Electricity Test

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Electricity Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

You are given $$5 m$$ length of heating wire, it has resistance of $$24 \Omega$$. It is cut into two and connected to $$110$$ volt line individually. The total power for the two half lengths is:

A
500 watts

B
1000 watts

C
2000 watts

D
2500 watts

Solution

Resistance of the full wire $$= 24 ohm$$
Since resistance is directly proportional to length.
$$\therefore$$ Resistance of the half wire $$= \displaystyle\frac{1}{2} \times 24 = 12 ohm$$
Power in one half $$= V. I.$$
and $$V=IR$$ $$I=\dfrac{V}{R}$$ $$V=110volts$$ $$R=12ohms$$
Power in one half= $$= 110 \times \displaystyle\frac{110}{12}$$
$$= 55 \times \displaystyle\frac{55}{3} watts$$
Total power of two half lengths $$= 2 \times 55 \times \displaystyle\frac{55}{3} = \displaystyle\frac{6050}{3}$$
$$= 2000 watts$$ (approximately)
Question 2
1 / -0

A $$100 watt$$ bulb is connected to $$200 volts$$ main. The resistance of the filament of the bulb is

A
$$400 ohm$$

B
$$100 ohm$$

C
$$200 ohm$$

D
$$50 ohm$$

Solution

Power in resistor is given by
$$P=\dfrac{V^2}{R}$$
$$P=100Watt$$ $$V=200Volts$$
$$\therefore R = \displaystyle\frac{{V}^{2}}{P} = \displaystyle\frac{200 \times 200}{100} = 400 ohm$$.
Question 3
1 / -0

Kilowatt hour (kWh) represents the unit of :

A
power

B
impulse

C
momentum

D
energy

Solution

1 kW h = 1 kW ×1 h = 1000 W × 3600 s = 3600000 J = 3.6 $$\ast$$ 10$$^{6}$$ J
The energy used in households, industries and commercial establishments are usually expressed in kilowatt hour.
Question 4
1 / -0

A lamp is marked 60 W, 220 V. If it operates at 200 V, the rate of consumption of energy will _____

A
decrease

B
increase

C
remain unchanged

D
first increase then decrease

Solution

Given power of bulb =60W
Voltage=220V
Power $$=\dfrac{V^2}{R}$$
$$60=\dfrac{220\times220}{R}$$
$$R=(\dfrac{220\times220}{60})\Omega$$
$$R=(\dfrac{220\times22}{6})\Omega$$ (1)
When voltage = 200V, resistance will be the same
$$P=\dfrac{200\times200}{R}$$
From 1,
$$=\dfrac { 200\times 200\times 6 }{ 220\times 22 } $$

$$\dfrac { 200\times 20\times 6 }{ 22\times 22 } $$

$$=49.5 W$$

We see that power gets decreased.
Power means rate of consumption of energy, so rate of consumption of energy decreases.
Option A is correct.
Question 5
1 / -0

The number of joules contained in $$1 kWh$$ is

A
$$36 \times {10}^{2}$$

B
$$36 \times {10}^{3}$$

C
$$36 \times {10}^{4}$$

D
$$3.6 \times {10}^{6}$$

Solution

1 KWh = 1000 $$\ast$$ joule/sec $$\ast$$ 60 $$\ast$$ 60 sec

1KWh = 1000 $$\ast$$ 60 $$\ast$$ 60joue/sec $$\ast$$ sec

1KWh = 1000 $$\ast$$ 60 $$\ast$$ 60 joule

1KWh = 36,00,000 joule = 3.6 $$\ast$$ 10$$^{6}$$
Question 6
1 / -0

What is the effective resistance between points P and Q?

A
$$\dfrac {r}{2}$$

B
$$\dfrac {r}{3}$$

C
$$\dfrac {2r}{3}$$

D
$$\dfrac {4r}{3}$$

Solution

In the given combination, point $$B$$ is equivalent to point $$P$$ as they are connected by a conducting wire. Similarly, point $$A$$ is equivalent to point $$Q$$ for the same reason.
So, the first resistance ($$2r$$) is connected between $$P$$ and $$A$$, i.e. $$P$$ and $$Q$$.
The second resistance ($$2r$$) is connected between $$A$$ and $$B$$, i.e. $$P$$ and $$Q$$.
the third resistance ($$r$$) is connected between $$B$$ and $$Q$$, i.e. $$P$$ and $$Q$$.
Therefore, $$\text{the three resistors are connected in parallel to each other.}$$

$$\therefore$$ Equivalent resistance between P and Q $$\dfrac{1}{R_{eq}} = \dfrac{1}{2r}+\dfrac{1}{2r}+\dfrac{1}{r}$$
$$\implies$$ $$R_{eq} = \dfrac{r}{2}$$
Question 7
1 / -0

An electric heater has a rating of 2 kW, 220 V. The cost of running the heater for 10 hours at the rate of Rs 350 per unit is Rs ____

A
216

B
165

C
209

D
105

Solution

Power is energy consumed per unit time.

Here, energy consumed$$=P=2kW=2000W$$ and time$$=t=10hr=36000s$$

$$P=\dfrac{E}{t}$$

$$\implies E=Pt=2kW\times 10h=20kWh=20units$$

Since $$1unit=1kWh$$

Cost$$=20\times 350=Rs 7000$$
Question 8
1 / -0

Calculate the effect the resistance when two resistors of resistances 10 $$\Omega$$ and 20 $$\Omega$$ are connected in parallel.

A
0.15 $$\Omega$$

B
6.66 $$\Omega$$

C
3.33 $$\Omega$$

D
0.25 $$\Omega$$

Solution

We know that in parallel connection of resistors $$R_1,R_2,R_3...R_n$$ the equivalent resistance is given by-

$$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+...+\dfrac{1}{R_n}$$

For two given resistances-

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

$$\implies R=\dfrac{R_1R_2}{R_1+R_2}$$

$$\implies R=\dfrac{10\times 20}{10+20}=\dfrac{200}{30}$$

$$\implies R=6.66\Omega$$

Answer-(B)
Question 9
1 / -0

A dynamo develops $$0.5 A$$ at $$6 V$$, the power delivered is ______.

A
$$30 W$$

B
$$3 W$$

C
$$1.5 W$$

D
$$2.5 W$$

Solution

Given,
$$i=0.5\ A\\V=6\ volt$$
Power is given by $$P=VI$$

$$\implies P=6\times 0.5$$

$$\implies P=3W$$

Answer-(B)
Question 10
1 / -0

Calculate the power of an electric bulb which consumes $$2400\ J$$ in a minute

A
$$80\ W$$

B
$$120\ W$$

C
$$60\ W$$

D
$$40\ W$$

Solution

Power is the energy consumed per unit of time.
Here, energy consumed,
$$E=2400J$$,
and time
$$t=1\ min=60s$$

$$P=\dfrac{E}{t}$$

$$\Rightarrow P=\dfrac{2400}{60}$$

$$\Rightarrow P=40W$$

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Electricity Test - 47

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Electricity Test - 47

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SHARING IS CARING

Question 1

500 watts

1000 watts

2000 watts

2500 watts

Solution

Question 2

$$400 ohm$$

$$100 ohm$$

$$200 ohm$$

$$50 ohm$$

Solution

Question 3

power

impulse

momentum

energy

Solution

Question 4

decrease

increase

remain unchanged

first increase then decrease

Solution

Question 5

$$36 \times {10}^{2}$$

$$36 \times {10}^{3}$$

$$36 \times {10}^{4}$$

$$3.6 \times {10}^{6}$$

Solution

Question 6

$$\dfrac {r}{2}$$

$$\dfrac {r}{3}$$

$$\dfrac {2r}{3}$$

$$\dfrac {4r}{3}$$

Solution

Question 7

216

165

209

105

Solution

Question 8

0.15 $$\Omega$$

6.66 $$\Omega$$

3.33 $$\Omega$$

0.25 $$\Omega$$

Solution

Question 9

$$30 W$$

$$3 W$$

$$1.5 W$$

$$2.5 W$$

Solution

Question 10

$$80\ W$$

$$120\ W$$

$$60\ W$$

$$40\ W$$

Solution

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