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| By Parveen Sharma
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Parveen Sharma
Community Contributor
Quizzes Created: 2 | Total Attempts: 308
Questions: 30 | Attempts: 127

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• 1.

To prevent a DC return between source and load, it is necessary to use

• A.

Resistor between source and load

• B.

Inductor between source and load

• C.

Capacitor between source and load

• D.

Either (a) or (b)

C. Capacitor between source and load
Explanation
To prevent a DC return between source and load, a capacitor is necessary to be used between them. A capacitor blocks the flow of direct current (DC) while allowing alternating current (AC) to pass through. By placing a capacitor between the source and load, it acts as an open circuit for DC signals, effectively preventing the return of DC current. This allows the AC signal to pass through to the load while blocking any DC component, ensuring the proper functioning of the circuit.

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• 2.

For a base current of 10 μA, what is the value of collector current in common emitter if βdc = 100

• A.

10 μA

• B.

100 μA

• C.

1 mA

• D.

10 mA

A. 10 μA
Explanation
The value of the collector current in a common emitter configuration is equal to the product of the base current and the DC current gain (Î²dc). In this case, the base current is given as 10 Î¼A and the Î²dc is given as 100. Therefore, the collector current can be calculated as 10 Î¼A x 100 = 1 mA.

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• 3.

To protect the diodes in a rectifier and capacitor input filter circuit it is necessary to use

• A.

Surge resistor

• B.

Surge inductor

• C.

Surge capacitor

• D.

Both (a) and (b)

A. Surge resistor
Explanation
To protect the diodes in a rectifier and capacitor input filter circuit, it is necessary to use a surge resistor. A surge resistor helps to limit the surge current that can flow through the diodes during power-up or power-down, preventing any damage to the diodes. It acts as a current-limiting device, absorbing the excess energy and dissipating it as heat. By using a surge resistor, the diodes are protected from the high current surges that can occur in the circuit, ensuring their longevity and reliability. A surge resistor is therefore essential in safeguarding the diodes in this type of circuit.

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• 4.

If the input to the ideal comparator shown in the figure is a sinusoidal signal of 8 V (peak to peak) without any DC component, then the output of the comparator has a duty cycle of

• A.

1/2

• B.

1/3

• C.

1/6

• D.

1/12

B. 1/3
Explanation
The ideal comparator compares the input signal to a reference voltage and produces a digital output based on whether the input signal is above or below the reference voltage. In this case, since the input signal is a sinusoidal signal without any DC component, the output of the comparator will be a square wave with a duty cycle of 1/3. This means that the signal will be high for one-third of the time and low for two-thirds of the time.

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• 5.

N a CE amplifier the input impedance is equal to the ratio of

• A.

Ac base voltage to ac base current

• B.

Ac base voltage to ac emitter current

• C.

Ac emitter voltage to ac collector current

• D.

Ac collector voltage to ac collector current

A. Ac base voltage to ac base current
Explanation
In a CE (common emitter) amplifier, the input impedance is equal to the ratio of the AC base voltage to the AC base current. This means that the input impedance is determined by the voltage applied to the base terminal divided by the current flowing into the base terminal. The base current is directly related to the input voltage, so by measuring the ratio of these two quantities, we can determine the input impedance of the amplifier.

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• 6.

For a system to work, as oscillator the total phase shift of the loop gain must be equal to

• A.

90°

• B.

45°

• C.

270°

• D.

360°

D. 360°
Explanation
For a system to work as an oscillator, the total phase shift of the loop gain must be 360Â°. This means that the output of the system must be in phase with the input after completing one full cycle. A phase shift of 360Â° indicates that the system is able to sustain oscillations and maintain a constant frequency.

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• 7.

The most widely used LC oscillator is

• A.

Hartley oscillator

• B.

Crystal oscillator

• C.

Colpitt's oscillator

• D.

Clapp's oscillator

A. Hartley oscillator
Explanation
The Hartley oscillator is the most widely used LC oscillator. It is a type of oscillator that uses an LC circuit to generate a stable frequency. The Hartley oscillator is known for its simplicity and stability, making it a popular choice in various electronic applications. It consists of an inductor and a capacitor connected in parallel, with a feedback loop that allows the oscillator to sustain oscillation. The Hartley oscillator is commonly used in radio transmitters, receivers, and other electronic devices that require a stable frequency signal.

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• 8.

"Half-splitting" is

• A.

A means for nuclear fuel cells to produce electricity

• B.

A means of isolating a problem in a circuit

• C.

A means for reducing a high ac voltage to a low dc voltage

• D.

A means of limiting current in a circuit

B. A means of isolating a problem in a circuit
Explanation
Half-splitting is a means of isolating a problem in a circuit. This technique involves dividing a circuit into two halves and testing each half separately to identify the faulty component or section. By isolating the problem, it becomes easier to troubleshoot and fix the issue, saving time and effort in the process.

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• 9.

If the resistance in a circuit with constant voltage increases, the current will

• A.

Increase

• B.

Decrease

• C.

Stay the same

• D.

None

B. Decrease
Explanation
When the resistance in a circuit with constant voltage increases, the current will decrease. This is because according to Ohm's Law, the current in a circuit is inversely proportional to the resistance. As the resistance increases, the flow of current is impeded, resulting in a decrease in the current.

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• 10.

• A.

3.6 W

• B.

35 W

• C.

175 W

• D.

245 W

B. 35 W
• 11.

A battery's capacity to deliver power is measured in

• A.

Kwh

• B.

Ah

• C.

Mh

• D.

Vh

B. Ah
Explanation
The capacity of a battery to deliver power is measured in ampere-hours (Ah). Ampere-hour is a unit of electrical charge, representing the amount of charge that can be delivered by a current of one ampere for one hour. This measurement is commonly used to determine the energy storage capacity of batteries, indicating how long a battery can provide a certain amount of current before it needs to be recharged. Other options such as kilowatt-hours (Kwh), milliampere-hours (Mh), and volt-hours (Vh) are not commonly used to measure battery capacity.

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• 12.

If current through a resistance is halved

• A.

The resistance is halved

• B.

The voltage is halved

• C.

The voltage doubles

• D.

None of the above

B. The voltage is halved
Explanation
When the current through a resistance is halved, according to Ohm's Law (V = IR), the voltage across the resistance will also be halved. This is because the voltage is directly proportional to the current flowing through the resistance. Therefore, if the current is reduced by half, the voltage will also be reduced by half.

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• 13.

Ohm's law is a relationship between

• A.

Voltage, current, and time

• B.

Power, current, and resistance

• C.

Resistance, time, and current

• D.

Voltage, current, and resistance

D. Voltage, current, and resistance
Explanation
Ohm's law states that the current flowing through a conductor is directly proportional to the voltage applied across it and inversely proportional to the resistance of the conductor. In other words, the relationship between voltage, current, and resistance can be expressed mathematically as V = I * R, where V is the voltage, I is the current, and R is the resistance. Therefore, the correct answer is voltage, current, and resistance.

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• 14.

The brake power of a diesel engine, keeping other parameters constant, can be increased by

• A.

Decreasing the density of intake air

• B.

Increasing the temperature of intake air

• C.

Increasing the pressure of intake air

• D.

Decreasing the pressure of intake air

C. Increasing the pressure of intake air
Explanation
Increasing the pressure of intake air can increase the brake power of a diesel engine. This is because when the pressure of intake air is increased, more air is forced into the combustion chamber, resulting in a higher oxygen concentration. This allows for a more efficient combustion process, leading to increased power output.

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• 15.

The expansion of fuel in a four stroke cycle diesel engine

• A.

Starts at 15° before top dead centre and ends at 30° after top dead centre

• B.

Starts at top dead centre and ends at 30° after top dead centre

• C.

Starts at 15° after top dead centre and ends at 30° before bottom dead centre

• D.

May start and end anywhere

C. Starts at 15° after top dead centre and ends at 30° before bottom dead centre
Explanation
The expansion of fuel in a four stroke cycle diesel engine starts at 15Â° after top dead centre and ends at 30Â° before bottom dead centre. This is the correct answer because in the expansion stroke, the piston moves from top dead centre to bottom dead centre, creating downward pressure and expanding the fuel-air mixture in the cylinder. The expansion stroke begins after the piston has passed top dead centre and continues until it reaches 30Â° before bottom dead centre, at which point the exhaust valve opens and the exhaust stroke begins.

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• 16.

Reactors for propulsion applications are designed for

• A.

Any form of uranium

• B.

Natura uranium

• C.

Enriched uranium

• D.

Plutonium

C. Enriched uranium
Explanation
Reactors for propulsion applications are designed to use enriched uranium. Enriched uranium is uranium that has a higher concentration of the isotope uranium-235, which is more fissile and can sustain a chain reaction. This is important for propulsion applications as it allows for a sustained release of energy, which is necessary for powering a reactor. Natural uranium, on the other hand, has a lower concentration of uranium-235 and would not be able to sustain a chain reaction efficiently. Plutonium is not commonly used in reactors for propulsion applications.

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• 17.

The primary fuel used in nuclear power plants is

• A.

U235

• B.

U238

• C.

Pu239

• D.

Pu233

A. U235
Explanation
The primary fuel used in nuclear power plants is U235. This is because U235 is a fissile isotope, meaning it can undergo nuclear fission, which releases a large amount of energy. U238 and Pu239 are also used as fuel in some reactors, but U235 is the most commonly used isotope due to its higher fissionability. Pu233 is not commonly used as a primary fuel in nuclear power plants.

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• 18.

In which of the following are operational amplifiers (op-amps) used?

• A.

Oscillators

• B.

Filters

• C.

Instrumentation circuits

• D.

All of the above

D. All of the above
Explanation
Operational amplifiers (op-amps) are versatile electronic components commonly used in various applications. They can be used in oscillators to generate continuous waveforms, in filters to modify the frequency response of a circuit, and in instrumentation circuits to amplify and process signals accurately. Therefore, the correct answer is "All of the above" as op-amps are used in all these applications.

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• 19.

For an op-amp having a slew rate SR = 5 V/ms, what is the maximum closed-loop voltage gain that can be used when the input signal varies by 0.2 V in 10 ms?

• A.

150

• B.

200

• C.

250

• D.

300

C. 250
Explanation
The maximum closed-loop voltage gain that can be used is determined by the slew rate of the op-amp and the rate at which the input signal is changing. The slew rate is given as 5 V/ms, which means that the op-amp can change its output voltage by a maximum of 5 V in 1 ms. The input signal varies by 0.2 V in 10 ms, so the maximum output voltage change required is 0.2 V in 10 ms. To ensure that the op-amp can handle this, the maximum closed-loop voltage gain can be calculated as the ratio of the maximum output voltage change to the maximum input voltage change, which is 0.2 V in 10 ms divided by 5 V/ms, resulting in a gain of 250.

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• 20.

What is the level of the voltage between the input terminals of an op-amp?

• A.

Virtually zero

• B.

5 V

• C.

18 V

• D.

22 V

A. Virtually zero
Explanation
The level of voltage between the input terminals of an op-amp is virtually zero. This is because an ideal op-amp has a very high input impedance, which means that it draws almost no current from the input source. As a result, the voltage drop across the input terminals is negligible, making the voltage level virtually zero.

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• 21.

A three-stage op-amp can have a maximum phase lag of ________°.

• A.

-180

• B.

-90

• C.

-270

• D.

None of these

C. -270
Explanation
A three-stage op-amp can have a maximum phase lag of -270Â°. This means that the output signal of the op-amp lags behind the input signal by 270 degrees at maximum. The negative sign indicates that the output signal is delayed compared to the input signal. This large phase lag can occur due to the internal circuitry and feedback mechanisms within the op-amp.

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• 22.

The summing amplifier contains an inverting amplifier.

• A.

True

• B.

False

A. True
Explanation
The summing amplifier is a type of operational amplifier circuit that combines multiple input voltages and produces an output voltage that is the sum of the input voltages. In order to achieve this, it uses an inverting amplifier configuration, where the input voltages are inverted and then summed. This is why the statement "The summing amplifier contains an inverting amplifier" is true.

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• 23.

Which of the following symptom is caused as a result of brake disc run out ?

• A.

Ineffectiveness of the brakes

• B.

Judder during braking

• C.

Localized wearing of the brake pads

• D.

Rapid wearing of the brake pads

B. Judder during braking
Explanation
Judder during braking is caused as a result of brake disc run out. When the brake disc is not properly aligned or has uneven thickness, it causes the brake pads to make uneven contact with the disc. This uneven contact leads to vibrations and juddering sensations when the brakes are applied. Therefore, judder during braking is a common symptom of brake disc run out.

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• 24.

The function of an alternator in an automobile is to

• A.

Supply electric power

• B.

Convert mechanical energy into electrical energy

• C.

Continually recharge the battery

• D.

Partly convert engine power into electric power

C. Continually recharge the battery
Explanation
The function of an alternator in an automobile is to continually recharge the battery. While it does supply electric power and convert mechanical energy into electrical energy, its primary purpose is to ensure that the battery remains charged. The alternator uses the engine's power to generate electricity, which is then used to recharge the battery and power the electrical systems of the vehicle. This ensures that the battery does not lose its charge and can continue to provide power to start the engine and run other electrical components.

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• 25.

In a unit type body (frameless body) design, the sheet metal parts are welded together, forming a frame work to which outer skin is attached.

• A.

True

• B.

False

A. True
Explanation
In a unit type body design, the sheet metal parts are welded together to create a framework. This framework serves as the support structure to which the outer skin is attached. This design eliminates the need for a separate frame, making the body more lightweight and efficient. Therefore, the statement that the sheet metal parts are welded together to form a framework to which the outer skin is attached is true.

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• 26.

The flywheel and the pressure plate bind the clutch disc between them so that the engine and the transmission can be engaged.

• A.

Yes

• B.

No

A. Yes
Explanation
The flywheel and pressure plate are two essential components of a clutch system in a vehicle. The flywheel is connected to the engine, while the pressure plate is connected to the transmission. When the clutch pedal is pressed, the pressure plate releases the clutch disc, allowing it to spin freely. When the pedal is released, the pressure plate applies pressure to the clutch disc, which then engages with the flywheel, connecting the engine and the transmission. Therefore, the statement is correct, as the flywheel and pressure plate indeed bind the clutch disc between them for the engine and transmission to be engaged.

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• 27.

The information provided by the oxygen (O2) sensor to the feedback control system is about the

• A.

Air-fuel ratio

• B.

Air temperature

• C.

Air flow speed

• D.

Exhaust gas volume

A. Air-fuel ratio
Explanation
The oxygen (O2) sensor provides information to the feedback control system about the air-fuel ratio. This ratio is important for the efficient combustion of fuel in the engine. The oxygen sensor measures the amount of oxygen in the exhaust gases and sends this information to the control system. Based on this data, the control system can adjust the fuel injection to maintain the optimal air-fuel ratio for combustion.

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• 28.

The most commonly used supplementary restraint system (SRS) component is

• A.

Seat belt

• B.

Brake

• C.

Airbag

• D.

Steering

C. Airbag
Explanation
The most commonly used supplementary restraint system (SRS) component is the airbag. Airbags are designed to rapidly inflate in the event of a collision, providing an additional layer of protection for the occupants of a vehicle. They work in conjunction with seat belts to help reduce the risk of injury during a crash. While brakes and steering are important components of a vehicle's safety system, the airbag is specifically designed as a supplementary restraint system to protect occupants in the event of a collision.

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• 29.

The air gap between the central electrode and ground (or side) electrode of a spark plug is around

• A.

0.2 mm

• B.

0.5 mm

• C.

1 mm

• D.

1.5 mm

C. 1 mm
Explanation
The air gap between the central electrode and ground electrode of a spark plug is typically around 1 mm. This gap is important because it allows for the formation of a spark to ignite the air-fuel mixture in the combustion chamber of an engine. A larger gap may result in a weaker spark, while a smaller gap may cause the spark to be too hot and potentially cause damage to the spark plug or engine components. Therefore, a 1 mm gap is a common and optimal choice for spark plug performance.

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• 30.

Which of the following indicates a multigrade oil ?

• A.

SAE 30

• B.

API SF

• C.

SAE 20 W-50

• D.

API 50

C. SAE 20 W-50
Explanation
SAE 20 W-50 indicates a multigrade oil because it has two viscosity ratings separated by a "W" (winter) symbol. The "20" indicates the oil's viscosity at low temperatures, making it suitable for winter use, while the "50" indicates its viscosity at high temperatures. This means that the oil can provide adequate lubrication in both cold and hot weather conditions, making it a multigrade oil.

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• Current Version
• Mar 20, 2023
Quiz Edited by
ProProfs Editorial Team
• Jun 10, 2019
Quiz Created by
Parveen Sharma

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