Electricity Test

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Electricity Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

An electric heater converts electrical energy into:

A
chemical energy

B
heat energy

C
muscular energy

D
none of the above

Solution

When a current passes through an electric heater, it becomes hot. Therefore, in an electric heater, electrical energy is converted into heat energy.
Question 2
1 / -0

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

A
electrical energy is converted into heat energy.

B
electrical energy is converted into mechanical energy.

C
mechanical energy is converted into. electrical energy.

D
heat energy is converted into electrical energy.

Solution

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.
Question 3
1 / -0

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

A
Copper coin

B
Steel spoon

C
Wooden ruler

D
Iron nail

Solution

Wood is a good insulator of electricity. Copper, steel, and iron are metals that are good conductors of electricity. Therefore, the wooden ruler is an electrical insulator, and all the other options in the question are electrical conductors.
Option C is correct.
Question 4
1 / -0

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

A
Iron

B
Tin

C
Glass

D
Steel

Solution

In the given group, iron, tin and steel are electric conductors while glass is an electric insulator.

Hence, glass does not belong to the group formed by others.

So, option C is correct.
Question 5
1 / -0

Which of the following is a conductor?

A
glass

B
tungsten

C
wood

D
plastic

Solution

Answer is B.

We know that metals are good conductors of heat and electricity. As tungsten is a metal, it is a good conductor of electricity. Also, it is very hard and has very high melting point. Because of its high thermal stability and good conductivity properties it has many uses in electronic devices. It is also used as filaments in bulbs and vacuum tubes.
Question 6
1 / -0

In a cell, electrons move from:

A
positive electrode to negative electrode.

B
negative electrode to positive electrode.

C
both A and B.

D
electrons do not move and only negative charge moves from one place to another place.

Solution

In cells, electrons move from negative electrode to positive electrode and an electron has negative charge. So, we can say that negative charge flows from negative electrode to positive electrode. But, by convention flow of charges are measured only through positive charges.When an electron moves from point A to point B,an equal amount of positive charge moves from point B to point A.
So, the direction of charge is from positive electrode to negative electrode. As unlike charges attract each other and like charges repel each other, thus electrons move from negative electrode to positive electrode inside a cell.
Question 7
1 / -0

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

A
4 times

B
8 times

C
double

D
half

Solution

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.
Question 8
1 / -0

A current of 300 mA is made to flow through a 6k$$\Omega$$ resistor. What is the potential difference across the resistor?

A
1800V

B
50V

C
0.02V

D
500V

Solution

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.
Question 9
1 / -0

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

A
1.2

B
12

C
0.3

D
3

Solution

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.
Question 10
1 / -0

Consider a circuit which contains a resistance free battery and a single resistance $$R$$ . If a second resistance is added in parallel with $$R$$ .

A
The voltage across $$R$$ will decrease

B
The current through $$R$$ will increase

C
The total current drawn from the battery will increases

D
The power dissipated in $$R$$ will decrease current drawn from battery will increase

Solution

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

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Re-Attempt Weekly Quiz Competition

Electricity Test - 23

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Electricity Test - 23

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SHARING IS CARING

Question 1

chemical energy

heat energy

muscular energy

none of the above

Solution

Question 2

electrical energy is converted into heat energy.

electrical energy is converted into mechanical energy.

mechanical energy is converted into. electrical energy.

heat energy is converted into electrical energy.

Solution

Question 3

Copper coin

Steel spoon

Wooden ruler

Iron nail

Solution

Question 4

Iron

Tin

Glass

Steel

Solution

Question 5

glass

tungsten

wood

plastic

Solution

Question 6

positive electrode to negative electrode.

negative electrode to positive electrode.

both A and B.

electrons do not move and only negative charge moves from one place to another place.

Solution

Question 7

4 times

8 times

double

half

Solution

Question 8

1800V

50V

0.02V

500V

Solution

Question 9

1.2

12

0.3

3

Solution

Question 10

The voltage across $$R$$ will decrease

The current through $$R$$ will increase

The total current drawn from the battery will increases

The power dissipated in $$R$$ will decrease current drawn from battery will increase

Solution

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