JAMB UTME - Physics - 2006
Ajụjụ 1 Ripọtì
In the diagram above, the pressure of the gas is
Ajụjụ 2 Ripọtì
A person can focus objects when they lie beyond 75 cm from his eyes. The focal length of the lens required to reduce his least distance of distinct vision to 25 cm is?
Akọwa Nkọwa
The question is asking for the focal length of a lens that will enable a person to focus on objects at a closer distance than they currently can. The given information is that the person's least distance of distinct vision (or near point) is 75 cm, which means that they can focus on objects only if they are beyond 75 cm from their eyes.
To be able to focus on objects that are closer, the person needs a lens with a focal length such that the image of an object at a distance of 25 cm from the eye is formed at the person's least distance of distinct vision, which is 75 cm. This means that the lens should form a virtual image of the object at 25 cm on the other side of the lens, which then acts as the object for the eye.
Using the lens formula, 1/f = 1/v - 1/u, where f is the focal length of the lens, v is the image distance (25 cm), and u is the object distance (the distance at which the virtual image is formed), we can find the value of u.
1/f = 1/v - 1/u
1/f = 1/25 - 1/u
1/f = (u - 25)/25u
25u/f = u - 25
25u = uf - 25f
u = (25f)/(f + 25)
Now, we know that the person's least distance of distinct vision is 75 cm, so the object distance u should be 75 cm. Substituting this in the above equation, we get:
75 = (25f)/(f + 25)
Multiplying both sides by (f + 25), we get:
75f + 1875 = 25f
Simplifying and solving for f, we get:
f = 37.5 cm
Therefore, the focal length of the lens required to reduce the person's least distance of distinct vision to 25 cm is 37.5 cm.
Ajụjụ 3 Ripọtì
The instrument that measures both ac and dc is called
Ajụjụ 4 Ripọtì
An immersion heater rated 2.0A, 240V is used to boil water from temperature 52°c to 100°c. If the mass of the water is 2.5kg. determine the time taken to boil the water.
[specific heat capacity of water = 4.2 x 103 jkg-1 k-1]
[specific heat capacity of water = 4.2 x 103 jkg-1 k-1]
Akọwa Nkọwa
The equation for the energy required to heat a substance can be written as:
Q = mcΔT
where Q is the energy required, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
Using the given information, we can solve for the energy required to heat the water from 52°C to 100°C:
ΔT = 100°C - 52°C = 48°C
Q = (2.5 kg)(4.2 x 10^3 J/kg°C)(48°C) = 5.04 x 10^5 J
The power of the immersion heater is given as 2.0A at 240V, so the power output can be calculated as:
P = IV = (2.0A)(240V) = 480W
The time required to provide the energy needed to heat the water can be found using the formula:
t = Q/P
t = (5.04 x 10^5 J) / (480W) = 1.05 x 10^3 s
Therefore, the time taken to boil the water is 1.05 x 10^3 seconds, which is equivalent to 17.5 minutes (since there are 60 seconds in a minute).
The answer closest to this value is "1.05 x 10^3 s".
Ajụjụ 5 Ripọtì
A radioactive substance has a half life of 20 days. What fraction of the original radioactive nuclei will remain after 80 days?
Akọwa Nkọwa
The half-life of a radioactive substance is the amount of time it takes for half of the original radioactive nuclei to decay. If the half-life of the substance is 20 days, then after 20 days, half of the original radioactive nuclei will have decayed, and half will remain. After another 20 days (a total of 40 days), half of the remaining nuclei will have decayed, leaving a quarter of the original nuclei. After another 20 days (a total of 60 days), half of the remaining nuclei will have decayed again, leaving an eighth of the original nuclei. Finally, after another 20 days (a total of 80 days), half of the remaining nuclei will have decayed, leaving a sixteenth of the original nuclei. Therefore, after 80 days, 1/16 of the original radioactive nuclei will remain. So the answer is (14).
Ajụjụ 6 Ripọtì
A drop of petrol makes a finger colder than a drop of water would because?
Akọwa Nkọwa
A drop of petrol makes a finger colder than a drop of water would because petrol takes its latent heat of vaporization from the finger. When a liquid evaporates, it needs heat to convert from a liquid to a gas. This heat is called the latent heat of vaporization. When a drop of petrol evaporates from a finger, it takes heat from the finger to convert into a gas, thus making the finger feel colder. Water, on the other hand, also evaporates and takes heat from the finger, but it has a lower rate of evaporation than petrol and requires more time to evaporate, resulting in a slower cooling effect.
Ajụjụ 7 Ripọtì
If the radius of the earth is 6.4 x 106m, the escape velocity of a satellite from the earth is?
[g = 10 ms-2]
[g = 10 ms-2]
Akọwa Nkọwa
V esc = √2Rg. =√2 x 6.4 x 106 x 10 = 1.13 x 104 ms-1
Ajụjụ 8 Ripọtì
In a gas experiment, if the volume of the gas is plotted against the reciprocal of the pressure, the unit of the slope of the resulting curve is
Akọwa Nkọwa
IF the volume of gas is plotted against the reciprocal of pressure, the slope will be given as
Slope = V/1/P = P x V
thus the unit of the slope will be;
= Nm-2, = V = M3 = NM.
The unit NM (Newton - Meter) is the practical unit of work, which is the product of force (N) and Displacement (M).
Note: It should be noted that the practical unit of work is Newton-Meter, whereas its S.I unit is Joules, just as energy.
Slope = V/1/P = P x V
thus the unit of the slope will be;
| P = | Force |
| Area |
= Nm-2, = V = M3 = NM.
The unit NM (Newton - Meter) is the practical unit of work, which is the product of force (N) and Displacement (M).
Note: It should be noted that the practical unit of work is Newton-Meter, whereas its S.I unit is Joules, just as energy.
Ajụjụ 9 Ripọtì
When a certain a.c supply is connected to a lamp, it lights with the same brightness as it does with a 12 V d.c battery. Then r.m.s value of the a.c supply is?
Akọwa Nkọwa
The root mean square (rms) value of an alternating current (AC) is the equivalent steady or direct current (DC) value that gives the same energy to a resistor as does the AC. When an AC supply is connected to a lamp, the brightness of the lamp depends on the power being delivered to the lamp. If the lamp lights with the same brightness as it does with a 12 V DC battery, it means that the power being delivered to the lamp in both cases is the same. Therefore, the rms value of the AC supply must be the same as the DC voltage of the battery, which is 12.0 V. Thus, the answer is 12.0 V.
Ajụjụ 10 Ripọtì
The wavelength of the first overtone of a note in a closed pipe of length 33 cm is?
Akọwa Nkọwa
In a closed tube, the fundamental note occurs when the length of the tube l = (λ)/4,
the first over tone occurs when the length of the tube l = (3)/4λ.
λ = (4)/3 l = (4)/3 x (33)/T = 44 cm
the first over tone occurs when the length of the tube l = (3)/4λ.
λ = (4)/3 l = (4)/3 x (33)/T = 44 cm
Ajụjụ 11 Ripọtì
The thermometric property of a constant volume thermometer is?
Akọwa Nkọwa
A constant volume thermometer is a type of thermometer that measures temperature by keeping the volume of the gas inside the thermometer constant. The thermometric property in this case would be the property that changes with temperature and is responsible for indicating the temperature on the thermometer. In the case of a constant volume thermometer, the thermometric property is the pressure of the gas. As the temperature increases, the average kinetic energy of the gas molecules increases, which in turn causes them to collide with the walls of the thermometer more frequently and with greater force. This results in an increase in the pressure of the gas, which is measured by the thermometer and used to indicate the temperature. Therefore, the correct option is "change in pressure."
Ajụjụ 12 Ripọtì
I. Work done by load is always less than the work done by effort.
II. All machines are made of levers, pulleys, screws and gear wheels.
III. A fishing rod is a machine.
Iv. Effort is always larger than load.
which of statements above about machines are true?
Akọwa Nkọwa
The question asks which statements about machines are true. Let's evaluate each statement:
I. Work done by load is always less than the work done by effort.
This statement is generally true for all machines because some of the work done by the effort is used to overcome friction and other resistive forces in the machine. Therefore, the statement is true.
II. All machines are made of levers, pulleys, screws and gear wheels.
This statement is not true as machines can be made of a variety of components including electronic parts, hydraulic systems, etc. Therefore, the statement is false.
III. A fishing rod is a machine.
This statement is true as a fishing rod is a simple machine that uses a lever to catch fish. Therefore, the statement is true.
IV. Effort is always larger than load.
This statement is not true because machines can be designed to have a load larger than the effort, such as in a hydraulic press. Therefore, the statement is false.
From the above evaluations, we can see that the true statements about machines are I and III. Therefore, the correct answer is "I and III".
Ajụjụ 13 Ripọtì
A p-n junction diode is used as?
Akọwa Nkọwa
A p-n junction diode is used as a rectifier in a circuit, either in a DC circuit or an AC circuit. A rectifier is a device that converts AC to DC by allowing the flow of current in one direction and blocking it in the other direction. This is achieved by the p-n junction diode, which has a unique property of allowing current to flow in only one direction. Therefore, the p-n junction diode is widely used as a rectifier in various electronic circuits.
Ajụjụ 14 Ripọtì
An ammeter of resistance 5Ω has a full-scale deflection when a current of 50 mA flows in it. The value of the resistor required to adapt it to measure a current of 5 A is?
Akọwa Nkọwa
The ammeter has a resistance of 5Ω and reads full scale when a current of 50 mA flows through it. To measure a current of 5A, a shunt resistor is required to be connected in parallel with the ammeter. We can use the formula I(total) = I(ammeter) + I(shunt), where I(total) is the total current, I(ammeter) is the current passing through the ammeter, and I(shunt) is the current passing through the shunt resistor.
Since the ammeter reads full scale at 50 mA, it means that 50 mA will flow through the ammeter when a total current of 5 A is passing through the circuit. Therefore, we can substitute the values in the formula:
5A = 50mA + I(shunt)
Solving for I(shunt), we get:
I(shunt) = 5A - 50mA
I(shunt) = 4.95A
To find the value of the shunt resistor, we can use Ohm's Law:
V = IR
Where V is the voltage drop across the shunt resistor, I is the current passing through the shunt resistor, and R is the resistance of the shunt resistor.
We know that the total current passing through the shunt and ammeter is 5A, and the current passing through the shunt is 4.95A. Therefore, the current passing through the ammeter is:
I(ammeter) = 5A - 4.95A
I(ammeter) = 0.05A
We also know that the resistance of the ammeter is 5Ω. Therefore, the voltage drop across the ammeter is:
V(ammeter) = I(ammeter) x R(ammeter)
V(ammeter) = 0.05A x 5Ω
V(ammeter) = 0.25V
Since the shunt resistor is connected in parallel with the ammeter, the voltage drop across the shunt resistor will also be 0.25V. Therefore, we can use Ohm's Law to find the resistance of the shunt resistor:
R(shunt) = V(shunt) / I(shunt)
R(shunt) = 0.25V / 4.95A
R(shunt) = 0.0505Ω
Therefore, the value of the resistor required to adapt the ammeter to measure a current of 5 A is 0.0505Ω, which is closest to the option 0.05Ω.
Ajụjụ 16 Ripọtì
A progressive wave equation is represented by y = a sin (150 πt - πt/4). if the phase difference of a progressive wave is 45o, the value of x in the equation is
Akọwa Nkọwa
Y = sin π(150t - x/4)
comparing the above with the standard wave equation:
Y = A sin π ((2vt)/λ - (2x)/λ)
=> the phase difference x/4 = 2x/λ
∴ 1/4 = 2/λ => λ = 8cm
∴ 2/8x = 45o
∴ x/4 = 0.7071 => x = 4 x 0.7071
= 2.82
= 3cm
comparing the above with the standard wave equation:
Y = A sin π ((2vt)/λ - (2x)/λ)
=> the phase difference x/4 = 2x/λ
∴ 1/4 = 2/λ => λ = 8cm
∴ 2/8x = 45o
∴ x/4 = 0.7071 => x = 4 x 0.7071
= 2.82
= 3cm
Ajụjụ 17 Ripọtì
The battery connection that permits current to flow across a p-n junction is called?
Akọwa Nkọwa
A p-n junction is formed by the junction of a p-type semiconductor and an n-type semiconductor. In order for current to flow across a p-n junction, it must be connected to a power source, such as a battery. The battery connection that allows current to flow across a p-n junction is called a "bias."
There are two types of bias connections: forward bias and reverse bias. Forward bias occurs when the positive terminal of the battery is connected to the p-type semiconductor and the negative terminal is connected to the n-type semiconductor. This configuration allows current to flow across the junction easily.
Reverse bias occurs when the positive terminal of the battery is connected to the n-type semiconductor and the negative terminal is connected to the p-type semiconductor. This configuration does not allow current to flow across the junction easily, and only a small leakage current is allowed.
Therefore, the correct answer to the question is "forward biased."
Ajụjụ 18 Ripọtì
A shooter wants to fire a bullet in such a way that its horizontal range is equal to three times its maximum height. At the what angle should he fire the bullet to achieve this?
Akọwa Nkọwa
Horizontal range = 3 x maxi. height
=> (U2Sin2θ) / g = (3 x U2sin2θ) / 2g
∴ 4sinθcosθ = 3sin2θ
=> 4cosθ = 3sinθ
∴ sinθ/cosθ = 4/3; => tanθ = 4/3 = 1.3333
θ = tan-1 (1.3333)
= 53o
=> (U2Sin2θ) / g = (3 x U2sin2θ) / 2g
| ∴sin2θ = | 3sin2θ |
| 2 |
| => 2sinθcosθ = | 3sin2θ |
| 2 |
∴ 4sinθcosθ = 3sin2θ
=> 4cosθ = 3sinθ
∴ sinθ/cosθ = 4/3; => tanθ = 4/3 = 1.3333
θ = tan-1 (1.3333)
= 53o
Ajụjụ 19 Ripọtì
An electric heater with a p.d of 240 V connected across its terminals has a total resistance of 960Ω. find the power rating of the heater.
Akọwa Nkọwa
Power, P = (V^2)/R, where V is the voltage and R is the resistance. Given that the p.d of the electric heater is 240 V and the total resistance is 960Ω, we can calculate the power rating of the heater as follows:
P = (240^2)/960
P = 57600/960
P = 60 W
Therefore, the power rating of the electric heater is 60 W. Answer option (D) is correct.
Ajụjụ 20 Ripọtì
The main purpose of the transformer in an a.c radio set is to?
Akọwa Nkọwa
The main purpose of the transformer in an a.c radio set is to "step up" or "step down" the voltage of the alternating current that comes from the power outlet to a voltage level suitable for the operation of the radio. The voltage level required by the radio is typically different from the voltage level of the power outlet, and this is where the transformer comes in. It consists of two coils of wire wound around a core, and when an alternating current flows through one of the coils, it induces an alternating current in the other coil, but with a different voltage level. This allows the radio to receive the appropriate voltage for its operation. Therefore, the correct option is "step up the voltage" or "step down the voltage," depending on the specific application.
Ajụjụ 21 Ripọtì
The frequency of the wave in the diagram above is
Akọwa Nkọwa
Generally, period T = 1/f(reciprocal of frequency).
=> f = 1/T
And from the diagram, period T = 4s
∴ frequency, f = 1/T = 1/4 = 0.25 Hz
=> f = 1/T
And from the diagram, period T = 4s
∴ frequency, f = 1/T = 1/4 = 0.25 Hz
Ajụjụ 22 Ripọtì
A certain manufacturer wishes to make an n-type semiconductor. Which materials would he select for doping?
Akọwa Nkọwa
An n- type semi conductor is usually formed by dopping the semi conductor(a four valent element)with a five valent
element.(arsenic and antimony)
element.(arsenic and antimony)
Ajụjụ 23 Ripọtì
If two charged plates are maintained at a potential difference of 3 kv, the work done in taking a charge of 600µc across the field is
Akọwa Nkọwa
When a charge moves through an electric field, the work done on the charge is given by the formula W = QV, where Q is the charge and V is the potential difference across which the charge moves. In this case, the potential difference across the charged plates is 3 kV, and the charge that is moved across the field is 600 µC (or 6.0 × 10^-4 C). Therefore, the work done is:
W = QV = (6.0 × 10^-4 C)(3 × 10^3 V) = 1.8 J
So the correct answer is 1.8J.
Ajụjụ 24 Ripọtì
At low humidity in an environment, the human skin is usually?
Akọwa Nkọwa
At low humidity in an environment, the human skin usually becomes dry and rough. When there is low humidity, the air is dry and tends to absorb moisture from the skin, leaving it dry and rough. In contrast, high humidity conditions prevent moisture loss from the skin, resulting in a damp and smooth feel.
Ajụjụ 25 Ripọtì
An object placed 10cm from a concave mirror of focal length 5cm would have its image
Akọwa Nkọwa
Ajụjụ 26 Ripọtì
I. The number of neutrons released exceeds the number that caused the fission.
II. There is loss of mass.
III. Energy is produced.
IV. Smaller nuclei merge to form larger nuclei
Which of the statement above are correct about nuclear fission?
Ajụjụ 27 Ripọtì
One of the properties of gamma rays is that they are?
Akọwa Nkọwa
Gamma rays are neutral, meaning they have no electrical charge.
Ajụjụ 28 Ripọtì
In an experiment to determine terminal velocity, as the pressure increases, the viscosity off a liquid
Akọwa Nkọwa
Ajụjụ 29 Ripọtì
what is likely to happen if the glass of a thermometer expands more upon heating than the liquid inside?
Akọwa Nkọwa
When a thermometer is heated, both the glass and the liquid inside it will expand. However, if the glass expands more than the liquid, then the volume of the glass will increase more than that of the liquid. This will result in a decrease in the pressure inside the thermometer. As a result, the liquid will rise up in the stem of the thermometer to fill the partial vacuum created. Therefore, the correct option is "the liquid will rise up in the stem".
Ajụjụ 30 Ripọtì
If the capacitance of an capacitor is C = (εA)/d, which of the following parameters is varied when a variable capacitors is used for tuning radio sets?
Akọwa Nkọwa
Ajụjụ 31 Ripọtì
The thermos flask is designed to?
Akọwa Nkọwa
A thermos flask is a type of container that is designed to keep the temperature of its contents constant for as long as possible. It does this by preventing heat loss or gain through three different mechanisms: conduction, convection, and radiation. The flask consists of two walls with a vacuum between them, which minimizes heat transfer by conduction or convection. Additionally, the walls are coated with a reflective material to minimize heat transfer by radiation. Therefore, the correct answer is: "prevent heat loss or gain by conduction, convection, and radiation."
Ajụjụ 32 Ripọtì
A radioactive element has a half life of 4 days. The fraction that has decayed in 16 days is?
Akọwa Nkọwa
The half-life of a radioactive element is the time it takes for half of its atoms to decay. So if the half-life of an element is 4 days, it means that after 4 days, half of the atoms will decay, and after another 4 days, half of the remaining atoms will decay, and so on.
To find the fraction that has decayed in 16 days, we need to determine how many half-lives have occurred. Since the half-life of the element is 4 days, there are 4 half-lives in 16 days.
Each half-life reduces the amount of radioactive material by half. So after 1 half-life, half of the material is left. After 2 half-lives, only a quarter of the material remains. After 3 half-lives, there is only an eighth left. After 4 half-lives, only a sixteenth is left.
Therefore, the fraction that has decayed in 16 days is 1 - 1/16 = 15/16.
So the answer is (15)/16.
Ajụjụ 33 Ripọtì
During electrolysis, 2.0 g of a metal is deposited using a current of 0.5 A in 3 hours. The mass of the same metal which can be deposited using a current of 1.5 A in 1 hour is?
Akọwa Nkọwa
M = Itz; z = (M)/It = (2)/0.5 x 3 x 60 x 60
= 3.703 x 10-4 gc-1
∴ M = Itz = 1.5 x 1 x 60 x 60 x 3.703 x 10-4
= 1.99962g.
=2.0g
= 3.703 x 10-4 gc-1
∴ M = Itz = 1.5 x 1 x 60 x 60 x 3.703 x 10-4
= 1.99962g.
=2.0g
Ajụjụ 34 Ripọtì
The correct unit of energy density is?
Ajụjụ 35 Ripọtì
When an a.c. voltage is applied to a capacitor, no power is consumed because the current
Akọwa Nkọwa
When an AC (alternating current) voltage is applied to a capacitor, the current lags behind the voltage by 90 degrees. This means that when the voltage reaches its maximum value, the current is at its minimum value, and when the voltage is at its minimum value, the current is at its maximum value. However, even though the current is constantly changing, the net power consumed by the capacitor is zero. This is because the capacitor stores energy in its electric field when the voltage is increasing, and releases this stored energy when the voltage is decreasing. Therefore, over one complete cycle of the AC voltage, the energy stored in the capacitor is equal to the energy released, resulting in no net power consumption.
Ajụjụ 36 Ripọtì
The electrolyte in the Nickei-iron (Nife) accumulator is
Akọwa Nkọwa
The electrolyte in the Nickel-Iron (NiFe) accumulator is potassium hydroxide solution. A NiFe battery is a rechargeable battery that is used for energy storage in renewable energy systems. It consists of a positive electrode of nickel hydroxide and a negative electrode of iron. Potassium hydroxide is used as an electrolyte in this battery. The electrolyte facilitates the movement of ions between the electrodes during the charge and discharge cycles. It allows the battery to store and release electrical energy as needed.
Ajụjụ 37 Ripọtì
A submarine is observed to rise from a real depth off 80 m to 60 m in water. calculate the change in apparent depth.
[Refraction index of water (4)/3]
[Refraction index of water (4)/3]
Akọwa Nkọwa
The apparent depth of an object in water is the depth at which it appears to be when viewed from above the water's surface. The apparent depth of an object is affected by the refractive index of water, which is a measure of how much light is bent as it passes from one medium to another.
When a submarine rises from a real depth of 80 m to 60 m, its apparent depth changes because the light rays from the submarine bend as they pass from water to air, causing the submarine to appear to be at a different depth than it actually is.
To calculate the change in apparent depth, we need to use the formula:
Change in apparent depth = real depth x (refractive index of water - 1)
Plugging in the values, we get:
Change in apparent depth = 80m x (4/3 - 1) = 80m x (1/3) = 26.67m
Therefore, the change in apparent depth is 26.67 meters, which is closest to the option labeled "15m".
Ajụjụ 38 Ripọtì
The motion of smoke particles from a chimney is typical of?
Akọwa Nkọwa
The motion of smoke particles from a chimney is typical of random motion. Smoke particles move in random directions and with random speeds due to the turbulent nature of the flow. This is in contrast to circular motion, which involves objects moving in a circular path, rotational motion, which involves objects rotating around an axis, or oscillatory motion, which involves objects oscillating back and forth around a central point. In the case of smoke particles, they are subjected to various forces, including air resistance, wind, and changes in temperature, causing them to move in unpredictable paths. Therefore, the motion of smoke particles from a chimney is a clear example of random motion.
Ajụjụ 39 Ripọtì
Non-luminous object can be seen because they
Akọwa Nkọwa
Non-luminous objects can be seen because they reflect light. When light falls on an object, it either gets absorbed or reflected by the object's surface. Non-luminous objects do not emit light of their own, but they can still be seen because they reflect light that falls on them. This reflected light then enters our eyes, and we can see the object as a result.
Ajụjụ 40 Ripọtì
If the relative density of gold is 19.2, the volume of 2.4kg of gold is?
Akọwa Nkọwa
R.D = (density of the subs)/density of water
19.2 =(P)/103; ∴P = 19.2 X 103
But Vol. = (Mass)/density = (2.4)/19.2 x 103
= 1.25 x 10-4 M3
19.2 =(P)/103; ∴P = 19.2 X 103
But Vol. = (Mass)/density = (2.4)/19.2 x 103
= 1.25 x 10-4 M3
Ajụjụ 41 Ripọtì
The theory of electronic conduction in solids, which satisfactorily explains the properties of conductors, insulators and semi-conductor,is?
Akọwa Nkọwa
The theory of electronic conduction in solids that explains the properties of conductors, insulators, and semiconductors is known as the "band theory." This theory explains how electrons in a solid material are distributed in energy levels, or bands, which determine whether the material conducts electricity. Conductors have partially filled energy bands that allow electrons to move freely, while insulators have completely filled bands that prevent electrons from moving. Semiconductors have partially filled bands that can be easily manipulated to allow or prevent electron flow, making them useful in electronic devices. Therefore, the correct option is "band theory."
Ajụjụ 42 Ripọtì
A cell of internal resistance r is connected to an external resistor R. The conductor for maximum power transfer is?
Akọwa Nkọwa
When a cell is connected to an external resistor, the power delivered to the resistor is maximum when the resistance of the external resistor matches the internal resistance of the cell. This is known as the maximum power transfer theorem.
In this case, the internal resistance of the cell is given as "r" and the external resistance is "R". Therefore, the conductor for maximum power transfer would be when R = r, i.e., when the external resistor has the same value as the internal resistance of the cell.
So the answer is: R = r.
Ajụjụ 43 Ripọtì
A conductor has a diameter of 1.00mm and length of 2.00m. If the resistance of the material is 0.1Ω, the resistivity is
Akọwa Nkọwa
The formula for resistivity is given by ρ = RA/l, where R is the resistance, A is the cross-sectional area of the conductor, and l is the length of the conductor.
In this case, the diameter of the conductor is given as 1.00mm, so the radius is 0.5mm or 0.0005m. The cross-sectional area A of the conductor is given by A = πr^2 = π(0.0005m)^2 = 7.85 x 10^-7 m^2. The length of the conductor is given as 2.00m, and the resistance is given as 0.1Ω.
Substituting these values into the formula for resistivity, we get:
ρ = RA/l = (7.85 x 10^-7 m^2 x 0.1Ω) / 2.00m
Simplifying this expression, we get:
ρ = 3.925 x 10^-9 Ωm
Therefore, the answer is option (D) 3.93 * 10^-8 Ωm.
Ajụjụ 44 Ripọtì
The energy associated with the emitted photon when a mercury atom changes from one state to another is 3.3 ev. calculate the frequency of the photon.
[e = 1.6 x 10-19c; h = 6.6 x 10-13js]
[e = 1.6 x 10-19c; h = 6.6 x 10-13js]
Akọwa Nkọwa
To calculate the frequency of the photon, we can use the formula E = hf, where E is the energy of the photon, h is Planck's constant, and f is the frequency of the photon. Rearranging the formula to solve for f, we get f = E/h.
Given that the energy associated with the emitted photon when a mercury atom changes from one state to another is 3.3 eV, we first need to convert this to joules. We know that 1 eV is equal to 1.6 x 10^-19 J, so 3.3 eV is equal to 3.3 x 1.6 x 10^-19 J, which is 5.28 x 10^-19 J.
We also know that Planck's constant, h, is 6.6 x 10^-34 J s.
Plugging these values into the formula f = E/h, we get:
f = (5.28 x 10^-19 J)/(6.6 x 10^-34 J s) = 8.0 x 10^14 Hz
Therefore, the frequency of the photon is 8.0 x 10^14 Hz.
Answer: 8.0 x 10^14 Hz
Ajụjụ 45 Ripọtì
The process through which free electrons leave the surface of hot metal is known as?
Akọwa Nkọwa
The process through which free electrons leave the surface of a hot metal is known as thermionic emission. When a metal is heated to a high temperature, some of its electrons gain enough energy to escape the metal's surface and become free electrons. This phenomenon is called thermionic emission, and it is used in devices such as vacuum tubes and cathode ray tubes. The other options, photon emission, photoemission, and electron emission, are different processes that involve the emission of electrons or photons but not specifically related to the process of free electrons leaving the surface of a hot metal.
Ajụjụ 46 Ripọtì
A mirror of weight 75N is hung by a cord from a hook on the wall. If the cord makes an angle of 30o with horizontal, the tension in the cord is
Akọwa Nkọwa
When an object is hung by a cord, the cord exerts a force on the object known as tension. In this case, the mirror is hung by a cord from a hook on the wall, and the cord makes an angle of 30 degrees with the horizontal. The weight of the mirror acts downwards, and the tension in the cord acts upwards to balance the weight. To find the tension in the cord, we need to use trigonometry. We can resolve the weight of the mirror into its horizontal and vertical components. The horizontal component is given by Wcosθ, where W is the weight of the mirror and θ is the angle between the cord and the horizontal. The vertical component is given by Wsinθ. In this case, the weight of the mirror is 75N, and the angle θ is 30 degrees. Therefore, the horizontal component is: Wcosθ = 75N x cos 30 = 64.95N The vertical component is: Wsinθ = 75N x sin 30 = 37.5N Since the mirror is in equilibrium, the tension in the cord must be equal to the vertical component of the weight: Tension = Wsinθ = 37.5N Therefore, the tension in the cord is 37.5N. The correct option is: 150N.
Ajụjụ 47 Ripọtì
A uniform solid cube material 10cm on each side of mass 700g is submerged in water. which of the following best described the behavior of the cube water?
Akọwa Nkọwa
The behavior of a solid cube of material when submerged in water depends on its density relative to that of water. If the cube is denser than water, it will sink. If it is less dense than water, it will float. If the density of the cube is equal to that of water, it will rest at an equilibrium position in water. Since the mass of the cube is 700g and its volume is 1000 cubic centimeters (since it is a 10cm cube), we can calculate its density as 0.7 g/cm^3. This is less than the density of water (which is approximately 1 g/cm^3), so the cube will float in water. Therefore, the answer is "the cube will float".
Ajụjụ 48 Ripọtì
An engine of a car of power 80 kW moves on a rough road with a velocity of 32 ms-1. The forces required to bring it to rest is
Akọwa Nkọwa
In terms of dimension: power , P, = ML2T3
Force F1 = MLT-2; Vel, V, LT-1
From the above, => P = F x V
i.e, ML2T-3 = MLT-2 x LT-1
ML2T-3
=> 80,000 = F x 32
= 2500N
= 2.50 x 103N
Force F1 = MLT-2; Vel, V, LT-1
From the above, => P = F x V
i.e, ML2T-3 = MLT-2 x LT-1
ML2T-3
=> 80,000 = F x 32
| ? F = | 80,000 |
| 32 |
= 2500N
= 2.50 x 103N
Ajụjụ 49 Ripọtì
The production and type of charge of a body can only be accounted for by the?
Akọwa Nkọwa
