JAMB UTME - Physics - 2007

Bayanin Amsa

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

Bayanin Amsa

Bayanin Amsa

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

Bayanin Amsa

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

Bayanin Amsa

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).

Bayanin Amsa

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

Bayanin Amsa

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.

Bayanin Amsa

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

Bayanin Amsa

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

Bayanin Amsa

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

Bayanin Amsa

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

Bayanin Amsa

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.

Bayanin Amsa

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".

Bayanin Amsa

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

Bayanin Amsa

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.

Bayanin Amsa

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

= force x displacement

Bayanin Amsa

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.

Bayanin Amsa

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.

Bayanin Amsa

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.

Bayanin Amsa

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

Bayanin Amsa

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.

Bayanin Amsa

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

Bayanin Amsa

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.

Bayanin Amsa

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

Bayanin Amsa

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.

Bayanin Amsa

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".

Bayanin Amsa

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.

Bayanin Amsa

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

Bayanin Amsa

The weight of the rod can be determined by balancing the moments on either side of the fulcrum, F. The moment of a force is its magnitude multiplied by the perpendicular distance from the fulcrum to the line of action of the force. Let the weight of the rod be W, and the distance from the left end of the rod to the fulcrum be x. Then, the distance from the right end of the rod to the fulcrum is (50 - x). Since the rod is in equilibrium, the total clockwise moment about F must be equal to the total anticlockwise moment. The weight W acts downwards at the centre of gravity of the rod, which is at its midpoint, so its moment about F is W(50/2 - x) = 25W - Wx. The force acting to the left at the left end of the rod is 40N, and its moment about F is 40x. The force acting to the right at the right end of the rod is unknown, and its moment about F is P(50 - x), where P is the magnitude of the force. Setting the clockwise moment equal to the anticlockwise moment gives: 25W - Wx = 40x + P(50 - x) Simplifying, we get: 25W = 41x + 50P We need to find W, so we need to find P. We can use the fact that the rod is in equilibrium to say that the total upward force must be equal to the total downward force. Upward force = weight of rod + 40N Downward force = P Therefore: W + 40 = P Substituting this into the previous equation, we get: 25W = 41x + 50(W + 40) Simplifying, we get: 25W = 41x + 50W + 2000 Solving for W, we get: W = 80N Therefore, the weight of the rod is 80N.

Bayanin Amsa

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.

Bayanin Amsa

In general, Power = IV;$⟹40KW=IV\text{Therefore}40000=1\times 800⟹I=\frac{40000}{800}=50A,\text{tune the current through}$

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

Bayanin Amsa

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

Bayanin Amsa

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.

Bayanin Amsa

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.

Bayanin Amsa

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.

Bayanin Amsa

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."

Bayanin Amsa

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.

Bayanin Amsa

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

Bayanin Amsa

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

Bayanin Amsa

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.

Bayanin Amsa

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

Bayanin Amsa

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.

Bayanin Amsa

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.

Bayanin Amsa

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.

Bayanin Amsa

Efficiency = $\frac{W\left(120N\right)INPUT}{WORKOUTPUT}\times \frac{100}{1}$

And for incline plane

V.R =$\frac{1}{Sin\theta }=V.R=\frac{1}{Sin30}=\frac{1}{0.5}=2\text{therefore}\frac{60}{120}=\frac{M.A}{2}=M.A=\frac{120}{100}=1.2\text{but M.A}=\frac{load}{effort}=\frac{120}{E}\text{therefore}\frac{120}{E}=1.2=E=\frac{120}{1.2}=100N$

Therefore Effort up the plane = 100N

Bayanin Amsa

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.

Bayanin Amsa

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.

Bayanin Amsa

E = $\frac{\text{F}}{\text{q}}$ and Q1 = 50μF, E1 = 360μF, E2 = ?, Q2 = 120μF
Using E = $\frac{\text{F}}{\text{q}}$.
So we have F = Q $\times$ E
F = 50 $\times$ 10${}^{-6}$ $\times$ 360 = 0.018N
When E2 = F/q
$\frac{0.018}{120\times {10}^{-6}}$ = 150NC${}^{-1}$

Bayanin Amsa

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

Bayanin Amsa

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.