The network of resistors is a balanced wheatstone bridge. The equivalent circuit is

$${\operatorname{R} _{eq}} = \frac{{15 \times 30}}{{15 + 30}} = 10\Omega \Rightarrow I = \frac{V}{R} = \frac{5}{{10}} = 0.5\,A$$

In steady state capacitor is fully charged hence no current will flow through line 2.
By simplyfing the circuit

Hence resultant potential difference across resistances will be $$8.0\,V.$$
Thus current $$I = \frac{V}{R} = \frac{{8.0}}{{3 + 9}} = \frac{8}{{12}}$$
$${\text{or,}}\,I = \frac{2}{3} = 0.67\,A$$

Statements I is false and Statement II is true
For ammeter, shunt resistance, $$S = \frac{{IgG}}{{I - Ig}}$$
Therefore for $$I$$ to increase, $$S$$ should decrease, So additional $$S$$ can be connected across it.

It follows from symmetry considerations that if we remove the first element from the circuit, the resistance of the remaining circuit between points $$C$$ and $$D$$ will be $${R_{CD}} = k{R_{AB}}.$$   Therefore, the equivalent circuit of the infinite chain will have the form shown in figure.

Applying to this circuit the formulas for the resistance of series and parallel resistors, we obtain
$${R_{AB}} = \frac{{{R_1} + {R_2}k{R_{AB}}}}{{{R_2} + k{R_{AB}}}}$$
Solving the quadratic equation for $${R_{AB}},$$  we obtain (in particular, for $$k = \frac{1}{2}$$ )
$${R_{AB}} = \frac{{{R_1} - {R_2} + \sqrt {R_1^2 + R_2^2 + 6{R_1}{R_2}} }}{2}$$

$$\eqalign{ & iR = 2 = 12 - 500i\,\,\therefore i = \frac{1}{{50}} \cr & \therefore \frac{1}{{50}} \times R = 2 \cr & \therefore R = 100\Omega \cr} $$

In ohm's law, we check $$V = IR$$   where $$I$$ is the current flowing through a resistor and $$V$$ is the potential difference across that resistor. Only option (A) fits the above criteria.
NOTE : Remember that ammeter is connected in series with resistance and voltmeter parallel with the net resistance.

$$\eqalign{ & S = {R_1} + {R_2}{\text{ and }}P = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} \cr & S = nP \Rightarrow {R_1} + {R_2} = \frac{{n\left( {{R_1}{R_2}} \right)}}{{\left( {{R_1} + {R_2}} \right)}} \cr & \Rightarrow {\left( {{R_1} + {R_2}} \right)^2} = n{R_1}{R_2} \Rightarrow n = \frac{{R_1^1 + R_2^2 + {R_1}{R_2}}}{{{R_1}{R_2}}} \cr & n = \frac{{{R_1}}}{{{R_2}}} + \frac{{{R_2}}}{{{R_1}}} + 2 \cr} $$
Arithmetic mean > Geometric mean
Minimum value of $$n$$ is 4

Power of bulb $$= 60\,W$$ (given)
Resistance of bulb = $$\frac{{120 \times 120}}{{60}} = 240\Omega \,\,\left[ {\because P = \frac{{{V^2}}}{R}} \right]$$
Power of heater $$= 240W$$  (given)
Resistance of heater = $$\frac{{120 \times 120}}{{240}} = 60\Omega $$
Voltage across bulb before heater is switched on, $${V_1} = \frac{{240}}{{246}} \times 120 = 117.73\,volt$$
Voltage across bulb after heater is switched on, $${V_2} = \frac{{48}}{{54}} \times 120 = 106.66\,volt$$
Hence decrease in voltage
$${V_1} - {V_2} = 117.73 - 106.66 = 10.04\,Volt\left( {{\text{approximately}}} \right)$$

Resistances $$4\,\Omega $$  and $$4\,\Omega $$  are connected in series, so their effective resistance is
$$R' = 4 + 4 = 8\,\Omega $$
Similarly, $$1\,\Omega $$  and $$3\,\Omega $$  are in series
So, $$R'' = 1 + 3 = 4\,\Omega $$
Now $${R'}$$ and $${R''}$$ will be in parallel, hence effective resistance
$$\eqalign{ & R = \frac{{R' \times R''}}{{R' + R''}} \cr & = \frac{{8 \times 4}}{{8 + 4}} = \frac{{32}}{{12}} = \frac{8}{3}\Omega \cr} $$
Current through the circuit, from Ohm's law
$$i = \frac{V}{R} = \frac{{3V}}{8}A$$
Let currents $${i_1}$$ and $${i_1}$$ flow in the branches as shown. As voltage remains same in parallel combination i.e. $${V_1} = {V_2}$$

$$\eqalign{ & \therefore 8{i_1} = 4{i_2} \cr & \Rightarrow {i_2} = 2{i_1}\,......\left( {\text{i}} \right) \cr & {\text{Also,}}\,\,i = {i_1} + {i_2} \cr & \Rightarrow \frac{{3V}}{8} = {i_1} + 2{i_1} \cr & \Rightarrow {i_1} = \frac{V}{8}A\,\,{\text{and}}\,\,{i_2} = \frac{V}{4}A \cr} $$
Potential drop at $$A,$$
$${V_A} = 4 \times {i_1} = \frac{{4V}}{8} = \frac{V}{2}$$
Potential drop at $$B,$$
$${V_B} = 1 \times {i_2} = 1 \times \frac{V}{4} = \frac{V}{4}$$
Since, drop of potential is greater in $$4\,\Omega $$  resistance so, it will be at lower potential than $$B.$$ Hence, on connecting wire between points $$A$$ and $$B$$ the current will flow from $$B$$ to $$A.$$

This is a balanced wheatstone bridge condition,
$$\eqalign{ & \frac{5}{R} = \frac{{{\ell _1}}}{{100 - {\ell _1}}}\,{\text{and}}\,\frac{5}{{\frac{R}{2}}} = \frac{{1.6{\ell _1}}}{{100 - 1.6{\ell _1}}} \cr & \Rightarrow R = 15\,\Omega \cr} $$