Electricity Test

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Electricity Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

In effective resistance between $$C$$ and $$B$$ is .

A
$$2 \Omega$$

B
$$3 \Omega$$

C
$$1 \Omega$$

D
$$0$$

Solution

Since all resistance between B and C are short circuit as a there is a direct wire available from B to C so all current will flow from that wire. So, Resistance is zero.
Therefore, D is correct option.
Question 2
1 / -0

The electricity tariff in a town is $$Rs.3.00$$ per unit. Calculate the cost of running an $$80 W$$ fan for $$10 h$$ a day for the month of June.

A
$$Rs. 12.00.$$

B
$$Rs. 52.00.$$

C
$$Rs. 72.00.$$

D
$$Rs. 92.00.$$

Solution

Total time for which the fan is used $$= (10 hours) \times (30 days) = 300$$ hours.
Energy consumed $$= (80 W) \times (300$$ hours$$)$$
$$= 24,000 Wh = 24 kWh.$$
But $$1 kWh =1$$ unit.
Therefore, $$24 kWh = 24$$ units
The cost of this power $$= 24$$ units $$\times Rs. 3= Rs. 72.00.$$
Question 3
1 / -0

Electric current cannot flow through

A
Silver

B
Wood

C
Copper

D
None of these

Solution

Wood is an insulator and hence does not allow electric current to pass through it.

Answer-(B)
Question 4
1 / -0

The amount of work done (or, energy spent) by a source of power $$1kW$$ in $$1h$$ time is:

A
$$1J$$

B
$$1Wh$$

C
$$1kWh$$

D
$$1calorie$$

Solution

$$Energy$$ $$spent$$ = $$Power \times time$$
$$E=1kWh \times 1h$$ = $$1kWh$$
$$1kWh$$ is the amount of work done (or, energy spent) by a source of $$1kW$$power in $$1h$$ time.
Question 5
1 / -0

What will be the current drawn by an electric fan of 805 W, when it is connected to a source of 230 V ?

A
$$3$$ A

B
$$4$$ A

C
$$3.5$$ A

D
$$4.5$$ A

Solution

Given:
Power of the electric fan $$P=805\ W$$
Voltage of the source $$V=230\ V$$

We know that, Power $$P=VI$$
$$\therefore Current\ I=\dfrac{P}{V}=\dfrac{805}{230}=3.5\ A$$
Question 6
1 / -0

Convert 1 kWh to SI unit of energy.

A
$$3.6\times 10^6J$$

B
$$3.6\times 10^8J$$

C
$$1.8\times 10^6J$$

D
$$1.8\times 10^8J$$

Solution

We know that,
$$1\ watt = 1\ joule/sec$$ and $$1\ kW=1000\ watt$$
Therefore,
$$1\ kWh$$ $$ = 1000\ watt\times 1\ hour = 1000\ joule/sec \times 3600\ sec$$ $$ = 3.6\times 10^6$$ $$J$$
Question 7
1 / -0

The substance which allows electric current to flow through it is called

A
Conductor

B
Insulator

C
Semiconductor

D
None of these

Solution

The substance which allows electric current to pass through it are called $$conductor.$$

Answer-(A)
Question 8
1 / -0

An experiment is being performed to test the conduction of electricity through a sample liquid as shown in the image, What happens when the current passing through the circuit is too weak?

A
The filament does not get heated sufficiently and the bulb does not glow.

B
The filament gets heated but the bulb does not glow.

C
The filament does not get heated but the bulb glows.

D
The filament gets heated and the bulb glows.

Solution

Electric heating is also used to produce light, as in an electric bulb. Here, the filament must retain as much of the heat generated as is possible, so that it gets very hot and emits light. If the sample liquid is a poor conductor, the current passing through the circuit will be weak. Therefore, the filament does not get heated sufficiently and the bulb does not glow.
Question 9
1 / -0

Two resistors are connected a) in series b) in parallel.
The equivalent resistance in two cases are $$9\Omega$$ and $$2\Omega$$ respectively. Then the resistance of the component resistors are.

A
$$2\Omega$$ and $$7\Omega$$

B
$$3\Omega$$ and $$6\Omega$$

C
$$3\Omega$$ and $$9\Omega$$

D
$$5\Omega$$ and $$4\Omega$$

Solution

$$\textbf{Step 1: Equivalent resistance for series connection}$$
Let the resistances be $$R_1$$ and $$R_2$$.
$$R_{s} = R_1+R_2$$
$$\Rightarrow$$ $$9 = R_1+R_2$$
$$\Rightarrow R_1 = 9-R_2$$    $$......(1)$$

$$\textbf{Step 2: Equivalent resistance for parallel connection} $$
$$R_{p} = \dfrac{R_1R_2}{R_1+R_2}$$
$$\Rightarrow$$ $$2 = \dfrac{R_1 R_2}{R_1 +R_2}$$   $$......(2)$$

$$\textbf{Step 3: Solving Equations}$$
Put the value of $$R_1$$ from $$(1)$$ in equation $$(2)$$
$$2 = \dfrac{(9-R_2) R_2}{9}$$

$$\Rightarrow R_2^2 - 9R_2 +18 = 0 $$
$$\Rightarrow (R_2-6)(R_2 -3)= 0$$
$$\Rightarrow R_2= 6\Omega $$ or $$ R_2 =3\Omega$$

From equation (1)
$$\Rightarrow R_1= 3\Omega $$ or $$ R_1 =6\Omega$$

Hence, option(B) is correct.
Question 10
1 / -0

Equivalent resistance of the above combination will be:

A
$$\displaystyle 5\Omega $$

B
$$\displaystyle 5/4\Omega $$

C
$$\displaystyle 10\Omega $$

D
$$\displaystyle 20\Omega $$

Solution

Equivalent resistance of resistors connected in series
$$R_{eq} = R_1+R_2+R_3+R_4$$

$$\therefore$$ $$R_{eq} = 5+5+5+5 =20\Omega$$

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Electricity Test - 49

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Electricity Test - 49

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Question 1

$$2 \Omega$$

$$3 \Omega$$

$$1 \Omega$$

$$0$$

Solution

Question 2

$$Rs. 12.00.$$

$$Rs. 52.00.$$

$$Rs. 72.00.$$

$$Rs. 92.00.$$

Solution

Question 3

Silver

Wood

Copper

None of these

Solution

Question 4

$$1J$$

$$1Wh$$

$$1kWh$$

$$1calorie$$

Solution

Question 5

$$3$$ A

$$4$$ A

$$3.5$$ A

$$4.5$$ A

Solution

Question 6

$$3.6\times 10^6J$$

$$3.6\times 10^8J$$

$$1.8\times 10^6J$$

$$1.8\times 10^8J$$

Solution

Question 7

Conductor

Insulator

Semiconductor

None of these

Solution

Question 8

The filament does not get heated sufficiently and the bulb does not glow.

The filament gets heated but the bulb does not glow.

The filament does not get heated but the bulb glows.

The filament gets heated and the bulb glows.

Solution

Question 9

$$2\Omega$$ and $$7\Omega$$

$$3\Omega$$ and $$6\Omega$$

$$3\Omega$$ and $$9\Omega$$

$$5\Omega$$ and $$4\Omega$$

Solution

Question 10

$$\displaystyle 5\Omega $$

$$\displaystyle 5/4\Omega $$

$$\displaystyle 10\Omega $$

$$\displaystyle 20\Omega $$

Solution

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