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Electricity Online Test 10th Science Lesson 4 Questions in English
Electricity Online Test 10th Science Lesson 4 Questions in English
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Question 1 |
The motion of what through a conductor (e.g., copper wire) will constitute an electric current?
Proton | |
Electron | |
Neutron | |
All the above |
Question 1 Explanation:
The motion of electric charges (electrons) through a conductor (e.g., copper wire) will constitute an electric current. This is similar to the flow of water through a channel or flow of air from a region of high pressure to a region of low pressure. In a similar manner, the electric current passes from the positive terminal (higher electric potential) of a battery to the negative terminal (lower electric potential) through a wire
Question 2 |
Which among the following is the symbol that represent the Electric current?
I | |
L | |
W | |
T |
Question 2 Explanation:
Electric current is often termed as ‘current’ and it is represented by the symbol ‘I’. It is defined as the rate of flow of charges in a conductor. This means that the electric current represents the amount of charges flowing in any cross section of a conductor (say a metal wire) in unit time.
Question 3 |
If a net charge ‘Q’ passes through any cross section of a conductor in time ‘t’, then the current flowing through the conductor is _____
I = Q / T | |
I = QT | |
I = T / Q | |
I = 1 / QT |
Question 3 Explanation:
If a net charge ‘Q’ passes through any cross section of a conductor in time ‘t’, then the current flowing through the conductor is I = Q / T.
Question 4 |
What is the SI unit of electric current?
Newton | |
Volt | |
Ampere | |
Joule |
Question 4 Explanation:
The SI unit of electric current is ampere (A). The current flowing through a conductor is said to be one ampere, when a charge of one coulomb flows across any cross-section of a conductor, in one second. Hence, 1 ampere = 1 coulomb / 1 second.
Question 5 |
A charge of 12 coulomb flows through a bulb in 5 second. What is the current through the bulb?
60 A | |
1.2 A | |
2.4 A | |
4.16 A |
Question 5 Explanation:
Charge Q = 12 C, Time t = 5 s. Therefore,
Current I = Q /T = 12 / 5 = 2.4 A.
Question 6 |
Which among the following component is used to fix the magnitude of the current through a circuit?
Resistor | |
A Diode | |
Galvanometer | |
Rheostat |
Question 6 Explanation:
Resistor is used to fix the magnitude of the current through a circuit.
Question 7 |
Which is used to select the magnitude of the current through a circuit?
Light Emitting Diode | |
A Diode | |
Galvanometer | |
Rheostat |
Question 7 Explanation:
Variable resistor or Rheostat is used to select the magnitude of the current through a circuit.
Question 8 |
Which among the following is used to detect the current and its direction?
Ammeter | |
Voltmeter | |
Galvanometer | |
Ground connection |
Question 8 Explanation:
Galvanometer is used to detect the current and its direction.
Question 9 |
Which among the following statement is correct
An electric circuit is an open conducting loop (or) path, which has a network of electrical components through which electrons and protons are able to flow. This path is made using electrical wires so as to connect an electric appliance to a source of electric charges (battery).
In this circuit, if the switch is ‘on’, the bulb glows. If it is switched off, the bulb does not glow. Therefore, the circuit must be closed in order that the current passes through it. The potential difference required for the flow of charges is provided by the battery. The electrons flow from the negative terminal to the positive terminal of the battery.
By convention, the direction of current is taken as the direction of flow of positive charge (or) opposite to the direction of flow of electrons. Thus, electric current passes in the circuit from the positive terminal to the negative terminal.
Both 1 and 2 | |
Both 1 and 3 | |
Both 2 and 3 | |
All 1, 2 and 3 |
Question 9 Explanation:
An electric circuit is a closed conducting loop (or) path, which has a network of electrical components through which electrons are able to flow. This path is made using electrical wires so as to connect an electric appliance to a source of electric charges (battery).
Question 10 |
Which among the following component is used to measure the potential difference?
Ammeter | |
Voltmeter | |
A diode | |
Ground connection |
Question 10 Explanation:
Voltmeter is used to measure the potential difference.
Question 11 |
Which among the following components is wrongly matched with its unit
- Ammeter – Used to measure the difference in current and voltage
- A diode – It is used in electronic devices
- Ground connection – Used to provide protection to the electrical components. It also serves as a reference point to measure the electric potential
Only 1 | |
Only 2 | |
Both 2 and 3 | |
None of the above |
Question 12 |
Which among the following statement is correct
- In the conductor, the charges will flow from a point in it, which is at a lower electric potential to a point, which is at a higher electric potential.
- In the conductor, the charges will flow from a point in it, which is at a higher electric potential to a point, which is at a lower electric potential.
- There must be a difference in temperature between two points in a solid for the heat to flow in it. Similarly, a difference in electric potential is needed for the flow of electric charges in a conductor.
Both 1 and 2 | |
Both 1 and 3 | |
Both 2 and 3 | |
All 1, 2 and 3 |
Question 12 Explanation:
There must be a difference in temperature between two points in a solid for the heat to flow in it. Similarly, a difference in electric potential is needed for the flow of electric charges in a conductor. In the conductor, the charges will flow from a point in it, which is at a higher electric potential to a point, which is at a lower electric potential.
Question 13 |
Which at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against the electric force?
Electric resistance | |
Electric conductance | |
Electric manual | |
Electric potential |
Question 13 Explanation:
The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against the electric force.
Question 14 |
Which among the following statement is correct
- The electric potential difference between two points is defined as the amount of work done in moving a unit positive charge from one point to another point against the electric force.
- Suppose, you have moved a charge Q from a point A to another point B. Let ‘W’ be the work done to move the charge from A to B. Then, the potential difference between the points A and B is given by the following expression: Potential Difference (V) = work done (W) / charge (Q).
- Potential difference is inversely proportion to the difference in the electric potential of these two points. If VA and VB represent the electric potential at the points A and B respectively, then, the potential difference between the points A and B is given by: V = VA – VB (if VA is less than VB) V = VB – VA (if VB is less than VA).
Both 1 and 2 | |
Both 1 and 3 | |
Both 2 and 3 | |
All 1, 2 and 3 |
Question 14 Explanation:
Potential difference is also equal to the difference in the electric potential of these two points. If VA and VB represent the electric potential at the points A and B respectively, then, the potential difference between the points A and B is given by: V = VA – VB (if VA is more than VB) V = VB – VA (if VB is more than VA).
Question 15 |
The SI unit of electric potential or potential difference is ___
Joule | |
Volt | |
Newton | |
Ohm |
Question 15 Explanation:
The SI unit of electric potential or potential difference is volt (V).
Question 16 |
Which among the following equation defines 1 Volt?
1 Joule / 1 Coulomb | |
1 Coulomb / 1 Joule | |
1 Joule × 1 Coulomb | |
1 Joule + 1 Coulomb |
Question 16 Explanation:
The potential difference between two points is one volt, if one joule of work is done in moving one coulomb of charge from one point to another against the electric force.
Question 17 |
Who among the following established the relation between the potential difference and current?
Michael Faraday | |
Gustav Kirchhoff | |
Georg Simon Ohm | |
Ernest Rutherford |
Question 17 Explanation:
A German physicist, Georg Simon Ohm established the relation between the potential difference and current, which is known as Ohm’s Law. This relationship can be understood from the following activity
Question 18 |
Which among the following statement is correct
- According to Ohm’s law, at a constant temperature, the steady current ‘I’ flowing through a conductor is directly proportional to the potential difference ‘V’ between the two ends of the conductor. I is proportion to V. Hence, I / V = constant. The value of this proportionality constant is found to be R. Therefore I = R V
- V = I R. Here, R is a constant for a given material (say Nichrome) at a given temperature and is known as the resistance of the material. Since, the potential difference V is proportional to the current I, the graph between V and I is a straight line for a conductor.
Only 1 | |
Only 2 | |
Both 1 and 2 | |
None of the above |
Question 18 Explanation:
According to Ohm’s law, at a constant temperature, the steady current ‘I’ flowing through a conductor is directly proportional to the potential difference ‘V’ between the two ends of the conductor. I is proportion to V. Hence, I / V = constant.
The value of this proportionality constant is found to be 1 / R. Therefore I = (1 / R) V.
Question 19 |
The work done in moving a charge of 10 C across two points in a circuit is 100 J. What is the potential difference between the points?
V = 0.1 volt | |
V = 110 volt | |
V = 1000 volt | |
V = 10 volt |
Question 19 Explanation:
Charge, Q = 10 C Work Done, W = 100 J
Potential Difference V = W / Q = 100 / 10
Therefore V = 10 volt.
Question 20 |
Which can be defined as the ratio between the potential difference across the ends of the conductor and the current flowing through it?
Inductance of a conductor | |
Resistance of a conductor | |
Capacity of conductor | |
None of the above |
Question 20 Explanation:
The resistance of a conductor can be defined as the ratio between the potential difference across the ends of the conductor and the current flowing through it.
Question 21 |
Which among the following statement is correct
- In this Figure, a Nichrome wire was connected between X and Y. If you replace the Nichrome wire with a copper wire and conduct the same experiment, you will notice a same current for the same value of the potential difference across the wire. If you again replace the copper wire with an aluminium wire, you will same value again for the current passing through it.
- From equation V = I R, you have learnt that V/I must be equal to the resistance of the conductor used. The variations in the current for the same values of potential difference indicate that the resistance of different materials is different.
- Resistance of a material is its property to oppose the flow of charges and hence the passage of current through it. It is different for different materials. From Ohm’s Law, V / I = R.
Both 1 and 2 | |
Both 1 and 3 | |
Both 2 and 3 | |
All 1, 2 and 3 |
Question 21 Explanation:
In this Figure, a Nichrome wire was connected between X and Y. If you replace the Nichrome wire with a copper wire and conduct the same experiment, you will notice a different current for the same value of the potential difference across the wire. If you again replace the copper wire with an aluminium wire, you will get another value for the current passing through it.
Question 22 |
What is the SI unit of resistance?
Newton | |
Ohm | |
Volt | |
Coulomb |
Question 22 Explanation:
The SI unit of resistance is ohm and it is represented by the symbol Ω.
Question 23 |
Which among the following equation denotes Resistance of a conductor is said to be one ohm if a current of one ampere flows through it when a potential difference of one volt is maintained across its ends?
1 ohm = 1 volt / 1 ampere | |
1 ohm = 1 ampere / 1 volt | |
1 ohm = 1 ampere × 1 volt | |
1 ohm = 1 ampere + 1 volt |
Question 23 Explanation:
Resistance of a conductor is said to be one ohm if a current of one ampere flows through it when a potential difference of one volt is maintained across its ends.
1 ohm = 1 volt / 1 ampere
Question 24 |
Which among the following is the unit of Electric resistivity?
Ohm | |
Ohm metre | |
Ohm-1 | |
Ohm metre-1 |
Question 24 Explanation:
The electrical resistivity of a material is defined as the resistance of a conductor of unit length and unit area of cross section. Its unit is ohm metre.
Question 25 |
Which among the following statement is correct
- You can verify electric resistivity by doing an experiment that the resistance of any conductor ‘R’ is directly proportional to the length of the conductor ‘L’ and is inversely proportional to its area of cross section ‘A’. R LA. Therefore R = ρ (L/A)
- ρ (rho) is a constant, called as electrical resistivity or specific resistance of the material of the conductor. ρ = (RA) / L. If L = 1 m, A = 1 m2 then, from the above equation ρ = R. Electrical resistivity of a conductor is a measure of the resisting power of a specified material to the passage of an electric current. It is a constant for a given material.
Only 1 | |
Only 2 | |
Both 1 and 2 | |
None of the above |
Question 26 |
Calculate the resistance of a conductor through which a current of 2 A passes, when the potential difference between its ends is 30 V?
10 Ω | |
15 Ω | |
30 Ω | |
60 Ω |
Question 26 Explanation:
Current through the conductor I = 2 A, Potential Difference V = 30 V
From Ohm’s Law: R = V / I
Therefore, R = 30 / 2 = 15 Ω.
Question 27 |
Which is a conductor with highest resistivity equal to 1.5 × 10-6 Ω m?
Aluminium | |
Copper | |
Silver | |
Nichrome |
Question 27 Explanation:
Nichrome is a conductor with highest resistivity equal to 1.5 × 10–6Ω m. Hence, it is used in making heating elements.
Question 28 |
The reciprocal of electrical resistivity of a material is called _____
Electrical Projective | |
Electrical Conductivity | |
Electrical Inductance | |
All the above |
Question 28 Explanation:
The reciprocal of electrical resistivity of a material is called its electrical conductivity. Conductance of a material is the property of a material to aid the flow of charges and hence, the passage of current in it. The conductance of a material is mathematically defined as the reciprocal of its resistance (R). Hence, the conductance ‘G’ of a conductor is given by G = 1 / R.
Question 29 |
The unit of Conductance is ____
Ohm | |
Ohm metre | |
Ohm-1 | |
Ohm metre-1 |
Question 29 Explanation:
The unit of conductance is Ohm-1. It is also represented as ‘mho’.
Question 30 |
Which among the following is the Insulator?
Copper | |
Aluminium | |
Rubber | |
Nickle |
Question 30 Explanation:
Electrical conductivity of a conductor is a measure of its ability to pass the current through it. Some materials are good conductors of electric current. Example: copper, aluminium, etc. While some other materials are non-conductors of electric current (insulators). Example: glass, wood, rubber, etc.
Question 31 |
What is the unit of electrical conductivity?
Ohm | |
Ohm metre | |
Ohm-1 | |
Ohm metre-1 |
Question 31 Explanation:
The unit of electrical conductivity is ohm-1 metre-1. It is also represented as mho metre-1. The conductivity is a constant for a given material.
Question 32 |
Which among the following material materials resistivity is wrongly matched
- Copper – 1.62 × 10-8
- Nickel – 10.14 × 10–8
- Chromium – 12.9 × 10–8
Only 1 | |
Only 2 | |
Both 1 and 3 | |
Both 2 and 3 |
Question 32 Explanation:
Copper – 1.62 × 10-8; Nickel – 6.84 × 10–8; Chromium – 12.9 × 10–8; Glass – 1010 to 1014; Rubber – 1013 to 1016.
Question 33 |
The combination of resistors is known as ______
System of resistors | |
Network of resistors | |
Bridge of resistors | |
Gate of resistors |
Question 33 Explanation:
The combination of resistors is known as ‘system of resistors’ or ‘grouping of resistors. Resistors can be connected in various combinations. The two basic methods of joining resistors together are: a) Resistors connected in series, and b) Resistors connected in parallel.
Question 34 |
The resistance of a wire of length 10 m is 2 ohms. If the area of cross section of the wire is 2×10-7 m2, determine its (i) resistivity (ii) conductance and (iii) conductivity.
ρ= 0.25× 10-8 Ω m; G=0.5 mho; = 4 × 108 mho m-1 | |
ρ= 0.5 × 10-8 Ω m;G=4 mho;= 0.25 × 108 mho m-1 | |
ρ= 0.25 × 10-8 Ω m;G=4 mho;= 0.5× 108 mho m-1 | |
ρ= 4 × 10-8 Ω m;G=0.5 mho;= 0.25 × 108 mho m-1 |
Question 34 Explanation:
Given: Length, L = 10 m, Resistance, R = 2 ohm and Area, A = 2 × 10–7 m2
Resistivity, ρ=RAL = (2×2×10-7)/10 = 4 × 10-8 Ω m
Conductance, G = 1R = 12= 0.5 mho
Conductivity, σ=1= 1/ (4 × 10-8) = 0.25 × 108 mho m–1.
Question 35 |
Which among the following statement is correct
- A series circuit connects the components one after the other to form a ‘multiple loop’. A series circuit has two or more loop through which current can pass. If the circuit is interrupted at any point in the loop, no current can pass through the circuit and hence no electric appliances connected in the circuit will work.
- Series circuits are commonly used in devices such as flashlights. Thus, if resistors are connected end to end, so that the same current passes through each of them, then they are said to be connected in series.
Only 1 | |
Only 2 | |
Both 1 and 2 | |
None of the above |
Question 35 Explanation:
A series circuit connects the components one after the other to form a ‘single loop’. A series circuit has only one loop through which current can pass. If the circuit is interrupted at any point in the loop, no current can pass through the circuit and hence no electric appliances connected in the circuit will work.
Question 36 |
Which among the following statement is correct
- Let, three resistances R1, R2 and R3 be connected in series (Figure 4.6). Let the current flowing through them be I. According to Joules Law, the potential differences V1, V2 and V3 across R1, R2 and R3 respectively, are given by V1 = IR1, V2 = IR2, V3 = IR3. The sum of the potential differences across the ends of each resistor is given by: V = V1 + V2 + V3 ---->>> 1
- Using equations, we get V = I R1 + I R2 + I R3. The effective resistor is a single resistor, which can replace the resistors effectively, so as to allow the same current through the electric circuit. Let, the effective resistance of the series-combination of the resistors, be RS. Then, V = I RS. --->>> 2
- Combining 1 and 2, I RS = I R1 + I R2 + I R3 i.e., RS = R1 + R2 + R3. Thus, you can understand that when a number of resistors are connected in series, their equivalent resistance or effective resistance is equal to the sum of the individual resistances. When ‘n’ resistors of equal resistance R are connected in series, the equivalent resistance is ‘n R’.
Both 1 and 2 | |
Both 1 and 3 | |
Both 2 and 3 | |
All 1, 2 and 3 |
Question 36 Explanation:
Let, three resistances R1, R2 and R3 be connected in series (Figure 4.6). Let the current flowing through them be I. According to Ohm’s Law, the potential differences V1, V2 and V3 across R1, R2 and R3 respectively, are given by V1 = IR1, V2 = IR2, V3 = IR3. The sum of the potential differences across the ends of each resistor is given by: V = V1 + V2 + V3.
Question 37 |
Three resistors of resistances 5 ohms, 3 ohm and 2 ohms are connected in series with 10 V battery. Calculate their effective resistance and the current flowing through the circuit?
RS = 15 Ω, I = 10A | |
RS = 10 Ω, I = 1A | |
RS = 15 Ω, I = 1A | |
RS = 10 Ω, I = 15A |
Question 37 Explanation:
R1 = 5 Ω, R2 = 3 Ω, R3 = 2 Ω, V = 10 V
Rs = R1 + R2 + R3, Rs = 5 + 3 + 2 = 10, hence
Rs = 10 Ω
The current, I = V / RS = 10 / 10 = 1 A.
Question 38 |
Which among the following statement is correct
- If you consider the connection of a set of parallel resistors that are connected in series, you get a series – parallel circuit. Let R1 and R2 be connected in parallel to give an effective resistance of RP1. Similarly, let R3 and R4 be connected in parallel to give an effective resistance of RP2. Then, both of these parallel segments are connected in series. 1 / Rp1 = (1 / R1)+(1 / R2); 1 / Rp2 = (1 / R3)+(1 / R4) Hence, RTotal = RP1 + RP2
- If you consider a connection of a set of series resistors connected in a parallel circuit, you get a parallel-series circuit. Let R1 and R2 be connected in series to give an effective resistance of RS1. Similarly, let R3 and R4 be connected in series to give an effective resistance of RS2. Then, both of these serial segments are connected in parallel.
Only 1 | |
Only 2 | |
Both 1 and 2 | |
None of the above |
Question 39 |
Which among the following is not the property of series circuit?
Equivalent resistance is more than the highest resistance | |
Current is more as effective resistance is more | |
If one appliance is disconnected, others also do not work | |
None of the above |
Question 39 Explanation:
Current is less as effective resistance is more.
Question 40 |
Which among the following is not the property of parallel circuit?
Equivalent resistance is less than the lowest resistance | |
Current is more as effective resistance is less | |
If one appliance is disconnected, others will work independently | |
None of the above |
Question 41 |
Which among the following statement is correct
- Generally, a source of electrical energy can develop a potential difference across a capacitor, which is connected to that source. This potential difference constitutes a current through the capacitor. For continuous drawing of current, the source has to continuously spend its energy.
- A part of the energy from the source can be converted into useful work and the rest will be converted into heat energy. Thus, the passage of electric current through a wire, results in the production of heat. This phenomenon is called heating effect of current. This heating effect of current is used in devices like electric heater, electric iron, etc.
Only 1 | |
Only 2 | |
Both 1 and 2 | |
None of the above |
Question 41 Explanation:
Generally, a source of electrical energy can develop a potential difference across a resistor, which is connected to that source. This potential difference constitutes a current through the resistor. For continuous drawing of current, the source has to continuously spend its energy.
Question 42 |
Let ‘I’ be the current flowing through a resistor of resistance ‘R’, and ‘V’ be the potential difference across the resistor. The charge flowing through the circuit for a time interval ‘t’ is ‘Q’. The work done in moving the charge Q across the ends of the resistor with a potential difference of V is VQ. What is heat produced in resistor?
H = R / I t | |
H = I2 R t | |
H = I2 / R t | |
H = I R t |
Question 42 Explanation:
This energy spent by the source gets dissipated in the resistor as heat. Thus, the heat produced in the resistor is: H = W = VQ
You know that the relation between the charge and current is Q = I t. Using this, you get H = V I t
From Ohm’s Law, V = I R. Hence, you have H = I2 R t.
This is known as Joule’s law of heating.
Question 43 |
Which among the following does not states the joules law of heating that the heat produced in any resistor is
directly proportional to the square of the current passing through the resistor | |
directly proportional to the resistance of the resistor. | |
directly proportional to the time for which the current is passing through the resistor. | |
directly proportional to the series of resistance connected. |
Question 43 Explanation:
Joule’s law of heating states that the heat produced in any resistor is: • directly proportional to the square of the current passing through the resistor. • directly proportional to the resistance of the resistor. • directly proportional to the time for which the current is passing through the resistor.
Question 44 |
Which among the following is not the property of Nichrome which is used as the heating element?
High resistivity | |
High melting point | |
Easily oxidized | |
None of the above |
Question 44 Explanation:
The heating effect of electric current is used in many home appliances such as electric iron, electric toaster, electric oven, electric heater, geyser, etc. In these appliances Nichrome, which is an alloy of Nickel and Chromium is used as the heating element. Because: (i) it has high resistivity, (ii) it has a high melting point, (iii) it is not easily oxidized.
Question 45 |
How fuse wire is connected in an electric circuit, which when a large current pass through the circuit, the fuse wire melts due to Joule’s heating effect and hence the circuit gets disconnected?
Parallel | |
Series | |
Series- parallel combination | |
Series-parallel combination |
Question 45 Explanation:
The fuse wire is connected in series, in an electric circuit. When a large current pass through the circuit, the fuse wire melts due to Joule’s heating effect and hence the circuit gets disconnected. Therefore, the circuit and the electric appliances are saved from any damage. The fuse wire is made up of a material whose melting point is relatively low.
Question 46 |
In electric bulbs, a small wire made up of a material whose melting point is very high is used, known as ______
Chock | |
Insulation | |
Stem | |
Filament |
Question 46 Explanation:
In electric bulbs, a small wire is used, known as filament. The filament is made up of a material whose melting point is very high. When current passes through this wire, heat is produced in the filament. When the filament is heated, it glows and gives out light. Tungsten is the commonly used material to make the filament in bulbs.
Question 47 |
An electric heater of resistance 5 Ω is connected to an electric source. If a current of 6 A flows through the heater, then find the amount of heat produced in 5 minutes.
H = 18000 J | |
H = 36000 J | |
H = 54000 J | |
H = 72000 J |
Question 47 Explanation:
Given resistance R = 5 Ω, Current I = 6 A, Time t = 5 minutes = 5 × 60 s = 300 s
Amount of heat produced, H = I2 Rt, H = 62 × 5 × 300. Hence, H = 54000 J
Question 48 |
In general, what is defined as the rate of doing work or rate of spending energy?
Power | |
Static | |
Modem | |
Wire |
Question 48 Explanation:
In general, power is defined as the rate of doing work or rate of spending energy. Similarly, the electric power is defined as the rate of consumption of electrical energy. It represents the rate at which the electrical energy is converted into some other form of energy.
Question 49 |
Suppose a current ‘I’ flows through a conductor of resistance ‘R’ for a time ‘t’, then the potential difference across the two ends of the conductor is ‘V’. The work done ‘W’ to move the charge across the ends of the conductor is given by what equation?
W = V I t | |
W = V I / t | |
W = V / I t | |
W = I t / V |
Question 49 Explanation:
The work done ‘W’ to move the charge across the ends of the conductor is given by the equation
W = V I t, Power P = work / Time = (V I t)/t = V I
Question 50 |
The SI unit of electric power is _____
Ohm | |
Newton | |
Watt | |
Coulomb |
Question 50 Explanation:
The SI unit of electric power is watt. When a current of 1 ampere passes across the ends of a conductor, which is at a potential difference of 1 volt, then the electric power is
P = 1 volt × 1 ampere = 1 watt. Thus, one watt is the power consumed when an electric device is operated at a potential difference of one volt and it carries a current of one ampere. A larger unit of power, which is more commonly used is kilowatt.
Question 51 |
Which is a unit in the foot-pound-second (fps) or English system, sometimes used to express the electric power?
Coulomb | |
Horse power | |
Newton metre | |
None of the above |
Question 51 Explanation:
The horse power (hp) is a unit in the foot-pound-second (fps) or English system, sometimes used to express the electric power.
Question 52 |
One horse power is equal to how many watts?
451 watts | |
572 watts | |
746 watts | |
824 watts |
Question 52 Explanation:
One horse power is equal to 746 watts
Question 53 |
Which among the following statement is correct
- Electricity is consumed both in houses and industries. Consumption of electricity is based on two factors: (i) Amount of electric power and (ii) Duration of usage. Electrical energy consumed is taken as the product of electric power and time of usage.
- Consumption of electrical energy is measured and expressed in watt hour, though its SI unit is watt second. In practice, a larger unit of electrical energy is needed. This larger unit is kilowatt hour (kWh).
- One kilowatt hour is otherwise known as one unit of electrical energy. One kilowatt hour means that an electric power of 100 watt has been utilized for an hour. Hence,
Both 1 and 2 | |
Both 1 and 3 | |
Both 2 and 3 | |
All 1, 2 and 3 |
Question 53 Explanation:
One kilowatt hour is otherwise known as one unit of electrical energy. One kilowatt hour means that an electric power of 1000 watt has been utilized for an hour. Hence,
1 kWh = 1000 watts hour = 1000 × (60 × 60) watt second = 3.6 × 106 J.
Question 54 |
Which among the following statement is correct
- The electricity produced in power stations is distributed to all the domestic and industrial consumers through overhead and underground cables. In our homes, electricity is distributed through the domestic electric circuits wired by the electricians.
- The first stage of the domestic circuit is to bring the power supply to the main-box from a distribution panel, such as a transformer. The important components of the main-box are: (i) a fuse box and (ii) a meter. The meter is used to record the consumption of electrical energy. The fuse box contains either a fuse wire or a miniature circuit breaker (MCB).
- The function of the fuse wire or a MCB is to protect the house hold electrical appliances from overloading due to excess current. An MCB is a switching device, which can be activated automatically as well as manually. It has a spring attached to the switch, which is attracted by an electromagnet when an excess current pass through the circuit. Hence, the circuit is broken and the protection of the appliance is ensured.
Both 1 and 2 | |
Both 1 and 3 | |
Both 2 and 3 | |
All 1, 2 and 3 |
Question 55 |
In India, domestic circuits are supplied with an alternating current of potential what?
110/120 V | |
220/230 V | |
310/320 V | |
420/430 V |
Question 55 Explanation:
In India, domestic circuits are supplied with an alternating current of potential 220/230V and frequency 50 Hz.
Question 56 |
In countries like USA and UK, domestic circuits are supplied with an alternating current of potential 110/120 V and frequency of what?
50 Hz | |
60 Hz | |
70 Hz | |
80 Hz |
Question 56 Explanation:
In countries like USA and UK, domestic circuits are supplied with an alternating current of potential 110/120 V and frequency 60 Hz.
Question 57 |
The electricity is brought to houses by how many insulated wires?
Two | |
Three | |
Five | |
Six |
Question 57 Explanation:
The electricity is brought to houses by two insulated wires.
Question 58 |
Out of the two Insulated wires brought to house, one wire has a red insulation and is called ____
Flaw wire | |
Neutral wire | |
Trip wire | |
Live wire |
Question 58 Explanation:
Out of the two Insulated wires brought to house, one wire has a red insulation and is called the ‘live wire’. The other wire has a black insulation and is called the ‘neutral wire’.
Question 59 |
Which among the following statement is correct
- The electricity supplied to your house is actually an direct current having an electric potential of 220 V. Both, the live wire and the neutral wire enter into a box where the main fuse is connected with the live wire.
- After the electricity meter, these wires enter into the main switch, which is used to discontinue the electricity supply whenever required. After the main switch, these wires are connected to live wires of two separate circuits. Out of these two circuits, one circuit is of a 5 A rating, which is used to run the electric appliances with a lower power rating, such as tube lights, bulbs and fans.
- The other circuit is of a 15 A rating, which is used to run electric appliances with a high-power rating, such as air-conditioners, refrigerators, electric iron and heaters. It should be noted that all the circuits in a house are connected in parallel, so that the disconnection of one circuit does not affect the other circuit.
Both 1 and 2 | |
Both 1 and 3 | |
Both 2 and 3 | |
All 1, 2 and 3 |
Question 59 Explanation:
The electricity supplied to your house is actually an alternating current having an electric potential of 220 V. Both, the live wire and the neutral wire enter into a box where the main fuse is connected with the live wire.
Question 60 |
In which among the following circuit, each electric appliance gets an equal voltage?
Series connection | |
Parallel connection | |
Series-parallel combination | |
Parallel-series combination |
Question 60 Explanation:
One more advantage of the parallel connection of circuits is that each electric appliance gets an equal voltage.
Question 61 |
The fuse wire or MCB will not disconnect the circuit in the event of what?
Short circuit | |
Overloading | |
Weathering | |
None of the above |
Question 61 Explanation:
The fuse wire or MCB will disconnect the circuit in the event of an overloading and short circuiting.
Question 62 |
When a live wire comes in contact with a neutral wire, it causes what?
Over loading | |
Short circuit | |
Transcription | |
All the above |
Question 62 Explanation:
When a live wire comes in contact with a neutral wire, it causes a ‘short circuit’. This happens when the insulation of the wires gets damaged due to temperature changes or some external force. Due to a short circuit, the effective resistance in the circuit becomes very small, which leads to the flow of a large current through the wires.
Question 63 |
What happens when a large number of appliances are connected in series to the same source of electric power?
Over loading | |
Transcription | |
Power fluctuation | |
Shortage |
Question 63 Explanation:
Over loading happens when a large number of appliances are connected in series to the same source of electric power. This leads to a flow of excess current in the electric circuit. When the amount of current passing through a wire exceeds the maximum permissible limit, the wires get heated to such an extent that a fire may be caused.
Question 64 |
In domestic circuit which wire is used as third wire having a green insulation which is usually connected to the body of the metallic electric?
Tear wire | |
Live wire | |
Sub wire | |
Earth wire |
Question 64 Explanation:
In domestic circuits, a third wire called the earth wire having a green insulation is usually connected to the body of the metallic electric appliance. The other end of the earth wire is connected to a metal tube or a metal electrode, which is buried into the Earth. This wire provides a low resistance path to the electric current. The earth wire sends the current from the body of the appliance to the Earth, whenever a live wire accidentally touches the body of the metallic electric appliance which saves us from electric shock.
Question 65 |
Two bulbs are having the ratings as 60 W, 220 V and 40 W, 220 V respectively. Which one has a greater resistance?
The bulb with 60 W, 220 V rating has a greater resistance | |
The bulb with 40 W, 220 V rating has a greater resistance | |
Both the bulb has the same resistance | |
None of the above |
Question 65 Explanation:
Electric power P = V2/ R
For the same value of V, R is inversely proportional to P.
Therefore, lesser the power, greater the resistance
Hence, the bulb with 40 W, 220 V rating has a greater resistance.
Question 66 |
With the help of which chemical compounds, the manufacturer can produce LED bulbs that radiates red, green, yellow and orange colours?
Cadmium Selenide | |
Silicon Carbide | |
Gallium Arsenide | |
Aluminium Nitride |
Question 66 Explanation:
An LED bulb is a semiconductor device that emits visible light when an electric current pass through it. The colour of the emitted light will depend on the type of materials used. With the help of the chemical compounds like Gallium Arsenide and Gallium Phosphide, the manufacturer can produce LED bulbs that radiates red, green, yellow and orange colours.
Question 67 |
Calculate the current and the resistance of a 100 W, 200 V electric bulb in an electric circuit?
I = 2 A; R = 100 Ω | |
I = 1.5 A; R = 300 Ω | |
I = 1 A; R = 200 Ω | |
I = 0.5 A; R = 400 Ω |
Question 67 Explanation:
Power P = 100 W and Voltage V = 200 V Power P = V I
So, Current, I = P / V = 100 / 200 = 0.5 A
Resistance, R = V / I = 200 / 0.5 = 400 Ω
Question 68 |
Which among the following is incorrect about LED bulb?
- As there is one large filament, there is less loss of energy in the form of heat. It is cooler than the incandescent bulb. It is not harmful to the environment. A wide range of colours is possible here.
- In comparison with the fluorescent light, the LED bulbs have significantly low power requirement. It is cost-efficient and energy efficient. Mercury and other toxic materials are not required.
Only 1 | |
Only 2 | |
Both 1 and 2 | |
None of the above |
Question 68 Explanation:
As there is no filament, there is no loss of energy in the form of heat. It is cooler than the incandescent bulb. It is not harmful to the environment. A wide range of colours is possible here.
Question 69 |
In the circuit diagram given below, three resistors R1, R2 and R3 of 5 Ω, 10 Ω and 20 Ω respectively are connected as shown. Calculate:
- Current through each resistor ii) Total current in the circuit iii) Total resistance in the circuit
R1 = 2A, R2 = 1A, R3 = 0.5A; I = 3.5A; RP = 2.857 Ω | |
R1 = 0.5A, R2 = 1A, R3 = 2A; I = 3.5A; RP = 2.857 Ω | |
R1 = 1.5A, R2 = 2A, R3 = 3A; I = 6.5A; RP = 2.857 Ω | |
R1 = 2A, R2 = 3A, R3 = 1A; I = 6A; RP = 2.857 Ω |
Question 69 Explanation:
Since the resistors are connected in parallel, the potential difference across each resistor is same (i.e., V=10V) Therefore, the current through R1 is I1 = V / R1 = 10 / 5 = 2A;
R2 = 10 / 10 = 1A;
R3 = 10 / 20 = 0.5A.
Total current in the circuit, I = I1 + I2 + I3 = = 2 + 1 + 0.5 = 3.5 A
Total resistance in the circuit 1 / RP = (1/R1) + (1/R2) + (1/R3) = (1/5) + (1/10) + (1/20)
= (4+2+1)/20 = 7/20. Hence, Rp = 20/7 = 2.857 Ω.
Question 70 |
LEDs emitting which colour light are used in monochrome TV?
Blue | |
Red | |
Black | |
White |
Question 70 Explanation:
LED Television is one of the most important applications of Light Emitting Diodes. An LED TV is actually an LCD TV (Liquid Crystal Display) with LED display. An LED display uses LEDs for backlight and an array of LEDs act as pixels. LEDs emitting white light are used in monochrome (black and white) TV.
Question 71 |
Which among the following colour LEDs are not used in colour television?
Red | |
Green | |
Violet | |
Blue |
Question 71 Explanation:
Red, Green and Blue (RGB) LEDs are used in colour television
Question 72 |
Which is the display device used to give an output in the form of numbers or text?
Seven Segment Display | |
Eight Segment Display | |
Nine Segment Display | |
Ten Segment Display |
Question 72 Explanation:
A ‘Seven Segment Display’ is the display device used to give an output in the form of numbers or text. It is used in digital meters, digital clocks, micro wave ovens, etc. It consists of 7 segments of LEDs in the form of the digit 8. These seven LEDs are named as a, b, c, d, e, f and g.
Question 73 |
The first LED television screen was developed by whom in 1977?
James P Mitchell | |
Ada Lovelace | |
William A Dubilier | |
Reginald Fessenden |
Question 73 Explanation:
The first LED television screen was developed by James P. Mitchell in 1977. It was a monochromatic display. But, after about three decades, in 2009, SONY introduced the first commercial LED Television.
Question 74 |
Three resistors of 1 Ω, 2 Ω and 4 Ω are connected in parallel in a circuit. If a 1 Ω resistor draws a current of 1 A, find the current through the other two resistors?
I2 = 0.25 A, I3 = 0.5 A | |
I2 = 0.5 A, I3 = 0.25 A | |
I2 = 1 A, I3 = 1.5 A | |
I2 = 0.25 A, I3 = 0.1 A |
Question 74 Explanation:
R1 = 1 Ω, R2 = 2 Ω, R3 = 4 Ω Current I1 = 1 A
The potential difference across the 1 Ω resistor = I1 R1 = 1 × 1 = 1 V
Since, the resistors are connected in parallel in the circuit, the same potential difference will exist across the other resistors also.
So, the current in the 2 Ω resistor, I2 = V / R2 = 1 / 2 = 0.5 A
Similarly, the current in the 4 Ω resistor, I3 = V / R3 = 1 / 4 = 0.25 A.
Question 75 |
Which among the following statement is correct
- A parallel circuit has two or more loops through which current can pass. If the circuit is disconnected in one of the loops, the current can still pass through the other loop(s). The wiring in a house consists of parallel circuits. Consider that three resistors R1, R2 and R3 are connected across two common points A and B.
- The potential difference across each resistance is the same and equal to the potential difference between A and B. This is measured using the voltmeter. The current I arriving at A divides into three branches I1, I2 and I3 passing through R1, R2 and R3 respectively. According to the Ohm’s law, you have, I1 = V / R1, I2 = V / R2, I3 = V / R3. The total current through the circuit is given by I = I1 + I2 + I3 = V / R1 + V / R2 + V / R3. --->>>
- Let the effective resistance of the parallel combination of resistors be Rp. Then, I = V / Rp. combining 1 and 2 we have V / Rp = (V / R1)+(V / R2)+(V / R3) → 1 / Rp = (1 / R1)+(1 / R2)+(1 / R3). Thus, when a number of resistors are connected in parallel, the sum of the reciprocals of the individual resistances is equal to the reciprocal of the effective or equivalent resistance. When ‘n’ resistors of equal resistances R are connected in parallel, the equivalent resistance is n / R. Hence, RP = n / R.
Both 1 and 2 | |
Both 1 and 3 | |
Both 2 and 3 | |
All 1, 2 and 3 |
Question 75 Explanation:
Let the effective resistance of the parallel combination of resistors be Rp. Then, I = V / Rp. combining 1 and 2 we have V / Rp = (V / R1)+(V / R2)+(V / R3) → 1 / Rp = (1 / R1)+(1 / R2)+(1 / R3). Thus, when a number of resistors are connected in parallel, the sum of the reciprocals of the individual resistances is equal to the reciprocal of the effective or equivalent resistance. When ‘n’ resistors of equal resistances R are connected in parallel, the equivalent resistance is R / n. Hence, Rp = R / n.
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