JAMB UTME - Physics - 1997
Ajụjụ 2 Ripọtì
Equal masses of copper and rubber are raised to the same temperature. After sometimes, the copper was observed to be at a lower temperature because
Akọwa Nkọwa
The correct answer is "the specific heat capacity of copper is lower than that of rubber." When two objects with the same mass are heated to the same temperature, they will absorb the same amount of heat energy. The specific heat capacity of a material is the amount of heat energy required to raise the temperature of one kilogram of that material by one degree Celsius. Copper has a lower specific heat capacity than rubber, which means that it requires less heat energy to raise its temperature by one degree Celsius. As a result, when the same amount of heat energy is transferred to both copper and rubber, the temperature of copper will increase more quickly than that of rubber. Therefore, when both copper and rubber are raised to the same temperature and left for some time, the copper will cool down faster than the rubber, and its temperature will become lower.
Ajụjụ 3 Ripọtì
An object of mass 100g projected vertically upwards from the ground level has a velocity of 20ms-1 at a height of 10m. calculate its initial kinetic energy at the ground level.[g = 10ms-2, neglect air resistance]
Akọwa Nkọwa
Using, v2 = u2 - 2gx
(20)2 = u21/2 x 10 x 10
u2 = 600(m/s)2
initial K.E = - mu2
= 12 x 600
= 30 J
Ajụjụ 4 Ripọtì
A heating coil rated at 100W is used to boil off 0.5kg of boiling water. The time taken to boil off the water is [specific latent heat of vaporization of water = 2.3 x 106Jkg-1]
Akọwa Nkọwa
Heat energy = Elect. energy
0.5 x 2.3 x 106 = 100t
t = 1.15 x 104s
Ajụjụ 5 Ripọtì
The power dissipated in an a.c. circuit with an r.m.s, current of 5A, r.m.s voltage of 10V and a phase angle of 60o is
Akọwa Nkọwa
The power dissipated in an AC circuit can be calculated using the formula P = Vrms × Irms × cos(θ), where Vrms and Irms are the root mean square values of voltage and current respectively, and θ is the phase angle between them. Here, Vrms = 10V and Irms = 5A, and the phase angle is 60°. So, the power dissipated is P = 10V × 5A × cos(60°) = 25W. Therefore, the answer is 25W.
Ajụjụ 6 Ripọtì
A uniform light beam XZ is Hinged at X and kept in equilibrium by the forces T and F as shown in the diagram above. if XO = 20m and OY = 30cm, express T in terms of F
Ajụjụ 7 Ripọtì
A cube of sides 0.1m hangs freely from a string. What is the upthrust on the cube when totally immersed in water? [density of water is 1000kgm-2, g = 10ms-2]
Akọwa Nkọwa
When an object is submerged in a fluid, it experiences an upward force called buoyant force or upthrust. The magnitude of this force is equal to the weight of the fluid displaced by the object. In this problem, the cube has a volume of (0.1m)^3 = 0.001m^3. When the cube is immersed in water, it displaces an equal volume of water. Using the density of water, we can calculate the mass of the displaced water as: mass = density x volume = 1000kg/m^3 x 0.001m^3 = 1kg The weight of the displaced water is: weight = mass x gravitational acceleration = 1kg x 10m/s^2 = 10N Therefore, the upthrust on the cube is 10N. Therefore, the correct option is 10 N.
Ajụjụ 8 Ripọtì
A boy on looking into a mirror observed that his face appeared to have grown bigger. The boy must have been looking at a
Akọwa Nkọwa
When an object is placed in front of a concave mirror, the image formed can either be real or virtual depending on the position of the object relative to the mirror. If the object is placed between the focus and the mirror, the image formed is virtual, erect and magnified. Therefore, the boy must have been looking at a concave mirror with his face between the focus and the mirror.
Ajụjụ 9 Ripọtì
Which of the following with respect to a body performing simple harmonic motion are in phase?
Akọwa Nkọwa
Ajụjụ 11 Ripọtì
Under constant tension and constant mass per unit length, the note produced by a plucked string is 500Hz when the length of the string is 0.90m. At what length is the frequency 150Hz?
Akọwa Nkọwa
F = 12L √Tm
F1L1 = F2L2
500 x 0.9 = 150 x L2
therefore, L2 = 3m
Ajụjụ 12 Ripọtì
A planet has mass m1 and is at a distance r, from the sun. A second planet has mass m2 = 10m1 and at a distance of r2 = 2r1 from the sun. Determine the ratio of the gravitational force experienced by the planets.
Akọwa Nkọwa
g = GMR2
G1M1r21 = G2M2r22
G1M1r21 = G2×10M1(2r1)2
G1G2 = 25
Ajụjụ 13 Ripọtì
An electric water pump rate 1.5 kW, lifts 200kg of water through a vertical height of 6 meters in seconds. What is the efficiency of the pump? [g = 10ms-2, neglecting air resistance]
Akọwa Nkọwa
Efficiency = output x 100
= 200×10×θ101500 x 1001
= 80%
Ajụjụ 14 Ripọtì
The length of mercury thread when it is at 0oC, 100oC and at an unknown temperature θ is 25mm, 225mm and 175mm respectively. The value of θ is
Ajụjụ 15 Ripọtì
A galvanometer of internal resistance 50Ω
has a full scale deflection for a current of 5mA. What is the resistance required to covert it to a voltmeter with full scale deflection of 10V?
Akọwa Nkọwa
Ajụjụ 16 Ripọtì
An arrow of mass 0.1kg moving with a horizontal velocity of 15ms-1 is shot into a wooden block of mass 0.4kg lying at rest on a smooth horizontal surface. Their common velocity after impact is
Akọwa Nkọwa
We can solve this problem using the law of conservation of momentum, which states that the total momentum of a closed system remains constant before and after an interaction. In this case, the arrow and the wooden block form a closed system. Let the velocity of the arrow after impact be v. We can set up the following equation: (mass of arrow * initial velocity of arrow) = (total mass * final velocity) 0.1 kg * 15 m/s = (0.1 kg + 0.4 kg) * v 1.5 kg m/s = 0.5 kg * v v = 3 m/s Therefore, the common velocity of the arrow and the wooden block after impact is 3 m/s. None of the answer options provided matches this result.
Ajụjụ 17 Ripọtì
Three electric cells each of e.m.f 1.5V and internal resistance of 1.0Ω are connected in parallel across an external resistance of 23 Ω . Calculate the value of the current in the resistor
Akọwa Nkọwa
E.m.f = 1.5
r = 1× 1×1(1× 1)+(1× 1)+(1× 1)
r = 13 Ω
I = E.m.fr+R
I = 1.513 +23
I = 1.5A
Ajụjụ 18 Ripọtì
The figure above shows a block of mass m sliding down a rough inclined plane QP at angle θ
. The forces acting on the block along QP are
Ajụjụ 19 Ripọtì
If 8 x 10-2J of work is required to move 100μ C of charge from a point X to a point Y in an electrical circuit, the potential difference between X and Y is
Akọwa Nkọwa
The potential difference between X and Y in an electrical circuit is given by the equation: V = W/Q where V is the potential difference, W is the work done, and Q is the charge. From the question, we know that W = 8 x 10^-2 J and Q = 100 x 10^-6 C. Substituting these values into the equation, we have: V = (8 x 10^-2)/(100 x 10^-6) = 8 x 10^2 V Therefore, the potential difference between X and Y in the electrical circuit is 8.0 x 10^2 V. Option C is the correct answer.
Ajụjụ 20 Ripọtì
A load of 20N on a wire of cross-sectional area 8 x 10-7m2, produces an extension of 10m2. calculate Young's modulus for the material of the wire if its length is 3m
Akọwa Nkọwa
Young's modulus is defined as the ratio of stress to strain in a material under tensile load. In this problem, the load on the wire is 20N, the extension produced is 10^(-2)m and the cross-sectional area is 8 x 10^(-7) m^2. From the given information, we can calculate the stress in the wire using the formula: stress = load / area Therefore, stress = 20N / (8 x 10^(-7) m^2) = 2.5 x 10^7 N/m^2 We can also calculate the strain using the formula: strain = extension / original length Here, the extension is 10^(-2)m and the original length is 3m. Therefore, the strain is: strain = 10^(-2)m / 3m = 3.33 x 10^(-3) Now, we can use the definition of Young's modulus: Young's modulus = stress / strain Substituting the values we calculated above, we get: Young's modulus = (2.5 x 10^7 N/m^2) / (3.33 x 10^(-3)) = 7.5 x 10^10 N/m^2 Therefore, the answer is:
7.5 x 1011Nm-2
Ajụjụ 21 Ripọtì
The distance xm travelled by a particle in time t seconds is described by the equation x = 10 + 12t2, Find the average speed of the particle between the time interval t = 2s and t = 5s
Akọwa Nkọwa
speed = dxdt
dxdt = 10 + 12t2
dxdt = 24t
dxdt =24 x 3
dxdt = 72 ms-1
Ajụjụ 22 Ripọtì
What happens when a, certain quantity of pure ice is completely changed to water at 0oC?
Akọwa Nkọwa
Ajụjụ 23 Ripọtì
Light of velocity 3 x 108ms-1 is incident on a material of reflective of index n. If the velocity of light is reduced to 2.4 x 108ms-1 in the material, what is n?
Akọwa Nkọwa
The question is asking about the index of refraction of a material that is being hit by light. The speed of light changes when it passes through a material, which is described by the index of refraction. The question gives the velocity of light in a vacuum and in the material, so we can use the equation n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the material, to find the index of refraction. Plugging in the given values, we get n = 3 x 10^8 m/s / (2.4 x 10^8 m/s) = 1.25. Therefore, the correct answer is.
Ajụjụ 25 Ripọtì
The graphite rods surrounding the uranium fuel rods in a nuclear reactor, are used to
Akọwa Nkọwa
The graphite rods in a nuclear reactor have the function of controlling the nuclear reaction by slowing down the neutrons produced by the fission process. This is because fast neutrons are less likely to be absorbed by uranium-235, which is the fuel used in nuclear reactors, whereas slow neutrons can be easily captured by it, thus sustaining the nuclear reaction. Therefore, the correct option is: "slowdown the neutrons and hence slow up the nuclear process".
Ajụjụ 26 Ripọtì
A body of mass 2kg moving vertically upwards has its velocity increased uniformly from 10ms-1 to 40ms-1 in 4s. Neglecting air resistance, calculate the upward vertical force acting on the body.[g = 10 ms-1]
Akọwa Nkọwa
F = mg + ma
F = 2 [10 + (40−10)4 ]
F = 35 N
Ajụjụ 28 Ripọtì
What happens when a gas expands at constant temperature?
Akọwa Nkọwa
When a gas expands at constant temperature, its pressure decreases. This is due to the fact that the gas molecules now have more space to move around, and as a result, they collide with the walls of the container less frequently, causing the pressure to decrease. However, the total momentum of the gas molecules remains constant. This is because momentum is a conserved quantity, and the gas molecules only transfer their momentum to other molecules or to the walls of the container. Therefore, the total momentum of the gas remains the same, even though the individual molecules may have different speeds and directions. Option (C) is correct, as it correctly identifies both of these phenomena. Option (D) is incorrect, as the total kinetic energy of the gas molecules does not necessarily decrease when the gas expands at constant temperature. While individual molecules may lose kinetic energy due to collisions, others may gain kinetic energy due to the same collisions.
Ajụjụ 29 Ripọtì
For a rough inclined plane on which lies a body of weight W, the angle θ in the diagram above becomes the angle of friction if
Akọwa Nkọwa
Ajụjụ 30 Ripọtì
Two bodies X and Y are projected on the same horizontal plane, with the same initial speed but at angles 30 and 60 respectively to the horizontal. Neglecting air resistance, the ratio of the range of X to that of Y is
Akọwa Nkọwa
Range = U2sin2θg
RxRy = U2xsin60U2ysin120
sin60o = sin120o
RxRy = 11
hence, = 1:1
Ajụjụ 31 Ripọtì
What is the total electrical energy consumed by using an electric cooker rated 1000W for 5 hrs?
Akọwa Nkọwa
To find the total electrical energy consumed by the electric cooker, we need to use the formula: Electrical Energy = Power x Time where Power is given in watts and Time is given in hours. In this case, the power rating of the electric cooker is 1000W and the time for which it is used is 5 hours. Therefore, Electrical Energy = 1000W x 5 hours = 5000 watt-hours However, we usually express electrical energy in joules. To convert watt-hours to joules, we need to multiply by 3600 (the number of seconds in an hour) and by the electrical energy conversion factor (1 watt = 1 joule per second). Therefore, Electrical Energy = 5000 watt-hours x 3600 seconds/hour x 1 joule/second = 18,000,000 joules = 1.8 x 10^7 joules Therefore, the answer is: 1.8 x 10^7 J
Ajụjụ 33 Ripọtì
The electromagnetic waves that are sensitive to temperature changes are
Akọwa Nkọwa
The electromagnetic waves that are sensitive to temperature changes are the infra-red rays. Infra-red radiation is a type of electromagnetic radiation that is invisible to the human eye but can be felt as heat. All objects with a temperature above absolute zero emit infrared radiation, and the amount of radiation increases as the temperature of the object increases. This property of infra-red radiation makes it useful for detecting temperature changes in a wide range of applications, including thermal imaging cameras, temperature sensors, and heat-seeking missiles.
Ajụjụ 34 Ripọtì
The diagram above shows two wave forms P and Q at a particular instant in time. the two waves will interface
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
If the load at the end of a sonometer wire is immersed in a bucket of water, the original fundamental frequency of the wire could be restored by
Ajụjụ 36 Ripọtì
For what value of 6 are the forces in the diagram above in equilib4ruim
Ajụjụ 37 Ripọtì
The colours seen in the thin films of oil on the road and in soap bubbles are due to
Ajụjụ 38 Ripọtì
A piece of wire gauze can be made to lie on water because
Akọwa Nkọwa
The reason a piece of wire gauze can be made to lie on water is that the water surface has the effect of an elastic thin skin. This effect is due to the cohesive forces between the water molecules that cause them to stick together and form a surface tension. The wire gauze is able to rest on the water surface because it is not heavy enough to break the surface tension, allowing it to be supported by the surface of the water. Therefore, the correct option is "the water surface has the effect of an elastic thin skin".
Ajụjụ 39 Ripọtì
At what respective values of X, Y and Z would the unit of force, the newton, be dimensionally equivalent to MxLyTz?
Akọwa Nkọwa
F = m (v-u)
= kg x m x S-2
= M1L1T-2
Ajụjụ 40 Ripọtì
Two 50 micro Farad parallel plate capacitors are connected in series. The combined capacitor is then connected across a 100-V battery. The charge on each plate of the
Akọwa Nkọwa
When capacitors are connected in series, the total capacitance decreases. The formula for calculating the total capacitance of capacitors connected in series is: 1/C_total = 1/C1 + 1/C2 + 1/C3 + ... where C1, C2, C3, ... are the capacitances of each capacitor. In this problem, two capacitors with capacitance 50 microFarads are connected in series. Using the formula above, we can calculate the total capacitance: 1/C_total = 1/C1 + 1/C2 1/C_total = 1/50 microF + 1/50 microF 1/C_total = 2/50 microF C_total = 25 microF Now, we can use the formula for capacitance to calculate the charge on each plate of the combined capacitor when it is connected to a 100-V battery: Q = C * V Q = 25 microF * 100 V Q = 2.5 x 10^-3 C Therefore, the charge on each plate of the combined capacitor is 2.5 x 10^-3 C. Option (B) is the closest answer, which is 2.50 x 10^-3 C.
Ajụjụ 41 Ripọtì
In a purely inductive circuit, the current
Akọwa Nkọwa
In a purely inductive circuit, the current lags behind the voltage in phase by 90 degrees. This means that the maximum value of the current occurs after the maximum value of the voltage. In simpler terms, when the voltage is at its highest point, the current is still increasing and has not yet reached its highest point. This is because in an inductive circuit, the current is impeded by the inductor, which resists changes in current. As a result, the current takes some time to catch up to the voltage.
Ajụjụ 42 Ripọtì
A mass of a liquid at 30oC is mixed with a mass of the same liquid at 70oC and the temperature of the mixture is 45oC. Find the ratio of the mass of the cold liquid to the mass of the other liquid.
Akọwa Nkọwa
We can solve this problem using the principle of mixtures. Let's assume that the masses of the cold and hot liquids are m1 and m2, respectively, and their specific heat capacities are the same. The temperature change of the cold liquid is (45-30)=15°C, and that of the hot liquid is (70-45)=25°C. Now, using the principle of mixtures, the total heat lost by the hot liquid = the total heat gained by the cold liquid, that is: m1*15 = m2*25 Simplifying the above equation, we get: m1/m2 = 25/15 = 5/3 Therefore, the ratio of the mass of the cold liquid to the mass of the other liquid is 5 : 3. So, is the correct answer.
Ajụjụ 43 Ripọtì
In a meter bridge experiment, two resistor 2Ω
and 3Ω
occupy the left and right gaps respectively. Find the balance point from the left side of the bridge
Akọwa Nkọwa
Let the point be x
2x = 3100?x
making x subject of formula,
2(100 - x) = 3x
200 - 2x = 3x
200 = 3x + 2x
200 = 5x
divide both side by 5
x = 40cm
Ajụjụ 44 Ripọtì
A 5kg block is released from rest on a smooth plane inclined at an angle of 30o to the horizontal. What is the acceleration down the plane? [g = 10ms-2]
Akọwa Nkọwa
The acceleration down the plane can be found using the formula: a = g sin(theta) where g is the acceleration due to gravity and theta is the angle of inclination of the plane. Given: mass of block, m = 5 kg angle of inclination, theta = 30 degrees acceleration due to gravity, g = 10 m/s^2 We can first find the component of gravity acting down the plane: force due to gravity along the plane = m g sin(theta) where sin(theta) = sin(30 degrees) = 0.5 force due to gravity along the plane = 5 x 10 x 0.5 = 25 N Since there are no other forces acting on the block, the force along the plane is equal to the force causing the acceleration of the block down the plane. Using Newton's second law of motion, we can find the acceleration down the plane: force along the plane = m a where a is the acceleration down the plane. Substituting the values, we get: 25 N = 5 kg x a a = 25 / 5 = 5 m/s^2 Therefore, the acceleration down the plane is 5 m/s^2. Looking at the options given, we can see that the correct answer is the first option, 5.0 ms^-2.
Ajụjụ 45 Ripọtì
A liquid of mass 1.0 x 103kg fills a rectangular tank of length 2.5m and width 2.0m. If the tank is 4m high, what is the pressure at the middle of the tank?[g = 10ms-2]
Akọwa Nkọwa
The pressure at the middle of the tank is given by the formula: pressure = (density of liquid) x (acceleration due to gravity) x (height of liquid column) Since the liquid fills the entire tank, the height of the liquid column is half the height of the tank, which is 2m. The density of the liquid is given as 1000kg/m³, and acceleration due to gravity is 10m/s². Therefore, the pressure at the middle of the tank is: pressure = 1000 kg/m³ x 10 m/s² x 2 m = 20,000 N/m² So the answer is: - 1.0 x 103Nm-2
Ajụjụ 46 Ripọtì
An empty 60 litre petrol tank has a mass of 10kg. Its mass when full of fuel of relative density 0.72 is
Akọwa Nkọwa
The mass of the tank when empty is 10 kg. The volume of the tank is given as 60 litres. When it is full of fuel of relative density 0.72, the mass of fuel it contains is equal to the volume of the fuel multiplied by its density, which is given by: Mass of fuel = Volume of fuel x Density of fuel Mass of fuel = 60 x 0.72 = 43.2 kg Therefore, the total mass of the tank when full of fuel is equal to the sum of the mass of the tank when empty and the mass of the fuel it contains, which is: Total mass = Mass of tank when empty + Mass of fuel Total mass = 10 kg + 43.2 kg = 53.2 kg Hence, the correct answer is 53.2kg.
