JAMB UTME - Physics - 2000

Akọwa Nkọwa

Akọwa Nkọwa

The question is asking which of the given statements are true for sliding friction. I. The statement that "The frictional force is independent of the area of the surfaces in conduct" is false for sliding friction because the frictional force depends on the area of surfaces in contact. II. The statement that "The frictional force depends on the nature of the surfaces in contact" is true for sliding friction because the amount of friction generated between two surfaces depends on their nature or texture. III. The statement that "The frictional force depends on the speed of sliding" is also true because the amount of friction changes with the speed of sliding. IV. The statement that "The frictional force is directly proportional to the normal reaction" is true because the amount of frictional force is directly proportional to the normal reaction between the two surfaces in contact. Therefore, the combination of statements that are true for sliding friction is: I, II and IV

Akọwa Nkọwa

When the tension in a sonometer wire is doubled while the length and the mass per unit length are kept constant, the new frequency of vibration is: f = (1/2L) * √(T/μ) Where: - f is the frequency of vibration - L is the length of the wire - T is the tension in the wire - μ is the mass per unit length of the wire Since the length and mass per unit length of the wire are kept constant, we can simplify the equation as follows: f2 = (1/2L) * √(2T/μ) where f2 is the new frequency of vibration after doubling the tension. Dividing the new frequency by the original frequency gives: f2/f0 = √(2T/μ) / √T/μ) = √2 Therefore, the new frequency of vibration is equal to the square root of two times the original frequency: f2 = f0√2 So the answer is: - f0√2

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Since capacitance C = Q/V, => Q = CV
∴ Qp = 5 x 10-b x 20
= 100 x 10-b or 100μC
Qp = 10 x 10-b x 70
= 200 x 10-b or 200μC
Correct option 100μC and 200μC

Akọwa Nkọwa

The reactance of the inductor is 100/π Ω. Reactance is the opposition that an inductor or a capacitor offers to the flow of alternating current. The reactance of an inductor is given by the equation: XL = 2πfL where XL is the inductive reactance, f is the frequency of the AC supply, and L is the inductance of the inductor. In this case, we are given that the inductance of the inductor is 2H and the frequency of the AC supply is 50/π Hz. Substituting the given values into the equation for inductive reactance, we get: XL = 2πfL = 2π(50/π)(2) = 200 Ω Therefore, the reactance of the inductor is 200 Ω. Note that in the given options, 200 Ω is one of the answer choices.

Akọwa Nkọwa

Efficiency = M.A/V.R
75% = M.A/V.R
75/100 = M.A/5
M.A = 75(5)/100 = 3.75

But M.A = Load/Effort
3.75 = 150/Effort
Effort = 150/3.75
Effort = 40N

Akọwa Nkọwa

When a charged ion passes through combined electric and magnetic fields, it experiences both electric force and magnetic force. The direction of electric force is in the direction of electric field while the direction of magnetic force is perpendicular to both the magnetic field and the velocity of the ion. The combined effect of these forces on the ion results in a curved path known as a helical path. The emergent velocity of the ion is the velocity of the ion at any given point along the helical path. To determine the emergent velocity, we need to find the balance between the electric and magnetic forces acting on the ion. This can be done by equating the electric force and the magnetic force: Eq1: Eq = Bqv where E is the electric field, B is the magnetic field, q is the charge of the ion, and v is the emergent velocity of the ion. Solving for v, we get: v = E/B Therefore, the resultant emergent velocity of the ion is given by the ratio of the electric field to the magnetic field, which is E/B. Hence, the correct option is "E/B".

Akọwa Nkọwa

The half-life of a radioactive material is defined as the time it takes for half of the original sample to decay. In this problem, we know that after 120 years, only 1/64th of the original sample is left. This means that the fraction of the original sample that has decayed is: 1 - 1/64 = 63/64 We want to know the half-life, which we can call "t". We know that after one half-life, half of the original sample will remain. So we can write: (1/2) = (63/64)^(120/t) To solve for t, we can take the natural logarithm of both sides and use the fact that ln(a^b) = b*ln(a): ln(1/2) = ln[(63/64)^(120/t)] ln(1/2) = (120/t)*ln(63/64) t = (120/ln(63/64))*ln(2) ≈ 20.2 years Therefore, the half-life of the radioactive material is approximately 20.2 years. The answer is closest to the option: 20 years.

Akọwa Nkọwa

Refractive Index = Real Depth/App. Depth
4/3 = 6/app. depth
App Depth = (3 x 6)/4 = 4.5m

Akọwa Nkọwa

Power is the rate at which work is done or energy is transferred. It is given by the formula P = Fv, where P is power, F is force and v is velocity. If both force and velocity are reduced by half, then the power is given by P = (F/2)(v/2) = (1/4)Fv. Therefore, the power of the system is reduced to a quarter of its original value. So, the option closest in meaning is "reduced to a quarter".

Akọwa Nkọwa

In a pure semiconductor, the number of electrons in the conduction band is equal to the number of holes in the valence band. A semiconductor has a valence band and a conduction band separated by a bandgap. The valence band is the highest energy band that is fully occupied by electrons, while the conduction band is the lowest energy band that is not occupied by electrons. In a pure semiconductor at absolute zero temperature, all electrons occupy the valence band and there are no electrons in the conduction band. As the temperature increases, some electrons gain enough energy to overcome the bandgap and transition from the valence band to the conduction band. This creates holes in the valence band, which are vacancies left by the electrons that have transitioned to the conduction band. The number of electrons in the conduction band is therefore equal to the number of holes in the valence band. Therefore, the correct option is: equal to the number of holes in the valence band.

Akọwa Nkọwa

Let the specific heat cap. of the metal = C
∴ Heat lost by the metal = heat gained by the Calorimeter + Water
∴ 0.5 x C x [100 - 25]
80 x [25 - 15] + 420 [25 - 15]
0.5 x C x 75 = 80 x 10 + 420 x 10
37.5C = 800 + 4200
C = 5000/37.5 = 133.3Jkg-1K-1

Akọwa Nkọwa

Total internal reflection occurs when light moves from a denser medium to a less dense medium. This happens when light strikes the boundary between the two media at an angle greater than the critical angle, which is the angle of incidence that produces an angle of refraction of 90 degrees. At this angle, the light is completely reflected back into the denser medium, and none of it is transmitted into the less dense medium. This phenomenon is used in various optical devices, such as fiber optics and prisms.

Akọwa Nkọwa

Impedance Z =
√ R2 x X2 =
√ 302 x 402 
∴Z =
√ 2500  = 50W

But current I = V/Z = 200/50 = 4.00 A

Akọwa Nkọwa

Akọwa Nkọwa

The given equation relates the velocity y of a particle to the time t. To find the instantaneous acceleration after 5 seconds, we need to take the derivative of the velocity function with respect to time: acceleration (a) = d(y)/dt = d/dt (10 + 2t^2) = 4t To find the instantaneous acceleration after 5 seconds, we substitute t=5 into the acceleration equation: a = 4t = 4(5) = 20 m/s^2 Therefore, the correct answer is 20 ms^-2, which is shown in.

Akọwa Nkọwa

V = 72m
U = Magnification = 24
But Magnification = V/U
∴ 24 = 72/U
U = 72/24
U = 3.0m

Akọwa Nkọwa

In thermonuclear reactions,
Energy = ∆mC2
∆m = Change in Mass
C = Speed of light
∴ 5.02 x 10-27 - 5.01 x 10-27 x [3 x 108]2
0.01 x 10-27 x 9 x 1016
E = 9.0 x 10-13J

Akọwa Nkọwa

The process described in the question is when the wire cuts through the ice block while the block remains solid behind the wire. This phenomenon is called "regelation". Regelation occurs when ice is subjected to pressure, causing it to melt and then refreeze as the pressure is released. In this case, the weight of the heavy weights attached to both ends of the thin wire creates enough pressure to melt the ice, which then refreezes behind the wire. Therefore, the correct answer is "regelation".

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

When the cells are connected in series, their voltages add up to give the total voltage of the battery. In this case, eight cells each of 2.0V connected in series will give a total voltage of 16V. If two cells are wrongly connected, the direction of current in the circuit will change, and the voltage of those two cells will be subtracted instead of being added. Therefore, the net voltage of the battery will be 16V - (2V + 2V) = 12V. Hence, the correct answer is 12V.

Akọwa Nkọwa

The optical instrument that can best be constructed with converging lenses of focal lengths 50cm and 5cm is an astronomical telescope. An astronomical telescope is used to observe distant objects in space, such as stars and planets. It uses two converging lenses - an objective lens and an eyepiece - to produce an enlarged image of the object being observed. The objective lens has a longer focal length, while the eyepiece has a shorter focal length. The objective lens collects and focuses the light from the object, forming a real image that is then magnified by the eyepiece. In this case, the 50cm focal length lens would be used as the objective lens, and the 5cm focal length lens would be used as the eyepiece.

Akọwa Nkọwa

Based on the information given, it is likely that the radioactive source emits particles that can be deflected by both magnetic and electric fields. This suggests that the particles are charged, and therefore, the answer is likely to be beta rays. Beta rays are high-energy, high-speed electrons or positrons that can be emitted from a radioactive nucleus during beta decay. They are charged particles and can be deflected by magnetic and electric fields. Gamma rays, X-rays, and ultraviolet rays, on the other hand, are not charged particles and are not deflected by electric or magnetic fields.

Akọwa Nkọwa

When a ship sails from salt water into fresh water, the fraction of its volume above the water surface will decrease. This is due to the principle of buoyancy, which states that the buoyant force acting on an object is equal to the weight of the fluid displaced by the object. In other words, the more fluid an object displaces, the greater the buoyant force it experiences. Salt water is denser than fresh water, which means that a ship will displace more water (and hence experience a greater buoyant force) in salt water than in fresh water. Therefore, when a ship sails from salt water into fresh water, the buoyant force acting on it decreases. However, the weight of the ship remains the same regardless of whether it is in salt water or fresh water. Therefore, when the buoyant force acting on the ship decreases, the ship sinks slightly deeper into the water, and the fraction of its volume above the water surface decreases. Therefore, the fraction of the ship's volume above the water surface will decrease when it sails from salt water into fresh water.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

When a vibrator is energized and attached to a long wire, it generates waves that propagate through the wire. The nature of these waves depends on the type of vibration produced by the vibrator. If the vibrator produces a transverse vibration, then the waves generated in the wire will also be transverse. If the vibrator produces a longitudinal vibration, then the waves generated in the wire will also be longitudinal. Now, if the vibrator is attached to one end of a long wire which is fixed at the other end, the waves generated in the wire will be progressive because they will travel through the wire in a particular direction. Therefore, the correct answer is "progressive and transverse".

Akọwa Nkọwa

Akọwa Nkọwa

The wet and dry bulb hygrometer is used to measure the relative humidity of the air. The dry bulb thermometer measures the actual temperature while the wet bulb thermometer is covered with a wet wick and measures the lowest temperature that can be achieved through evaporative cooling. The greater the difference between the two temperatures, the drier the air is. In the given readings, the dry bulb temperature decreases gradually from 30°C to 25°C while the wet bulb temperature remains constant at 22°C for all three days. This means that the difference between the dry and wet bulb temperatures is decreasing, indicating an increase in the relative humidity of the air over the three days. Therefore, the correct answer is "increased steadily."

Akọwa Nkọwa

Power input = 3KW = 3000W
Voltage = 1.5KV = 1500V
Current (I) = P/V = 3000/1500 = 2Amps.
∴ Power loss in the cables = I2R
= 2 x 2 x 20
= 80W

Akọwa Nkọwa

Akọwa Nkọwa

Angle of deviation = 180 - [i + v]
= 180 - 2i [since i = v]
∴ 120 = 180 - 2i
2i = 180 - 120
= 60
∴ i = 30o

Akọwa Nkọwa

For a photoemitted electron,
eV = K.E.max, where V = stopping potential and K.E. = maximum kinetic energy of the photo emitted electron given by K.E. = Energy of incident rad - Work - function
= 5ev - 3ev = 2ev
∴ eV = 2eV
V = 2eV/e = 2V
∴ Stopping Potential = 2.0V

Akọwa Nkọwa

When a charged particle is accelerated across a region of p.d. V Volt, the work done is the product of the charge Q and the p.d. V

i.e. Work done = Q x V

But this work done is equal to the K.E. imparted to the charge given by K.E = 1/2MV2, where V is the velocity of the charge = Q x V = 1/2MV2

∴ p.d V = (1/2MV2)/Q
= 1/2 x 9.1-31 x [4 x 107]2/1.6 x 10-19
= 8 x 9.1 x 102/1.6
= 4550V
= 4.55 x 103V

Akọwa Nkọwa

For a simple pendulum, the period T = 2π
√  1/g 
Since g, 2 and π are constant, => T ∝ √l
∴ T = K√l
Let the original length = l
thus for the length l, period T = 17s
∴ 17 = K√1 .................. [1]
Again, for the new length [1 - 1.5]
the period T = 8.5s
∴ 8.5 = k
√ [1-1.5] ...............[2]

thus dividing eqn [1] by eqn[2] we have
17/8.5 = √(l/l-1.5)
∴2 = √(l/l-1.5)
=>4 = (l/l-1.5)
∴l = 4[l-1.5]
= 4l - 6.0
∴4l - l = 6.0
3l = 6.0
l = 2.0m

Akọwa Nkọwa

The pressure at any point in a liquid depends on the depth of the point and the density of the liquid. At a fixed point below the liquid surface, the pressure downwards is P1 and pressure upwards is P2. It can be deduced that P1 > P2 because the pressure increases with depth due to the weight of the liquid above. Therefore, the correct option is: P1 > P2.

Akọwa Nkọwa

The speed of propagation of a wave on a string is given by the formula: v = √(T/μ), where T is the tension in the string and μ is the mass per unit length of the string. Substituting the given values, we get: v = √(12 N / 3 × 10^(-2) kg/m) v = √(4 × 10^2) v = 20 m/s Therefore, the speed of propagation of the wave is 20 m/s, which corresponds to option (B).

Akọwa Nkọwa

A gas would serve as an electrical conductor under reduced pressure and high current. In general, gases are poor conductors of electricity because they are made up of molecules that are widely spaced and have few free electrons. However, under certain conditions, some gases can become ionized and form plasma, which can conduct electricity. When the pressure of a gas is reduced, the distance between its molecules increases, and the probability of electron collisions with gas molecules decreases. This increases the likelihood that electrons can move freely through the gas and form plasma. Additionally, when a high current is passed through a gas, it can cause the gas molecules to ionize and form plasma, which can then conduct electricity. The high current provides enough energy to remove electrons from gas molecules and create a large number of free electrons, which can then move through the gas and conduct electricity. Therefore, a gas would serve as an electrical conductor under reduced pressure and high current. Note that in the given options, reduced pressure and high current is one of the answer choices.

Akọwa Nkọwa

In the diagram shown above, the voltage division rule can be applied to determine the voltage values. According to the voltage division rule, the voltage across a resistor in a series circuit is proportional to the resistance of the resistor. Here, the total resistance of the circuit is 3 ohms (1 ohm + 2 ohms), and the voltage applied is 3 volts. So, using the voltage division rule, we can find the voltage V1 as follows: V1 = (1 ohm / 3 ohms) x 3 volts = 1 volt Similarly, we can find the voltage V2 as: V2 = (2 ohms / 3 ohms) x 3 volts = 2 volts Therefore, the values of V1 and V2 are 1V and 2/3V, respectively. Hence, the correct option is: "1V and 2/3V".

Akọwa Nkọwa

When a beam of white light passes from a rare medium (e.g. air) to a dense medium (e.g. glass), it bends or refracts towards the normal (the line perpendicular to the surface). The degree of bending depends on the wavelength (or color) of the light. The shorter the wavelength, the more the light bends. In the given diagram, the red ray (a) is the least bent, while the blue ray (c) is the most bent. The green ray (b) is somewhere in between. This means that the colors are separated out as they pass through the medium, with the shorter wavelengths (violet, blue, and green) bending more than the longer wavelengths (yellow, orange, and red). Therefore, the correct answer is: red, green, and blue, as shown in.

Akọwa Nkọwa

The force acting on a current-carrying conductor in a magnetic field is given by the formula F = BILsinθ, where B is the magnetic flux density, 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. Since the conductor is perpendicular to the magnetic field, sinθ = 1. Therefore, the maximum force on the conductor is F = BIL = (0.5T)(0.8A)(2m) = 0.8N. Hence, the correct option is 0.8N.

Akọwa Nkọwa

The value of F is 2.0 x 10^3 N. To solve this problem, we can use the principle of conservation of energy. The potential energy stored in the catapult is converted into the kinetic energy of the stone as it leaves the catapult. The potential energy stored in the catapult is given by the equation: PE = 0.5kx^2 where PE is the potential energy, k is the spring constant of the catapult, and x is the extension of the catapult. We can rearrange this equation to solve for the spring constant: k = (2PE) / x^2 The force applied to the catapult is given by: F = kx To solve for F, we need to first calculate the spring constant k. We know that the extension of the catapult is 20 cm, or 0.2 m. We also know that the mass of the stone is 500 g, or 0.5 kg. The potential energy stored in the catapult is equal to the work done in extending the catapult, which is given by: PE = Fx where F is the force applied to the catapult. Substituting the given values, we get: PE = Fx = (F)(0.2) At the moment the stone leaves the catapult, all of the potential energy stored in the catapult is converted into the kinetic energy of the stone. The kinetic energy of the stone is given by: KE = 0.5mv^2 where KE is the kinetic energy, m is the mass of the stone, and v is the velocity of the stone. Substituting the given values, we get: KE = 0.5(0.5)(40)^2 = 400 J Since the potential energy stored in the catapult is equal to the kinetic energy of the stone, we have: PE = KE = 400 J Substituting this into the equation for the spring constant, we get: k = (2PE) / x^2 = (2)(400) / (0.2)^2 = 40000 N/m Finally, we can solve for the force F applied to the catapult: F = kx = (40000)(0.2) = 8000 N Therefore, the value of F is 2.0 x 10^3 N.

Akọwa Nkọwa

Let the velocity of the boat = Vm/s
0.75/V = adj/hyp = Cost30°
0.75/V = 0.8660
∴ V = 0.75/0.8660
V = 0.86-1

Akọwa Nkọwa

Energy imparted by the accelerating system = e x V, where V is the accelerating potential; and this energy is equal to the energy of the X-ray, give by hf

e x V = hf
∴ V = hf/e
(6.63 x 10-34 x 1.6 x 10-16)/1.6 x 10-19

= 6.63 x 10 = 66.3V

Akọwa Nkọwa

The efficiency of a cell is defined as the ratio of the power output to the power input. In this case, the power output is the power dissipated in the external resistor, and the power input is the power supplied by the cell. First, we can calculate the current in the circuit using Ohm's law: I = V / R where V is the voltage supplied by the cell, and R is the total resistance of the circuit. In this case, R is the sum of the internal resistance of the cell (1Ω) and the external resistor (3Ω): R = 1Ω + 3Ω = 4Ω Now, we can calculate the current: I = V / R = V / 4Ω Next, we can calculate the power output: Pout = I^2 * Rext where Rext is the resistance of the external resistor. In this case, Rext is 3Ω, so: Pout = I^2 * 3Ω Finally, we can calculate the power input: Pin = I^2 * R where R is the total resistance of the circuit, including the internal resistance of the cell. In this case, R is 4Ω, so: Pin = I^2 * 4Ω Now, we can calculate the efficiency: efficiency = Pout / Pin Substituting the values we calculated earlier: efficiency = (I^2 * 3Ω) / (I^2 * 4Ω) = 3/4 = 0.75 = 75% Therefore, the efficiency of the cell is 75%.

Akọwa Nkọwa

The equation of a travelling wave is given by: y = A sin (kx - ωt + φ) where A is the amplitude, k is the wave number, ω is the angular frequency, t is time, x is position, and φ is the phase constant. Given the amplitude, frequency, and wavelength of the wave, we can find the wave number and angular frequency using the following equations: k = 2π/λ ω = 2πf Substituting the values, we get: k = 2π/0.01 = 200π ω = 2π × 550 = 1100π The phase constant φ is usually assumed to be zero. Hence, the equation describing the wave is: y = 0.15 sin (200πx - 1100πt) which can be simplified as: y = 0.15 sin 200π(x - 5.5t) Therefore, the correct option is: y = 0.15 sin 200π(x - 5.5t)