Given, $$e = 2\;V$$  and $$l = 4\,m$$
Potential drop per unit length
$$\phi = \frac{e}{l} = \frac{2}{4} = 0.5\;V/m$$
I
$$e' = \phi {l_1}\,.......\left( {\text{i}} \right)\,\,\,\left( {e' = {\text{emf of the given cell}}} \right)$$
II
$$V = \phi {l_2}\,.......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii),
$$\eqalign{ & \frac{{e'}}{V} = \frac{{{l_1}}}{{{l_2}}} \cr & \therefore e' = I\left( {r + R} \right) \cr} $$
and $$V = IR$$   for the second case
$$\eqalign{ & \frac{{I\left( {r + R} \right)}}{{IR}} = \frac{{{l_1}}}{{{l_2}}} \cr & {\text{So,}}\,\,r = R\left( {\frac{{{l_1}}}{{{l_2}}} - 1} \right) \cr & = 9.5\left( {\frac{3}{{2.85}} - 1} \right) \cr & = 9.5\left( {1.05 - 1} \right) \cr & = 9.5 \times 0.05 = 0.475 \simeq 0.5\,\Omega \cr} $$

$$\eqalign{ & R = \frac{{\rho \ell }}{{\left( {\frac{{\pi {D^2}}}{4}} \right)}}\,{\text{or}}\,R \propto \frac{\ell }{{{D^2}}}. \cr & \frac{{{R_x}}}{{{R_y}}} = \frac{{{\ell _x}}}{{D_x^2}} \times \frac{{D_y^2}}{{{\ell _y}}} = \frac{{\frac{{{\ell _y}}}{2}}}{{{{\left( {\frac{{{D_y}}}{2}} \right)}^2}}} \times \frac{{D_y^2}}{{{\ell _y}}} = \frac{2}{1} \cr} $$

Since, power rating of bulb in both the countries India and USA is same, so
$$\eqalign{ & \frac{{V_1^2}}{{{R_1}}} = \frac{{V_2^2}}{{{R_2}}}\,\,\left[ {P = \frac{{{V^2}}}{R}} \right] \cr & \Rightarrow \frac{{220 \times 220}}{{{R_1}}} = \frac{{110 \times 110}}{{{R_2}}} \cr & \Rightarrow \frac{{{R_2}}}{{{R_1}}} = \frac{{110 \times 110}}{{220 \times 220}} \cr & \Rightarrow {R_2} = \frac{R}{4}\,\,\left( {\because {R_1} = R} \right) \cr} $$

Using formula, internal resistance,
$$r = \left( {\frac{{{l_1} - {l_2}}}{{{l_2}}}} \right)s = \left( {\frac{{52 - 40}}{{40}}} \right) \times 5 = 1.5\Omega $$

$$\eqalign{ & \frac{{\left( {\frac{I}{4} \times 12000} \right)}}{{6000}} = 2000, \cr & I = \frac{{220}}{{9000 + 2000}}\;A \cr} $$
$$\therefore $$ Potential difference across voltmeter $$= 40\,V.$$

Kirchhoff’s junction rule states that the algebraic sum of all currents into and out of any branch point is zero: $$\Sigma I = 0.$$  By convention, the sign of current entering a junction is positive and current leaving a junction is negative.
$$4A + 5A - 6A + {I_{AB}} = 0,$$     therefore $${I_{AB}} = - 3A.$$   The wire between points $$A$$ and $$B$$ carries a current of $$3A$$ away from the junction.

Length of each wire $$ = \ell ;$$  Area of thick wire $$\left( {{A_1}} \right) = 3A;$$   Area of thin wire $$\left( {{A_2}} \right) = A$$   and resistance of thick wire $$\left( {{R_1}} \right) = 10\,\Omega .$$
Resistance $$\left( R \right) = \rho \frac{\ell }{A} \propto \frac{1}{A}\left( {{\text{if}}\,\ell \,{\text{is constant}}} \right)$$
$$\eqalign{ & \therefore \frac{{{R_1}}}{{{R_2}}} = \frac{{{A_2}}}{{{A_1}}} = \frac{A}{{3A}} = \frac{1}{3} \cr & {\text{or,}}\,{R_2} = 3{R_1} = 3 \times 10 = 30\,\Omega \cr} $$
The equivalent resistance of these two resistors in series
$$ = {R_1} + {R_2} = 30 + 10 = 40\,\Omega .$$

$$\eqalign{ & \Delta Q = mC \times \Delta T \cr & = 1 \times 4180 \times \left( {40 - 10} \right) = 4180 \times 30 \cr} $$
($$\therefore $$ $$\Delta Q =$$  heat supplied in time $$t$$ for heating $$1L$$ water from $${10^ \circ }C$$  to $${40^ \circ }C$$ )
$${\text{also}}\,\Delta Q = 836 \times t \Rightarrow t = \frac{{4180 \times 30}}{{836}} = 150s$$

Total current drawn from the battery
$$i = \frac{E}{{R + r}} = \frac{6}{{100 + 0}} = 0.06\,A$$
Resistance of $$50\,cm$$  wire is
$$R' = \frac{{\rho l'}}{A} = \left( {\frac{\rho }{A}} \right)l' = \left( {\frac{R}{l}} \right)l'\left( {\because R = \frac{{\rho l}}{A}} \right)$$
$$ = \frac{{100}}{{300}} \times 50$$
So, $$R' = \frac{{50}}{3}\Omega $$
Hence, the potential difference between two points on the wire separated by a distance $${l'}$$ is given by Ohm's law
i.e. $$V = iR' = 0.06 \times \frac{{50}}{3} = 1\,V$$

Let $$R$$ be the resistance of wire.

Energy released in $$t$$ second $$ = \frac{{{{\left( {3V} \right)}^2}}}{R} \times t$$
$$\eqalign{ & \therefore Q = \frac{{9{V^2}}}{R} \times t \cr & \therefore mc\Delta T = \frac{{9{V^2}}}{R} \times t\,\,\left( {\because But\,Q = mc\Delta T} \right)\,......\left( {\text{i}} \right) \cr} $$
Let $$R'$$ be the resistance of the second wire

$$ \Rightarrow R' = 2R\,\,\,\,\,\,\left( {\because {\text{length}}\,{\text{is}}\,{\text{twice}}} \right)$$
∴ Energy released in $$t$$-seconds $$ = \frac{{{{\left( {NV} \right)}^2}}}{{2R}} \times t$$
$$\eqalign{ & {\text{Also}}\,Q' = m'c\Delta T = \left( {2m} \right)C\Delta T \cr & 2mc\Delta T = \frac{{{N^2}{V^2}}}{{2R}} \times t\,......\left( {{\text{ii}}} \right) \cr} $$
Dividing (i) by (ii)
$$\eqalign{ & \frac{{mc\Delta T}}{{2mc\Delta T}} = \frac{{\frac{{9{V^2} \times t}}{R}}}{{\frac{{{N^2}{V^2} \times t}}{{2R}}}}\,\,\,{\text{or,}}\,\,\,\,\frac{1}{2} = \frac{{9 \times 2}}{{{N^2}}} \cr & {\text{or,}}\,{N^2} = 18 \times 2\,\,\,\,\,\therefore N = 6 \cr} $$