Current Electricity
- Two metal wires of identical dimension are connected in series. If σ1 and σ2 are the conductivities of the metal wires respectively, the effective conductivity of the combination is :
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In figure, two metal wires of identical dimension are connected in series
Correct Option: D
In figure, two metal wires of identical dimension are connected in series
- A, B and C are voltmeters of resistance R, 1.5 R and 3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are VA, VB and VC respectively. Then
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Effective resistance of B and C
= RB . RC = 1.5 R × 3R = 4.5R² = R RB + RC 1.5R +3R 4.5R
i.e., equal to resistance of voltmeter A.
In parallel potential difference is same so, VB = VC and in series current is same
So, VA = VB = VCCorrect Option: D
Effective resistance of B and C
= RB . RC = 1.5 R × 3R = 4.5R² = R RB + RC 1.5R +3R 4.5R
i.e., equal to resistance of voltmeter A.
In parallel potential difference is same so, VB = VC and in series current is same
So, VA = VB = VC
- The masses of the three wires of copper are in the ratio of 1 : 3 : 5 and their lengths are in the ratio of 5 : 3 : 1. The ratio of their electrical resistance is
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R = ρ l . But m = πr²ld ∴ πr² = m πr² ld ∴ R = ρl²d ,R1 = ρl1²d ,R2 = ρl2²d m m1 m2 R3 = ρl1²d m3 R1 : R2 : R3 = 25 : 9 : 1 = 125 : 15 :1 1 3 5
Correct Option: D
R = ρ l . But m = πr²ld ∴ πr² = m πr² ld ∴ R = ρl²d ,R1 = ρl1²d ,R2 = ρl2²d m m1 m2 R3 = ρl1²d m3 R1 : R2 : R3 = 25 : 9 : 1 = 125 : 15 :1 1 3 5
- The velocity of charge carriers of current (about 1 amp) in a metal under normal conditions is of the order of
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NA
Correct Option: A
NA
- If a wire of resistance R is melted and recasted to half of its length, then the new resistance of the wire will be
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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, thenR ∝ 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, thenR ∝ 1 (Area of cross-section)²