The resistance of a wire is \(R\) ohm. If it is melted and stretched to \(n\) times its original length, its new resistance will be:

1. \(nR\) 2. \(\frac{R}{n}\)
3. \(n^2R\) 4. \(\frac{R}{n^2}\)

Subtopic:  Derivation of Ohm's Law |
 83%
From NCERT
NEET - 2017
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A potentiometer is an accurate and versatile device to make electrical measurement of EMF because the method involves

1. cells

2. potential gradients 

3. a condition of no current flow through the galvanometer 

4. a combination of cells, galvanometer, and resistances 

Subtopic:  Meter Bridge & Potentiometer |
 71%
From NCERT
NEET - 2017
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The potential difference VA-VB between points A and B in the given figure is if the current is flowing from A to B:

1. - 3 V                   

2. +3 V

3. +6 V                 

4. +9 V

Subtopic:  Kirchoff's Voltage Law |
 51%
From NCERT
NEET - 2016
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A potentiometer wire is 100 cm long and a constant potential is maintained across it. Two cells are connected in series first to support one another and then in the opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf is

1. 5:4             
2. 3:4
3. 3:2             
4. 5:1  

Subtopic:  Meter Bridge & Potentiometer |
 65%
From NCERT
NEET - 2016
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A potentiometer wire has a length 4 m and resistance 8Ω. The resistance that must be connected in series with the wire and an accumulator of emf 2V, so as to get a potential gradient 1mV per cm of the wire is 

(1)32Ω

(2)40Ω

(3)44Ω

(4)48Ω 

Subtopic:  Meter Bridge & Potentiometer |
 67%
From NCERT
NEET - 2015
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A, B, and C are voltmeters of resistance R, 1.5R, 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,


1. VA=VB=VC

2. VA≠VB=VC

3. VA=VB≠VC

4. VA≠VB≠VC

Subtopic:  EMF & Terminal Voltage |
 60%
From NCERT
NEET - 2015
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A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. Eo and a resistance r1. An unknown e.m.f. is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by

(1)EorLr1l

(2)Eorlr+r1L

(3)Eol/L

(4)EorLr+r1l

Subtopic:  Meter Bridge & Potentiometer |
 73%
From NCERT
NEET - 2015
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Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8V and the  average resistance per km is 0.5 Ω. The power loss in the wire is

(1) 19.2W

(2) 19.2kW

(3) 19.2J

(4) 12.2kW 

Subtopic:  Heating Effects of Current |
 82%
From NCERT
NEET - 2014
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The resistances in the two arms of the meter bridge are 5 and R , respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6l1. The resistance R, is


(1)10Ω

(2)15Ω

(3)20Ω

(4)25Ω 


Subtopic:  Kirchoff's Voltage Law |
 71%
From NCERT
NEET - 2014
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A potentiometer circuit has been setup for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance, R, connected across the given cell, has values of

(i)infinity

(ii)9.5Ω 

the 'balancing lengths', on the potentiometer wire are found to be 3m and 2.85m, respectively. The value of internal resistance of the cell is

(1) 0.25Ω 

(2) 0.95Ω 

(3) 0.5Ω 

(4) 0.75Ω  

Subtopic:  Meter Bridge & Potentiometer |
 68%
From NCERT
NEET - 2014
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