Electricity Test

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Electricity Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

A block of carbon, $$1.0 cm$$ by $$2.0 cm$$ by $$5.0 cm$$, has a resistance of $$0.015 \ \Omega$$ between its two smaller faces. What is the resistivity of carbon?

A
$$1 \times 10 ^{-8} \ \Omega m$$

B
$$2 \times 10 ^{-8} \ \Omega m$$

C
$$4 \times 10 ^{-5} \ \Omega m$$

D
$$6 \times 10 ^{-8} \ \Omega m$$

Solution

$$R = ρ\dfrac { l }{ A } \quad $$

where $$R$$ is resistance $$l$$ is length of wire $$A$$ is the cross-section area.
As the question says the resistance is $$ 0.015Ω $$ along its smaller faces, this implies,

$$l = 5cm=0.05m\quad $$

$$A = 1\times 2{ cm }^{ 2 }=0.0002{ m }^{ 2 }\quad $$

$$R = ρ\dfrac { l }{ A } \quad $$

$$ 0.015=ρ\times \dfrac { 0.05 }{ 0.0002 } \quad $$

So , $$ρ = 6\times { 10 }^{ -5 } \ \Omega m $$
Question 2
1 / -0

A wire of uniform cross-section has a resistance of $$R$$ . If it is drawn to three times the length, but the volume remains constant, what will be its resistance?

A
$$3 R \ \Omega$$

B
$$5 R \ \Omega$$

C
$$7 R \ \Omega$$

D
$$9 R \ \Omega$$

Solution

Wire of uniform cross-section has a resistance of $$R$$.

Let the length of the wire be $$l$$.
Cross sectional Area be $$A$$:

And volume, $$V=Al$$

Now the resistance of a wire can be expressed as: $$R=\rho\dfrac{l}{A}$$
Where resistivity is $$\rho$$.

Now the wire is stretched such that final length ($$l_1$$) becomes three times the initial length (l), but the volume (V) remains constant.

$$Al=3A_1l$$
$$A_1=\dfrac{A}{3}$$

The new resistance will be:
$$R_1=\rho\dfrac{l_1}{A_1}$$

Putting the value of $$l_1=3l$$ and $$A_1$$=$$\dfrac{A}{3}$$

We have, $$R_1=\rho\dfrac{3l}{\dfrac{A}{3}}=9\rho\dfrac{l}{A}$$
$$ R_1=9R$$ ohm
Question 3
1 / -0

Human body is a

A
good conductor of electricity

B
poor conductor of electricity

C
semiconductor

D
good insulator

Solution

Human body is a good conductor of electricity which allow electric current to flow.

Answer-(A)
Question 4
1 / -0

What length of German silver wire, diameter $$0.050 cm$$, is needed to make a $$28 \Omega$$ resistor, if the resistivity of German silver is $$2.2 \times 10^{-7} \Omega m$$?

A
$$20 m $$

B
$$25 m $$

C
$$30 m $$

D
$$35 m $$

Solution

Given :
Resistance R = 28$$\Omega $$
Using formula, $$ R=ρ\dfrac { l }{ A } \quad $$,
where
$$l$$ is length of wire
$$A$$ is the cross-section area and,
$$\rho $$ is resistivity of the wire.

To find length $$l$$:
$$ A=\pi r^2$$
Radius $$r=\dfrac{d}{2}=\dfrac{0.050}{2}cm=0.00025m$$
Area $$A= \pi { (0.00025) }^{ 2 }m^2=625\pi \times { 10 }^{ -10 } m^2 \quad $$

$$ 28=2.2\times { 10 }^{ -7 }\dfrac { (l) }{ A } \quad $$
So,
$$l=\dfrac{28A}{2.2 \times 10^{-7}}$$
$$l=\dfrac{28\times625\pi \times 10^{-3}}{2.2}=24.98m \approx 25m $$
Question 5
1 / -0

A wire of uniform cross-section has a resistance of $$R \ \ \Omega$$. What would be the resistance of a similar wire, made of the same material, but twice as long and of twice the diameter?

A
$$\dfrac{3}{2} R$$

B
$$\dfrac{5}{2} R$$

C
$$\dfrac{1}{2} R$$

D
$$\dfrac{1}{4} R$$

Solution

Wire of uniform cross-section has a resistance of $$R$$.
let the diameter be $$d$$.
Let the length of the wire be $$l$$.
Cross sectional Area be $$A$$:
$$A=\pi\dfrac{d^2}{4}$$
Now the resistance of a wire can be expressed as:
$$R=\rho\dfrac{l}{A}$$
Where resistivity is $$\rho$$.
Now a new wire is taken made of the same material, but twice as long and of twice the diameter.
Let final length be $$l_1$$, diameter be $$d_1$$ and cross section Area be $$A_1$$.
$$A_1=\pi\dfrac{d_1^2}{4}=\pi\dfrac{4d^2}{4}=4A$$
$$R_1=\rho\dfrac{l_1}{A_1}$$

Putting the value of $$l_1=2l$$ and $$A_1=4A$$
We have, $$R_1=\rho\dfrac{2l}{4A}=\rho\dfrac{l}{2A}$$
$$ R_1=\dfrac{R}{2}$$
Question 6
1 / -0

Four bulbs marked $$40W, \ 250V$$ are connected in series with $$250\ V$$ mains, the total power consumed is

A
$$10W$$

B
$$40W$$

C
$$320W$$

D
$$160W$$

Solution

Power of each bulb,
$$P = 40 W$$
Total power consumed when four bulbs are connected in series
$$\dfrac{1}{P_s} = \dfrac{1}{P_1}+$$ $$\dfrac{1}{P_2} + \dfrac{1}{P_3}+\dfrac{1}{P_4}$$
$$\therefore$$ $$\dfrac{1}{P_s} = \dfrac{1}{P}+$$ $$\dfrac{1}{P} + \dfrac{1}{P}+\dfrac{1}{P}$$
We get
$$P_s =\dfrac{P}{4}$$
$$\Rightarrow P_s = \dfrac{40}{4}$$
$$\Rightarrow P_s =10 W $$
Question 7
1 / -0

What is the minimum number of bulbs, each marked $$60W, \ 40V,$$ that can work safely when connected in series with a $$240V$$ mains supply?

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

Resistance of each bulb
$$R = \dfrac{V^2}{P}$$
$$\Rightarrow R = \dfrac{(40)^2}{60}$$
$$\Rightarrow R = 26.67\Omega$$
Maximum current that can flow through the bulb without damaging it
$$I_{max}= \dfrac{P}{V}$$
$$\Rightarrow I_{max} = \dfrac{60}{40}$$
$$\Rightarrow I_{max} = 1.5\ A$$
Thus minimum resistance required in the circuit
$$R_{min} = \dfrac{V_s}{I_{max}}$$

$$\Rightarrow R_{min} = \dfrac{240}{1.5}$$
$$\Rightarrow R_{min} = 160 \Omega$$

Let $$n$$ number of bulbs are connected in series in the circuit.
Equivalent resistance of $$n$$ bulbs
$$R_{eq} = nR = 26.67 n$$
But,
$$26.67 n \leq 160$$
$$\Rightarrow n \leq 6$$
Thus minimum number of bulbs required is $$6$$.
Question 8
1 / -0

Two wire of same metal have same length but their cross-sections are in the ratio $$3:1$$. They are joined in series. The resistance of thick wire is $$10$$. The total resistance of combination will be

A
$$10 \Omega$$

B
$$20 \Omega$$

C
$$40 \Omega$$

D
$$100 \Omega$$

Solution

Resistance of the wire $$R = \dfrac{\rho L}{A}$$
where $$\rho$$ is the resistivity and $$L$$ and $$A$$ is the length and cross-section area of the wire, respectively.
$$\implies$$ $$R\propto \dfrac{1}{A}$$ (for same $$L$$ and $$\rho$$)
Ratio of cross-sectional area of thick wire to thin wire $$\dfrac{A_1}{A_2} = \dfrac{3}{1}$$
Thus ratio of resistance of thin wire to thick wire $$\dfrac{R_2}{R_1} = \dfrac{A_1}{A_2} = \dfrac{3}{1}$$
Or $$\dfrac{R_2}{10}= \dfrac{3}{1}$$
$$\implies$$ $$R_2 = 30\Omega$$
Total resistance of the (series) combination $$R_{eq} = R_1 + R_2$$
$$\implies$$ $$R_{eq} = 10 + 30 = 40 \Omega$$
Question 9
1 / -0

A $$100W, 200V$$ bulb is connected to a $$160V$$ supply. The actual power consumption would be

A
$$185W$$

B
$$100W$$

C
$$54W$$

D
$$64W$$

Solution

Resistance of the bulb $$R = \dfrac{V^2}{P}$$
where $$P = 100 W$$ and $$V = 200$$ V.
$$\therefore$$ $$R = \dfrac{(200)^2}{100} = 400\Omega$$
Actual power consumption $$P' = \dfrac{V_1^2}{R}$$
where $$V_1 = 160$$ V
$$\therefore$$ $$P' = \dfrac{(160)^2}{400} = 64 W$$
Question 10
1 / -0

$$1$$ kilowatt hr $$=$$ __________ joules.

A
$$10.6 \times {10}^{6}$$

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

C
$$30.6 \times {10}^{6}$$

D
$$3.6 \times {10}^{5}$$

Solution

$$kWh$$ is the commercial unit of energy.
$$1\ kWh =1\times 1000\times 3600 = 3.6\times 10^6$$ joules

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Electricity Test - 57

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Electricity Test - 57

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Question 1

$$1 \times 10 ^{-8} \ \Omega m$$

$$2 \times 10 ^{-8} \ \Omega m$$

$$4 \times 10 ^{-5} \ \Omega m$$

$$6 \times 10 ^{-8} \ \Omega m$$

Solution

Question 2

$$3 R \ \Omega$$

$$5 R \ \Omega$$

$$7 R \ \Omega$$

$$9 R \ \Omega$$

Solution

Question 3

good conductor of electricity

poor conductor of electricity

semiconductor

good insulator

Solution

Question 4

$$20 m $$

$$25 m $$

$$30 m $$

$$35 m $$

Solution

Question 5

$$\dfrac{3}{2} R$$

$$\dfrac{5}{2} R$$

$$\dfrac{1}{2} R$$

$$\dfrac{1}{4} R$$

Solution

Question 6

$$10W$$

$$40W$$

$$320W$$

$$160W$$

Solution

Question 7

$$2$$

$$4$$

$$6$$

$$8$$

Solution

Question 8

$$10 \Omega$$

$$20 \Omega$$

$$40 \Omega$$

$$100 \Omega$$

Solution

Question 9

$$185W$$

$$100W$$

$$54W$$

$$64W$$

Solution

Question 10

$$10.6 \times {10}^{6}$$

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

$$30.6 \times {10}^{6}$$

$$3.6 \times {10}^{5}$$

Solution

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