JAMB UTME - Physics - 2004
Swali 1 Ripoti
For semi conductor to have negative temperature coefficient of resistance implies that
Maelezo ya Majibu
Temperature coefficient of resistance is given by
| α = | Increase in resistance per oC rise in temp. |
| __resistance at 0oC |
And for metals where resistance increase with temperature rise, temperature coefficient of resistance is positive; on the other hand in semi conductors, where resistance decrease with temperature rise, temperature coefficient of resistance is negative.
Swali 3 Ripoti
In which of the following material media will sound travel fastest?
Maelezo ya Majibu
Sound travels fastest through solids, such as metals, because the particles are closest together, making it easier for the sound wave to propagate. Therefore, the option that best answers the question is "metal".
Swali 5 Ripoti
The diagram above shows forces 4N, 6N, 10N and 8N which act at a point O in the directions indicated. the net horizontal force is
Maelezo ya Majibu
Component of 8N along the horizontal = 8 cos 60o = 8 x 0.5
= 4N
Similarity Component of 6N along the horizontal = 6 cos 60o = 3N
∴ Total force towards the right = 10 + 4 + 3
= 17N
Again total force toward the left = 4N
∴ Net horizontal force (towards the right) = 17-4
= 13N
Swali 6 Ripoti
A radioisotope has a decay constant of 10-7s-1. The average life of the radioisotope is
Maelezo ya Majibu
Average life / Half life (Tl/2) = ln2/λ = 0.693/10-7
= 0.693 x 107
= 6.93 x 108
Swali 7 Ripoti
A machine whose efficiency is 60% has a velocity ratio of 5. If a force of 500N is applied to lift a load P, what is the magnitude of P?
Maelezo ya Majibu
efficiency = (M.A / V.R) x 100/1 =>60/100 = M.A/5
∴M.A = (60 X 5) / 100 = 3
But M.A = Load/Effort; => 3 = Load/500
∴Load = 500 x 3
= 1500N
Swali 8 Ripoti
A body of mass 4kg is acted on by a constant force of 12N for 3 seconds. The kinetic energy gained by the body at the end of the time is
Maelezo ya Majibu
The kinetic energy gained by a body is equal to the work done on it by a force, as given by the formula: Kinetic energy gained = work done = force x distance In this problem, the force acting on the body is constant at 12N, and it acts for a time of 3 seconds. We can find the distance moved by the body using the formula for distance travelled by an object under constant acceleration: Distance = (1/2) x acceleration x time^2 Since the force on the body is constant, we can use Newton's second law to find its acceleration: Force = mass x acceleration Acceleration = Force / mass = 12N / 4kg = 3 m/s^2 Using this acceleration, we can find the distance travelled by the body: Distance = (1/2) x 3 m/s^2 x (3 s)^2 = 13.5 m Now, we can use the formula for work done to find the kinetic energy gained by the body: Work done = force x distance = 12N x 13.5m = 162 J Therefore, the correct answer is 162J.
Swali 9 Ripoti
When left in a freezer, a bottle full of water cracks on freezing into ice because of the
Maelezo ya Majibu
Swali 10 Ripoti
The energy stored in an inductor of inductance 5mH when a current of 6A flows through it is
Maelezo ya Majibu
The energy stored in an inductor can be calculated using the formula: E = 0.5 * L * I^2 where E is the energy stored, L is the inductance, and I is the current flowing through the inductor. Substituting the given values, we get: E = 0.5 * 5mH * (6A)^2 First, we need to convert millihenry (mH) to henry (H): 1mH = 0.001H Therefore, E = 0.5 * 0.005H * (6A)^2 Simplifying, E = 0.09J So the energy stored in the inductor is 0.09J. Therefore, the correct option is: 9.0 x 10-2J.
Swali 11 Ripoti
As the pressure of a fluid increases, its viscosity
Maelezo ya Majibu
As the pressure of a fluid increases, its viscosity increases. Viscosity is the measure of a fluid's resistance to flow. When pressure is increased, the molecules in the fluid come closer together, leading to stronger intermolecular forces and increased friction between them. This results in an increase in viscosity, making it more difficult for the fluid to flow. Therefore, is correct.
Swali 12 Ripoti
Transverse waves can be distinguished from longitudinal waves using the characteristic of
Maelezo ya Majibu
Transverse waves can be distinguished from longitudinal waves using the characteristic of polarization. Polarization is a property of transverse waves, which describes the direction of the vibrations of the wave in relation to the direction of wave propagation. In a transverse wave, the vibrations are perpendicular to the direction of wave propagation, whereas in a longitudinal wave, the vibrations are parallel to the direction of wave propagation. Therefore, polarization is a key characteristic that distinguishes transverse waves from longitudinal waves. Diffraction, refraction, and reflection are properties that apply to both transverse and longitudinal waves.
Swali 13 Ripoti
Calculate the angle of minimum deviation for a ray which is refracted through an equiangular prism of refractive index 1.4
Maelezo ya Majibu
The formula for the angle of minimum deviation for an equiangular prism is given as: $$ \delta_{\text{min}} = \frac{A}{2} + \text{arcsin}\left(\frac{n}{\sqrt{2}}\sin\frac{A}{2}\right) - 90^\circ $$ where $A$ is the angle of the prism and $n$ is the refractive index of the prism. For an equiangular prism, $A = 60^\circ$. Substituting $A$ and $n=1.4$ into the formula, we get: $$ \delta_{\text{min}} = \frac{60^\circ}{2} + \text{arcsin}\left(\frac{1.4}{\sqrt{2}}\sin\frac{60^\circ}{2}\right) - 90^\circ \approx 29^\circ $$ Therefore, the angle of minimum deviation for a ray refracted through an equiangular prism of refractive index 1.4 is approximately 29 degrees. Thus, the correct option is 290.
Swali 14 Ripoti
The earth is four times the size of the moon and the acceleration due to gravity on the earth is 80 times that on the moon. The ratio of the mass of the moon to that of the earth is
Maelezo ya Majibu
NOTE: Firstly, it should be noted that the mass, as well as the weight of a body is independent of its size.
∴ Let the weight of the moon = Mm x gm
weight of the earth = Me x ge
But ge = 80gm
∴ weight of the earth = Me x 80gm
=> weight of the moon / weight of the earth = Mm x gm / Me x 80gm
thus Mm/Me ∝ gm/80gm ∝ 1/80
=> Mm:Me ∝ 1:80
Swali 15 Ripoti
Given three capacitors 0.3μF, 0.5μF and 0.2μF, the joint capacitance when arranged to give minimum capacitance is
Maelezo ya Majibu
When capacitors are arranged in parallel, the equivalent capacitance is given by the sum of individual capacitances. When capacitors are arranged in series, the inverse of the equivalent capacitance is given by the sum of the inverses of individual capacitances. To obtain the minimum capacitance, the capacitors should be arranged in series. Therefore, the equivalent capacitance will be: 1/C = 1/C1 + 1/C2 + 1/C3 Substituting the given values, we get: 1/C = 1/0.3μF + 1/0.5μF + 1/0.2μF Solving the above equation, we get: C = 0.086μF Therefore, the equivalent capacitance is approximately 0.1μF, which is the first option.
Swali 16 Ripoti
Electrical appliances at home are normally earthed so that
Maelezo ya Majibu
Electrical appliances at home are normally earthed so that a person touching the appliances is safe from electric shock. When an appliance is not earthed and there is a fault, it can become live and carry an electric current. If a person touches the live appliance, the current can flow through their body to the earth, causing electric shock. However, when an appliance is earthed, any fault current flows to the earth instead of flowing through the body of the person touching the appliance, making it safer.
Swali 17 Ripoti
In a hydraulic press , a force of 40N is applied on the effort piston of area 0.4m2. If the force exerted on the load piston is 400N, the area of the large piston is
Maelezo ya Majibu
Pressure developed in the effort piston = F/a
∴F/a = 40/0.4 = 400/4 = 100Nm2
this pressure is transmitted to the larger piston of area A; => P = F/A
∴ 100 = 400/A
=> A = 400/100 = 4m2
Swali 18 Ripoti
The refractive index of the medium M in the diagram above is
Maelezo ya Majibu
In the diagram given, the incident ray AO is perpendicular to the surface AB. Therefore, it will pass straight through without deviation. When it reaches the surface BC, it will refract according to Snell's law. We can write: sin(i)/sin(r) = n where i is the angle of incidence, r is the angle of refraction, and n is the refractive index of the medium. Since the incident ray is perpendicular to AB, i = 0. We can then simplify the equation to: sin(r) = 0 This means that the refracted ray is parallel to the surface BC. When it reaches the surface CD, it will once again refract according to Snell's law. We can write: sin(i)/sin(r) = n where i is the angle of incidence and r is the angle of refraction. The incident angle i is equal to the angle of reflection, which is 60 degrees in this case. We can then simplify the equation to: sin(60)/sin(r) = n Solving for n, we get: n = sin(60)/sin(r) The angle of minimum deviation occurs when the refracted ray is symmetric with respect to the incident ray. In other words, the angles of incidence and refraction are equal. Since the incident angle is 60 degrees, the angle of refraction at CD is also 60 degrees. Therefore, we can substitute r = 60 degrees into the equation above to get: n = sin(60)/sin(60) = 1 The refractive index of the medium M is therefore 1. Answer option (2√3) is incorrect because it is greater than 1, and answer options (1/√3) and (2/√3) are incorrect because they are less than 1. The correct answer is option ( √3).
Swali 19 Ripoti
A person can focus an object only when it lies within 200cm from him. Which spectacles should be used to increase his maximum distance of distinct vision infinity
Maelezo ya Majibu
A person who can only focus on an object within 200cm is nearsighted or myopic. To increase their maximum distance of distinct vision to infinity, they would need a lens that can correct their nearsightedness. A convex lens is used to correct nearsightedness by diverging the light rays before they enter the eye, allowing for proper focus on distant objects. Therefore, the spectacles that should be used to increase the person's maximum distance of distinct vision to infinity is a convex lens.
Swali 20 Ripoti
What types of mirror are capable of producing parallel beams of light such as those arising from the headlamps of a car?
Maelezo ya Majibu
Parabolic mirrors are capable of producing parallel beams of light such as those arising from the headlamps of a car. This is because parabolic mirrors have a unique shape that can reflect light rays that come in parallel to the axis of the mirror, back out as parallel beams. Plane mirrors, cylindrical mirrors, and spherical mirrors do not have this property, and so they cannot produce parallel beams of light in the same way that parabolic mirrors can.
Swali 21 Ripoti
Flourescent tubes produces light by the
Maelezo ya Majibu
Fluorescent tubes produce light by the excitation of gas molecules. Fluorescent tubes have a gas mixture containing low-pressure mercury vapor and noble gases like argon, neon, or krypton. The gas inside the tube gets excited by electricity and emits ultraviolet (UV) light. The UV light excites the phosphor coating on the inside of the glass tube, causing it to emit visible light. The gas molecules act as a medium for the transfer of energy between the electrical source and the phosphor coating.
Swali 22 Ripoti
A generator manufacturing company was contracted to produce an a.c dynamo but inadvertently produced a d.c dynamo. To correct this error, the
Maelezo ya Majibu
An a.c dynamo and a d.c dynamo are two different types of generators that produce different types of currents. An a.c dynamo produces alternating current while a d.c dynamo produces direct current. To correct the error of producing a d.c dynamo instead of an a.c dynamo, the commutator should be replaced with slip rings. The commutator is a device that converts the alternating current produced by the dynamo into direct current. Since the dynamo was inadvertently made to produce direct current, there is no need for a commutator, and it should be replaced with slip rings which allow the output to be collected as alternating current.
Swali 23 Ripoti
A steady current of 2A flows in a coil of emf 12V for 0.4s. A back emf of 3V was induced during this period. The stored energy in the loop that can be utilized is
Maelezo ya Majibu
Energy stored = Ivt = current x back e.m.f x time
= 2 x 3 x 0.4
= 2.4J
Swali 24 Ripoti
The phase difference between waves P and q in the diagram above is
Maelezo ya Majibu
From the given diagram, it can be observed that wave P reaches its maximum amplitude at time t=0, whereas wave Q reaches its maximum amplitude at time t=T/4. Since the time difference between the maximum points of the two waves is T/4, where T is the time period of the waves, we can say that the phase difference between waves P and Q is π/2 or 90 degrees. Therefore, the correct option is: π/2.
Swali 25 Ripoti
In a reverse biased junction diode, current flows in by
Maelezo ya Majibu
In a reverse biased junction diode, current flows in by minority carriers. When a junction diode is reverse biased, the P-type region is connected to the negative terminal of the battery and the N-type region is connected to the positive terminal. In this condition, the holes in the P-type region and the electrons in the N-type region are both moving away from the junction, creating a depletion region with very few charge carriers. However, due to the presence of a small number of impurities, there are still some charge carriers present in this region, known as minority carriers. These minority carriers can contribute to the reverse current that flows through the diode when a voltage is applied in the reverse direction. Therefore, the correct option is "minority carriers".
Swali 26 Ripoti
In a tuned radio receiver R, L, C series circuit for resonance, the inductive and capacitive reactance XL and XC respectively are related as
Maelezo ya Majibu
In a tuned radio receiver R, L, C series circuit for resonance, the inductive and capacitive reactance X_L and X_C respectively are related as X_L = 1/(2πfC) and X_C = 1/(2πfL), where f is the frequency of the applied voltage. Therefore, substituting X_C into X_L, we get X_L = X_C. Hence, the correct answer is X_L = X_C.
Swali 27 Ripoti
The particle and wave nature of matter are demonstrated in the equation
Maelezo ya Majibu
The equation that demonstrates the particle and wave nature of matter is the second option: λ = h/P. This equation is known as the de Broglie wavelength equation, and it relates the wavelength of a particle to its momentum. This equation implies that particles, like waves, can have a wavelength and frequency, which is a property of waves. In other words, it shows that particles have a dual nature, they can behave as both particles and waves. The first option, λ = hc/E, is the equation for the energy of a photon, which demonstrates the wave nature of light. The third option, λ = c/f, is the equation for the wavelength of light, which also demonstrates the wave nature of light. The fourth option, λ = 2dsinθ, is the equation for the diffraction of waves, which is also a property of waves.
Swali 29 Ripoti
| The binding energy of helium | 4 | He is |
| 2 |
[atomic mass of proton = 1.00783U,
Atomic mass of neutron = 1.00867U]
Maelezo ya Majibu
The binding energy of helium refers to the amount of energy required to separate the two protons and two neutrons in the helium nucleus. The table provided shows the atomic masses of helium with two different numbers of neutrons. By subtracting the atomic mass of the helium nucleus with two neutrons from the one with four neutrons, we can find the mass defect, which is the mass that is converted to energy in the process of binding the nucleus. Using Einstein's famous equation E=mc², we can calculate the binding energy from the mass defect. The mass defect can be calculated as: mass defect = (atomic mass of helium with 2 neutrons) + 2 × (atomic mass of a proton) + 2 × (atomic mass of a neutron) - atomic mass of helium with 4 neutrons Substituting the given values, we get: mass defect = 2.014101u + 2 × 1.00783u + 2 × 1.00867u - 4.002603u mass defect = 0.030379u Using E=mc², we can find the binding energy, where c is the speed of light in a vacuum (299,792,458 m/s): binding energy = mass defect × c² binding energy = 0.030379u × (299,792,458 m/s)² binding energy ≈ 2.73 × 10⁻¹² J Therefore, the answer is 0.033U.
Swali 30 Ripoti
In the diagram above, the ratio of he electric power dissipated in the 6Ω and the 3Ω resistor respectively is
Maelezo ya Majibu
To solve the problem, we can use the formula for electric power, which is P = VI, where P is power, V is voltage, and I is current. Using Ohm's law, we can also write V = IR, where R is resistance. First, we need to find the current flowing through the circuit. Since the resistors are in parallel, the voltage across them is the same, and we can use Ohm's law to find the current through each resistor. For the 6Ω resistor, V = IR = (12V)/(6Ω) = 2A For the 3Ω resistor, V = IR = (12V)/(3Ω) = 4A Now, we can calculate the power dissipated in each resistor using the formula P = VI. For the 6Ω resistor, P = VI = (2A)(12V) = 24W For the 3Ω resistor, P = VI = (4A)(12V) = 48W The ratio of the power dissipated in the 6Ω resistor to that in the 3Ω resistor is 24W:48W, which simplifies to 1:2. Therefore, the answer is option (D) 1:2.
Swali 31 Ripoti
X rays can be used in the study of crystal structure because they
Maelezo ya Majibu
X-rays can be used in the study of crystal structure because they have an extremely short wavelength. This allows them to diffract off the atoms in the crystal lattice, producing a unique pattern that can be used to determine the arrangement of atoms within the crystal. The short wavelength of X-rays allows them to interact with the closely spaced atoms in the crystal, giving rise to the diffraction pattern. The other options are incorrect: X-rays are not long-reaching, they are not invisible (they can be detected using photographic film or electronic detectors), and their speed is not a factor in their usefulness for crystallography.
Swali 32 Ripoti
A 100kg box is pushed along a road with force of 500N. If the box moves with a uniform velocity, the coefficient of friction between the box and the road is
Maelezo ya Majibu
To find the coefficient of friction between the box and the road, we need to use the formula for frictional force, which is given by: f_friction = coefficient of friction x normal force. Since the box is moving with a uniform velocity, we know that the net force acting on the box is zero. The force pushing the box forward is balanced by the force of friction acting in the opposite direction. Therefore, we can set up an equation to find the force of friction acting on the box: f_push - f_friction = 0 where f_push is the force pushing the box forward, and f_friction is the force of friction acting in the opposite direction. Substituting the values given in the problem, we get: 500N - f_friction = 0 Solving for f_friction, we get: f_friction = 500N Now, we can use the formula for frictional force to find the coefficient of friction: f_friction = coefficient of friction x normal force The normal force is the force exerted by the road on the box, which is equal to the weight of the box (since the box is not accelerating): normal force = weight of box = mg = 100kg x 9.8 m/s^2 = 980N Substituting this value and the value we found for the force of friction into the formula for frictional force, we get: 500N = coefficient of friction x 980N Solving for the coefficient of friction, we get: coefficient of friction = 500N / 980N = 0.51 Therefore, the coefficient of friction between the box and the road is approximately 0.5, which is the fourth option.
Swali 33 Ripoti
A 50W electric heater is used to heat a metal block of mass 5kg. If in 10 minutes a temperature rise of 12oC is achieved, the specific heat capacity of the metal is
Maelezo ya Majibu
The specific heat capacity of a substance is the amount of energy required to raise the temperature of 1 kilogram of that substance by 1 degree Celsius. In this problem, we have a 50W electric heater that is used to heat a metal block of mass 5kg. We are also given that the temperature of the block increases by 12°C in 10 minutes. To find the specific heat capacity of the metal, we can use the formula: Q = mcΔT where Q is the amount of heat energy transferred, m is the mass of the metal block, c is the specific heat capacity of the metal, and ΔT is the change in temperature of the metal block. We know that the electric heater is supplying 50J of heat energy per second (since 1W = 1J/s), and the heater is operating for 10 minutes, or 600 seconds. Therefore, the total amount of heat energy supplied to the metal block is: Q = (50 J/s) x (600 s) = 30,000 J We also know that the mass of the metal block is 5kg, and the temperature of the block increases by 12°C. Therefore, we can rearrange the formula above to solve for the specific heat capacity: c = Q/(mΔT) = 30,000 J/(5 kg x 12°C) = 500 J/(kg·°C) Therefore, the specific heat capacity of the metal is 500 J/(kg·°C). So, the correct option is: 500 J kg-1K-1.
Swali 34 Ripoti
The pitch of a sound note depends on
Maelezo ya Majibu
The pitch of a sound note depends on its frequency. Pitch refers to the perceived highness or lowness of a sound, and is determined by the frequency of the sound wave. The higher the frequency, the higher the pitch of the sound. Therefore, among the given options, the correct answer is "frequency". Timbre, harmonics, and quality also affect the perceived sound, but they do not directly determine the pitch of the sound.
Swali 35 Ripoti
Metal rods of length 20m each are laid end to end to form a bridge at 25oC. What gap will be provided between consecutive rails for the bridge to withstand 75oC?
[Linear expansivity of the material = 2.0 x 10-5K-1]
Maelezo ya Majibu
The problem involves finding the gap to be provided between the consecutive metal rods, which are laid end to end to form a bridge at 25°C, so that the bridge can withstand a temperature of 75°C. The linear expansivity of the material is given as 2.0 x 10^-5 K^-1. When the temperature of a material increases, it expands in length. The amount of expansion is given by the coefficient of linear expansion, which is the increase in length per unit length per unit increase in temperature. Let x be the gap to be provided between consecutive rails at 25°C. When the temperature increases to 75°C, each metal rod will expand by a length of (20m x 2.0 x 10^-5 K^-1 x 50K) = 0.02m. Therefore, the total increase in length for two consecutive rods and the gap between them will be 0.04m. To ensure that the bridge can withstand the increased temperature, the gap between the consecutive rails should be equal to the total increase in length, which is 0.04m. Since there are two gaps between three rods, the gap to be provided between consecutive rails is half of 0.04m, which is equal to 0.02m. Therefore, the answer is option D, 0.02m.
Swali 36 Ripoti
A moving coil galvanometer has a full-scale deflection of 3A equivalent to 30o deflection. The sensitivity of the instrument is
Maelezo ya Majibu
The sensitivity of a moving coil galvanometer is the deflection produced per unit current passing through it. In this question, we are given that the full-scale deflection of the galvanometer is 30° which is equivalent to 3A. So, the sensitivity of the galvanometer can be calculated as follows: Sensitivity = (Full-scale deflection) / (Current producing it) Sensitivity = 30° / 3A = 10°/A Therefore, the sensitivity of the instrument is 10.0. Option (D) is the correct answer.
Swali 37 Ripoti
The change in volume when 450kg of ice is completely melted is
[density of ice = 900 kgm-3, density of water = 100 kgm-3]
Maelezo ya Majibu
Vol. of ice = M/?D = 450/900 = 0.500m3
vol. of water = M/D" = 450/1000 = 0.450m3
∴ Difference in vol. = 0.500 - 0.450
= 0.05m3
Swali 38 Ripoti
If u is the object distance and v the image distance, which of the following expressions gives the linear magnification produced by a convex lens of focal length f?
Maelezo ya Majibu
Magnification, M = v/u = v (1/u)
And for the Lens formula 1/u + 1/v = 1/f
=>1/u = 1/f - 1/v
∴M = V(1/u)
= V(1/f - 1/v) = v/f - v/v = v/f - 1
∴M = v/f - 1
Swali 39 Ripoti
An a.c circuit of e.m.f 12V has a resistor of resistance 8Ω connected in series to an inductor of inductive reactance 16Ω and a capacitor capacitive reactance 10Ω. The current flow in the circuit is
Maelezo ya Majibu
In this AC circuit, there is a resistor, an inductor, and a capacitor connected in series. The current flow in the circuit is given by the equation I = V/Z, where V is the voltage, and Z is the impedance of the circuit. The impedance of a series circuit is given by the equation Z = sqrt(R^2 + (Xl - Xc)^2), where R is the resistance, Xl is the inductive reactance, and Xc is the capacitive reactance. Substituting the given values, we have: Z = sqrt(8^2 + (16 - 10)^2) Z = sqrt(64 + 36) Z = sqrt(100) Z = 10Ω The voltage is given as 12V. Therefore, the current flow in the circuit is: I = V/Z I = 12/10 I = 1.2A Hence, the answer is 1.2A.
Swali 40 Ripoti
If a force of 50N stretches a wire from 20m to 20.01m, what is the amount of force required to stretch the same material from 20m to 20.05m?
Maelezo ya Majibu
To solve this problem, we need to use Hooke's law, which states that the force required to stretch or compress a spring is directly proportional to the extension or compression of the spring. We can represent this law as: F = kx where F is the force applied, x is the extension or compression of the spring, and k is a constant known as the spring constant. We can use this formula to solve the problem by first finding the spring constant of the wire. To do this, we can use the given force and extension: 50N = k(20.01m - 20m) k = 50N / 0.01m k = 5000 N/m Now we can use this spring constant to find the force required to stretch the wire from 20m to 20.05m: F = kx F = 5000 N/m * (20.05m - 20m) F = 250N Therefore, the amount of force required to stretch the wire from 20m to 20.05m is 250N. Answer: 250N.
Swali 41 Ripoti
The instrument used for securing a large number of similar charges by induction is called
Maelezo ya Majibu
The instrument used for securing a large number of similar charges by induction is called an electrophorus. An electrophorus consists of two parts: a metal plate and a non-conducting plate, usually made of plastic. The metal plate is charged by rubbing it with a fur or other material, which causes a buildup of negative charge on the plate. When the non-conducting plate is placed on top of the metal plate, the negative charge on the metal plate induces a positive charge on the top of the non-conducting plate. This process can be repeated multiple times, resulting in a large number of charges that can be used for various experiments or applications.
Swali 42 Ripoti
Tea pots are often silver-coated to prevent heat loss by
Maelezo ya Majibu
Tea pots are often silver-coated to prevent heat loss by radiation only. Silver is a good reflector of heat, and the shiny surface of the silver-coated tea pot reflects the heat back into the pot, preventing it from radiating out into the surrounding environment.
Swali 43 Ripoti
To protect a material from the influnce of an external magnetic field, the materials should be kept
Maelezo ya Majibu
In order to protect a material from the influence of an external magnetic field, the material is placed in a soft iron ring, usually referred to as magnetic screening.
Swali 44 Ripoti
A ray of light makes an angle of 35o with a plane mirror. What is the angle of reflection?
Maelezo ya Majibu
35o =35o = 70o
180 - 70 = 110
110/2 = 55o
Swali 45 Ripoti
I. Wavelength
II. Medium of propagation
III. Wave velocity
IV. Frequency
V. Energy
Which of the above are used for characterizing waves?
Maelezo ya Majibu
Swali 46 Ripoti
I. Jet propelled aircraft
II. Rocket propulsion
III. The recoil of a gun
IV. A person walking
Which of the above is based on Newton's third law of motion
Maelezo ya Majibu
Newton's third law of motion states that for every action, there is an equal and opposite reaction. Among the given options, the following satisfy this law: - III. The recoil of a gun: The action is the explosion of gas in the gunpowder which creates a forward force on the bullet and an equal and opposite force on the gun, causing the gun to recoil. - I. Jet propelled aircraft: The action is the expulsion of gases from the back of the jet engine, which creates a forward force on the aircraft, and an equal and opposite force in the opposite direction. - II. Rocket propulsion: The action is the ejection of exhaust gases from the nozzle of the rocket, which creates a forward force on the rocket, and an equal and opposite force in the opposite direction. Therefore, the answer is: I, II, and III only.
Swali 47 Ripoti
If 16mA of current flows through a conductor in one second, the number of electron transported per seconds is
[1 electronic charge = 1.6 x 10-19C]
Maelezo ya Majibu
Current (I) = q/t → q = i x t
= 16 x 10-3 x 1
= 1.6 x 10-2C
But 1.6 x 10-19C = I electronic charge
1 coulomb = 1 / (1.6x10-19)
∴ 1.6 x 10-2 coulombs = 1 / (1.6x10-19) x 1.6x10-2 / 1
= 1.0 x 10-2 x 1019
= 1.0 x 1017
Swali 48 Ripoti
A cell whose internal resistance is 0.5Ω delivers a current of 4A to an external resistor,The loss voltage of the cell is
Maelezo ya Majibu
When a cell delivers a current to an external resistor, there is a drop in voltage across the internal resistance of the cell due to the resistance of the cell. This is known as the "loss voltage" of the cell. We can use Ohm's law to calculate the loss voltage of the cell. Ohm's law states that V = IR, where V is voltage, I is current, and R is resistance. In this case, the internal resistance of the cell is given as 0.5Ω and the current delivered by the cell is 4A. Thus, the loss voltage of the cell is: V = IR = (4A)(0.5Ω) = 2V Therefore, the answer is 2.000V.
Swali 49 Ripoti
If the angle between two vector P and Q is Oo, the vectors are said to
Maelezo ya Majibu
When the angle between two vectors P and Q is 0, it means that they are pointing in the same direction, i.e., they are parallel. Therefore, the correct option is "be parallel."
Swali 50 Ripoti
A man exerts a pressure of 2.8 x 103Nm-2 on the ground and has 4 x 10-2m2 of his feet in contact with the ground. The weight of the man is
Maelezo ya Majibu
The pressure exerted by a person is defined as the force per unit area, where force is the weight of the person and area is the surface area of the contact between the person and the ground. Here, the pressure exerted by the man is given as 2.8 x 10^3 Nm^-2 and the surface area of his feet in contact with the ground is 4 x 10^-2 m^2. To find the weight of the man, we can use the formula: pressure = force/area Rearranging the formula, we get: force = pressure x area Substituting the given values, we get: force = (2.8 x 10^3 Nm^-2) x (4 x 10^-2 m^2) = 112 N Therefore, the weight of the man is 112N. Hence, the answer is 112N.
