The resistance of length $$2\pi R$$  is $$12\Omega .$$  Hence the resistance of length $$\pi R$$ is $$6\Omega .$$  Thus two resistances of $$6\Omega $$  can be represented as shown in fig. 2.
$$\therefore $$ Equivalent resistance $$R = \frac{{6 \times 6}}{{12}} = 3\Omega $$

$$\eqalign{ & H = \frac{{{V^2}}}{R} \times 15 \times 60 = \frac{{{V^2}}}{{\left( {\frac{2}{3}} \right)R}} \times t \cr & {\text{or}}\,t = \frac{2}{3} \times 15 \times 60 = 600s = 10\,{\text{minutes}}{\text{.}} \cr} $$

The current through the resistance $$R$$
$$I = \left( {\frac{\varepsilon }{{R + r}}} \right)$$
The potential difference across $$R,$$
$$V = IR = \left( {\frac{\varepsilon }{{R + r}}} \right)R$$

$$V = \frac{\varepsilon }{{\left( {1 + \frac{r}{R}} \right)}};$$   when $$R = 0,V = 0,R = \infty ,V = \varepsilon $$
Thus $$V$$ increases as $$R$$ increases upto certain limit, but it does not increase further.

$$\eqalign{ & I = neA{v_d} \cr & = 2 \times {10^{21}} \times 1.6 \times {10^{ - 19}} \times 10 \times 0.25 \times {10^{ - 3}} \cr & = 2 \times 1.6 \times 0.25 = \frac{8}{{10}} = 0.8\,A \cr} $$

$$\eqalign{ & \frac{P}{Q} = \frac{l}{{\left( {100 - l} \right)}} \cr & {\text{or}}\,P = \frac{l}{{100 - l}} \times Q = \frac{{20}}{{80}} \times 1 = 0.25\Omega . \cr} $$

$$\eqalign{ & i = neA{V_d}\,{\text{and}}\,{V_d} \propto \sqrt E \,\left( {{\text{Given}}} \right) \cr & {\text{or,}}\,i \propto \sqrt E \cr & {i^2} \propto E \cr & {i^2} \propto V \cr} $$
Hence graph (C) correctly depicts the $$V-I$$  graph for a wire made of such type of material.

For the same length and same material,
$$\eqalign{ & \frac{{{R_2}}}{{{R_1}}} = \frac{{{A_1}}}{{{A_2}}} = \frac{3}{1} \cr & {\text{or}}\,\,{R_2} = 3{R_1} \cr} $$
The resistance of thick wire,
$${R_1} = 10\,\Omega $$
The resistance of thin wire $$ = 3{R_1} = 3 \times 10 = 30\,\Omega $$
Total resistance $$ = 10 + 30 = 40\,\Omega $$

The following points should be considered while making the circuit :
(i) An ammeter is made by connecting a low resistance $${R_2}$$ in parallel with the galvanometer $${G_2}.$$

(ii) A voltmeter is made by connecting a high resistance $${R_1}$$ in series with the galvanometer $${G_1}.$$

(iii) Voltmeter is connected in parallel with the test resistor $${R_T}.$$
(iv) Ammeter is connected in series with the test resistor $${R_T}.$$
(v) A variable voltage source $$V$$ is connected in series with the test resistor $${R_T}.$$

Power, $$\left( P \right) = \frac{{{V^2}}}{R}$$
For small variation,
$$\frac{{\Delta P}}{P} \times 100 = 2 \times \frac{{\Delta V}}{V} \times 100 = 2 \times 2.5 = 5\% $$
Therefore, power would decrease by $$5\% .$$

$$\eqalign{ & {R_p} = \frac{{3 \times 6}}{{3 + 6}} = \frac{{18}}{9} = 2\Omega \cr & \therefore V = IR\,\,\,\,\, \Rightarrow I = \frac{V}{R} = \frac{3}{2} = 1.5A \cr} $$