Electricity Test

Result

Electricity Test - 39

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

An electric heater, operating at $$220\ V$$, boils $$5\ l$$ of water in $$5\ \text{minutes}$$. If it is used on a $$110\ V$$ line, it will boil the same amount of water in:

A
$$10\ \text{minutes}$$

B
$$20\ \text{minutes}$$

C
$$5\ \text{minutes}$$

D
$$1\ \text{minute}$$

Solution

Heat required to boil the water will be same in the two cases.
Heat produced by heater is:
$$H = \cfrac{V^2}{R} t$$
$$V^2 = \cfrac{HR}{t}$$
Hence, other things remaining constant, $$V^2 \propto \dfrac{1}{t}$$

Let the first case be labeled by subscript 1 and the second by subscript 2. Then,

$$\cfrac{t_2}{t_1} = \cfrac{V_1^2}{V_2^2}$$
$$t_2 = \dfrac{220^2 \times 5}{110^2}$$
$$t_2 = 20\ min$$
Question 2
1 / -0

An electric bulb is rated '100 W, 250 V' . What is the resistance of bulb ?

A
2.5 ohm

B
625 ohm

C
25000 ohm

D
1500 ohm

Solution

Generally electrical appliances such as an electric bulb, geyser, heater, etc. are rated with power and voltage. If a bulb is rated 100W - 250V, it means that when a bulb is lit on a 250V supply, it consumes 100 W of electric power. This means that, 100 Joules of electrical energy gets converted into heat and light energy in 1 second.
From this rating we can calculate the resistance of the filament of the appliance while it is glowing and also the current which flows through it.
$$Since\quad P=\frac { { V }^{ 2 } }{ R } ,\quad R=\frac { { V }^{ 2 } }{ P } .$$
For the bulb rated as 100 W - 250 V, the resistance of its filament
$$R=\frac { { V }^{ 2 } }{ P } =\frac { 250^{ 2 } }{ 100 } =625\Omega .$$
The resistance of the filament of a bulb is much less than this value when it is not lit because the resistance of a filament increases with the increasing temperature.
Question 3
1 / -0

A given wire of resistance $$1\,\Omega$$ is stretched to double its length. What will be its new resistance?

A
$$1\, \Omega$$.

B
$$8\, \Omega$$.

C
$$2\, \Omega$$.

D
$$4\, \Omega$$.

Solution

When we stretch a wire to make its length double or whatever the size, to conserve volume, its cross-sectional area must decrease. For a common observation, when we stretch something its length increases but thickness decreases. Now the tricky part is in the question we are being told the new length but not the cross-sectional area, so we have to find it first.
Let's say a wire having length L cross-sectional area A has resistance R. If it is so stretched that its length becomes 2L so new length L' = 2L and when the length is doubled by stretching the cross-sectional area decreases by half.
The resistance of the wire is expressed as $$R=\dfrac { \rho L }{ A } $$where,
rho is the resistivity of the wire,

L is the length of the wire,

A is the area of the cross-section of the wire.

The new resistance is calculated as follows.

$$R=\dfrac { \rho 2L }{ \dfrac { A }{ 2 } } \quad =\dfrac { 4\rho L }{ A } \quad $$.

That is, the new resistance is four times the old resistance.

Hence, the new resistance is $$1\times 4=4\quad ohms$$.
Question 4
1 / -0

Choose the correct option.

A
$$kW=$$ unit of power and $$kWh =$$ unit of energy.

B
$$kW=$$unit of energy and $$kWh =$$ unit of power.

C
$$kW$$ and $$kWh$$ are both of unit of energy.

D
$$kW$$ and $$kWh$$ are both of unit of power.

Solution

Electrical Power is measured in watts ($$W$$), kilo-watts ($$kW$$), mega-watts ($$MW$$), and giga-watts ($$GW$$).

Electrical Energy is measured in kilowatt-hours ($$kWh$$), megawatt-hours, and sometimes (less commonly) in watt-hours or watt-seconds.
Question 5
1 / -0

A wire of 3 ohm resistance and 10 cm length is stretched to 30 cm length. Assuming that it has a uniform cross section, what will be its new resistance?

A
$$27\, \Omega$$.

B
$$9\, \Omega$$.

C
$$3\, \Omega$$.

D
$$30\, \Omega$$.

Solution

On stretching a wire, both its area of cross-section and length changes but its volume remains constant.

$$R_1=3\Omega$$ , $$l_1=10cm$$ , $$l_2=30cm=3l_1$$

And, $$A=\dfrac{V}{l}$$

Now, $$R=\rho\dfrac{l}{A}$$

$$\implies R=\rho\dfrac{l^2}{V}$$

$$\implies R\propto l^2$$

On increasing length to 3 times of initial value.
Hence, $$\dfrac{R_2}{R_1}=\dfrac{l_2 ^2}{l_1 ^2}=9$$

$$\implies R_2=9R_1=9\times 3\Omega$$

$$R_2=27\Omega$$

Answer-(A)
Question 6
1 / -0

A wire of 9 ohm resistance having 30 cm length is tripled on itself. What is its new resistance?

A
$$81\, \Omega$$

B
$$9\, \Omega$$

C
$$3\, \Omega$$

D
$$18\, \Omega$$

Solution

When we stretch a wire to make its length double or whatever the size, to conserve volume, its cross sectional area must decrease. For a common observation, when we stretch something its length increases but thickness decreases. Now the tricky part is in the question we are being told the new length but not the cross sectional area so we have to find it first.
Let's say a wire having length L cross sectional area A have resistance R. If it is so stretched that its length becomes 3L so new length L' = 3L and when length is tripled by stretching the cross sectional area decreases to one-third.
The resistance of the wire is expressed as $$R=\dfrac { \rho L }{ A } $$where,
rho is the resistivity of the wire,
L is the length of the wire,

A is the area of coss section of the wire.

The new resistance is calculated as follows.

$$R=\dfrac { \rho 3L }{ \dfrac { A }{ 3 } } \quad =\dfrac { 9\rho L }{ A } \quad $$.

That is, the new resistance is nine times the old resistance.

Hence, the new resistance is $$9\times 9=81\quad ohms$$.
Question 7
1 / -0

A car bulb connected to a 12 volt battery draw 2 A current when glowing. What is the resistance of the filament of the bulb?

A
$$3\, \Omega$$

B
$$6\, \Omega$$

C
$$9\, \Omega$$

D
$$24\, \Omega$$

Solution

Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Introducing the constant of proportionality, the resistance, one arrives at the usual mathematical equation that describes this relationship, I = V/R.
Rewriting this relation we arrive at $$R=\dfrac { V }{ I } =\dfrac { 12\ V }{ 2\ A } =6\quad ohms$$.
Hence the resistance of the filament of the car bulb is 6 ohms.
There will be less resistance in the car bulb if the current or voltage does not flow in the circuit.
Question 8
1 / -0

The rating of an electrical appliance is $$60 W$$, $$220 V$$. What is the safe limit of current in it, while in use ?

A
$$0.20 A$$

B
$$0.27 A$$

C
$$0.50 A$$

D
$$1.2 A$$

Solution

Given : $$P =60$$ W $$V = 220$$ Volts
Safe limit of current $$I_{max} =\dfrac{P}{V} =\dfrac{60}{220} =0.27$$ A
Question 9
1 / -0

An electric bulb draws $$1.2$$ $$A$$ current at $$6.0\ V$$. The resistance of filament of bulb will be

A
$$5\, \Omega$$

B
$$7.2\, \Omega$$

C
$$6\, \Omega$$

D
$$10\, \Omega$$

Solution

Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Introducing the constant of proportionality, the resistance, one arrives at the usual mathematical equation that describes this relationship, I = V/R.

Rewriting this relation we arrive at $$R=\dfrac { V }{ I } =\dfrac { 6\quad V }{ 1.2\quad A } =5\quad ohms$$.

Hence the resistance of the filament of the bulb is 5 ohms.
Question 10
1 / -0

In our household applications, commercial unit of electricity is used. One unit is equal to:

A
$$1\ kW$$

B
$$1\ kWh$$

C
$$1\ kW/h$$

D
$$1\ Wh$$

Solution

Electrical Energy is measured in kilowatt-hours (kWh). A unit is the primary measurement used by the utility company to determine your bill each month (although demand and power factor are also sometimes used). Unlike power, energy does not change quickly, but instead accumulates gradually.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Electricity Test - 39

Result

Electricity Test - 39

Score

-

Rank

-

SHARING IS CARING

Question 1

$$10\ \text{minutes}$$

$$20\ \text{minutes}$$

$$5\ \text{minutes}$$

$$1\ \text{minute}$$

Solution

Question 2

2.5 ohm

625 ohm

25000 ohm

1500 ohm

Solution

Question 3

$$1\, \Omega$$.

$$8\, \Omega$$.

$$2\, \Omega$$.

$$4\, \Omega$$.

Solution

Question 4

$$kW=$$ unit of power and $$kWh =$$ unit of energy.

$$kW=$$unit of energy and $$kWh =$$ unit of power.

$$kW$$ and $$kWh$$ are both of unit of energy.

$$kW$$ and $$kWh$$ are both of unit of power.

Solution

Question 5

$$27\, \Omega$$.

$$9\, \Omega$$.

$$3\, \Omega$$.

$$30\, \Omega$$.

Solution

Question 6

$$81\, \Omega$$

$$9\, \Omega$$

$$3\, \Omega$$

$$18\, \Omega$$

Solution

Question 7

$$3\, \Omega$$

$$6\, \Omega$$

$$9\, \Omega$$

$$24\, \Omega$$

Solution

Question 8

$$0.20 A$$

$$0.27 A$$

$$0.50 A$$

$$1.2 A$$

Solution

Question 9

$$5\, \Omega$$

$$7.2\, \Omega$$

$$6\, \Omega$$

$$10\, \Omega$$

Solution

Question 10

$$1\ kW$$

$$1\ kWh$$

$$1\ kW/h$$

$$1\ Wh$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.