Total resistance $$R = \left( {0.5\,\Omega /km} \right) \times \left( {150\,km} \right) = 75\,\Omega $$
Total voltage drop $$ = \left( {8\,V/km} \right) \times \left( {150\,km} \right) = 1200\,V$$
Power loss $$ = \frac{{{{\left( {\Delta V} \right)}^2}}}{R} = \frac{{{{\left( {1200} \right)}^2}}}{{75}}W$$
$$\eqalign{ & = 19200\,W \cr & = 19.2\,kW \cr} $$

$$\eqalign{ & {i_1}{R_1} = {i_2}{R_2}\,\,\,\,\,\,\,\,\left( {{\text{same potential difference}}} \right) \cr & \therefore \frac{{{i_1}}}{{{i_2}}} = \frac{{{R_2}}}{{{R_1}}} = \frac{{{\ell _2}}}{{{\ell _1}}} \times \frac{{r_1^2}}{{r_2^2}} = \frac{3}{4} \times \frac{4}{9} \times \frac{1}{3}\left( {{\text{same}}\,\rho } \right) \cr} $$

The resistance of $$6\,\Omega $$  and $$3\,\Omega $$  are in parallel in the given circuit, their equivalent resistance is
$$\eqalign{ & \frac{1}{{{R_1}}} = \frac{1}{6} + \frac{1}{3} = \frac{{1 + 2}}{6} = \frac{1}{2} \cr & {\text{or}}\,\,{R_1} = 2\,\Omega \cr} $$
Again, $${R_1}$$ is in series with $$4\,\Omega $$  resistance, hence
$$R = {R_1} + 4 = 2 + 4 = 6\,\Omega $$
Thus, the total power dissipated in the circuit
$$P = \frac{{{V^2}}}{R}$$
Here, $$V = 18\,V,R = 6\,\Omega $$
Thus, $$P = \frac{{{{\left( {18} \right)}^2}}}{6} = 54\,W$$

$$\eqalign{ & P = Vi = \frac{{{V^2}}}{R} \cr & {R_{{\text{hot}}}} = \frac{{{V^2}}}{P} = \frac{{200 \times 200}}{{100}} = 400\Omega \cr & {R_{{\text{cold}}}} = \frac{{400}}{{10}} = 40\Omega \cr} $$

$$\eqalign{ & \left[ {{\text{Hint}} \Rightarrow {R_t} = {R_0}\left( {1 + \alpha t} \right)} \right] \cr & 5\Omega = {R_0}\left( {1 + \alpha \times 50} \right)\,{\text{and}}\,7\Omega = {R_0}\left( {1 + \alpha \times 100} \right) \cr & {\text{or}}\,\frac{5}{7} = \frac{{1 + 50\alpha }}{{1 + 100\alpha }} \cr & {\text{or}}\,\alpha = \frac{2}{{150}} = 0.0133{/^ \circ }C \cr} $$

We know that,
$${\text{Work done}}\left( W \right) = q\Delta V$$
$$\Delta V$$ is same in all the cases. So, work done will be same in the all cases.

Let initial resistance of wire be $$R,$$ its initial length is $${l_1}$$ and final length is $${l_2}.$$ According to problem $${l_2} = 0.5\,{l_1},$$   volume of the wire is $$= lA.$$  Since, the volume of wire remains the same after recasting, therefore
$$\eqalign{ & {l_1}{A_1} = {l_2}{A_2} \cr & \therefore \frac{{{l_1}}}{{{l_2}}} = \frac{{{A_2}}}{{{A_1}}}\,\,{\text{or}}\,\,\frac{{{l_1}}}{{0.5\,{l_1}}} = \frac{{{A_2}}}{{{A_1}}} \cr & \therefore \frac{{{A_2}}}{{{A_1}}} = 2 \cr} $$
As resistance of wire, $$R \propto \frac{l}{A}$$
$$\eqalign{ & \therefore \frac{{{R_1}}}{{{R_2}}} = \frac{{{l_1}}}{{{l_2}}} \times \frac{{{A_2}}}{{{A_1}}} = \frac{{{l_1}}}{{0.5\,{l_1}}} \times 2 = 4 \cr & {\text{or}}\,\,{R_2} = \frac{{{R_1}}}{4} = \frac{R}{4} \cr} $$
NOTE
As we know the resistance $$R$$ of wire of length $$l$$ and area of cross-section, $$A$$ is given by $$R = \int {\frac{l}{A}} $$

Required resistance to convert a galvanometer into voltmeter of $$30\,V$$  is given by
$$R = \frac{V}{{{i_g}}} - G$$
Symbols have their equal meaning
$$ = \frac{{30}}{{30 \times {{10}^{ - 3}}}} - 100 = 900\,\Omega $$

$$\frac{F}{\ell } = \frac{{{\mu _0}{i_1}{i_2}}}{{2\pi d}} = \frac{{{\mu _0}{i^2}}}{{2\pi d}}$$
(attractive as current is in the same direction)

Resistance of a given conducting wire is given by
$$R = \rho \cdot \frac{l}{A}$$
where, $$\rho $$ is the specific resistance of the material of the conductor.
Here, $$l = 15\,m$$
$$\eqalign{ & A = 6 \times {10^{ - 7}}{m^2} \cr & R = 5\,\Omega ,\rho = ? \cr & \therefore \rho = \frac{{RA}}{l} \cr & = \frac{{5 \times 6 \times {{10}^{ - 7}}}}{{15}} \cr & = 2 \times {10^{ - 7}}\Omega - m \cr} $$