The second major problem is that the capacitor will take a huge current if the circuit is turned on near the peak of the mains cycle. You should add a 1k resistor in series with the supply to limit the turn-on surge.
Let's do a quick calculation. This circuit consists of a capacitor connected in series with a diode arrangement and a battery and then connected to 220 VAC (presumably @ 50 Hz in New Delhi). The diodes and battery part of the circuit will drop a few volts; let's say 3 volts. So what we have essentially is a capacitor connected to 220-3 = 217 volts. Now let's add a 1k resistor in series with that. Let the capacitor be 10 µF, which would be typical for a charger like this; it would give us something like 1 amp charging current. Now we have 1k in series with the reactance of the capacitor which is 1/(2*Pi*50*10E-6) = 318.3 ohms. The total impedance will be 1000 - j318.3; divide 217 VAC by that value and we get a current in the capacitor and in the 1k resistor of .2068 amps. The dissipation in the 1k resistor will be (.2068)^2*1000 = 42.76 watts. This would not be a good choice for the series current limiting resistance.
The worst case initial surge current which would occur if the circuit were suddenly connected to the grid when the grid sine wave is at its peak won't be a problem. The surge will be very short and will be limited by the leakage inductance of the distribution transformer on the pole outside your house. I'm in the U.S. where the voltage at the plug in the house is 120 VAC @ 60 Hz rather than the 220 VAC @ 50Hz in India. But the surge should be similar.
Just to get a feel for what could be expected I performed an experiment. I connected a short line cord in series with a 50 µF 370 VAC rated motor run capacitor with a measured ESR of 18 milliohms, and also in series with a 1 milliohm current shunt. Using an oscilloscope with isolated inputs, I monitored the voltage across the capacitor (which is essentially the grid voltage minus the few millivolts across the shunt), and the voltage across the shunt. The shunt voltage is 1 millivolt per amp. I then repeatedly plugged this arrangement into an outlet (discharging the capacitor before each insertion) until I caught the grid voltage at a peak when I plugged it in.
The first image shows the grid voltage in green and the capacitor current (also the grid current) in purple. The scale is 50 amps/cm for the current and 50 volts/cm for the voltage. The second image shows the steady state current with the capacitor connected to the grid. Notice how the current in the second image is not very sinusoidal. A capacitor connected to the grid draws a current waveform which is the differentiated version of the voltage waveform.
The current surge shows some oscillations, but its average value is about 70 amps for about 100 µs. Note that the capacitor voltage is almost a linear rising ramp for that 100 µs; the leakage inductance of the distribution transformer is what limits the rate of rise of the current. This measurement was made on a 120 VAC winding of the distribution transformer. Had I used the 220 VAC winding, the leakage inductance would be about 4 times as much.
A typical 1N400x series rectifier is rated to withstand a surge of 30 amps for 8.3 milliseconds. A surge of 70 amps lasting for only 100 µS is well within its ratings, and that's with a 50 µF capacitor. A more typical 10 µF capacitor would produce a "surge" for substantially less than 100 µS.
The capacitor, which should be rated for connection to the grid, will easily be able to withstand this pulse of current without damage. It's just not necessary to try to reduce this <100 µS "surge"; it won't cause any damage to the diodes or the capacitor.
Consider the possible value we might use for a series surge limiting resistor. The steady state current for charging the battery might be on the order of 1 amp. If we place a 1 ohm resistor in series with the capacitor, it will dissipate 1 watt, so we should probably use a 2 watt resistor. Any higher value than 1 Ω will dissipate even more power and would be undesirable. The current surge of 70 amps when the peak voltage of 169 volts is applied implies an apparent "impedance" of 169/70 = 2.4 ohms. We would need to use a surge limiting resistor of this value, or more, to achieve a substantial reduction in the "surge" current. Such a high value resistor will get much too hot for a 2 watt size resistor.
The whole point of this (unsafe for a hobbyist to build) charger circuit is to place a current limiting impedance in series with the grid voltage and then place the battery in series with that. The current in the battery is limited by the series impedance. That series impedance could be a resistor, but if we used a resistor for that purpose, the resistor would get very hot. If we wanted a battery current of 1 amp (RMS), we would need to put a resistance of (220-3)/1 = 217 ohms in series with the 220 VAC; the resistor would then dissipate 217 watts--not a good thing. By using a capacitor as the current limiting element, we avoid the waste of power that would occur if we used a resistor (we could also use an inductor for this purpose).
By the way, when this series capacitor, line connected, charger is used on the output of a non-true-sinewave inverter (sometimes called "modified sine wave") inverter, the fast rising edges of the voltage waveform cause the current drawn to be excessive.
I once had to determine why an inverter was blowing the internal fuse in a charger for a Black & Decker battery operated drill. It turned out that the reason was because the charger was just this sort of series capacitor, directly off the grid circuit.