A cell can be balanced against 100 cm and 110 cm of potentiometer wire, respectively with and without being short-circuited through a resistance of 10 Ω. lts internal resistance is :

1.  1.0 Ω

2.  0.5 Ω

3.  2.0 Ω

4.  zero

In the circuit shown cells, \(A\) and \(B\) have negligible resistance. For \(V_A =12 ~\text{V}\), \(R_1 = 500 ~\Omega \), and $\mathrm{}$\(R = 100 ~\Omega \) the galvanometer \((\text{G}) \) shows no deflection. The value of \(V_B\) is: 
     

1. \(4\) V
2. \(2\) V
3. \(12\) V
4. \(6\) V

A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. $E_0$and resistance $r_1$. 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. $\frac{LE{ }_{0}r}{l{r}_1}$

2. $\frac{E_{0}r}{(r+r_1)}·\frac{l}{L}$

3. $\frac{E_{0}l}{L}$

4. $\frac{L{E}_{0}r}{(r+r_1)1}$

A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is :

1.  230 $\mathrm{\Omega }$

2.  46 $\mathrm{\Omega }$

3.  26 $\mathrm{\Omega }$

4.  13 $\mathrm{\Omega }$

A potentiometer circuit has been set up 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) 9.5, the 'balancing lengths, on the potentiometer wire, are found to be 3m and 2.85 m, respectively. 
The value of internal resistance of the cell is (in ohm) :

1. 0.25

2. 0.95

3. 0.5

4. 0.75

A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn becomes 10I. The value of 'n' is:-

1. 10

2. 11

3. 20

4. 9