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 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 

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 

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  

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%

The power dissipated in the circuit shown in

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

(1) $20\mathrm{\Omega }$

(2) $15\mathrm{\Omega }$

(3) $10\mathrm{\Omega }$

(4) $30\mathrm{\Omega }$

If power dissipated in the 9$\mathrm{\Omega }$ resistor in the circuit 

shown is 36 W, the potential difference across 

the 2$\mathrm{\Omega }$ resistor is

(a) 8 V

(b) 10 V

(c) 2 V

(d) 4 V