Current Electricity Easy Questions and Answers | Page - 19

Current Electricity

If a negligibly small current is passed through a wire of length 15 m and of resistance 5Ω having uniform cross-section of 6 × 10–7 m2 , then coefficient of resistivity of material, is

1. 1 × 10^–7 Ω-m
1. 2 × 10^–7 Ω-m
1. 3 × 10^–7 Ω-m
1. 4 × 10^–7 Ω-m

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Given : Length of wire (l) = 15m
Area (A) = 6 × 10^–7 m²
Resistance (R) = 5Ω.
We know that resistance of the wire material
R = ρ l
A

⇒ 5 = ρ × 15 = 2.5 × 10^-7ρ
6 × 10^-7

⇒ ρ = 5 = 2.5 × 10^-7Ω - m
2.5 × 10^-7

[where ρ = coefficient of resistivity]

Correct Option: B

Given : Length of wire (l) = 15m
Area (A) = 6 × 10^–7 m²
Resistance (R) = 5Ω.
We know that resistance of the wire material
R = ρ l
A

⇒ 5 = ρ × 15 = 2.5 × 10^-7ρ
6 × 10^-7

⇒ ρ = 5 = 2.5 × 10^-7Ω - m
2.5 × 10^-7

[where ρ = coefficient of resistivity]

If the resistance of a conductor is 5 Ω at 50ºC and 7Ω at 100ºC, then the mean temperature coefficient of resistance (of the material) is

1. 0.001/ºC
1. 0.004/ºC
1. 0.006/ºC
1. 0.008/ºC

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As we know that resistance varies with temperature as
R = R₀ [1 + αt]
Ist Case : 5 = R₀ [1 + α(50)] ....(I)
IInd Case : 7 = R₀ [1 + α(100)] .....(II)
Divided (I) by (II), 5 = 1 + 50 α
7 1 + 100 α

5 + 500α = 7 + 350α
150α = 2 ⇒ α - 2 - 0.001 /°C
150

Correct Option: A

As we know that resistance varies with temperature as
R = R₀ [1 + αt]
Ist Case : 5 = R₀ [1 + α(50)] ....(I)
IInd Case : 7 = R₀ [1 + α(100)] .....(II)
Divided (I) by (II), 5 = 1 + 50 α
7 1 + 100 α

5 + 500α = 7 + 350α
150α = 2 ⇒ α - 2 - 0.001 /°C
150

There are three copper wires of length and cross sectional area (L, A),[2L,(1/2)A] ,[(1/2)L, 2A] . In which case is the resistance minimum?

1. It is the same in all three cases
1. Wire of cross-sectional area 2A
1. Wire of cross-sectional area A
1. Wire of cross-sectional area 1/2 A

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R - ρ L
A

R₁ = ρ L ...(i)
A

R₂ = ρ 2L × 2 ...(ii)
A

R ₃ = ρ L = ρL ...(iii)
2.2A 4A

⇒ R₃ < R₁ < R₂

Correct Option: B

R - ρ L
A

R₁ = ρ L ...(i)
A

R₂ = ρ 2L × 2 ...(ii)
A

R ₃ = ρ L = ρL ...(iii)
2.2A 4A

⇒ R₃ < R₁ < R₂

The resistance of a discharge tube is

1. zero
1. ohmic
1. non-ohmic
1. infinity

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In discharge tube the current is due to flow of positive ions and electrons. Moreover, secondary emission of electrons is also possible. So V-I curve is non-linear; hence resistance is non-ohmic.

Correct Option: C

In discharge tube the current is due to flow of positive ions and electrons. Moreover, secondary emission of electrons is also possible. So V-I curve is non-linear; hence resistance is non-ohmic.

A wire has a resistance of 3.1Ω at 30ºC and a resistance 4.5Ω at 100ºC. The temperature coefficient of resistance of the wire

1. 0.0064ºC^–1
1. 0.0034ºC^–1
1. 0.0025ºC^–1
1. 0.0012ºC^–1

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R₁ = 3.1 Ω at t = 30°C
R₂ = 4.5 Ω at t = 100°C
We have, R = R₀ (1 + αt)
∴ R₁ = R₀ [1 + α (30)]
R₂ = R₀ [1 + α (100)]
⇒ R₁ = 1 + 30α
R₂ 1 + 100α

⇒ 3.1 = 1 + 30α ⇒ α = 0.0064 °C^-1
4.5 1 + 100α

Correct Option: A

R₁ = 3.1 Ω at t = 30°C
R₂ = 4.5 Ω at t = 100°C
We have, R = R₀ (1 + αt)
∴ R₁ = R₀ [1 + α (30)]
R₂ = R₀ [1 + α (100)]
⇒ R₁ = 1 + 30α
R₂ 1 + 100α

⇒ 3.1 = 1 + 30α ⇒ α = 0.0064 °C^-1
4.5 1 + 100α