Let the effective internal resistance of the battery is $R_{\text {eff }}$, the effective external resistance $R$ and the effective voltage of the battery is $V_{\text {eff }}$.

Applying Ohm's law,

Then current through $R$ is given by

$$I=\frac{V_{\text {eff }}}{R_{\text {eff }}+R}$$

If all the resistances and the effective voltage are increased $n$-times, then we have

$$\begin{aligned} V_{\text {eff }}^{\text {new }} & =n V_{\text {eff }}, R_{\text {eff }}^{\text {new }}=n R_{\text {eff }} \\ \text{and}\quad R^{\text {new }} & =n R \end{aligned}$$

Then, the new current is given by

$$I^{\prime}=\frac{n V_{\text {eff }}}{n R_{\text {eff }}+n R}=\frac{n\left(V_{\text {eff }}\right)}{n\left(R_{\text {eff }}+R\right)}=\frac{\left(V_{\text {eff }}\right)}{\left(R_{\text {eff }}+R\right)}=I$$

Thus, current remains the same.

Applying Kirchhoff's junction rule, $\quad I_1=I+I_2$

Applying Kirchhoff's II law / loop rule applied in outer loop containing 10V cell and resistance $R$, we have

$$10=I R+10 I_1\quad\text{.... (i)}$$

Applying Kirchhoff II law / loop rule applied in outer loop containing 2 V cell and resistance $R$, we have

$$\begin{aligned} 2 & =5 I_2-R I=5\left(I_1-I\right)-R I \\ \text{or}\quad 4 & =10 I_1-10 I-2 R I\quad \text{... (ii)} \end{aligned}$$

$$\begin{aligned} &\text { Solving Eqs. (i) and (ii), gives }\\ &\begin{aligned} & \Rightarrow \quad 6=3 R I+10 I \\ & 2=I\left(R+\frac{10}{3}\right) \end{aligned} \end{aligned}$$

Also, the external resistance is $R$. The Ohm's law states that

$$V=I\left(R+R_{\text {eff }}\right)$$

On comparing, we have $V=2 V$ and effective internal resistance

$$\left(R_{\text {eff }}\right)=\left(\frac{10}{3}\right) \Omega$$

Since, the effective internal resistance $\left(R_{\text {eff }}\right)$ of two cells is $\left(\frac{10}{3}\right) \Omega$, being the parallel combination of $5 \Omega$ and $10 \Omega$. The equivalent circuit is given below

Power consumption in a day i.e., in $5=10$ units

Or power consumption per hour $=2$ units

Or power consumption $=2$ units $=2 \mathrm{~kW}=2000 \mathrm{~J} / \mathrm{s}$

Also, we know that power consumption in resistor,

$$\begin{array}{rlrl} P =V \times l \\ \Rightarrow \quad 2000 \mathrm{~W} & =220 \mathrm{~V} \times l \text { or } l \approx 9 \mathrm{~A} \end{array}$$

Now, the resistance of wire is given by $R=\rho \frac{l}{A}$ where, $A$ is cross-sectional area of conductor.

Power consumption in first current carrying wire is given by

$$\begin{aligned} P & =l^2 R \\ \rho \frac{l}{A} l^2 & =1.7 \times 10^{-8} \times \frac{10}{\pi \times 10^{-6}} \times 81 \mathrm{~J} / \mathrm{s} \approx 4 \mathrm{~J} / \mathrm{s} \end{aligned}$$

The fractional loss due to the joule heating in first wire $=\frac{4}{2000} \times 100=0.2 \%$

Power loss in Al wire $=4 \frac{\rho_{\mathrm{Al}}}{\rho_{\mathrm{Cu}}}=1.6 \times 4=6.4 \mathrm{~J} / \mathrm{s}$

The fractional loss due to the joule heating in second wire $=\frac{6.4}{2000} \times 100=0.32 \%$

Let $R^{\prime}$ be the resistance of the potentiometer wire.

Effective resistance of potentiometer and variable resistor $(R=50 \Omega)$ is given by $=50 \Omega+R^{\prime}$

Effective voltage applied across potentiometer $=10 \mathrm{~V}$.

The current through the main circuit,

$$I=\frac{V}{50 \Omega+R}=\frac{10}{50 \Omega+R}$$

Potential difference across wire of potentiometer,

$$I R^{\prime}=\frac{10 R^{\prime}}{50 \Omega+R}$$

Since with $50 \Omega$ resistor, null point is not obtained it's possible only when

$$\begin{aligned} \frac{10 \times R^{\prime}}{50+R} & <8 \\ 10 R^{\prime} & <400+8 R^{\prime} \\ \Rightarrow\quad 2 R^{\prime}<400 \text { or } R^{\prime} & <200 \Omega . \end{aligned}$$

Similarly with $10 \Omega$ resistor, null point is obtained its possible only when

$$\frac{10 \times R^{\prime}}{10+R^{\prime}}>8$$

$$\begin{array}{llr} \Rightarrow & 2 R^{\prime}> 80 \\ \Rightarrow & R^{\prime}> 40 \\ & \frac{10 \times \frac{3}{4} R^{\prime}}{10+R^{\prime}} <8 \\ \Rightarrow & 7.5 R^{\prime} < 80+8 R^{\prime} \\ \Rightarrow & R^{\prime} > 160 \\ & 160 < R^{\prime}<200 . \end{array}$$

Any $R^{\prime}$ between $160 \Omega$ and $200 \Omega$ wlll achieve.

Since, the null point on the last (4th) segment of the potentiometer, therefore potential drop across 400 cm of wire $>8 \mathrm{~V}$.

This imply that potential gradient

$$\begin{aligned} k \times 400 \mathrm{~cm} & >8 \mathrm{~V} \\ \text{or}\quad k \times 4 \mathrm{~m} & >8 \mathrm{~V} \\ k & >2 \mathrm{~V} / \mathrm{m} \end{aligned}$$

Similarly, potential drop across 300 cm wire $<8 \mathrm{~V}$.

$$\begin{aligned} k \times 300 \mathrm{~cm} & <8 \mathrm{~V} \\ \text{or}\quad \mathrm{k} \times 3 \mathrm{~m}<8 \mathrm{~V}, k & <2 \frac{2}{3} \mathrm{~V} / \mathrm{m} \\ \text{Thus,}\quad 2 \frac{2}{3} \mathrm{~V} / \mathrm{m} & >k>2 \mathrm{~V} / \mathrm{m} \end{aligned}$$

(a) By Ohm's law, current $I$ is given by

$$\begin{aligned} & I=6 \mathrm{~V} / 6 \Omega=1 \mathrm{~A} \\ \text{But,}\quad & I=n e t ~ A v_d \text { or } v_d=\frac{i}{n e A} \end{aligned}$$

On substituting the values

For, $$\quad n=\text { number of electron/volume }=10^{29} / \mathrm{m}^3$$

$$\begin{aligned} \text { length of circuit } & =10 \mathrm{~cm}, \text { cross-section }=A=(1 \mathrm{~mm})^2 \\ v_d & =\frac{1}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-6}} \\ & =\frac{1}{1.6} \times 10^{-4} \mathrm{~m} / \mathrm{s} \end{aligned}$$

Therefore, the energy absorbed in the form of KE is given by

$$\begin{aligned} \mathrm{KE} & =\frac{1}{2} m_e v_d^2 \times n A I \\ & =\frac{1}{2} \times 9.1 \times 10^{31} \times \frac{1}{2.56} \times 10^{20} \times 10^8 \times 10^6 \times 10^1 \\ & =2 \times 10^{-17} \mathrm{~J} \end{aligned}$$

(b) Power loss is given by $P=I^2 R=6 \times 1^2=6 \mathrm{~W}=6 \mathrm{~J} / \mathrm{s}$

Since, $$P=\frac{E}{t}$$

Therefore, $$E=P \times t$$

or $$t=\frac{E}{P}=\frac{2 \times 10^{-17}}{6} \approx 10^{-17} \mathrm{~s}$$