 4A x 240V = 960W Since the time taken to toast the bread is given as 1 minute, which is 60 seconds, we can now calculate the energy consumed: E = 960W x 60s = 57,600 J Therefore, the energy consumed by the toaster is 5.76 x 10^4 J (option A).
Frage 13 Bericht
The magnetic force on a charged particle moving with velocity v is
Antwortdetails
The magnetic force on a charged particle moving with velocity v is proportional to both the magnitude of the charge and the velocity v. This means that the greater the magnitude of the charge on the particle and the faster it is moving, the stronger the magnetic force will be. The magnetic force on a charged particle is given by the equation F = qvBsinθ, where q is the magnitude of the charge, v is the velocity of the particle, B is the strength of the magnetic field, and θ is the angle between the velocity vector of the particle and the magnetic field vector. Therefore, the magnetic force is dependent on both the charge and the velocity of the particle, and it is not solely proportional to either the magnitude of the charge or the velocity alone.
Frage 14 Bericht
A transistor function mainly as a
Antwortdetails
A transistor is an electronic device that has multiple uses in electronic circuits. Its main function is to amplify and switch electronic signals. Therefore, the option that best describes the function of a transistor is "switch and amplifier."
Frage 15 Bericht
The diagram above shows a closed side 0.5m in a uniform electric field E in the direction shown by the arrows. What is the flux Φ for the box?
Antwortdetails
0.0 E since one of the electric fields points in the opposite direction (left)
Frage 16 Bericht
The primary coil of a transformer has N turns and is connected to a 120 V a.c. power line. If the secondary coils has 1 000 turns and a terminal voltage of 1 200 volts, what is the value of N?
Antwortdetails
In a transformer, the ratio of the number of turns in the primary coil to that in the secondary coil is equal to the ratio of the primary voltage to the secondary voltage. This is given by the equation: N1/N2 = V1/V2 where N1 and N2 are the number of turns in the primary and secondary coils, respectively, and V1 and V2 are the voltages across the primary and secondary coils, respectively. In this question, we are given that the primary coil has N turns and is connected to a 120 V AC power line. We are also given that the secondary coil has 1,000 turns and a terminal voltage of 1,200 volts. Using the above equation, we can rearrange it to solve for N1: N1 = (V1/V2) * N2 Substituting the given values, we get: N1 = (120/1200) * 1000 = 100 Therefore, the value of N is 100.
Frage 17 Bericht
The terrestrial telescope has one extra lens more than the astronomical telescope. The extra lens is for
Antwortdetails
The extra lens in a terrestrial telescope is used for erecting the image. The terrestrial telescope has one extra lens more than the astronomical telescope to provide an upright image, which is necessary for terrestrial viewing. Without this extra lens, the image would be inverted, making it difficult to use the telescope for ground observations. Therefore, the extra lens in the terrestrial telescope helps to erect the image and make it easier for the observer to view the object in its true orientation.
Frage 18 Bericht
Find the frequencies of the first three harmonics of a piano string of length 1.5m, if the velocity of the wave on the string is 120ms-1
Antwortdetails
The frequency of a wave on a string is related to its length, tension, and mass per unit length by the equation f= (1/2L) * sqrt(T/μ), where L is the length of the string, T is the tension in the string, μ is the mass per unit length, and f is the frequency of the wave. In this question, the length of the string is given as 1.5m and the velocity of the wave on the string is given as 120ms^-1. We need to first calculate the tension in the string. The velocity of a wave on a string is also related to tension and mass per unit length by the equation v= sqrt(T/μ), where v is the velocity of the wave. Rearranging this equation, we get T= μv^2. Substituting the given values, T= (1/120)^2= 6.94 N. Now we can use the equation for frequency to calculate the frequencies of the first three harmonics. The first harmonic has one antinode (at the middle of the string), the second harmonic has two antinodes (one at the middle and one at a quarter of the length of the string from one end), and the third harmonic has three antinodes (at the middle, a quarter, and three-quarters of the length of the string from one end). The formula for frequency for a string fixed at both ends is f= (n/2L) * v, where n is the harmonic number. Therefore, for the first three harmonics, we have: f1= (1/2*1.5) * 120= 40Hz f2= (2/2*1.5) * 120= 80Hz f3= (3/2*1.5) * 120= 120Hz Therefore, the correct option is 40Hz, 80Hz, 120Hz.
Frage 19 Bericht
In the circuit diagram above , the ammeter reads a current of 3A when R is 5Ω and 6A when R is 2Ω. Determine the value of x
Antwortdetails
for a parallel R and X, when R = 5Ω,
| effective resistance R = | 5X |
| 5+X |
and when R = 2Ω
| effetive resistance R = | 2X |
| 2+X |
Thus since E = I(R+r), and if r = 0 then E = IR
∴(i) E = 3(5X/(5+x)); (ii) E = 6(2X(2+x))
∴3(5X/(5+X)) = 6(2X/(2+X))
=> (5X/(5+X)) = 2(2x/(2+X))
∴ (5X/(5+X)) = (4X/(2+X))
5x(2+x) = 4x(5+x)
10x + 5x2 = 20x + 4x2
5x2 - 4x2 = 20x - 10x
=>x2 = 10x
∴x2 - 10x = 0
x(x - 10) = 0
∴ x = 0 or 10
∴ x = 10Ω
Frage 20 Bericht
A working electric motor takes a current of 1.5 A when the p.d across it is 250V. If its efficiency is 80%, the power output is
Antwortdetails
The power output of the electric motor is given by the formula: Power output = Efficiency x Power input The power input can be calculated using the formula: Power input = Voltage x Current Substituting the given values in the above formula: Power input = 250V x 1.5A = 375W Now, using the efficiency of 80%, we can calculate the power output: Power output = 0.80 x 375W = 300W Therefore, the power output of the electric motor is 300.0W. So, the correct option is: 300.0W.
Frage 21 Bericht
If a ray traveling in air is incident on a transparent medium as shown in the diagram, the refractive index of the medium is given as
Frage 22 Bericht
A plane sound wave of frequency 85.5Hz and velocity 342ms-1 is reflected from a vertical wall. At what distance from the wall does the wave have antinode?
Antwortdetails
If F = 85.5Hz, V = 342ms-1,
Then wavelength = V/F = 342/85.5 = 4m
Then if wavelength = 4m
=> wavelength/2 = 4/2 = 2m = NN,
Thus NA = wavelength/4 = 4/4 = 1m.
Frage 23 Bericht
A cell can supply current of 0.4 A and 0.2 A through a 4.0Ω and 10.0Ω resistors respectively.
This internal resistance of the cell is
Antwortdetails
To find the internal resistance of the cell, we can use the formula: Internal resistance (r) = (EMF - V) / I where EMF is the electromotive force of the cell, V is the potential difference across the external resistor, and I is the current flowing through the circuit. First, let's find the EMF of the cell. We know that the current through the 4.0Ω resistor is 0.4 A, so the potential difference across it is: V = IR = 0.4 A x 4.0Ω = 1.6 V Similarly, for the 10.0Ω resistor, we have: V = IR = 0.2 A x 10.0Ω = 2.0 V Now we can find the EMF by adding these potential differences: EMF = V + IR = 1.6 V + 2.0 V = 3.6 V Finally, we can use the formula above to find the internal resistance: r = (EMF - V) / I = (3.6 V - 1.6 V) / 0.4 A = 2.0 Ω Therefore, the internal resistance of the cell is 2.0Ω. Answer option A, "2.0Ω", is correct.
Frage 24 Bericht
The diagram above is a block and tackle pulley system in which an effort of 80N is use to lift a load of 240N. The efficiency of the machine is
Antwortdetails
ince the number of pulleys present is 6, it implies that the velocity ratio VR = 6
| ∴ efficiency = | M.A |
|
|||
| V.R |
= Load/Effort x 1/V.R x 100/1
= 240/80 x 1/6 x 100/1
= 50%
∴ Efficiency = 50%
Frage 25 Bericht
Which of the following metals will provide the greatest shield against ionizing radiation
Antwortdetails
Ionizing radiation is high-energy radiation that can cause ionization of atoms and molecules. It is harmful to living organisms, and shielding is often used to protect against it. The effectiveness of shielding depends on the material used, with denser materials generally providing better protection. Among the given options, lead is the densest material and is therefore the most effective at shielding against ionizing radiation. Therefore, lead would provide the greatest shield against ionizing radiation.
Frage 26 Bericht
The pressure of a given mass of a gas changes from 200Nm-2 to 120Nm-2 while the temperature drops from 127°C to -73°C. The ratio of the final volume to the initial volume is
Antwortdetails
P1 = 300Nm-2; P2 = 120Nm-2
T1 = 127 + 273; T2 = -73 + 273
= 400K; = 200K
Let the initial volume V1; and the final volume = V2
? P1V1/T1 = P2V2/T2
P1V1T2 = P2V2T1
? V2/V1 = P1T2/P2T1
? V2/V1 = (300 x 200)/(120 x 400) = 5/4
? V2 : V1 = 5 : 4
Frage 27 Bericht
I. The earth is not spherical but elliptical in shape
II. Variation in latitude and longitude
III. Rotation of the earth on its axis
IV. Variation in the density of the earthOn which combination of the above does the weight of an object vary on the earth's surface?
Antwortdetails
Frage 28 Bericht
A student is at a height 4m above the ground during a thunderstorm. Give that the potential difference between the thundercloud and the ground is 107v, the electric field created by the storm is
Antwortdetails
The electric field created by the storm can be calculated using the formula E = V/d, where E is the electric field, V is the potential difference, and d is the distance between the thundercloud and the ground. In this case, the potential difference is given as 10^7 V, and the distance is the height of the student above the ground, which is 4m. Therefore, E = V/d = (10^7 V)/(4m) E = 2.5 * 10^6 NC^-1 So the electric field created by the storm is 2.5 * 10^6 NC^-1. Therefore, the answer is: 2.5 * 106 NC -1.
Frage 30 Bericht
The driving mirror of a car has a radius of curvature of 1m. A vehicle behind the car is 4m from the mirror. Find the image distance behind the mirror.
Antwortdetails
A driving mirror is always a convex mirror which is always of negative focal length and always forms a virtual image
Thus if r = 1m = 100cm
f = 1/2 = -50
∴ if U = 4cm = 400cm
f = -50cm
∴1/u + 1/v = 1/f
1/400 + 1/v = -1/50
∴ -1/v = 1/400 + 1/50
-1/v = (1 + 8)/400 = 9/400
∴ v = -400/9
v = 4/9
Frage 31 Bericht
The diagram above shows the capacitors C1, C2 and C3 2μF, 6μF and 3μF respectively. The potential difference across C1, C2 and C3 respectively are
Antwortdetails
C = Q/V => Q = CV : ∴V = Q/C
∴V1 = Q/C1; V2 = Q/C;
V3 = Q/C3
for series
1/C = 1/C1 + 1/C2 + 1/C3
= 1/2 + 1/6 + 1/3
| = | 3+1+2 |
| _6 |
= 6/6 = 1
∴ C = 1μF = 1.0 x 10-6F
∴ Q = CV = 1.0 x 10-6 x 12
= 12.0 x 10-6C
∴V1 = Q/C1
| = | 12.0x10-6 |
| 2 x 10-6 |
= 6V
V2 = Q/C2
| = | 12.0x10-6 |
| 6 x 10-6 |
= 2V
V3 = Q/C3
| = | 12.0x10-6 |
| 3 x 10-6 |
∴ 6V, 2V and 4V
Frage 32 Bericht
A capacitor of 20 x 10-12F and an inductor are joined in series. The value of the inductance that will give the circuit a resonance frequency of 200kHz is
Antwortdetails
Resonant, frequency in an a.c. circuit is given by F = I/2π√LC
therefore 200 * 103 = I/2 * π√l * 20 * 10-12
therefore (2.0 * 105)2 = I/4 * π2 * L * 20 * 10-12
THEREFORE L = I/4.0 * 1010 * 4 * 10 * 20 * 10-12
(if π2 = 10)
= I/16 * 1011 * 2.0 * 10-11
Therefore L = I/32 H
Frage 33 Bericht
The force on a current carrying conductor in a magnetic field is greatest when the
Antwortdetails
The force on a current-carrying conductor in a magnetic field is given by the formula F = BIL sinθ, where B is the magnetic field strength, I is the current, L is the length of the conductor and θ is the angle between the direction of the current and the magnetic field. The force is greatest when sinθ is at its maximum value of 1, which occurs when the conductor is perpendicular (i.e., at a right angle) to the magnetic field. Therefore, the answer is "conductor is at right angle with the field".
Frage 34 Bericht
At reasonance, the phase angle in an a.c. circuit is
Antwortdetails
In an AC circuit, the current and voltage are constantly changing direction and magnitude. The phase angle between the current and voltage in an AC circuit indicates how much the current leads or lags the voltage. At resonance, the impedance of the circuit is at its minimum and the phase angle between the current and voltage is zero degrees (0°). Therefore, the correct answer is "0°".
Frage 35 Bericht
A ray of light strikes a plane mirror at an angle of incidence of 35°. If the mirror is rotated through 10°, through what angle is the reflected ray rotated?
Antwortdetails
When ever a ray of light is incident on a plane mirror at an angle of incidence l, if the mirror is then rotated through an angle θ, then the reflected ray is rotated through an angle twice θ
So if θ = 35°, then the reflected ray is rotated through 2 x 10o = 20°
Frage 36 Bericht
A resistance R is connected across the terminal of an electric cell of internal resistance 2Ω and the voltage was reduced to 3/5 of its nominal value.
The value of R is
Antwortdetails
1 = V/R = E/R+r , if r = 2Ω
V = 3/5 E
=> 3E/5R = E/R+2
i.e 3E(R+2) = 5RE
3RE + 6E = 5RE
i.e 3R + 6 = 5R
6 = 5R - 3R
=> 2r = 6 i.e R = 6/2 = 3Ω
Frage 37 Bericht
A gas with initial volume of 2 X 10-6m3 is allowed to expand to six times its initial value at constant pressure of 2 x 105Nm-2. The work done is
Antwortdetails
P = F/A => F = PA = PV/L
W = FD = (PV/L) x L = PV
= 2 x 105
(2 x 10-6 x 6)
= 2 x 105 x 1.2 x 10-5
= 2.4 J
Frage 38 Bericht
The diagram above shows two capacitors P and Q of capacitance 2μF and 4μF respectively connected to a d.c source. The ratio of energy stored in P to Q is
Antwortdetails
The energy stored in a capacitor is given by the formula 1/2CV^2, where C is the capacitance and V is the voltage across the capacitor. Since the capacitors P and Q are connected in parallel, they have the same voltage across them. Therefore, the ratio of energy stored in P to Q is the same as the ratio of their capacitances, which is 2:4 or 1:2. Therefore, the answer is option (D) 1:2.
Frage 39 Bericht
I. Low pressure.
II. High pressure.
III. High p.d.
IV. Low p.d.
Which combination of the above is true of the conduction of electricity through gases?
Antwortdetails
Electricity conduction through gases is usually done under low pressure (I) and high voltage (III). Thus /B/
Frage 40 Bericht
The current through a resistor in an a.c circuit is given as 2 sin wt. Determine the d.c. equivalent of the current
Antwortdetails
If I = 2 Sin wt; but in a.c. circuit current I = I0 Sin wt; thus comparing the two equations, it implies that I0 = 2A; Again, the root means square (I(rms)) of an a.c. current is the value of the d.c. current that will dissipate the same amount of heat in a given resistance as the a.c.
and I(rms) = I0/√2 = 2.0A/√2
Frage 41 Bericht
In the diagram above , the current I is
Antwortdetails
In order to determine the current I in the circuit, we need to apply Kirchhoff's Current Law (KCL), which states that the sum of currents entering a node must be equal to the sum of currents leaving the node. Looking at the circuit diagram, we see that there are two nodes: the one at the top, where the current source and resistor meet, and the one at the bottom, where the two resistors meet. Let's start with the top node. We know that the current entering the node is equal to the current leaving the node, since there are no other connections to this node. Therefore, we have: I = I1 Next, let's move to the bottom node. Here, we know that the current leaving the node is equal to the current entering the node. Therefore, we have: I1 = I2 + I3 Now, we can use Ohm's Law to relate the currents to the resistances and voltages: I1 = (V1 - V2) / R1 I2 = V2 / R2 I3 = V2 / R3 Substituting these expressions into the equation for the bottom node, we get: (V1 - V2) / R1 = V2 / R2 + V2 / R3 Multiplying both sides by R1R2R3 and simplifying, we get: R2R3(V1 - V2) = R1R3V2 + R1R2V2 Expanding and simplifying further, we get: (V1 - V2) / 8 = (2V2) / 9 Solving for V2, we get: V2 = V1 / 3 Substituting this value back into our expressions for I1, I2, and I3, we get: I1 = 2V1 / 27 I2 = V1 / 27 I3 = V1 / 9 Finally, substituting these values back into our equation for the top node, we get: I = I1 = 2V1 / 27 = 2/27 * 24 = 16/9 Therefore, the current I is 16/9 or 1.78, which corresponds to option (B) 9/11.
Frage 42 Bericht
In a Daniel cell, the depolarizer, positive and negative electrodes are respectively
Antwortdetails
In the Daniel cell,
i. Depolarize = CuSO4 solution
ii. Positive electrode = Copper container
iii Negative electrode = zinc rod/plate
Frage 43 Bericht
The particles emitted when 3919K decay to 3919K is
Antwortdetails
3919K 3919K.
Since neither the mass number (39), nor the atomic number (19) is affected by the emission,the particles emitted can only be gamma ray, which is not a charge particle, but simply a radiation.
Frage 44 Bericht
the ratio of electrostatic force FE to gravitational force FG between two protons each of charge e and mass m, at a distance d is
Antwortdetails
The electrostatic force between two protons is given by Coulomb's law, FE = (1/4πε0) (e2/d2), where ε0 is the permittivity of free space, e is the charge on each proton and d is the distance between them. The gravitational force between the two protons is given by the law of gravitation, FG = Gm2/d2, where G is the universal gravitational constant and m is the mass of each proton. Thus, the ratio of electrostatic force to gravitational force between the two protons is (FE/FG) = [(1/4πε0) (e2/d2)]/[Gm2/d2]. Simplifying this expression, we get e2 / (4πε0Gm2). Therefore, the correct option is e2/4πԐoeGm2.
Frage 45 Bericht
In the diagram above, determine the r.m.s current
Antwortdetails
Ir.m.s = Vr.m.s/ZBut Z =
√ R2 + XL2 =
√ 42 + 32 =
√ 16 + 9 =
√ 25
∴Ir.m.s = 240/5
= 48.0A
Frage 47 Bericht
A string is fastened tightly between two walls 24cm apart. The wavelength of the second overtone is
Antwortdetails
For a string fixed at both ends and plucked at the middle, the fundamental frequency is given
by f0 = v/λ, where v = velocity of sound, λ = wave length.
At its fundamental, note, the the length of the string is given l = λ/2, => λ0 = 2l, where λ0 = wave length fundamental note. for the first over tone, λ1 = l.
For the second over tone, λ2 = 21/3 = (2 x 24)/3 = 16cm.
i.e. λ2 = 16cm.
Frage 48 Bericht
A cell of internal resistance r supplies current to a 6.0Ω resistor and its efficiency is 75%. Find the value of r.
Antwortdetails
Efficiency = power out put/power input * 100/1
= power out put/ power + that due * 100/1
to internal resistance
=> 75/1 = 12R *100/1
12(R+r)
i.e 75/100 = 6/6+r => 75 (6+r) = 100/*6
i.e 450 + 75r = 600
75r =600 - 450
i.e r = 150/75 = 2Ω
i.e r = 2Ω
Frage 49 Bericht
Energy losses through eddy currents are reduced by using
Antwortdetails
Eddy currents are loops of electric current induced within conductors by a changing magnetic field in the conductor. These currents can cause energy losses in electrical devices. To reduce these losses, one can use insulated soft iron wires. Soft iron has low coercivity which means it can be easily magnetized and demagnetized, making it useful for reducing eddy currents. Insulating the soft iron wires prevents them from creating unwanted electrical paths and keeps the eddy currents localized, reducing the overall energy losses. Therefore, the correct option is "insulated soft iron wires".
