The charge following through a resistance R varies with time $t as Q= at-b{t}^2,$ where a and b are positive constants. The total heat produced in R is

(1) $\frac{a^{3}R}{3b}$               

(2) $\frac{a^{3}R}{2b}$

(3) $\frac{a^{3}R}{b}$               

(4) $\frac{a^{3}R}{6b}$

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 

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

Across a metallic conductor of non-uniform cross-section, a constant potential difference is applied. The quantity which remains constant along the conductor is :

1. current density
2. current
3. drift velocity
4. electric field

A circuit contains an ammeter, a battery of 30 V and a resistance 40.8Ω all connected in series. If the ammeter has a coil of resistance 480Ω  and a shunt of 20Ω then reading in the ammeter will be :

(1) 0.5A

(2) 0.25A

(3) 2A

(4) 1A

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