The formula for resistance of wire is
$$R = \frac{{\rho l}}{A}$$
where, $$\rho = $$  specific resistance of the wire
$$\eqalign{ & \Rightarrow R \propto \frac{l}{A} \cr & \Rightarrow R \propto \frac{l}{{{r^2}}}\,\,\left( {\because A = \pi {r^2}} \right) \cr & \therefore \frac{{{R_1}}}{{{R_2}}} = \frac{{{l_1}}}{{{l_2}}} \times \frac{{r_2^2}}{{r_1^2}}\,......\left( {\text{i}} \right) \cr} $$
Given, $${l_1} = l,{l_2} = 2l,{r_1} = r,{r_2} = 2r,{R_1} = R.$$
Substituting these values in Eq. (i), we have
$$\eqalign{ & \frac{{{R_1}}}{{{R_2}}} = \frac{l}{{2l}} \times \frac{{{{\left( {2r} \right)}^2}}}{{{r^2}}} \cr & \frac{{{R_1}}}{{{R_2}}} = \frac{l}{{2l}} \times \frac{{{{\left( {2r} \right)}^2}}}{{{r^2}}} \cr & \frac{{{R_1}}}{{{R_2}}} = 2 \cr & \Rightarrow {R_2} = \frac{R}{2} \cr} $$
Therefore, resistance will be halved.
Now, the specific resistance of the wire does not depend on the geometry of the wire. Hence, it remains unchanged.

Power $$P = 100\,W,$$   Voltage, $$V = 220\,V$$
$$\eqalign{ & P = \frac{{{V^2}}}{R} \cr & \therefore R = \frac{{{V^2}}}{P} = \frac{{{{\left( {220} \right)}^2}}}{{100}} = \frac{{220 \times 220}}{{100}}\Omega \cr} $$
As both the bulb have same voltage and power so, resistance of bulbs will also be same.
Case I
When two bulbs are connected in series.

In series, $${R_{{\text{eq}}}} = {R_1} + {R_2}$$
$$ = \left( {\frac{{220 \times 220}}{{100}}} \right) \times 2$$
Hence, $${P_{{\text{eq}}}} = \frac{{{V^2}}}{{{R_{{\text{eq}}}}}} = \frac{{220 \times 220}}{{\left( {\frac{{220 \times 220}}{{100}} \times 2} \right)}}$$
$$ = \frac{{100}}{2} = 50\,W$$
Case II
When two bulbs are connected in parallel.

In parallel, $${R_{{\text{eq}}}} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \frac{{{{\left( {\frac{{220 \times 220}}{{100}}} \right)}^2}}}{{\frac{{220 \times 220}}{{100}} \times 2}}$$
$${R_{{\text{eq}}}} = \frac{{220 \times 220}}{{100}} \times \frac{1}{2}$$
Hence, $${P_{{\text{eq}}}} = \frac{{{V^2}}}{{{R_{{\text{eq}}}}}} = \frac{{220 \times 220}}{{\frac{{220 \times 220}}{{100}} \times \frac{1}{2}}}$$
$$ = 200\,W$$
Alternative
For series, $${P_{{\text{eq}}}} = \frac{{{P_1}{P_2}}}{{{P_1} + {P_2}}}$$
$$ = \frac{{100 \times 100}}{{200}} = 50\,W$$
For parallel, $${P_{{\text{eq}}}} = {P_1} + {P_2}$$
$$ = 100 + 100 = 200\,W$$
NOTE
Power equivalent of two or more resistance in series is given by $$\frac{1}{{{P_{{\text{eq}}}}}} = \frac{1}{{{P_1}}} + \frac{1}{{{P_2}}}$$    and for parallel combination $${P_{{\text{eq}}}} = {P_1} + {P_2}$$

Required ratio
$$ = \frac{{{\text{Energy stored in capacitor}}}}{{{\text{Workdone by the battery}}}} = \frac{{\frac{1}{2}C{V^2}}}{{C{e^2}}}$$
where $$C =$$  Capacitance of capacitor
$$V =$$  Potential difference,
$$e =$$  emf of battery
$$\frac{{\frac{1}{2}C{e^2}}}{{C{e^2}}} = \frac{1}{2}\,\,\,\,\,\,\,\,\left( {\because V = e} \right)$$

In case of a meter bridge
$$\eqalign{ & \frac{R}{l} = \frac{X}{{100 - l}}\,{\text{Here}}\,X = 90\Omega ,l = 40.0\;cm \cr & \therefore R = \frac{{Xl}}{{100 - l}} \cr} $$
For finding the value of $$R$$
$$R = \frac{{90 \times 40}}{{60}} = 60\Omega $$
For finding the value of $$\Delta R$$
$$\eqalign{ & \frac{{\Delta R}}{R} = \frac{{\Delta l}}{l} + \frac{{\Delta \left( {100 - l} \right)}}{{100 - l}} \cr & \therefore \frac{{\Delta R}}{{60}} = \frac{{0.1}}{{40}} + \frac{{0.1}}{{60}} \cr & \therefore \Delta R = 0.25\Omega \cr} $$
Therefore, $$R = \left( {60 \pm 0.25} \right)\Omega $$

We know that $$P = \frac{{{V^2}}}{R}$$
For a given potential difference at a particular temperature
$$P \propto \frac{1}{R}$$
It is given that the powers of the bulbs are in the order
$$100W > 60W > 40W$$
$$\therefore \frac{1}{{{R_{100}}}} > \frac{1}{{{R_{60}}}} > \frac{1}{{{R_{40}}}}$$

Kirchhoff's first law is based on conservation of charge and Kirchhoff's second law is based on conservation of energy.

$$\left( A \right)\,\frac{{{V_{AB}}}}{{{V_{BC}}}} = \frac{{{I_{AB}}{R_{AB}}}}{{{I_{BC}}{R_{BC}}}} = \frac{{{R_{AB}}}}{{{R_{BC}}}} = \frac{{\rho \frac{\ell }{{2\left[ {\pi \times 4{r^2}} \right]}}}}{{\rho \frac{\ell }{{2\left[ {\pi {r^2}} \right]}}}} = \frac{1}{4}$$
$$\left[ {{I_{AB}} = {I_{BC}},\,{\text{wire is of same material}}} \right]$$
Therefore option (A) is incorrect.
$$\left( B \right)\,\frac{{{P_{BC}}}}{{{P_{AB}}}} = \frac{{{I^2}{R_{BC}}}}{{{I^2}{R_{AB}}}} = \frac{{\rho \frac{\ell }{{2\left[ {\pi \times 4{r^2}} \right]}}}}{{\rho \frac{\ell }{{2\left[ {\pi {r^2}} \right]}}}} = \frac{1}{4}$$
$$\therefore {P_{AB}} = 4{P_{BC}};$$    Therefore (B) is correct.
$$\left( C \right)\,\frac{{{J_{AB}}}}{{{J_{BC}}}} = \frac{{\frac{I}{{\pi \times 4{r^2}}}}}{{\frac{I}{{\pi \times {r^2}}}}} = \frac{1}{4};$$     Therefore (C) is incorrect.
$$\left( D \right)\,\frac{{{E_{AB}}}}{{{E_{BC}}}} = \frac{{\left[ {\frac{{{V_{AB}}}}{{\frac{\ell }{2}}}} \right]}}{{\left[ {\frac{{{V_{BC}}}}{{\frac{\ell }{2}}}} \right]}} = \frac{1}{4};$$     Therefore (D) is incorrect.

$$\eqalign{ & r = \frac{{{\ell _1} - {\ell _2}}}{{{\ell _2}}} \times R\Omega \cr & {\text{Here,}}\,{\ell _1} = 125\,cm,{\ell _2} = 100\,cm,R = 2\Omega . \cr & \therefore r = 0.5\Omega \cr} $$

The effective circuit is shown in figure.
Thus, $${R_{AB}} = 5\,\Omega .$$

The equivalent resistance is $${R_{eq}} = \frac{{2 \times R}}{{2 + R}}$$
$$\eqalign{ & \therefore {\text{Power dissipation }}P = \frac{{{V^2}}}{{{R_{eq}}}} \cr & \therefore 150 = \frac{{15 \times 15}}{{{R_{eq}}}}\,\,\therefore {R_{eq}} = \frac{{15}}{{10}} = \frac{3}{2} \cr & \Rightarrow \frac{{2R}}{{2 + R}} = \frac{3}{2} \Rightarrow 4R = 6 + 3R \Rightarrow R = 6\Omega \cr} $$