The potential difference in each loop is zero.
∴ No current will flow or current in each resistance is Zero.

The resistance of the bulb is $$R = \frac{{{V^2}}}{P} = \frac{{{{\left( {220} \right)}^2}}}{{100}}$$
The power consumed when operated at $$110\,V$$  is
$$P = \frac{{{{\left( {110} \right)}^2}}}{{\frac{{{{\left( {220} \right)}^2}}}{{100}}}} = \frac{{100}}{4} = 25W$$

According to the condition of balancing
$$\frac{{55}}{{20}} = \frac{R}{{80}} \Rightarrow R = 220\Omega $$

Current in the circuit,
$$I = \frac{\varepsilon }{{R + r}}$$
Potential difference across $$R,$$
$$\eqalign{ & V = IR = \left( {\frac{\varepsilon }{{R + r}}} \right)R = \frac{\varepsilon }{{1 + \frac{r}{R}}} \cr & {\text{When}}\,R = 0,V = 0 \cr & R = \infty ,V = \varepsilon \cr} $$

The equivalent circuit is as shown in figure. The resistance of arm $$AOD\left( { = R + R} \right)$$    is in parallel to the resistance $$R$$ of arm $$AD.$$
Their effective resistance

$${R_1} = \frac{{2R \times R}}{{2R + R}} = \frac{2}{3}R$$
The resistance of arms $$AB,BC$$   and $$CD$$  is
$${R_2} = R + \frac{2}{3}R + R = \frac{8}{3}R$$
The resistance $${R_1}$$ and $${R_2}$$ are in parallel. The effective resistance between $$A$$ and $$D$$ is
$${R_3} = \frac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}} = \frac{{\frac{2}{3}R \times \frac{8}{3}R}}{{\frac{2}{3}R + \frac{8}{3}R}} = \frac{8}{{15}}R.$$

$$\eqalign{ & {R_1} = {R_0}\left[ {1 + \alpha \times 100} \right] = 100\,......\left( 1 \right) \cr & {R_2} = {R_0}\left[ {1 + \alpha \times T} \right] = 200\,......\left( 2 \right) \cr} $$
On dividing we get
$$\frac{{200}}{{100}} = \frac{{1 + \alpha T}}{{1 + 100\alpha }} \Rightarrow 2 = \frac{{1 + 0.005T}}{{1 + 100 \times 0.005}} \Rightarrow T = {400^ \circ }C$$
NOTE : We may use this expression as an approximation because the difference in the answers is appreciable. For accurate results one should use $$R = {R_0}{e^{\alpha \Delta T}}$$

There is no change in null point, if the cell and the galvanometer are exchanged in a balanced wheatstone bridge.

On balancing condition $$\frac{{{R_1}}}{{{R_2}}} = \frac{{{R_3}}}{R}$$

Mass of the substance deposited at the cathode is given by
$$m = Zit$$    ($$Z =$$  electrochemical equivalent)
$$\eqalign{ & = Z\left( {\frac{W}{V}} \right)t = 0.367 \times {10^{ - 6}} \times \frac{{100 \times {{10}^3}}}{{125}} \times 60 \cr & = 17.6 \times {10^{ - 3}}kg \cr} $$

Given: $$V = 7\,V$$   and $$r = 5\Omega $$
$$\eqalign{ & {R_{{\text{eq}}}} = \frac{{40 \times 120}}{{40 + 120}}\Omega \cr & I = \frac{V}{R} = \frac{7}{{5 + \frac{{40 \times 120}}{{40 + 120}}}} = 0.2\,A. \cr} $$

As the ring has no resistance, the three resistances of $$3R$$ each are in parallel.
$$\eqalign{ & \Rightarrow \frac{1}{{R'}} = \frac{1}{{3R}} + \frac{1}{{3R}} + \frac{1}{{3R}} = \frac{1}{R} \cr & \Rightarrow R' = R \cr} $$
$$\therefore $$ between point $$A$$ and $$B$$ equivalent resistance $$= R + R = 2R.$$