JAMB UTME - Physics - 1994 | Green Bridge CBT

Question 1 Report

A spring of length 25cm is extended to 30cm by a load of 150N attached to one of its end, what is the energy stored in the spring?

3750J

2500J

3.75J

2.50J

Elasticity

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Question 2 Report

The mass of a specific gravity bottle is 15.2g when it is empty. it is 24.8g when filled with kerosene and 27.2g when filled with distilled water. Calculate the relative density of kerosene

1.25

1.10

0.90

0.80

Answer Details

To calculate the relative density of kerosene, we need to first find the mass of kerosene that fills the specific gravity bottle. We can do this by subtracting the mass of the empty bottle from the mass of the bottle filled with kerosene. Mass of kerosene = Mass of filled bottle - Mass of empty bottle Mass of kerosene = 24.8g - 15.2g Mass of kerosene = 9.6g Next, we need to find the mass of an equal volume of water that fills the specific gravity bottle. We can do this by subtracting the mass of the empty bottle from the mass of the bottle filled with water. Mass of water = Mass of filled bottle with water - Mass of empty bottle Mass of water = 27.2g - 15.2g Mass of water = 12g The relative density of kerosene is then calculated by dividing the density of kerosene by the density of water. Relative density of kerosene = (mass of kerosene / volume of kerosene) / (mass of water / volume of water) We don't have the volumes of the kerosene and water, but since they were both measured using the same specific gravity bottle, their volumes are equal. Therefore, we can simplify the equation to: Relative density of kerosene = (mass of kerosene) / (mass of water) Relative density of kerosene = 9.6g / 12g Relative density of kerosene = 0.8 Therefore, the relative density of kerosene is 0.8, which corresponds to the option: 0.80.

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Question 3 Report

Which of the following are found in the receiver but not in the microphone of a telephone handset? i. Diaphragm. ii. soft iron pole pieces. iii. permanent magnet. iv.carbon blocks

i and ii

i and iv

ii and iii

iii and iv

Answer Details

The receiver and microphone are two essential parts of a telephone handset. The microphone converts sound waves into electrical signals, while the receiver converts electrical signals into sound waves. In the microphone, sound waves cause a diaphragm to vibrate, which then changes the pressure on carbon blocks, varying the electrical resistance of the blocks. This variation in electrical resistance produces electrical signals that correspond to the sound waves. In the receiver, electrical signals produced by the microphone are transmitted to a coil of wire that surrounds a magnet. The electrical signals cause the coil to create a magnetic field that interacts with the permanent magnet. This interaction causes a diaphragm to vibrate, which produces sound waves that correspond to the original sound. From this explanation, we can see that the diaphragm is found in both the receiver and the microphone. The permanent magnet is found in the receiver, but not in the microphone. The soft iron pole pieces are also found in the receiver, but not in the microphone. The only component that is found in the microphone but not in the receiver is the carbon blocks. Therefore, the answer is option (ii) i and iv.

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Question 4 Report

If two parallel conductors carry currents flowing in the same direction, the conducts will

attract each other

repel each other

both move in the same direction

have no effect on each other

Answer Details

Two parallel conductors carrying currents flowing in the same direction will attract each other. This is due to the magnetic fields produced by the currents, which interact with each other to create a force. The direction of the force is determined by the direction of the current and the orientation of the conductors. When the currents are flowing in the same direction, the magnetic fields around the conductors will interact in a way that produces an attractive force between them. This phenomenon is known as the Ampere's law, which states that two parallel conductors carrying currents in the same direction will experience a magnetic attraction force between them.

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Electrostatics

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Question 5 Report

A force of 15N stretches a spring to a total length of 30cm. An additional force of 10N stretches the spring 5cm further. Find the natural length of the spring

25.0cm

22.5cm

20.0cm

15.0cm

Answer Details

We can use Hooke's Law, which states that the force exerted by a spring is proportional to its extension, to solve this problem. Let the natural length of the spring be x. When a force of 15N is applied, the spring is stretched by 30cm - x. Therefore, we have: 15N = k(30cm - x) where k is the spring constant. When an additional force of 10N is applied, the spring is stretched by an additional 5cm. Therefore, we have: 25N = k(35cm - x) Dividing the second equation by the first, we get: 25N/15N = (35cm - x)/(30cm - x) Simplifying this equation, we get: 5/3 = (35cm - x)/(30cm - x) Multiplying both sides by 30cm - x, we get: 5(30cm - x) = 3(35cm - x) 150cm - 5x = 105cm - 3x 2x = 45cm x = 22.5cm Therefore, the natural length of the spring is 22.5cm. Answer: 22.5cm.

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Elasticity

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Question 6 Report

A pipe of length 45cm is closed at the end. Calculate the fundamental frequency of the sound wave generated in the pipe if the velocity of sound in air is 360ms-1 [Neglect end correction]

5.5Hz

148.5Hz

200.0Hz

550.0Hz

Answer Details

The formula for calculating the fundamental frequency of a closed-end air column is given by: f = v/4L Where f is the frequency, v is the velocity of sound in air, and L is the length of the air column. Substituting the given values: f = 360/4 x 0.45 f = 200.0 Hz Therefore, the fundamental frequency of the sound wave generated in the pipe is 200.0 Hz.

Read lesson note on Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

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Question 7 Report

A capacitor $8\mu\text{F}$ is charged to a potential difference of 100V. The energy stored by the capacitor is

1.0 x 104J

8.0 x 10J

1.25 x 10J

4.0 x 10-2J

Answer Details

Energy =

\frac{1}{2}

CV2

= $\frac{1}{2}$ x 8 x 10-6 x (100)2

= 4 x 10-2J

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Question 8 Report

When 100g of liquids L1 at 78oC was mixed with Xg of liquid L2 at 50oC, the final temperature was 60oC. Given that the specific heat capacity of L2is half that of L1, find x

50g

100g

150g

200g

Answer Details

Heat lost by L1 = heat gained by L2

cL1 = 2cL2

100 x 2cL2 x (78 - 60)

= xg x cL2 x (60 - 50)

xg = 150g

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Temperature And Its Measurement

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Question 9 Report

An object is projected with a velocity of 80ms-1 at an angle of 30o to the horizontal.The maximum height reached is

20m

80m

160m

320m

Answer Details

The problem involves finding the maximum height reached by an object that is projected at a given velocity and angle to the horizontal. To solve this, we can use the equations of motion for projectile motion. Firstly, we need to break down the initial velocity into its horizontal and vertical components. The horizontal component remains constant throughout the motion and is given by: vx = v * cos(theta) where v is the initial velocity and theta is the angle of projection. The vertical component changes due to the force of gravity and is given by: vy = v * sin(theta) - gt where g is the acceleration due to gravity and t is the time elapsed. At the maximum height, the vertical component of the velocity becomes zero. We can use this fact to determine the time of flight (the time it takes for the object to reach its maximum height and then return to the same height) as follows: 0 = v * sin(theta) - gt_max where t_max is the time taken to reach the maximum height. Rearranging this equation gives: t_max = v * sin(theta) / g Using this value of t_max, we can determine the maximum height reached by the object using the equation: h_max = v * sin(theta) * t_max - 0.5 * g * t_max^2 Substituting the given values in these equations, we get: vx = 80 * cos(30) = 69.28 m/s vy = 80 * sin(30) - 9.81 * t 0 = 80 * sin(30) - 9.81 * t_max t_max = (80 * sin(30)) / 9.81 = 4.08 s h_max = 80 * sin(30) * t_max - 0.5 * 9.81 * t_max^2 = 80 * sin(30) * 4.08 - 0.5 * 9.81 * 4.08^2 = 80 m Therefore, the maximum height reached by the object is 80 meters.

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Motion

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Question 10 Report

A cone in an unstable equilibrium has its potential energy

decreased

increased

unchanged

oscillating

Equilibrium Of Forces

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Question 11 Report

A motor vehicle is brought to rest from a speed of 15ms-1 in 20 seconds. Calculate the retardation.

0.75ms-1

1.33ms-2

5.00ms-2

7.50ms-2

Answer Details

Retardation is the rate at which an object slows down. It is also known as deceleration or negative acceleration. To calculate the retardation of the motor vehicle, we can use the following formula: Retardation = (Final Velocity - Initial Velocity) / Time Taken Here, the final velocity is 0 m/s (since the vehicle is brought to rest), the initial velocity is 15 m/s (given in the question), and the time taken is 20 seconds (also given in the question). So, plugging in the values in the formula, we get: Retardation = (0 - 15) / 20 Retardation = -0.75 m/s^2 Therefore, the retardation of the motor vehicle is 0.75 m/s^2 (option A).

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Equilibrium Of Forces

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Question 12 Report

The figure above shows a uniform wood of weight 200N and length 50m. It is pivoted at one end and suspended by a cord at the other end at an angle of 30 degrees to the wood. Calculate the tension in the cord if the wood is horizontal.

10N

20N

100N

200N

Equilibrium Of Forces

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Question 13 Report

An object of mass 50kg is released from a height of 2m.Find the kinetic energy just before it strikes the ground

250J

1000J

10 000J

100 000J

Answer Details

When an object is lifted to a height h, it gains potential energy given by: P.E = mgh, where m is the mass of the object, g is the acceleration due to gravity and h is the height. When the object is released, it starts to fall and its potential energy is converted into kinetic energy, which is given by: K.E = (1/2)mv^2, where v is the velocity of the object just before it strikes the ground. At the highest point, the potential energy is maximum and the kinetic energy is minimum. At the instant just before the object strikes the ground, the potential energy is zero and the kinetic energy is maximum. Therefore, the potential energy at the highest point is: P.E = mgh = 50 x 9.81 x 2 = 981 J Since the potential energy is converted to kinetic energy just before the object strikes the ground, the kinetic energy at that point is equal to the potential energy at the highest point: K.E = 981 J. Therefore, the correct option is: 1000J.

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Work, Energy And Power

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Question 14 Report

Which of these statements are TRUE of pressure in liquids? i. pressure acts equally in all direction. ii. pressure decreases with depth. iii. pressure at the same level of a liquid is the same. iv pressure is dependent on the cross-sectional area of the barometer tube

i, iii only

i, ii, iii only

i, iii, and iv only

i, ii, iii and iv

Pressure

Liquids At Rest

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Question 15 Report

In the circuit diagram above, calculate the current in the 12 Ω resistor if the cell has an emf of 12V and an internal resistance of 1Ω

0.8A

1.0A

1.6A

2.4A

Current Electricity

Simple A.C Circuits

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Question 16 Report

If a room is saturated with water vapour, the temperature of the room must be

at 0oC

above the dew point

at 100oC

below or at the dew point

Answer Details

When air holds the maximum amount of water vapor that it can at a given temperature and pressure, it is said to be saturated. The maximum amount of water vapor that air can hold increases with temperature. Thus, if a room is saturated with water vapor, the temperature of the room must be at or below the dew point temperature, which is the temperature at which water vapor in the air begins to condense into liquid. This means that the answer is "below or at the dew point."

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Temperature And Its Measurement

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Question 17 Report

In the figure above if the atmospheric pressure is 760mm, the pressure of the chamber G is

660mm

690mm

830mm

860mm

Pressure

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Question 18 Report

In the curve above, PQRST represents an alternating voltage of frequency 50Hz. The time interval between points P and R on the graph is

25s

1/50s

1/100s

1/200s

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Question 19 Report

If the Nigeria flag (green, white, green) is viewed in pure yellow light, which of the following set of colours would be observed on the flag?

Green, yellow, green

Red, yellow, red

black, yellow, black

Green, white, green

Structure Of The Nucleus

Structure Of The Atom (Nigeria Only)

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Question 20 Report

Calculate the velocity ratio of a screw jack of pitch 0.3cm if the length of the tommy bar is 21cm

\frac{1}{140}

π

14

π

70

π

140

π

Answer Details

V.R =

\frac{2 π L}{p}

= $\frac{2 \times π \times 2}{0.3}$

V.R = 140 $π$

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Simple Machines

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Question 21 Report

When a biro pen rubbed on a dry silk cloth is moved very close to a piece of paper on a dry table, the pen is found to pick up the paper. This is because

both the pen and the cloth are magnetized

the pen is magnetized but the cloth is not

the pen is charged while the cloth is magnetized

both the pen and the cloth are charged

Answer Details

They are all charged either positively or negatively

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Electrostatics

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Question 22 Report

In the figure above, XY is of length 1m, The value of R at balance point Z is

Options

3.0Ω

13.3Ω

15.0Ω

30.0Ω

Scalars And Vectors

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Question 23 Report

Two similar kettles containing equal masses of boiling water are placed on a table. If the surface of one is highly polished and the surface of the other is covered with soot, Which of the following observations is correct

The two kettles will cool down at the same rate

the polished kettle cools down more quickly by conduction

the kettle covered with soot cools down more quickly because it is a good radiator

the kettle covered with soot cools down more quickly by the process of heat convection

Answer Details

The correct observation is: the kettle covered with soot cools down more quickly because it is a good radiator. The process by which a body loses heat to its surroundings is known as radiation, convection or conduction. The surface of the kettle covered with soot is rough and black, and it has a high emissivity that allows it to radiate more heat to its surroundings compared to the highly polished surface of the other kettle. This causes the kettle covered with soot to cool down faster than the polished one.

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Question 24 Report

A certain radioisotope of $ {}^{235}_{92}\mathrm{U} $ emits four alpha particles and three beta particles. The mass number and the atomic number of the resulting elements respectively are

219 and 87

84 and 223

223 and 87

219 and 81

Answer Details

_{92}^{235} U

\to

4

(_{2}^{4} α)

+ 3

(_{- 1}^{0} e)

+

_{a}^{b} y

a + (4 + 2) + (3x - 1) = 92 $\Rightarrow$ a = 87

b + (4 x 4) + (3 x 0) = 235 $\Rightarrow$ b = 219

$_{a}^{b} Y$ = $_{87}^{219} Y$

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Structure Of The Atom (Nigeria Only)

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Question 25 Report

In a projection lantern of focal length f, the object distance u, is such that

u > 2f > f

u < f < 2f

u = f < 2f

f < u < 2f

Reflection Of Light At Plane And Curved Surfaces

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Question 26 Report

What is the reading of the vernier calipers shown above?

1.79cm

1.73cm

1.39cm

1.34cm

Answer Details

The main scale reading on the vernier caliper is 1.3 cm, which is the closest value to the left of the zero mark on the vernier scale. The first mark on the vernier scale that aligns with a mark on the main scale is the 9th mark, which corresponds to 0.09 cm. The total reading is the sum of the main scale and the vernier scale readings, so: Total reading = 1.3 cm + 0.09 cm = 1.39 cm Therefore, the reading of the vernier calipers shown above is 1.39 cm.

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Measurements And Units

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Question 27 Report

If it takes 5.0 hrs to drain a container of 540.0cm3 of water, what is the flow rate of water from the container in kg-1? [Density of water = 1000kgm-3]

32.5kg-1 s

31.5kg-1 s

30.8kg-1 s

30.0kg-1 s

Answer Details

The flow rate of water is the volume of water drained per unit time. To calculate it, we first need to find the volume of water drained, which is given as 540.0 cm^3. To convert this to liters, we divide by 1000: 540.0 cm^3 = 0.54 L We are given the time taken to drain this volume of water as 5.0 hours. To convert this to seconds, we multiply by 3600: 5.0 hours = 18000 s The flow rate of water is therefore: Flow rate = Volume / Time = 0.54 L / 18000 s Now, we are given the density of water as 1000 kg/m^3. To convert the volume of water to mass, we multiply by the density and convert the units from liters to kilograms: 0.54 L × 1000 kg/m^3 × 10^-3 L/kg = 0.54 kg Putting it all together, the flow rate of water is: Flow rate = Volume / Time = 0.54 kg / 18000 s = 0.03 × 10^-3 kg/s = 30.0 kg^-1 s^-1 Therefore, the answer is option D, 30.0 kg^-1 s^-1.

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Measurements And Units

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Question 28 Report

The sound from a source travelled to the bottom of the sea and the echo was heard 4s later. If the speed of sound in sea water is 1500ms-1 the depth of the sea is

6000m

3000m

1500m

375m

Answer Details

t = 4s ; v = 150ms-1

depth = X ; but 2X = vt

X = $\frac{v t}{2}$ = $\frac{1500 \times 4}{2}$

X = 3000m

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Current Electricity

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Question 29 Report

If light with photon energy 2eV is incident suitably on the surface of a metal with work function 3eV, then

no electron will be emitted

the few electrons emitted will have a maximum kinetic energy of 1eV

the few electrons emitted will have a maximum kinetic energy of 3eV

many electrons will be emitted

Answer Details

An electron can only be emitted if photon energy is equal or greater than work function of the plate

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Electromagnetic Field (Part 2)

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Question 30 Report

Which of the following is TRUE of a particle moving in a horizontal circle with constant angular velocity?

The energy is costant but the linear momentum varies

The linear momentum is conatant but the energy varies

both energy and linear momentum are constant

The speed and the linear velocity are both constant

Newton’s Laws Of Motion

Simple A.C Circuits

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Question 31 Report

Light of frequency 6.0 x 1014 Hz travelling in air is transmitted through glass or refractive index 1.5. Calculate the frequency of the light in the glass

4.0 x 1014Hz

6.0 x 1014Hz

7.5 x 1014Hz

9.0 x 1014Hz

Answer Details

1.5 =

\frac{6 \times 10^{14}}{F}

F = 4 x 1014Hz

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Optical Instruments

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Question 32 Report

Calculate the magnitude of the force required to just move a 20kg object along a horizontal surface if the coefficient of friction is 0.2. [g = 10ms-2]

400.0N

40.0N

4.0N

0.4N

Answer Details

The force required to move an object along a surface is equal to the product of the coefficient of friction and the normal force acting on the object. The normal force is equal to the weight of the object, which is given by the product of the mass and the acceleration due to gravity. Therefore, the force required to move the 20kg object along a horizontal surface is: Force = coefficient of friction * normal force Normal force = mass * acceleration due to gravity = 20kg * 10ms^-2 = 200N Force = 0.2 * 200N = 40N Therefore, the magnitude of the force required to just move the 20kg object along a horizontal surface is 40N. The correct option is: 40.0N

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Friction

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Question 33 Report

If a solid X floats in liquids P of relative density 2.0 and in liquid Q of relative density 1.5, it can be inferred that the

weight of P displaced is greater than that of Q

weight of P displaced is less than that of Q

volume of P displaced is greater than that of Q

volume of P displaced is less than that of Q

Answer Details

R.d

α

weight of liquid displaced

$\frac{R.d of P}{R.d of Q}$ = $\frac{weight of p displaced}{weight of Q displaced}$

$\frac{2}{1.5}$ = $\frac{weight of p displaced}{weight of Q displaced}$

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Measurements And Units

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Question 34 Report

Which of the following has the lowest internal resistance when new?

leclanche cell

Daniel cell

Accumulator

Torch battery

Electric Cells

Characteristics Of Sound Waves

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Question 35 Report

The note produced by a stretched string has a fundamental frequency of 400Hz. If the length of the string is doubled while the tension in the string is increased by a factor of 4, the frequency is

200Hz

400Hz

800Hz

1600Hz

Answer Details

F =

\frac{1}{2 L}

\sqrt{\frac{T}{m}}

T2 = 4T1 ; L2 = 2L1

$\frac{2 F_{1} L_{1}}{2 F_{2} L_{2}}$ = $\sqrt{\frac{T_{1}}{T_{2}}}$

$\frac{400 \times L_{1}}{F \times L_{1}}$ = $\sqrt{\frac{T_{1}}{4 T_{1}}}$

F = 400Hz

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Elasticity

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Question 36 Report

A Motor Tyre is inflated to pressure of 2.0 x 105Nm-2 when the temperature of air is 27oC. What will be the pressure in it at 87oC assuming that the volume of the Tyre does not change?

2.6 x 105Nm-2

2.4 x 105Nm-2

2.2 x 105Nm-2

1.3 x 105Nm-2

Answer Details

The pressure inside a motor tire is directly proportional to the temperature of the air inside it. According to the Gay-Lussac's Law of pressure-temperature, when the temperature of gas increases, its pressure also increases if the volume is constant. Here, the volume of the tire is assumed to be constant. Using the formula P1/T1 = P2/T2, where P1 is the initial pressure, T1 is the initial temperature, P2 is the final pressure, and T2 is the final temperature, we can find the final pressure. Substituting the given values, we get: P1 = 2.0 x 10^5 N/m^2 T1 = 27°C + 273 = 300K T2 = 87°C + 273 = 360K So, P2 = (P1 x T2) / T1 = (2.0 x 10^5 N/m^2 x 360K) / 300K = 2.4 x 10^5 N/m^2. Therefore, the pressure inside the motor tire at 87°C is 2.4 x 10^5 N/m^2. So, the answer is 2.4 x 10^5 N/m^2.

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Pressure

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Question 37 Report

A boat travels due east at a speed 40ms-1s across a river flow due south at 30ms-1s. What is the resultant speed of the boat?

1.3ms-1

10.0ms-1

50.0ms-1

70.0ms-1

Answer Details

To find the resultant speed of the boat, we need to use the Pythagorean theorem which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In this case, the boat's velocity and the river's velocity are perpendicular to each other, so we can form a right-angled triangle. The speed of the boat is given as 40 ms^-1 due east, and the speed of the river's flow is 30 ms^-1 due south. Using Pythagoras theorem, we have: Resultant speed = √(40^2 + 30^2) = √(1600 + 900) = √2500 Resultant speed = 50 ms^-1 Therefore, the correct option is 50.0ms^-1.

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Motion

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Question 38 Report

Two capacitors $R_1 = 4\,\Omega$ and $R_2 = 5\,\Omega$ are connected in parallel across a potential difference. If $P_1$ and $P_2$ represent power dissipated in $R_1$ and $R_2$ respectively, then the ratio $P_1 : P_2$ is

4 : 5

5 : 4

16 : 25

25 : 16

Answer Details

P1 =

\frac{V_{1}^{2}}{R_{1}}

and P2 =

\frac{V_{2}^{2}}{R_{2}}

P1R1 = P2R2

4P1 = 5P2

$\frac{p_{1}}{p_{2}}$ = $\frac{5}{4}$

= 5 : 4

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Capacitors

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Question 39 Report

To produce a parallel beam of light from a concave mirror, the distance at a which the lamp should be placed from the mirror is equal to

the focal length

two times the focal length

the distance of the image

two times the radius of curvature

Answer Details

To produce a parallel beam of light from a concave mirror, the distance at which the lamp should be placed from the mirror is equal to the focal length. This is because a concave mirror has a focal point, which is the point where all incoming parallel rays of light converge after reflection. By placing the lamp at a distance equal to the focal length, the rays of light will reflect off the mirror and converge at the focal point, resulting in a parallel beam of light. Placing the lamp at any other distance would result in the rays of light either converging or diverging after reflection, which would not produce a parallel beam. Therefore, option A, the focal length, is the correct answer.

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Reflection Of Light At Plane And Curved Surfaces

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Question 40 Report

The melting point of naphthalene is 78oC. What is this temperature in kelvin

100k

351k

378k

444k

Answer Details

To convert a temperature from Celsius to Kelvin, we add 273.15 to the Celsius temperature. In this question, the melting point of naphthalene is given in Celsius which is 78°C. To convert it to Kelvin, we add 273.15 to 78°C: 78°C + 273.15 = 351.15 K Therefore, the temperature of the melting point of naphthalene in Kelvin is 351 K. The correct option is: 351k.

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Temperature And Its Measurement

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Question 41 Report

An object is placed directly below a glass block of thickness 3.0cm. Calculate the lateral displacement of the object if the refractive index of the glass is 1.5.

1.0cm

1.5cm

2.0cm

2.5cm

Answer Details

1.5 =

\frac{3}{apparent displacement}

Apparent depth = 2cm

displacement = 3 - 2

= 1cm

Read lesson note on Refraction Of Light Through Plane And Curved Surfaces (JAMB)

Refraction Of Light Through Plane And Curved Surfaces

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Question 42 Report

Heat is supplied to a test tube containing 100g of ice at its melting point. The ice melts completely in 1 min.What is the power rating of the source of heat? [Latent heat of fusion of ice = 336Jg-1]

336W

450W

560W

600W

Answer Details

When heat is supplied to the ice, the ice will absorb heat and its temperature will rise until it reaches its melting point. At this point, the ice will start melting, and the temperature will remain constant until all the ice has melted. The energy required to melt 1 gram of ice at its melting point is called the latent heat of fusion of ice and is given as 336J/g. In this question, we are given that 100g of ice melts completely in 1 min. Therefore, the total energy required to melt the ice can be calculated as follows: Total energy = mass x latent heat of fusion of ice Total energy = 100g x 336J/g Total energy = 33600J Since this energy is supplied in 1 minute, we can calculate the power rating of the source of heat as follows: Power = Total energy supplied / time taken Power = 33600J / 60s Power = 560W Therefore, the power rating of the source of heat is 560W. Option (C) is the correct answer.

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Heat Energy

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Question 43 Report

The equation of transverse wave traveling along a string is given by Y = 0.3 sin(0.5x - 50t) where Y and X are in cm and t is in seconds.Find the maximum displacement of the particles from the equilibrium position

50.0cm

2.5 cm

0.5cm

0.3cm

Answer Details

The equation of the transverse wave traveling along a string is given by Y = 0.3 sin(0.5x - 50t), where Y and X are in cm and t is in seconds. The general form of a sine wave equation is Y = A sin(kx - ωt + φ), where A is the amplitude, k is the wave number, ω is the angular frequency, t is time, x is distance and φ is the phase constant. Comparing the given equation with the general equation, we can see that the amplitude is 0.3. The maximum displacement of particles from the equilibrium position is equal to the amplitude. Hence, the maximum displacement of particles from the equilibrium position is 0.3 cm. Therefore, the correct option is 0.3cm.

Read lesson note on Conduction Of Electricity Through (JAMB)

Conduction Of Electricity Through

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Question 44 Report

The fundamental frequency of vibration of a sonometer wire may be halved by

doubling the length of the wire

doubling the mass of the wire

reducing the tension by half

reducing the absolute temperature by half

Answer Details

The fundamental frequency of a sonometer wire is related to its length, tension, and mass. To halve the fundamental frequency, we need to identify which factor(s) we can change to reduce the frequency by a factor of 2. - Doubling the length of the wire will reduce the frequency by a factor of 2, as the frequency is inversely proportional to the length of the wire. - Doubling the mass of the wire will not affect the frequency, as the frequency is independent of the mass of the wire. - Reducing the tension by half will reduce the frequency by a factor of sqrt(2), not by a factor of 2. - Reducing the absolute temperature by half will not affect the frequency of the wire. Therefore, the correct answer is: "doubling the length of the wire".

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Motion

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Question 45 Report

An object is placed in front of a converging lens of a focal length 20cm. The image is virtual and has a magnification of 2. What is the distance of the object from the lens?

5cm

10cm

30cm

40cm

Answer Details

V = -2u

F = $\frac{u v}{u + v}$

20 = $\frac{u v - 2 u}{u - 2 u}$

u = 10cm

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Electrostatics

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