MCQ 11 Mark
The unit for expressing electric power is:
- AVolt.
- BJoule.
- CCoulomb.
- ✓Watt.
Answer
View full question & answer→Correct option: D.
Watt.
Watt is the unit of electric power.
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57 questions · 56 auto-graded MCQ + 1 self-marked written.
MCQ 21 Mark
When a $4 \Omega $ resistor is connected across the terminals of a $12V$ battery, the number of coulombs passing through the resistor per second is:
- A$0.3$
- ✓$3$
- C$4$
- D$12$
Answer
View full question & answer→Correct option: B.
$3$
The number of coulombs passing through the resistor is the current passing through it.
$\text{Current}=\frac{\text{Voltage}}{\text{Resistance}}$
$\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=\frac{12}{4}$
$\text{I}={3\text{A}}$
Thus, when a $4 \Omega $ resistor is connected across the terminals of a $12V$ battery,
the number of coulombs passing through the resistor per second will be $3$.
$\text{Current}=\frac{\text{Voltage}}{\text{Resistance}}$
$\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=\frac{12}{4}$
$\text{I}={3\text{A}}$
Thus, when a $4 \Omega $ resistor is connected across the terminals of a $12V$ battery,
the number of coulombs passing through the resistor per second will be $3$.
MCQ 31 Mark
In the circuit shown below:
The potential difference across the $3 \Omega $ resistor is:
The potential difference across the $3 \Omega $ resistor is:
- A$\frac{1}{9}\text{V}$
- B$\frac{1}{2}\text{V}$
- ✓$1V$
- D$2V$
Answer
View full question & answer→Correct option: C.
$1V$
The resistors of $1 \Omega , 2 \Omega $ and $3 \Omega $ are connected in series.
Therefore, the net resistance,
$\text{R} = \text{R}_1 + \text{R}_2 +\text{R}_3$
$\text{R}=1\Omega+2\Omega+3\Omega=6\Omega$
Current in the circuit will be,
$\text{I}=\frac{\text{V}}{\text{R}}$
or $\text{I}=\frac{2}{6}=\frac{1}{3}\text{A}$
Current $=\frac{1}{3}\text{A}$
Therefore, the voltage across the $3I$ resistor,
$V = IR$
or $\text{V}=\frac{1}{3}\times3=1\text{V}$
Therefore, the net resistance,
$\text{R} = \text{R}_1 + \text{R}_2 +\text{R}_3$
$\text{R}=1\Omega+2\Omega+3\Omega=6\Omega$
Current in the circuit will be,
$\text{I}=\frac{\text{V}}{\text{R}}$
or $\text{I}=\frac{2}{6}=\frac{1}{3}\text{A}$
Current $=\frac{1}{3}\text{A}$
Therefore, the voltage across the $3I$ resistor,
$V = IR$
or $\text{V}=\frac{1}{3}\times3=1\text{V}$
MCQ 41 Mark
The current passing through an electric kettle has been doubled. The heat produced will become:
- AHalf.
- BDouble.
- CFour time.
- ✓One$-$fourth.
Answer
View full question & answer→Correct option: D.
One$-$fourth.
Heat produced is directly proportional to the square of the current.
MCQ 51 Mark
A battery and three lamps are connected as shown? Which of the following statements about the currents at $X, Y$ and $Z$ is correct?
- AThe current at $Z$ is greater than that at $Y.$
- ✓The current at $Y$ is greater than that at $Z.$
- CThe current at $X$ equals the current at $Y.$
- DThe current at $X$ equals the current at $Z.$
Answer
View full question & answer→Correct option: B.
The current at $Y$ is greater than that at $Z.$
This is so because $Y$ alone will offer less resistance.
MCQ 61 Mark
Which of the following characteristic is not suitable for a fuse wire?
- AThin and short.
- ✓Thick and short.
- CLow melting point.
- DHigher resistance than rest of wiring.
Answer
View full question & answer→Correct option: B.
Thick and short.
This is so because in this case, the resistance of the wire will be low.
MCQ 71 Mark
The lamps in a house hold circuit are connected in parallel because:
- AThis way they required less current.
- ✓If one lamp fails the others remain lit.
- CThis way they require less power.
- DIf one lamp fails the other also fail.
Answer
View full question & answer→Correct option: B.
If one lamp fails the others remain lit.
This is so because in a parallel circuit, if one device fails, it does not affect the working of the other devices.
MCQ 81 Mark
Keeping the $p.d.$ constant, the resistance of a circuit is halved. The current will become:
- AOne$-$fourth.
- BFour time.
- CHalf.
- ✓Double.
Answer
View full question & answer→Correct option: D.
Double.
As we know from Ohm’s law:
Voltage $=$ Current $\times$ Resistance
$V = IR$
If the voltage is constant, the resistance of the circuit is halved.
That is, it becomes $\frac{\text{R}}{2}$
Current, $\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=\frac{\frac{\text{V}}{\text{R}}}{2}=2\text{I}$
Thus by keeping the $p.d.$ constant, the resistance of a circuit is halved and the current is doubled.
Voltage $=$ Current $\times$ Resistance
$V = IR$
If the voltage is constant, the resistance of the circuit is halved.
That is, it becomes $\frac{\text{R}}{2}$
Current, $\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=\frac{\frac{\text{V}}{\text{R}}}{2}=2\text{I}$
Thus by keeping the $p.d.$ constant, the resistance of a circuit is halved and the current is doubled.
MCQ 91 Mark
Which of the following statements correctly defines a volt?
- AA volt is a joule per ampere.
- ✓A volt is a joule per coulomb.
- C$A$ and $B$ both
- DNone of these
Answer
View full question & answer→Correct option: B.
A volt is a joule per coulomb.
A volt is a joule per coulomb.
MCQ 101 Mark
When the diameter of a wire is doubled, its resistance becomes:
- ADouble.
- BFour times.
- COne$-$half.
- ✓One$-$fourth.
Answer
View full question & answer→Correct option: D.
One$-$fourth.
$\text{R}=\rho\frac{\text{I}}{\text{A}}$
When the diameter is doubled, $d = 2d$
Radius, $r\ ' = 2r$
Area of cross section, $\text{A}'=\pi\text{r}'^2=\pi(2\text{r})=4\pi\text{r}^2=4\text{A}$
The area of cross$-$section will increase by four times.
Then the new resistance, $\text{R}'=\frac{\rho\text{I}}{\text{A}'}$
$\text{R}'=\frac{\rho\text{I}}{4\text{A}}$
$\text{R}'=\frac{\text{R}}{4}$
Thus, the resistance will get reduced by four times.
When the diameter is doubled, $d = 2d$
Radius, $r\ ' = 2r$
Area of cross section, $\text{A}'=\pi\text{r}'^2=\pi(2\text{r})=4\pi\text{r}^2=4\text{A}$
The area of cross$-$section will increase by four times.
Then the new resistance, $\text{R}'=\frac{\rho\text{I}}{\text{A}'}$
$\text{R}'=\frac{\rho\text{I}}{4\text{A}}$
$\text{R}'=\frac{\text{R}}{4}$
Thus, the resistance will get reduced by four times.
MCQ 111 Mark
Which of the following is likely to be the correct wattage for an electric iron used in our homes?
- A$60W.$
- B$250W.$
- ✓$850W.$
- D$2000W.$
Answer
View full question & answer→Correct option: C.
$850W.$
An electric iron may use a power of $850$ watts.
MCQ 121 Mark
A car headlight bulb working on a $12V$ car battery draws a current of $0.5A.$ The resistance of the light bulb is:
- A$0.5\Omega $
- B$6\Omega $
- C$12\Omega $
- ✓$24\Omega $
Answer
View full question & answer→Correct option: D.
$24\Omega $
Voltage, $V = 12V$
Current, $I = 0.5A$
Resistance, $\text{R}=\frac{\text{Voltage}}{\text{Current}}$
$\text{R}=\frac{12}{0.5}$
$\text{R} = 24\Omega $
Thus, if a car's headlight bulb working on a $12V$ car battery draws a current of $0.5A,$
the resistance of the light bulb will be $ 24\Omega $
Current, $I = 0.5A$
Resistance, $\text{R}=\frac{\text{Voltage}}{\text{Current}}$
$\text{R}=\frac{12}{0.5}$
$\text{R} = 24\Omega $
Thus, if a car's headlight bulb working on a $12V$ car battery draws a current of $0.5A,$
the resistance of the light bulb will be $ 24\Omega $
MCQ 131 Mark
The commercial unit of energy is:
- AWatt.
- BWatt$-$hour.
- ✓Kilowatt$-$hour.
- DKilo$-$joule.
Answer
View full question & answer→Correct option: C.
Kilowatt$-$hour.
Kilowatt$-$hour is the commercial unit of electrical energy.
MCQ 141 Mark
One coulomb charge is equivalent to the charge contained in:
- A$2.6 \times 10^{19}$ electrons.
- B$6.2 \times 10^{19}$ electrons.
- C$2.65 \times 10^{18}$ electrons.
- ✓$6.25 \times 10^{18}$ electrons.
Answer
View full question & answer→Correct option: D.
$6.25 \times 10^{18}$ electrons.
The coulomb is the $SI$ unit of electric charge.
It is the charge transported by a constant current of one ampere in one second.
It is equivalent to the charge of approximately $6.25 \times 10^{18}$ protons or electrons.
It is the charge transported by a constant current of one ampere in one second.
It is equivalent to the charge of approximately $6.25 \times 10^{18}$ protons or electrons.
MCQ 151 Mark
A wire of resistance $R_1$ is cut into five equal pieces. These five pieces of wire are then connected in parallel. If the resultant resistance of this combination be $R_2$, then the ratio $\frac{\text{R}_1}{\text{R}_2}$ is:
- A$\frac{1}{25}$
- B$\frac{1}{5}$
- C$5$
- ✓$25$
Answer
View full question & answer→Correct option: D.
$25$
If the resistance wire is cut into five pieces, the resistance of each wire is $\frac{\text{R}}{5}.$
If we connected the pieces in parallel,
we will get the net resistance as $\frac{\text{R}}{25}$
Therefore, the ratio will be $25.$
If we connected the pieces in parallel,
we will get the net resistance as $\frac{\text{R}}{25}$
Therefore, the ratio will be $25.$
MCQ 161 Mark
The resistivity of copper metal depends on only one of the following factors. This factor is:
- ALength.
- BThickness.
- ✓Temperature.
- DArea of cross$-$section.
Answer
View full question & answer→Correct option: C.
Temperature.
The resistivity of copper depends only on temperature.
MCQ 171 Mark
If two resistors of $25 \Omega $ and $15 \Omega $ are joined together in series and then placed in parallel with a $15 \Omega $ resistor, the effective resistance of the combination is:
- A$0.1\Omega$
- B$10\Omega$
- ✓$20\Omega$
- D$40\Omega$
Answer
View full question & answer→Correct option: C.
$20\Omega$
When $25 \Omega $ and $15 \Omega $ are connected in series, then:
Total resistance, $R = 25 + 15 = 40I$
This $40I$ is connected in parallel with the $40I$ resistor.
Therefore, the net resistance $=\frac{40}{2}=20\text{I}$
Total resistance, $R = 25 + 15 = 40I$
This $40I$ is connected in parallel with the $40I$ resistor.
Therefore, the net resistance $=\frac{40}{2}=20\text{I}$
MCQ 181 Mark
The $SI$ unit of energy is:
- ✓Joule.
- BCoulomb.
- CWatt.
- DOhm$-$metre.
Answer
View full question & answer→Correct option: A.
Joule.
Joule is the $SI$ unit of energy.
MCQ 191 Mark
At a given time, a house is supplied with $100A$ at $220V.$ How many $75W, 220V$ light bulb could be switched on in the house at the same time $($if they are all connected in parallel$)?$
- A$93$
- B$193$
- ✓$293$
- D$393$
Answer
View full question & answer→Correct option: C.
$293$
$El$ Total power provided in the house, $P = VI = 100A \times 220V = 22000W$
This is the maximum power that can be drawn in the house.
Total power drawn by $x$ bulbs $= 75 \times W$
Maximum power $x$ bulbs can draw $= 22000W.$
$\Rightarrow75\text{x}=22000$
$\Rightarrow \text{x}=\frac{22000}{75}\approx293$
This is the maximum power that can be drawn in the house.
Total power drawn by $x$ bulbs $= 75 \times W$
Maximum power $x$ bulbs can draw $= 22000W.$
$\Rightarrow75\text{x}=22000$
$\Rightarrow \text{x}=\frac{22000}{75}\approx293$
MCQ 201 Mark
Ohm's law gives a relationship between:
- ACurrent and resistance.
- BResistance and potential difference.
- CPotential difference and electric charge.
- ✓Current and potential difference.
Answer
View full question & answer→Correct option: D.
Current and potential difference.
Ohm's law gives the relationship between current and potential difference.
MCQ 211 Mark
Keeping the potential difference constant, the resistance of a circuit is doubled. The current will become:
- ADouble.
- ✓Half.
- COne$-$fourth.
- DFour times.
Answer
View full question & answer→Correct option: B.
Half.
As we know from Ohm’s law:
Voltage $=$ Current $\times$ Resistance
$V = IR$
If the voltage is constant, the resistance is doubled and the current becomes half.
Voltage $=$ Current $\times$ Resistance
$V = IR$
If the voltage is constant, the resistance is doubled and the current becomes half.
Question 221 Mark
$V _1, V_2$ and $V _3$ are the p.ds. across the $1 \Omega, 2 \Omega$ and $3 \Omega$ resistors in the following diagram, and the current is 5 A .
| $V_1$ | $V_2$ | $V_3$ | |
| (a) | 1.0 | 2.0 | 3.0 |
| (b) | 5.0 | 10.0 | 15 |
| (c) | 5.0 | 2.5 | 1.6 |
| (d) | 4.0 | 3.0 | 2.0 |
Answer
View full question & answer→$V_1=5, V_2=10$ and $V_3=15$
Because V = IR, the net voltage can be obtained by multiplying current with resistance.
Because V = IR, the net voltage can be obtained by multiplying current with resistance.
MCQ 231 Mark
The $p.d.$ across a $ 3\Omega $ resistor is $6V.$ The current flowing in the resistor will be:
- A$\frac{1}{2}\text{A}$
- B$1A.$
- ✓$2A.$
- D$6A.$
Answer
View full question & answer→Correct option: C.
$2A.$
If the $p.d.$ across a $3\Omega$ resistor is $6V,$ the current flowing in the resistor will be $2A$ as current $(I)$ is given by the equation, $\text{I}=\frac{\text{V} (\text{Voltage})}{\text{R}(\text{Resistance})}.$
or $\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=6\ \frac{\text{V}}{3}\ \Omega $
$\text{I} = 2\text{A.}$
or $\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=6\ \frac{\text{V}}{3}\ \Omega $
$\text{I} = 2\text{A.}$
MCQ 241 Mark
An electrical appliance has a resistance of $25 \Omega .$ When this electrical appliance is connected to a $230V$ supply line, the current passing through it will be:
- A$0.92A.$
- B$2.9A.$
- ✓$9.2A.$
- D$92A.$
Answer
View full question & answer→Correct option: C.
$9.2A.$
Resistance $=25\Omega $
Voltage $= 230V$
$\text{Current}=\frac{\text{Voltage}}{\text{Resistance}}$
$\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=\frac{230}{25}$
$\text{I}=9.2\text{A}$
Thus, if an electrical appliance has a resistance of $25\Omega $ and when that electrical appliance is connected to a $230V$ supply line, the current passing through it will be $9.2A.$
Voltage $= 230V$
$\text{Current}=\frac{\text{Voltage}}{\text{Resistance}}$
$\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=\frac{230}{25}$
$\text{I}=9.2\text{A}$
Thus, if an electrical appliance has a resistance of $25\Omega $ and when that electrical appliance is connected to a $230V$ supply line, the current passing through it will be $9.2A.$
MCQ 251 Mark
The $SI$ unit of energy is:
- ✓Joule.
- BCoulomb.
- CWatt.
- DOhm$-$metre.
Answer
View full question & answer→Correct option: A.
Joule.
Joule is the $SI$ unit of energy.
MCQ 261 Mark
The device used for measuring potential difference is known as:
- APotentiometer.
- BAmmeter.
- CGalvanometer.
- ✓Voltmeter.
Answer
View full question & answer→Correct option: D.
Voltmeter.
A voltage meter is an instrument used for measuring the potential difference, or voltage, between two points in an electrical or electronic circuit.
MCQ 271 Mark
An electric heater is rated at $2 \ \ce{KW}.$ Electrical energy costs $Rs. 4$ per $\ce{KWh}.$ What is the cost of using the heater for $3$ hours?
- A$Rs. 12.$
- ✓$Rs. 24.$
- C$Rs. 36.$
- D$Rs. 48.$
Answer
View full question & answer→Correct option: B.
$Rs. 24.$
Electrical energy consumed in three hours $= P \times t = 2 \times 3 = 6$
Unit cost $= Rs. 4 =$ per
Therefore, the cost of energy used for three hours $= 4 \times 6 = Rs. 24$
Unit cost $= Rs. 4 =$ per
Therefore, the cost of energy used for three hours $= 4 \times 6 = Rs. 24$
MCQ 281 Mark
If the current flowing through a fixed resistor is halved, the heat produced in it will become:
- ADouble.
- BOne$-$half.
- ✓One$-$fourth.
- DFour times.
Answer
View full question & answer→Correct option: C.
One$-$fourth.
This is so because heat produced is directly proportional to the square of the current passing through the resistor.
MCQ 291 Mark
If the area of cross$-$section of a resistance wire is halved, then its resistance becomes:
- AOne$-$half.
- ✓$2$ times.
- COne$-$fourth.
- D$4$ times.
Answer
View full question & answer→Correct option: B.
$2$ times.
The resistance of a conductor is inversely proportional to its area of cross$-$section, i.e. $R \ \alpha \ 1/ A.$
So, when the area of cross$-$section of a resistance wire is halved, its resistance will increase by two times.
So, when the area of cross$-$section of a resistance wire is halved, its resistance will increase by two times.
MCQ 301 Mark
The figure given below shows three resistors? Their combined resistance is:
- A$1\frac{5}{7}\Omega$
- B$14\Omega$
- C$6\frac{2}{3}\Omega$
- ✓$7\frac{1}{2}\Omega$
Answer
View full question & answer→Correct option: D.
$7\frac{1}{2}\Omega$
The resistors of $6I$ and $2I$ are connected in parallel.
$\therefore\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$ Here,
$\text{R}_1=6\Omega$
$\text{R}_2=2\Omega\ \frac{1}{\ \text{R}}=\frac{1}{6}+\frac{1}{2}\frac{1}{\text{R}}=\frac{4}{6}\ \text{R}=\frac{6}{4}$
This arrangement is further connected in series with the $6I$ resistor.
$\therefore$ Net resistance $=\frac{6}{4}+6=7\frac{1}{2}\Omega$
$\therefore\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$ Here,
$\text{R}_1=6\Omega$
$\text{R}_2=2\Omega\ \frac{1}{\ \text{R}}=\frac{1}{6}+\frac{1}{2}\frac{1}{\text{R}}=\frac{4}{6}\ \text{R}=\frac{6}{4}$
This arrangement is further connected in series with the $6I$ resistor.
$\therefore$ Net resistance $=\frac{6}{4}+6=7\frac{1}{2}\Omega$
MCQ 311 Mark
If the potential difference between the ends of a fixed resistor is halved, the electric power will become:
- ADouble.
- BHalf.
- CFour times.
- ✓One$-$fourth.
Answer
View full question & answer→Correct option: D.
One$-$fourth.
Since $\text{P}=\frac{\text{V}^2}{\text{R}}, $ when the potential difference is halved, the power becomes one$-$fourth.
MCQ 321 Mark
The substance having infinitely high electrical resistance is called:
- AConductor.
- BResistor.
- CSuperconductor.
- ✓Insulator.
Answer
View full question & answer→Correct option: D.
Insulator.
A substance having infinitely high electrical resistance is called an insulator.
MCQ 331 Mark
When the area of cross$-$section of a conductor is doubled, its resistance becomes:
- ADouble.
- ✓Half.
- CFour time.
- DOne$-$fourth.
Answer
View full question & answer→Correct option: B.
Half.
We know that the resistance of a conductor is given by:
$\text{R}=\rho\frac{\text{I}}{\text{A}}$
where $\rho=\text{resistivity}$
$l =$ length of the conductor
$A =$ area of the cross$-$section of the conductor
Let the new resistance be $R\ '$ when the area of cross$-$section of the conductor is doubled
$\text{R}'=\rho\frac{\text{I}}{2\text{A}}$
$\text{R}=\rho\frac{\text{I}}{\text{A}}$
where $\rho=\text{resistivity}$
$l =$ length of the conductor
$A =$ area of the cross$-$section of the conductor
Let the new resistance be $R\ '$ when the area of cross$-$section of the conductor is doubled
$\text{R}'=\rho\frac{\text{I}}{2\text{A}}$
MCQ 341 Mark
An electric fuse works on the:
- AChemical effect of current.
- BMagnetic effect of current.
- CLighting effect of current.
- ✓Heating effect of current.
Answer
View full question & answer→Correct option: D.
Heating effect of current.
An electric fuse works on the heating effect of current.
MCQ 351 Mark
The heat produced by passing an electric current through a fixed resistor is proportional to the square of:
- AMagnitude of resistance of the resistor.
- BTemperature of the resistor.
- ✓Magnitude of current.
- DTime for which current is passed.
Answer
View full question & answer→Correct option: C.
Magnitude of current.
We know that:
$H = I^2Rt$
It shows that the heat produced is proportional to the square of the current.
$H = I^2Rt$
It shows that the heat produced is proportional to the square of the current.
MCQ 361 Mark
How much energy does a $100W$ electric bulb transfer in $1$ minute?
- A$100J.$
- ✓$600J.$
- C$3600J.$
- D$6000J.$
Answer
View full question & answer→Correct option: B.
$600J.$
Energy transferred, $E =$ power $\times$ time $= 100W \times 60s = 6000J$
MCQ 371 Mark
The other name of potential difference is:
- AAmpereage.
- BWattage.
- ✓Voltage.
- DPotential energy.
Answer
View full question & answer→Correct option: C.
Voltage.
The term voltage came from volt, which is the $SI$ unit of potential difference.
MCQ 381 Mark
Which of the following is the most likely temperature of the filament of an electric light bulb when it is working on the normal $220V$ supply line?
- A$500^\circ C$
- B$1500^\circ C$
- ✓$2500^\circ C$
- D$4500^\circ C $
Answer
View full question & answer→Correct option: C.
$2500^\circ C$
It is the temperature of an electric bulb filament when it glows.
MCQ 391 Mark
The resistivity of a certain material is $0.6 \Omega \ \text{m}. $ The material is most likely to be:
- AAn insulator.
- BA superconductor.
- CA conductor.
- ✓A semiconductor.
Answer
View full question & answer→Correct option: D.
A semiconductor.
The resistivity of a certain material is $0.6 \Omega \ \text{m}. $ The material is most likely to be a semiconductor because it has moderate resistivity.
MCQ 401 Mark
The diagram below shows part of a circuit:
If this arrangement of three resistors was to be replaced by a single resistor, its resistance should be:
If this arrangement of three resistors was to be replaced by a single resistor, its resistance should be:
- ✓$9\Omega$
- B$4\Omega$
- C$6\Omega$
- D$18\Omega$
Answer
View full question & answer→Correct option: A.
$9\Omega$
The two resistors of $6I$ are connected in parallel with each other.
So, their net resistance $3I$ is connected in series with a resistance of $6I.$
So, the net resistance of the complete arrangement is $9 \Omega .$
So, their net resistance $3I$ is connected in series with a resistance of $6I.$
So, the net resistance of the complete arrangement is $9 \Omega .$
MCQ 411 Mark
In a filament type light bulb, most of the electric power consumed appears as:
- AVisible light.
- ✓Infra$-$red$-$rays.
- CUltraviolet rays.
- DFluorescent light.
Answer
View full question & answer→Correct option: B.
Infra$-$red$-$rays.
This is so because of the heat produced when a filament type light bulb glows.
MCQ 421 Mark
When an electric lamp is connected to $12V$ battery, it draws a current of $0.5A.$ The Power of the lamp is:
- A$0.5W.$
- ✓$6W.$
- C$12W.$
- D$24W.$
Answer
View full question & answer→Correct option: B.
$6W.$
Power, $P = VI = 12V \times 0.5A = 6W$
MCQ 431 Mark
Which statement$/$ statement is$/$ are correct?
$1.$ An ammeter is connected in series in a circuit and a voltmeter is connected in parallel.
$2.$ An ammeter has a high resistance.
$3.$ A voltmeter has a low resistance.
$1.$ An ammeter is connected in series in a circuit and a voltmeter is connected in parallel.
$2.$ An ammeter has a high resistance.
$3.$ A voltmeter has a low resistance.
- A$1, 2, 3$
- B$1, 2$
- C$2, 3.$
- ✓$1.$
Answer
View full question & answer→Correct option: D.
$1.$
An ammeter is connected in series in a circuit and voltmeter is connected in parallel.
MCQ 441 Mark
The unit for measuring potential difference is:
- AWatt.
- B$\ce{Ohm}.$
- ✓Volt.
- D$\ce{KWh}.$
Answer
View full question & answer→Correct option: C.
Volt.
It is the $SI$ unit of potential difference.
MCQ 451 Mark
If the current through a floodlamp is $5A,$ what charge passed in $10$ seconds?
- A$0.5C.$
- B$2C.$
- C$5C.$
- ✓$50C.$
Answer
View full question & answer→Correct option: D.
$50C.$
Here, current, $I = 5A$
Time, $t = 10s$
So, using the relation, $\text{I} =\frac{\text{Q}}{\text{t}}$
$Q = I \times t$
Charge, $Q = 5 \times 10 = 50C$
Thus, a charge of $50C$ is passed in $10s.$
Time, $t = 10s$
So, using the relation, $\text{I} =\frac{\text{Q}}{\text{t}}$
$Q = I \times t$
Charge, $Q = 5 \times 10 = 50C$
Thus, a charge of $50C$ is passed in $10s.$
MCQ 461 Mark
The work done in moving a unit charge across two points in an electric circuit is a measure of:
- ACurrent.
- ✓Potential difference.
- CResistance.
- DPower.
Answer
View full question & answer→Correct option: B.
Potential difference.
The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.
MCQ 471 Mark
An electric kettle for use on a $230V$ supply is rated $3000W.$ For safe working, the cable connected to it should be able to carry at least:
- A$2A.$
- B$5A.$
- C$10A.$
- ✓$15A.$
Answer
View full question & answer→Correct option: D.
$15A.$
Current drawn by the kettle, $\text{I}=\frac{\text{P}}{\text{V}}=\frac{3000}{230}=13\text{A}$
So the cable should be able to carry more current than this and hence the answer is $15A.$
So the cable should be able to carry more current than this and hence the answer is $15A.$
MCQ 481 Mark
The unit of electrical resistance is:
- AAmpere.
- BVolt.
- CCoulomb.
- ✓$\ce{Ohm}.$
Answer
View full question & answer→Correct option: D.
$\ce{Ohm}.$
The unit of electrical resistance is $\ce{Ohm}.$
MCQ 491 Mark
The elements of electrical heating devices are usually made of:
- ATungsten.
- BBronze.
- ✓Nichrome.
- DArgon.
Answer
View full question & answer→Correct option: C.
Nichrome.
Nichrome has a high resistance.
MCQ 501 Mark
If the diameter of a resistance wire is halved, then its resistance becomes:
- AFour times.
- BHalf.
- ✓One$-$fourth.
- DTwo times.
Answer
View full question & answer→Correct option: C.
One$-$fourth.
Resistance of the wire is given by:
$\text{R}=\rho\frac{\text{I}}{\text{A}}$
When the diameter is halved:
$\text{d}'=\frac{\text{d }}{2}$
Radius:
$\text{r}'=\frac{\text{r}}{2}$
Area of cross$-$section:
$\text{A}'=\pi\text{r}'^2=\pi\Big(\frac{\text{r}}{2}\Big)^2=\frac{\pi\text{r}}{4}=\frac{\text{A}}{4}$
Area of cross$-$section will get reduced by four times.
Then the new resistance:
$\text{R}'=\frac{\rho\text{I}}{\text{A}}$
$\text{R}'=\frac{4\rho\text{I}}{\text{A}}$
$\text{R}'=4\text{R}$
Thus, the resistance will increase by four times.
$\text{R}=\rho\frac{\text{I}}{\text{A}}$
When the diameter is halved:
$\text{d}'=\frac{\text{d }}{2}$
Radius:
$\text{r}'=\frac{\text{r}}{2}$
Area of cross$-$section:
$\text{A}'=\pi\text{r}'^2=\pi\Big(\frac{\text{r}}{2}\Big)^2=\frac{\pi\text{r}}{4}=\frac{\text{A}}{4}$
Area of cross$-$section will get reduced by four times.
Then the new resistance:
$\text{R}'=\frac{\rho\text{I}}{\text{A}}$
$\text{R}'=\frac{4\rho\text{I}}{\text{A}}$
$\text{R}'=4\text{R}$
Thus, the resistance will increase by four times.
MCQ 511 Mark
Which of the following units could be used to measure electric charge?
- AAmpere.
- BJoule.
- CVolt.
- ✓Coulomb.
Answer
View full question & answer→Correct option: D.
Coulomb.
The coulomb $($symbol$: C)$ is the International System of Units $(SI)$ unit of electric charge. It is the charge $($symbol$: Q$ or $q)$ transported by a constant current of one ampere in one second.
MCQ 521 Mark
Which unit could be used to measure current?
- AWatt.
- BCoulomb.
- CVolt.
- ✓Ampere.
Answer
View full question & answer→Correct option: D.
Ampere.
It is the $SI$ unit of electric current.
MCQ 531 Mark
The resistance of a wire of length $300m$ and cross$-$section area $1.0\ mm^2$ made of material of resistivity $1.0\times10−7 Ω\ \text{m}$ is:
- A$2 Ω$
- B$3 Ω$
- C$20 Ω$
- ✓$30 Ω$
Answer
View full question & answer→Correct option: D.
$30 Ω$
Resistance, $\text{R}=\frac{\rho\text{I}}{\text{A}}$
Length, $I = 300m$
Cross section area,$A = 1.0mm^2 = 10^{-6} m^2$
Resistivity, $\rho=1.0\times10^{-7}\Omega\ \text{m}$
Resistance, $\text{R}=\frac{10^{-7}\times300}{10^{-6}}$
$\text{R}=30\Omega$
Length, $I = 300m$
Cross section area,$A = 1.0mm^2 = 10^{-6} m^2$
Resistivity, $\rho=1.0\times10^{-7}\Omega\ \text{m}$
Resistance, $\text{R}=\frac{10^{-7}\times300}{10^{-6}}$
$\text{R}=30\Omega$
MCQ 541 Mark
How many joules of electrical energy are transferred per second by a $6V; 0.5A$ lamp?
- ✓$30\ \ce{J/s}$
- B$12\ \ce{J/s}$
- C$0.83\ \ce{J/s}$
- D$3\ \ce{J/s}$
Answer
View full question & answer→Correct option: A.
$30\ \ce{J/s}$
Electrical energy transferred per second, $E = V \times I \times t = 6V \times 0.5A \times 1s \ 6V \times 0.5A \times 1s = 3\ \ce{J/s}$
MCQ 551 Mark
If the resistance of a certain copper wire is $1 \Omega ,$ then the resistance of a similar nichrome wire will be about:
- A$25 \Omega $
- B$30 \Omega $
- ✓$60 \Omega $
- D$45 \Omega $
Answer
View full question & answer→Correct option: C.
$60 \Omega $
If the resistance of a certain copper wire is $1 \Omega ,$ the resistance of a similar nichrome wire will be about $60 \Omega $ because the resistivity of nichrome is $60$ times the resistivity of copper.
MCQ 561 Mark
$f$ the amount of electric charge passing through a conductor in $10 $ minutes is $300C,$ the current flowing is:
- A$30A.$
- B$0.3A.$
- ✓$0.5A.$
- D$5A.$
Answer
View full question & answer→Correct option: C.
$0.5A.$
Here, charge, $Q = 300C$
Time, $t = 10$ minutes $= (10 \times 60)s = 600s$
Then, current, $\text{I}=\frac{\text{Q}}{\text{t}}$
$=\frac{300}{600}=0.5\text{A}.$
Time, $t = 10$ minutes $= (10 \times 60)s = 600s$
Then, current, $\text{I}=\frac{\text{Q}}{\text{t}}$
$=\frac{300}{600}=0.5\text{A}.$
MCQ 571 Mark
The heat produced in a wire of resistance $'x\ '$ when a current $'y\ '$ flows through it in time $'z\ '$ is given by:
- A$x^2 \times y \times z.$
- ✓$x \times z \times y^2.$
- C$y \times z^2 \times x.$
- D$y \times z \times x.$
Answer
View full question & answer→Correct option: B.
$x \times z \times y^2.$
This is so because heat, $H = I^2Rt.$