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


(a)10Ω

(b)15Ω

(c)20Ω

(d)25Ω 


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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

(a) 0.25Ω 

(b) 0.95Ω 

(c) 0.5Ω 

(d) 0.75Ω  

Concept Questions :-

Meter bridge and potentiometer
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A wire of resistance 4Ω is stretched to twice its original length. The resistance of a stretched wire would be :

(a)2Ω

(b)4Ω

(c)8Ω

(d)16Ω 

Concept Questions :-

Combination of resistors
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The internal resistance of a 2.1V cell which gives a current of 0.2A through a resistance of 10Ω is :

(a)0.2Ω

(b)0.5Ω

(c)0.8Ω

(d)1.0Ω

Concept Questions :-

Emf and terminal voltage
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A millivoltmeter of 25 mV range is to be converted into an ammeter of 25 A range. The value (in ohm) of necessary shunt will be

(a)0.001                                     (b)0.01

(c)1                                            (d)0.05

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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

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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

(a)20%                                             (b)2.5%

(c)5%                                               (d)10%

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A ring is made of a wire having a resistane 

R0=12Ω. Find the points A and B, as shown

in the figure, at which a current carrying 

conductor should be connected so that the 

resistance R of the sub circuit between these

points is equal to 8/3Ω

1. l1l2=58                    2. l1l2=13

3. l1l2=38                    4. l1l2=12

 

Concept Questions :-

Combination of resistors
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The power dissipated in the circuit shown in

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

(a) 20Ω

(b) 15Ω

(c) 10Ω

(d) 30Ω

Concept Questions :-

Heating effect of current
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A cell having an emf ε internal resistance r

is connected across a variable external 

resistance R. As the resistance R is increased,

the plot of potential difference V across R is given 

by

(a)          

(b) 

(c) 

(d) 

 

Concept Questions :-

Emf and terminal voltage
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Difficulty Level: