JAMB UTME - Physics - 2001

Akọwa Nkọwa

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.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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°

Akọwa Nkọwa

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

Akọwa Nkọwa

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

Akọwa Nkọwa

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.

Akọwa Nkọwa

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.

Akọwa Nkọwa

Ir.m.s = Vr.m.s/ZBut Z =
√ R2 + XL2 =
√ 42 + 32 =
√ 16 + 9 =
√ 25 
∴Ir.m.s = 240/5
= 48.0A

Akọwa Nkọwa

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

Akọwa Nkọwa

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.

Akọwa Nkọwa

Electricity conduction through gases is usually done under low pressure (I) and high voltage (III). Thus /B/

Akọwa Nkọwa

for a parallel R and X, when R = 5Ω,

Akọwa Nkọwa

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.

Akọwa Nkọwa

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

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

(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

Akọwa Nkọwa

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.

Akọwa Nkọwa

0.0 E since one of the electric fields points in the opposite direction (left)

Akọwa Nkọwa

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.

Akọwa Nkọwa

In the Daniel cell,
i. Depolarize = CuSO4 solution
ii. Positive electrode = Copper container
iii Negative electrode = zinc rod/plate

Akọwa Nkọwa

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°".

Akọwa Nkọwa

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.

Akọwa Nkọwa

The electrostatic force between two protons is given by Coulomb's law, F_E = (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, F_G = 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 (F_E/F_G) = [(1/4πε₀) (e²/d²)]/[Gm²/d²]. Simplifying this expression, we get e² / (4πε₀Gm²). Therefore, the correct option is e2/4πԐoeGm2.

Akọwa Nkọwa

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."

Akọwa Nkọwa

The pendulum that will resonate with P is T. being of the same length and therefore of the same frequency.

Akọwa Nkọwa

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.

Akọwa Nkọwa

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.

Akọwa Nkọwa

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.

Akọwa Nkọwa

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".

Akọwa Nkọwa

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.

Akọwa Nkọwa

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).

Akọwa Nkọwa

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

Akọwa Nkọwa

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.

Akọwa Nkọwa

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.

Akọwa Nkọwa

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.

Akọwa Nkọwa

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

Akọwa Nkọwa

ince the number of pulleys present is 6, it implies that the velocity ratio VR = 6

Akọwa Nkọwa

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.

Akọwa Nkọwa

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

Akọwa Nkọwa

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

Akọwa Nkọwa

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".

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

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