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 V_A, V_B, and V_C respectively. Then,


1. V_A=V_B=V_C

2. V_A≠V_B=V_C

3. V_A=V_B≠V_C

4. V_A≠V_B≠V_C

A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. $E_o$ 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)$\frac{E_{o}rL}{\left(r_1\right)l}$

(2)$\frac{E_{o}rl}{\left(r+r_1\right)L}$

(3)E_ol/L

(4)$\frac{E_{o}rL}{\left(r+r_1\right)l}$

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 

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 1.6$l_1$. The resistance R, is


(1)10Ω

(2)15Ω

(3)20Ω

(4)25Ω 

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  

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

(1)2Ω

(2)4Ω

(3)8Ω

(4)16Ω 

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

(1)0.2Ω

(2)0.5Ω

(3)0.8Ω

(4)1.0Ω

The resistances of the four arms P, Q,R and S in a Wheatstone's bridge are 10Ω ,30Ω ,30Ω and 90Ω, respectively. The emf and internal resistance of the cell are 7 V and 5 Ω respectively. If the galvanometer resistance is 50Ω, the current drawn from the cell will be

(1) 1.0A

(2) 0.2A

(3) 0.1A

(4) 2.0A

In the circuit shown the cells A and B have negligible resistances. For $V_A=12V, R_1=500\Omega and R=100\Omega$ the galvanometer (g) shown no deflection.The value of $V_B$ is 

(a)4V                                         (b)2V

(c)12V                                       (d)6V

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%