# Class 10 : Electricity Quiz

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| By Bhudev Singh
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Bhudev Singh
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Quizzes Created: 2 | Total Attempts: 1,555
Questions: 15 | Attempts: 978

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• 1.

### Unit of electric current :

• A.

Joule/sec

• B.

Coulomb

• C.

Ampere

• D.

None

C. Ampere
Explanation
The SI unit of electric current is the ampere, or amp, which is the flow of electric charge across a surface at the rate of one coulomb per second.

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• 2.

### Unit of conductance :

• A.

Volt/ampere

• B.

Ohm

• C.

Ohm-1

• D.

Both a & b

C. Ohm-1
Explanation
The unit of conductance is ohm-1. Conductance is the measure of how easily an electric current can flow through a material. It is the reciprocal of resistance, which is measured in ohms. Therefore, conductance is measured in ohm-1, indicating the number of ohms of resistance per unit of current. This unit represents the inverse relationship between resistance and conductance, where higher conductance values indicate easier flow of current.

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• 3.

### Which of the following permit the current flowing in circuit ?

• A.

Conductance

• B.

Resistance

• C.

Both a & b

• D.

None of these

A. Conductance
Explanation
Conductance is the measure of how easily an electric current can flow through a material. It is the reciprocal of resistance. Therefore, conductance allows the current to flow in a circuit. Resistance, on the other hand, opposes the flow of current. So, both conductance and resistance affect the current flowing in a circuit.

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• 4.

### Ohm-meter is the unit of :

• A.

Conductivity

• B.

Electric potential

• C.

Resistivity

• D.

None

C. Resistivity
Explanation
An ohm-meter is a unit used to measure resistivity, which is the inherent property of a material to resist the flow of electric current. It is a measure of how strongly a material opposes the passage of electric current through it. The higher the resistivity, the greater the resistance to the flow of current. Therefore, the correct answer is resistivity.

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• 5.

### 1 micro ampere = ________ Ampere

10^-6
Explanation
The given question is asking for the conversion of 1 micro ampere to amperes. The correct answer is 10^-6. This is because the prefix "micro" represents a factor of 10^-6, which means that 1 micro ampere is equal to 10^-6 amperes.

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• 6.

### If length and area of conductor both are double. Then, the resistance will be :

• A.

Increase 4 times

• B.

Halved

• C.

Remain same

• D.

Increase 2 times

C. Remain same
Explanation
When the length and area of a conductor both double, the resistance of the conductor remains the same. This is because resistance is directly proportional to the length of the conductor and inversely proportional to the cross-sectional area of the conductor. So, when both the length and area double, the increase in resistance due to the longer length is offset by the decrease in resistance due to the larger area, resulting in the resistance remaining unchanged.

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• 7.

### Resistivity of conductor depends on :

• A.

Nature

• B.

Length

• C.

Temperature

• D.

A & c

D. A & c
Explanation
The resistivity of a conductor depends on its nature, which refers to the type of material it is made of. Different materials have different resistivities due to variations in their atomic structure and electron mobility. Additionally, the resistivity of a conductor also depends on temperature. As temperature increases, the resistivity generally increases as well. This is because higher temperatures cause more collisions between the electrons and atoms in the conductor, impeding the flow of electric current. Therefore, the correct answer is a & c, which means that the resistivity of a conductor depends on both its nature and temperature.

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• 8.

### What is the power of conductor of resistance 5 ohm and current of 25 A ? Give the ans. in HP(Horse Power).   HINT: 1HP = 746 watt

• A.

3125/746 HP

• B.

3065/746 HP

• C.

4125/746 HP

• D.

2815/746 HP

A. 3125/746 HP
Explanation
The power of a conductor can be calculated using the formula P = I^2 * R, where P is the power, I is the current, and R is the resistance. In this case, the current is 25 A and the resistance is 5 ohm. Plugging these values into the formula, we get P = (25^2) * 5 = 3125 W. To convert this to HP, we divide by 746, so the answer is 3125/746 HP.

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• 9.

### Property of Series Combination -

• A.

Current same & p.d. is different

• B.

Current different & p.d. is same

• C.

Both same

• D.

Both different

A. Current same & p.d. is different
Explanation
In a series combination of resistors, the current flowing through each resistor is the same. This is because the current has only one path to flow through and it experiences the same resistance in each resistor. However, the potential difference (p.d.) across each resistor can be different. This is because the p.d. depends on the resistance of each resistor and the current flowing through it. If the resistors have different resistances, then the p.d. across each resistor will be different. Therefore, the correct answer is "Current same & p.d. is different".

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• 10.

### 1 kWh = ________ Joule and ________ Unit

3.6*10^6
1
Explanation
The given answer states that 1 kWh is equal to 3.6*10^6 Joules and 1 Unit. This means that 1 kilowatt-hour (kWh) of energy is equivalent to 3.6*10^6 Joules and also equal to 1 Unit.

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• 11.

### An electric heater is connected in a circuit of 400 volt and 4 A current for 2 hours. Find the cost of bill at the rate of 5 Rs. per unit.

• A.

32 Rs.

• B.

64 Rs.

• C.

16 Rs.

• D.

48 Rs.

C. 16 Rs.
Explanation
The cost of the bill can be calculated by multiplying the power consumed by the time it was used and the rate per unit. The power consumed can be calculated by multiplying the voltage and current. In this case, the power consumed is 400V * 4A = 1600W. Since the heater was used for 2 hours, the total energy consumed is 1600W * 2 hours = 3200Wh. Converting this to units, we divide by 1000 to get 3.2 units. Finally, multiplying the units by the rate of 5 Rs. per unit gives us a total cost of 3.2 units * 5 Rs. per unit = 16 Rs.

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• 12.

### Which of the following is not the formula of Power ?

• A.

Energy/Time

• B.

I2R

• C.

V*I

• D.

V*I*T

D. V*I*T
Explanation
The formula V*I*T represents the product of voltage, current, and time, which gives the total energy consumed in a circuit. Power is defined as the rate at which energy is transferred or converted, and it is calculated using the formula P = V*I. Therefore, V*I*T is not the formula for power.

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• 13.

### Heating Effect Law is used in :

• A.

Celling Fan

• B.

Electric Heater

• C.

Electric Iron

• D.

Both b & c

D. Both b & c
Explanation
The heating effect law is used in both electric heaters and electric irons. When electric current flows through a conductor, it produces heat due to the resistance of the material. This principle is utilized in electric heaters to generate heat for warming up a space. Similarly, electric irons use the heating effect law to produce heat that helps in ironing clothes effectively. Therefore, both electric heaters and electric irons utilize the heating effect law.

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• 14.

### Electric Iron work in heating effect law due to presence of ________ wire.

nichrome
Explanation
Electric irons work on the principle of heating effect of electric current. The presence of nichrome wire in the iron is crucial for this process. Nichrome is an alloy made of nickel and chromium, which has high electrical resistance. When an electric current passes through the nichrome wire, it encounters resistance, causing the wire to heat up. This heat is then transferred to the soleplate of the iron, allowing it to reach high temperatures and effectively iron clothes. Therefore, the presence of nichrome wire enables the electric iron to function properly.

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• 15.

### Voltmeter is used for

• A.

Measuring current

• B.

Measuring electric potential

B. Measuring electric potential
Explanation
A voltmeter is used to measure the electric potential, also known as voltage, across a circuit or a specific component. It is designed to measure the potential difference between two points in an electrical circuit. By connecting the voltmeter in parallel with the component or across the circuit, it can provide an accurate reading of the electric potential at that point. This allows for the measurement and monitoring of voltage levels, ensuring that the electrical system is functioning correctly and within the desired range.

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• Current Version
• Oct 16, 2023
Quiz Edited by
ProProfs Editorial Team
• Jun 12, 2020
Quiz Created by
Bhudev Singh

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