According to maximum power theorem, the power in the circuit is maximum if the value of external resistance is equal to the internal resistance of battery.

Let the resistance of half the turn be $$R.$$ Then in the former case, we have fifteen resistors of resistance $$R$$ connected in parallel, the total resistance being $$\frac{R}{{15}}.$$
In the latter case, we have the same fifteen resistors connected in series, the total resistance being $$15R.$$  Therefore, as a result of unwinding, the resistance of the wire will increase by a factor of 225.

The full scale deflection current
$${i_g} = \frac{{25\,mV}}{G}A$$
where, $$G$$ is the resistance of the meter.
The value of shunt required for converting it into ammeter of range $$25\,A$$  is $$S = \frac{{{i_g}G}}{{i - {i_g}}}$$
$$ \Rightarrow S = {i_g}\frac{G}{i}\,\,\,\left( {{\text{as}}\,i > > {i_g}} \right)$$
So that, $$S \approx \frac{{25\,mV}}{G} \cdot \frac{G}{i} = \frac{{25\,mV}}{{25}} = 0.001\,\Omega $$

In parallel arrangement of $$n$$ rods Equivalent thermal conductivity is given by
$${K_{{\text{eq}}}} = \frac{{{K_1}{A_1} + {K_2}{A_2} + \ldots + {K_n}{A_n}}}{{{A_1} + {A_2} + \ldots + {A_n}}}$$
If rods are of same area, then
$${K_{{\text{eq}}}} = \frac{{{K_1} + {K_2} + \ldots + {K_n}}}{n}$$
Now, in the question, it is not given that rods are of same area. But we can judge that from given diagram.
∴ Equivalent thermal conductivity of the system of two rods
$$ \Rightarrow {K_{{\text{eq}}}} = \frac{{{K_1} + {K_2}}}{2}$$

When wire is bent to form a complete circle, then Total circumference, $$2\pi r = $$  Total length, $$R$$
$$ \Rightarrow r = \frac{R}{{2\pi }}$$
Resistance of each semicircle $$ = \pi r$$
$$ = \frac{{\pi R}}{{2\pi }} = \frac{R}{2}$$

Thus, net resistance in parallel combination of two semicircular resistances
$$R' = \frac{{\frac{R}{2} \times \frac{R}{2}}}{{\frac{R}{2} + \frac{R}{2}}} = \frac{{\frac{{{R^2}}}{4}}}{R} = \frac{R}{4}$$

In the given circuit, resistances $${R_B}$$ and $${R_C}$$ are in series order, so their effective resistance,
$$\eqalign{ & R' = {R_B} + {R_C} \cr & = 6 + 6 = 12\,\Omega \cr} $$
Now, $${R_A}$$ and $${R'}$$ are in parallel order, hence net resistance of the circuit
$$R = \frac{{R' \times {R_A}}}{{R' + {R_A}}} = \frac{{12 \times 3}}{{12 + 3}} = \frac{{36}}{{15}}\Omega $$
The current flowing in the circuit,
$$i = \frac{V}{R} = 4.8 \times \frac{{15}}{{36}} = 2\,A$$

The equivalent resistance between $$Q$$ and $$S$$ is given by

$$\eqalign{ & \frac{1}{{R'}} = \frac{1}{{1.5R}} + \frac{1}{{3R}} = \frac{{2 + 1}}{{3R}} \cr & \Rightarrow R' = R \cr} $$
Now, $${V_{PQ}} = {V_A} = IR$$
Also, $${V_{QS}} = {V_B} = {V_C} = IR$$
Hence, $${V_A} = {V_B} = {V_C}$$

Net resistance of the circuit $$ = {r_1} + {r_2} + R$$
Net emf in series $$ = E + E = 2E$$

Therefore, from Ohm’s law, current in the circuit
$$\eqalign{ & i = \frac{{{\text{Net emf}}}}{{{\text{Net resistance}}}} \cr & \Rightarrow i = \frac{{2E}}{{{r_1} + {r_2} + R}}\,.......\left( {\text{i}} \right) \cr} $$
It is given that, as circuit is closed, potential difference across the first cell is zero.
i.e., $$V = E - i{r_1} = 0$$
$$ \Rightarrow i = \frac{E}{{{r_1}}}\,......\left( {{\text{ii}}} \right)$$
Equating Eqs. (i) and (ii), we get
$$\eqalign{ & \frac{E}{{{r_1}}} = \frac{{2E}}{{{r_1} + {r_2} + R}} \Rightarrow 2{r_1} = {r_1} + {r_2} + R \cr & \therefore R = {\text{external resistance}} \cr & \Rightarrow R = {r_1} - {r_2} \cr} $$
NOTE
The question is wrong as the statement is when the circuit is closed, the potential difference across the first cell is zero which implies that in a series circuit, one part cannot conduct current which is wrong, Kirchhoff 's law is violated. The question must have been modified.

Power consumed, $$P = \frac{{{V^2}}}{R}$$
$$\eqalign{ & {\text{and}}\,\,P' = \frac{{{{V'}^2}}}{R}\,\,\left( {{\text{As resistance is same for both the cases}}} \right) \cr & {\text{Given,}}\,\,V = 200\,V \cr & V' = 160\,V,P = 100\,W \cr} $$
Comparing two different cases of voltage supplied
$$\eqalign{ & {\text{So,}}\,\,\frac{{P'}}{P} = {\left( {\frac{{V'}}{V}} \right)^2} \cr & = {\left( {\frac{{160}}{{200}}} \right)^2} = {\left( {0.8} \right)^2} \cr & {\text{or}}\,\,P' = {\left( {0.8} \right)^2}P \cr & = 0.64 \times 100 \cr & = 64\,W \cr} $$

$$\eqalign{ & {R_1} = {R_0}\left[ {1 + {\alpha _1}\Delta t} \right];\,\,\,\,\,\,{R_2} = {R_0}\left[ {1 + {\alpha _2}\Delta t} \right] \cr & {\text{In Series,}}\,R = {R_1} + {R_2} \cr & = {R_0}\left[ {2 + \left( {{\alpha _1} + {\alpha _2}} \right)\Delta t} \right] = 2{R_0}\left[ {1 + \left( {\frac{{{\alpha _1} + {\alpha _2}}}{2}} \right)\Delta t} \right] \cr & \therefore {\alpha _{eq}} = \frac{{{\alpha _1} + {\alpha _2}}}{2} \cr & {\text{In Parallel}},\,\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} = \frac{1}{{{R_0}\left[ {1 + {\alpha _1}\Delta t} \right]}} + \frac{1}{{{R_0}\left[ {1 + {\alpha _2}\Delta t} \right]}} \cr & \Rightarrow \frac{1}{{\frac{{{R_0}}}{2}\left( {1 + {\alpha _{eq}}\Delta t} \right)}} = \frac{1}{{{R_0}\left( {1 + {\alpha _1}\Delta t} \right)}} + \frac{1}{{{R_0}\left( {1 + {\alpha _2}\Delta t} \right)}} \cr & 2\left( {1 - {\alpha _{eq}}\Delta t} \right) = \left( {1 - {\alpha _1}\Delta t} \right)\left( {1 - {\alpha _2}\Delta t} \right)\therefore {\alpha _{eq}} = \frac{{{\alpha _1} + {\alpha _2}}}{2} \cr} $$