Potential gradient = Potential fall per unit length $$ = \frac{V}{l}$$
$$\eqalign{ & = {\text{current}}\, \times {\text{resistance per unit length}} \cr & = i \times \frac{R}{l} \cr} $$
but $$R = \frac{{\rho l}}{A} \Rightarrow \frac{R}{l} = \frac{\rho }{A}$$
where, symbols have their usual meanings.
∴ Potential gradient $$ = i \times \frac{\rho }{A}$$
Here, $$\rho = {10^{ - 7}}\Omega {\text{ - }}m,\,i = 0.1\,A$$
and $$A = {10^{ - 6}}{m^2}$$
Hence, potential gradient $$ = 0.1 \times \frac{{{{10}^{ - 7}}}}{{{{10}^{ - 6}}}} = \frac{{0.1}}{{10}}$$
$$ = 0.01 = {10^{ - 2}}\,V/m$$

Heat supplied $$ = \frac{{{V^2}}}{R} \times t$$
$$\eqalign{ & \Rightarrow \frac{{{t_1}}}{{{R_1}}} = \frac{{{t_2}}}{{{R_2}}} \cr & \Rightarrow \frac{6}{{{R_1}}} = \frac{8}{{{R_2}}} \cr & \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{3}{4} \cr} $$

As the resistance of stretched wire to length $$n$$ times of original length is
$$R' = {n^2}R = {2^2} \times 4 = 4 \times 4 = 16\,\Omega $$
where, $$R =$$  original resistance
$${R'} =$$  final resistance

Total charge flowing through electrolyte,
$$\eqalign{ & q = it \cr & = 50 \times 20 \times 60\,\,\left[ {_{t\, = 20 \times 60\,s}^{{\text{as}}\,i = \,50\,A}} \right] \cr & = 6 \times {10^4}C \cr & \therefore {10^5}C\,{\text{release}} = 9{\text{ }}g{\text{ of }}Al \cr & \therefore 6 \times {10^4}C\,{\text{would}}\,{\text{release}} \cr & = \frac{{9 \times 6 \times {{10}^4}}}{{{{10}^5}}}g \cr & = 5.4g\,{\text{of}}\,Al \cr} $$

Given, charge $$Q = at - b{t^2}\,.......\left( {\text{i}} \right)$$
$$\because $$ We know that current, $$I = \frac{{d\theta }}{{dt}}$$
So, eq. (i) can be written as
$$I = \frac{d}{{dt}}\left( {at - b{t^2}} \right) \Rightarrow I = a - 2bt\,......\left( {{\text{ii}}} \right)$$
For maximum value of $$t,$$ till the current exist is given by
$$\eqalign{ & \Rightarrow a - 2bt = 0 \cr & \therefore t = \frac{a}{{2b}}\,......\left( {{\text{iii}}} \right) \cr} $$
$$\because $$ The total heat produced $$\left( H \right)$$ can be given as
$$\eqalign{ & H = \int_0^t {{I^2}Rdt} \cr & = \int_0^{\frac{a}{{2b}}} {{{\left( {a - 2bt} \right)}^2}R.dt\,\left\{ {\because t = \frac{a}{{2b}}} \right\}} \cr & = \int_0^{\frac{a}{{2b}}} {\left( {{a^2} + 4{b^2}{t^2} - 4abt} \right)Rdt} \cr & H = \left[ {{a^2}t + 4{b^2}\frac{{{t^3}}}{3} - \frac{{4ab{t^2}}}{2}} \right]_0^{\frac{a}{{2b}}}R \cr} $$
Solving above equation, we get
$$ \Rightarrow H = \frac{{{a^3}R}}{{6b}}$$

KEY CONCEPT : $$R = \frac{{{V^2}}}{P}$$
$$\therefore {R_1} = \frac{{{V^2}}}{{100}},{R_2} = \frac{{{V^2}}}{{60}} = {R_3};$$

$$\eqalign{ & {W_1} = \frac{{V_1^2}}{{{R_1}}} = \frac{{{V^2}{R_1}}}{{{{\left( {{R_1} + {R_2}} \right)}^2}}},\,{W_2} = \frac{{V_2^2}}{{{R_2}}} = \frac{{{V^2}{R_2}}}{{{{\left( {{R_1} + {R_2}} \right)}^2}}} \cr & {\text{and }}{W_3} = \frac{{{V^2}}}{{{R_3}}} \cr & {W_3}:{W_2}:{W_1} = \frac{{{{\left( {250} \right)}^2}}}{{{R_3}}}:\frac{{{{\left( {250} \right)}^2}}}{{{{\left( {{R_1} + {R_2}} \right)}^2}}}{R_2}:\frac{{{{\left( {250} \right)}^2}}}{{{{\left( {{R_1} + {R_2}} \right)}^2}}}{R_1} \cr & {\text{or }}{W_3}:{W_2}:{W_1} = \frac{{{{\left( {250} \right)}^2}}}{{{V^2}}} \times 60:\frac{{{{\left( {250} \right)}^2}}}{{{{\left[ {\frac{1}{{100}} + \frac{1}{{60}}} \right]}^2}{V^4}}} \times \frac{{{V^2}}}{{60}}:\frac{{{{\left( {250} \right)}^2}{V^2}}}{{{{\left[ {\frac{1}{{100}} + \frac{1}{{60}}} \right]}^2}{V^4} \times 1000}} \cr & {\text{or }}{W_3}:{W_2}:{W_1} = 60:\frac{{100 \times 100 \times 60 \times 60}}{{160 \times 160 \times 60}}:\frac{{100 \times 100 \times 60 \times 60}}{{160 \times 160 \times 100}} \cr & = 64:25:15 \cr} $$

Potential drop between two cities is $$ = 150 \times 8 = 1200\,V$$
Average resistance of total wire $$ = 0.5 \times 150 = 75\,\Omega $$
So, power loss $$P = \frac{{{V^2}}}{R} = \frac{{1200 \times 1200}}{{75}} = 19200\,W$$
$$ = 19.2\,kW$$

The bridge circuit can be shown as

The balanced condition of bridge circuit is given by
$$\frac{P}{Q} = \frac{3}{4},\frac{R}{S} = \frac{6}{8} = \frac{3}{4} \Rightarrow \frac{P}{Q} = \frac{R}{S}$$
Thus, it is balanced Wheatstone bridge, so potential at $$F$$ is equal to potential at $$H.$$ Therefore, no current will flow through $$7\,\Omega $$  resistance. So, circuit can be redrawn as

$$P$$ and $$Q$$ are in series, so their equivalent resistance
$$ = 3 + 4 = 7\,\Omega $$
$$R$$ and $$S$$ are also in series, so their equivalent resistance $$ = 6 + 8 = 14\,\Omega $$
Now, $$7\,\Omega $$  and $$14\,\Omega $$  resistances are in parallel, so
$${R_{AB}} = \frac{{7 \times 14}}{{7 + 14}} = \frac{{7 \times 14}}{{21}} = \frac{{14}}{3}\Omega $$
NOTE
Normally, in Wheatstone bridge in middle arm galvanometer must be connected. In Wheatstone bridge, cell and galvanometer arms are interchangeable. In both the cases, condition of balanced bridge is
$$\frac{P}{Q} = \frac{R}{S}$$

$$I = \frac{{2\varepsilon }}{{R + {R_1} + {R_2}}}$$
Potential difference across second cell $$ = V = \varepsilon - I{R_2} = 0$$
$$\eqalign{ & \varepsilon - \frac{{2\varepsilon }}{{R + {R_1} + {R_2}}}.{R_2} = 0 \cr & R + {R_1} + {R_2} - 2{R_2} = 0 \cr & R + {R_1} - {R_2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\therefore R = {R_2} - {R_1} \cr} $$

Let $${R_1}$$ and $${R_2}$$ be the resistances of the coils, $$V$$ be the supply voltage, $$H$$ be the heat required to boil the water.
For first coil, $$H = \frac{{{V^2}{t_1}}}{{{R_1}}}\,......\left( {\text{i}} \right)$$
For second coil, $$H = \frac{{{V^2}{t_2}}}{{{R_2}}}\,......\left( {{\text{ii}}} \right)$$
Equating Eqs. (i) and (ii), we get
$$\frac{{{t_1}}}{{{R_1}}} = \frac{{{t_2}}}{{{R_2}}}$$
\[{\rm{i}}{\rm{.e}}{\rm{.}}\,\,\frac{{{R_2}}}{{{R_1}}} = \frac{{{t_2}}}{{{t_1}}} = \frac{{40}}{{10}} = 4\,\,\left[ {\begin{array}{*{20}{l}} {{t_1} = 10\,\min }\\ {{t_2} = 40\,\min } \end{array}} \right]\]
$$ \Rightarrow {R_2} = 4{R_1}\,......\left( {{\text{iii}}} \right)$$
When the two heating coils are in parallel, equivalent resistance is given by
$$\eqalign{ & R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \frac{{{R_1} \times 4{R_1}}}{{{R_1} + 4{R_1}}} = \frac{{4{R_1}}}{5} \cr & {\text{and}}\,\,H = \frac{{{V^2}t}}{R}\,.......\left( {{\text{iv}}} \right) \cr} $$
Comparing Eqs. (i) and (iv), we get
$$\eqalign{ & \frac{{{V^2}{t_1}}}{{{R_1}}} = \frac{{{V^2}t}}{R} \cr & \Rightarrow t = \frac{R}{{{R_1}}} \times {t_1}\,\,\left[ {{t_1} = 10\,\min } \right] \cr & \therefore t = \frac{4}{5} \times 10 = 8\,\min \cr} $$