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 |
 66%
From NCERT
NEET - 2016
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The charge following through a resistance R varies with time t as Q= at-bt2, where a and b are positive constants. The total heat produced in R is

(1) a3R3b               

(2) a3R2b

(3) a3Rb               

(4) a3R6b

Subtopic:  Heating Effects of Current |
 53%
From NCERT
NEET - 2016
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A potentiometer wire has a length 4 m and resistance 8 Ω. The resistance that must be connected in series with the wire and an energy source of emf 2 V, so as to get a potential gradient 1 mV per cm of the wire is:
1. 32 Ω
2. 40 Ω
3. 44 Ω
4. 48 Ω 

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:  Combination of Resistors |
 61%
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 |
 75%
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 |
 85%
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 |
 66%
From NCERT
NEET - 2014
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In the circuit shown the cells A and B have negligible resistances. For VA=12V, R1=500Ω and R=100Ω the galvanometer (g) shown no deflection.The value of VB is 

(a)4V                                         (b)2V

(c)12V                                       (d)6V

Subtopic:  Kirchoff's Voltage Law |
 73%
From NCERT
NEET - 2012
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If voltage across a bulb rated 220 V-100 W drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is

(1)20%                                             

(2)2.5%

(3)5%                                               

(4)10%

Subtopic:  Heating Effects of Current |
 78%
From NCERT
NEET - 2012
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The power dissipated in the circuit shown in

the figure is 30 Watt. The value of R is :

(1) 20Ω

(2) 15Ω

(3) 10Ω

(4) 30Ω

Subtopic:  Heating Effects of Current |
 87%
From NCERT
NEET - 2012
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