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 

1. 4V                                         
2. 2V
3. 12V                                       
4. 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%

A ring is made of a wire having a resistance 

${\mathrm{R}}_0=12\mathrm{\Omega }$. 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 8/3$\mathrm{\Omega }$

1. $\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}=\frac{5}{8}$                   
2. $\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}=\frac{1}{3}$
3. $\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}=\frac{3}{8}$                   
4. $\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}=\frac{1}{2}$

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. $30\mathrm{\Omega }$

A cell having an emf $\mathrm{\epsilon }$ 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

1.    

2. 

3. 

4. 

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

1. 8 V
2. 10 V
3. 2 V
4. 4 V

A current of 2A flows through a 2$\mathrm{\Omega }$ resistor

when connected across a battery. The same

battery supplies a current of 0.5 A when

connected across a 9$\mathrm{\Omega }$ resistor. The internal

resistance of the battery is 

1. $1/3\mathrm{\Omega }$                   

2. $1/4\mathrm{\Omega }$

3. $1\mathrm{\Omega }$                       

4. $0.5\mathrm{\Omega }$

In the circuit in the figure, if the potential at point A is taken to be zero, the potential at point B is :

1. -1V                                          2. +2V

3. -2V                                          4. +1V