Conduct electricity only at low temperatures

Conduct electricity at room temperature

MCQ Questions for Class 10 Physics Electricity Quiz 2

An electric heater is rated

$1500 \ W$ . How much energy does it use in

$10 \ hours$ ?

0%
$25 \: kWh$
0%
$10 \: kWh$
0%
$15 \: kWh$
0%
$12 \: kWh$

Explanation

Correction option: C

Hint: Energy consumed = Power × Time

Explanation for correct option:

Step 1: Find the energy consumes by the electric heater.

Power rating of the heater (P) = 1500

P= 1.5 kW = 1500 W

Time for which the heater has operated (T) = 10 h

Work done = Energy consumed by the heater

∴ Energy consumed = Power × Time

Energy consumed = 1500 × 10

Energy consumed = 15000 W

Hence, the energy consumed by the heater in $\bf{10 h}$ is $\bf{15 kWh}$ .

Electricians wear rubber gloves because:

0%
it is soft
0%
it is water proof
0%
it is an insulator
0%
it is inexpensive

Which is/are the insulators ?

0%
wax
0%
silver
0%
ebonite
0%
gold

A material that lets electricity to pass through it is called a/an:

0%
electrical conductor
0%
electrical insulator
0%
electrical appliance
0%
key

An electric heater converts electrical energy into:

0%
chemical energy
0%
heat energy
0%
muscular energy
0%
none of the above

Which of the following does not belong to the group formed by the others?

0%
Copper coin
0%
Steel spoon
0%
Wooden ruler
0%
Iron nail

When a 1 ohm resistor is connected across the potential difference of 20 V, what will be power dissipation in it?

0%
$200 W$
0%
$20 W$
0%
$400 W$
0%
$10 W$

Explanation

Given :

$V = 20$ volts

$R =1\Omega$

Power dissipated in it

$P =\dfrac{V^2}{R} = \dfrac{20^2}{1} = 400$ W

Find the energy in

$kWh$ consumed in

$10$ hours by four devices of power

$500\ W$ each.

0%
$30 \:kWh$
0%
$25 \:kWh$
0%
$20 \:kWh$
0%
$35 \:kWh$

Explanation

Hint: Calculate the energy consumed by each device.

Correct answer is an option (C).

Step 1: Calculate energy consumed by each device

The power of each device is

$500W$ .

The time for which each device is used is

$10hr$

Energy consumed by each device

$=500\times 10$ =

$5000Wh$

$5kWh$

Step 2: Calculate energy consumed by four devices.

Energy consumed by one device

$=5kWh$

Energy consumed by four devices

$=5\times 4$

$=20kWh$

Hence, the energy consumed by four devices in $kWh$ is $20kWh$ .

The correct answer is an option (C).

A material that does not allow electricity to flow through it is called a/an:

0%
Electrical conductor
0%
Electrical insulator
0%
Electrical appliance
0%
Key

Metals are good conductors because:

0%
outer electrons are strongly bound to the atom
0%
outer electrons are loosely bound to the atom
0%
inner electrons are loosely bound to the atom
0%
protons can detach from the nucleus and conduct electricity

Electric current flows through:

0%
a conductor
0%
an insulator
0%
free space
0%
none of these

Insulators:

0%
Conduct electricity
0%
Do not conduct electricity
0%
Conduct electricity only at low temperatures
0%
Conduct electricity at room temperature

Explanation

$\textbf{Correct option: Option (B).}$

$\textbf{Explanation:}$

Insulators have tightly bound electrons which cannot move freely and hence cannot conduct electricity.

$\textbf{Thus, option (B) is correct.}$

Consider a circuit which contains a resistance free battery and a single resistance

$R$ . If a second resistance is added in parallel with

$R$ .

0%
The voltage across $R$ will decrease
0%
The current through $R$ will increase
0%
The total current drawn from the battery will increases
0%
The power dissipated in $R$ will decrease current drawn from battery will increase

Explanation

From given question, A new resistance has been adding to the older resistance circuit. Because this connection is making in parallel so potential difference will be same in each resistors which is equal to initial value. Both resistance will draw equal current that is equal to

$V/R$ and equal power that is equal to

$V^2/R$ but total current from the battery will be sum of currents in individual resistance that is equal to

$2V/R$ . Hence The total current drawn from the battery will increases.

Hence Option C

A body which cannot conduct electricity is

0%
an insulator
0%
a conductor
0%
can be an insulator or conductor
0%
none of these

When electric current is flown through a conductor, some amount of:

0%
electrical energy is converted into heat energy.
0%
electrical energy is converted into mechanical energy.
0%
mechanical energy is converted into. electrical energy.
0%
heat energy is converted into electrical energy.

Explanation

A conductor has its own resistance. Due to the flowing current, certain heat is produced in the conductor according to the Joule's heating effect i.e

$H = I^2Rt$ .

Thus electrical energy is converted into heat energy when a current flows through the conductor.

When 1 coulomb of charge flows through a conductor in 1 sec, then the current passing through the conductor is?

0%
1Hp
0%
1A
0%
1C
0%
1P

Explanation

Given :

$q =1$ C

$t = 1$ s

$\therefore$ Current

$I = \dfrac{q}{t} = 1$ s

Thus 1 Ampere is the current is defined as flow of 1 coulomb charge through a conductor in 1 second.

In a cell, electrons move from:

0%
positive electrode to negative electrode.
0%
negative electrode to positive electrode.
0%
both A and B.
0%
electrons do not move and only negative charge moves from one place to another place.

If the resistance of a circuit is halved and the potential difference is kept constant, then the current will become

0%
4 times
0%
8 times
0%
double
0%
half

Explanation

Answer is C.

According to Ohm's law, V = IR. That is, I = V/R.
Now, if the resistance is halved and the voltage remains the same, the new value of current becomes,

$I'=\dfrac { V }{ \dfrac { R }{ 2 } } =\dfrac { 2V }{ R } .\quad That\quad is,\quad I'=2(\dfrac { V }{ R } )=2I$ .
Hence, the current will become double.

A current of 300 mA is made to flow through a 6k

$\Omega$ resistor. What is the potential difference across the resistor?

0%
1800V
0%
50V
0%
0.02V
0%
500V

Explanation

Answer is A.

According to Ohm's law, V = IR.
Here, I = 300 mA = 0.3 A and R = 6000 ohms.
Therefore,

$V = 0.3\times 6000 = 1800 V.$
Hence, the p.d. across the resistor is 1800 V.

An electrical appliance has a resistance of 4 ohms. When it is connected to a 12 V battery, the number of coulombs passing through it per second is

0%
1.2
0%
12
0%
0.3
0%
3

Explanation

Answer is D.

The coulomb is defined in terms of the ampere and second:

$1 C = 1 A \times 1 s$ .
In this case, the resistance

$4$ ohms and voltage

$V = 12 V.$

According to Ohm's law,

$V = IR$

That is,

$I = \dfrac{V}{R}$

So,

$I = \dfrac{12}{4} = 3 A$

Hence, 3 coulomb of charge flows through the electrical appliance in 1 second.

A 4

$\Omega$ (or 4 ohm) wire and a 2

$\Omega$ (or 2 ohm)wire are connected in parallel. A current of 3A passes through the wires. How much current passes through the 2

$\Omega$ (or 2 ohm) wire?

0%
2A
0%
4A
0%
6A
0%
8A

Explanation

Answer is A.

Components connected in parallel are connected so the same voltage is applied to each component. In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents through each component. To find the total resistance of all components, the reciprocals of the resistances of each component is added and the reciprocal of the sum is taken. Hence, the Total resistance will always be less than the value of the smallest resistance. That is,

$\frac { 1 }{ { R }_{ total } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } ...\frac { 1 }{ { R }_{ n } } ,\quad R={ { R }_{ total } }^{ -1 }$ .

In this case, the 4 ohms and 2 ohms resistances are connected in parallel. So the total resistance will be

$\frac { 1 }{ 4 } +\frac { 1 }{ 2 } =\frac { 1 }{ 0.75 } =1.33\quad ohms$ .
As it is a parallel connection, the voltage across each of the components is the same, and the total current is the sum of the currents through each component.
Total voltage in the circuit is given as V = IR = 1.33 * 3 = 4 V.
Now, the current across the 2 ohms wire is given as I=V/R = 4/2 = 2A.
Hence, 2A of current passes through the 2 ohms wire.

The potential difference across a

$6\Omega$ resistor is

$12V$ . The current flowing in the resistor will be

0%
4A
0%
2A
0%
0.5A
0%
1A

A radio-set of resistance 20

$\Omega$ and a resistor of 4

$\Omega$ are connected in series with a 6V battery. What is the potential difference across the radio-set?

0%
24V
0%
120V
0%
80V
0%
5V

Explanation

Answer is D.
In series.

${ R }_{ total }={ R }_{ 1 }+{ R }_{ 2 }...{ R }_{ n }$ .
In this case, the total resistance is 20+4 = 24 ohms.
The total current in the circuit is calculated from the relation V = IR, that is, I=V/R.
I = 6 V/24 ohms = 0.25 A.
As it is a series connection, the current through each of the components is the same, the voltage across the circuit is the sum of the voltages across each component.
So, the voltage across the radio set is calculated as follows.
The current is 0.25 A and the resistance of the radio set is 20 ohms.
Therefore, V=IR = 0.25 * 20 = 5 V.
Hence, the potential difference across the radio-set is 5 V.

The potential difference between the terminals of an electric heater is 220 V and the current is 5A. What is the resistance of the heater?

0%
22 ohm
0%
11 ohm
0%
44 ohm
0%
1100 ohm

Two resistors of

$30 \Omega$ and

$20 \Omega$ are joined together in series and then placed in parallel with a

$50 \Omega$ resistor. What is the effective resistance of the combination?

0%
$25 \Omega$
0%
$50 \Omega$
0%
$5 \Omega$
0%
$10 \Omega$

If the current passing through an electric bell is doubled, the heat produced will become

0%
four times
0%
double
0%
one-fourth
0%
half

Explanation

Answer is A.

Energy (

$power\times time$ ) is measured in Joules and by including time (t) in the power formulae, the energy dissipated by a component or circuit can be calculated.
Energy dissipated = Pt or

$VI\quad \times \quad t\quad as\quad P=VI$ .
It can also be written as

${ I }^{ 2 }\times Rt\quad as\quad V=IR.$
Hence, the expression for Power in terms of current is given as

$P={ I }^{ 2 }Rt\quad J$ .
Therefore, if I becomes 2I, then

$P'=(2{ I) }^{ 2 }R\times t\quad =\quad 4({ I }^{ 2 }R\times t)$ .
Hence, if the current passing through an electric bell is doubled, the heat produced will become four times more.

Which one of the following substances has infinitely high electrical resistance?

0%
conductor
0%
insulator
0%
super conductor
0%
resistor

What is the effective resistance between points A and B?

0%
$\dfrac {R}{3}$
0%
$\dfrac {4R}{3}$
0%
$\dfrac {2R}{3}$
0%
$\dfrac {R}{2}$

Explanation

The above figure shows that the three resistors are connected in parallel to each other.

$\therefore$ Equivalent resistance between A and B

$\dfrac{1}{R_{eq}} = \dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}$

$\implies$

$R_{eq} =\dfrac{R}{3}$

The amount of heat produced in a wire depends on its material-

0%
Length
0%
Thickness
0%
Length and thickness
0%
None of these

Explanation

Amount of heat produced in wire =

$I^2R$

Where

$R$ = Resistance of wire

$I$ = Current through wire

As resistance of wire

$R \propto \dfrac{L}{A}$

So heat produced depends on Length as well as thickness.

Electric current is a ________.

0%
Scalar quantity
0%
Vector quantity
0%
Vector quantity when it is due to the flow of positive charges
0%
Vector quantity when it is due to electrons

The current carriers can have:

0%
Positive charge only
0%
Negative charge only
0%
Positive or negative charge
0%
Neither positive nor negative charge

The effective resistance of a circuit containing resistances in parallel is.

0%
equal to the sum of the individual resistances.
0%
smaller than any of the individual resistances.
0%
greater than any of the individual resistances.
0%
sometimes greater and sometimes smaller than the individual resistances.

Explanation

The effective resistance of a circuit containing resistances in parallel is smaller than any of the individual resistances.

For example :

$R_1 = 2\Omega$ and

$R_2 = 3\Omega$

$\therefore$ Effective resistance

$\dfrac{1}{R_p} = \dfrac{1}{R_1}+\dfrac{1}{R_2}$

$\dfrac{1}{R_p} = \dfrac{1}{2}+\dfrac{1}{3}$

$R_p = 1.2 \Omega < 2\Omega$

Power dissipated in the resistor is

$P=$ ________.

0%
$VI$
0%
$I^2R$
0%
$\displaystyle \frac {V^2}{R}$
0%
Any of these

Explanation

Power dissipated by a device that carries current

$I$ when operated at a potential difference

$V$ is given by

$$P=VI$$

If the device has a resistance

$R$ then we can use Ohm's law (

$V=IR$ )

$P=(IR)I=I^2R$

$P=V \left( \dfrac{V}{R} \right)=\dfrac{V^2}{R}$

So all the options are correct. Option (D) is the answer.

The maximum number of 40 W bulbs connected in parallel which can be lighted safely using a 240 V supply with a 5A fuse is.

0%
30
0%
45
0%
8
0%
44

Which among the following is the commercial unit of electrical energy?

0%
Volt
0%
Watt
0%
Ampere
0%
KWh

When a current flows through a conductor, its temperature

0%
increases
0%
decreases
0%
remains same
0%
may increase or decrease

Explanation

As we know,

$Power=I^2R$

Therefore, this power is released in the form of heat in the conductor and this heat increases the temperature of the conductor.

Hence, temperature increases.

A man has five resistors each of value

$\frac {1}{5}\Omega$ . What is the maximum resistance he can obtain by connecting them?

0%
${1} \Omega$
0%
$5 \Omega$
0%
$\dfrac {1}{2} \Omega$
0%
$\dfrac {2}{5} \Omega$

Explanation

Maximum resistance is obtained when all the resistors are connected in series.

Effective resistance

$R_{max} = R+R+R+R+R = 5R$

$\therefore$

$R_{max} = 5\times \dfrac{1}{5} =1\Omega$

A material that does not allow electricity to flow through it is called a/an-

0%
Electrical appliance
0%
Electrical conductor
0%
Electrical Insulator
0%
None of these

Most of the electricity carrying wires near highways do not have a plastic covering like the wires used at homes. But there is no substantial loss of energy because

0%
Air is a bad insulator of electricity
0%
Air is a bad conductor of electricity
0%
Electricity flows very fast near highways
0%
Both B & C

A material that lets electricity to pass through it is called a/an-

0%
Electrical appliances
0%
Electrical conductor
0%
Electrical insulator
0%
None of these

Heat produced in a resistor when current is passed through it depends upon-

0%
Current only
0%
Time only
0%
Current and time
0%
None of these

Explanation

$Power=i^2R$

and

$Energy=Heat=Power\times time=I^2Rt$

Which one of the following substances are used to make electric switches and switch boards?

0%
Conductors
0%
Insulators
0%
Semiconductors
0%
None of the above

Look at the given figure.
It consists of a cell, a bulb with the two terminals X and Y and wires with ends P & Q and S & R. The direction of current will be :

0%
PQRS
0%
SRQP
0%
PRQS
0%
SQRP

Explanation

$P$ is the positive terminal of the cell and

$S$ is the negative terminal of the cell. We know that the current flows from the positive end to the negative end. Therefore, the direction of the current flow will be

$PQRS$ .

Electric heaters used for cooking have the filament or the heating coil on a plate made up of clay because

0%
Clay is a bad conductor of heat
0%
Clay is a bad conductor of electricity
0%
Both A & B
0%
None of these

Which of the following statements is true?

0%
Electricity can be created
0%
Electricity flows in a circuit with gaps
0%
Electricity is the flow of negative charge
0%
All of the above

Electric current can flow through-

0%
Silver
0%
Copper
0%
Iron
0%
All of these

Find odd one out:

Copper, Salt water, Cotton, Human body.

0%
Human body
0%
Cotton
0%
Salt water
0%
Copper

State whether given statement is True or False
Glass is a good insulator.

0%
True
0%
False

Find the odd man out.

Copper, Aluminium, Iron, Wood, Brass.

0%
Wood
0%
Aluminium
0%
Iron
0%
Brass

Our body can conduct electric current.

0%
True
0%
False

CBSE Questions for Class 10 Physics Electricity Quiz 2 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

CBSE Questions for Class 10 Physics Electricity Quiz 2 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

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