Current Electricity Easy Questions and Answers | Page - 10

Current Electricity

A 12 cm wire is given a shape of a right angled triangle ABC having sides 3 cm, 4 cm and 5 cm as shown in the figure. The resistance between two ends (AB, BC, CA) of the respective sides are measured one by one by a multi-meter. The resistances will be in the ratio of

1. 3 : 4 : 5
1. 9 : 16 : 25
1. 27 : 32 : 35
1. 21 : 24 : 25

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Resistance is directly proportional to length
1 = 1 = 1 = (4 + 5) + 3
R_AB 3 4 + 5 (3)(4 + 5)

R_AB = 3 × (4 + 5) = 27
3 + (4 + 5) 12

Similarly,
R_BC = 4 × (3 + 5) = 32
4 + (3 + 5) 12

R_AC = 5 × (3 + 4) = 35
5 + (3 + 4) 12

∴ R_AB : R_BC : R_AC = 27 : 32 : 35

Correct Option: C

Resistance is directly proportional to length
1 = 1 = 1 = (4 + 5) + 3
R_AB 3 4 + 5 (3)(4 + 5)

R_AB = 3 × (4 + 5) = 27
3 + (4 + 5) 12

Similarly,
R_BC = 4 × (3 + 5) = 32
4 + (3 + 5) 12

R_AC = 5 × (3 + 4) = 35
5 + (3 + 4) 12

∴ R_AB : R_BC : R_AC = 27 : 32 : 35

The velocity of charge carriers of current (about 1 amp) in a metal under normal conditions is of the order of

1. a fraction of mm/sec
1. velocity of light
1. several thousand metres/second
1. a few hundred metres per second

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NA

Correct Option: A

NA

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 a wire of resistance R is melted and recasted to half of its length, then the new resistance of the wire will be

1. R / 4
1. R / 2
1. R
1. 2R

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Initial resistance (R₁) = R; Initial length l₁is 1and final length (l₂) = 0.5 l . Volume of a wire = l . A. Since the volume of the wire remains the same after recasting, therefore l₁. A₁ = l₂ A₂

We also know that resistance of a wire (R)
R = ρ × l .; R ∝ l
A A

∴ R₁ = l₁ × A₂ = l × 2 = 4
R₂ l₂ A₁ 0.5l

or, R₂ = R₁ = R
4 4

[Alt : When wires are drawn from same volume but with different area of cross-section, then
R ∝ 1
(Area of cross-section)²

Correct Option: A

Initial resistance (R₁) = R; Initial length l₁is 1and final length (l₂) = 0.5 l . Volume of a wire = l . A. Since the volume of the wire remains the same after recasting, therefore l₁. A₁ = l₂ A₂

We also know that resistance of a wire (R)
R = ρ × l .; R ∝ l
A A

∴ R₁ = l₁ × A₂ = l × 2 = 4
R₂ l₂ A₁ 0.5l

or, R₂ = R₁ = R
4 4

[Alt : When wires are drawn from same volume but with different area of cross-section, then
R ∝ 1
(Area of cross-section)²

Two batteries, one of emf 18 volt and internal

resistance 2Ω and the other of emf 12 volt and internal resistance 1Ω, are connected as shown. The voltmeter V will record a reading of

1. 30 volt
1. 18 volt
1. 15 volt
1. 14 volt

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(Since the cells are in parallel).

Correct Option: D

(Since the cells are in parallel).