I am very confused about voltage in short circuit.
Let's suppose there is a 6 volt battery and both ends of it are connected with the same wire and it's short-circuited.
Then because resistance of wire is 0 (negligible) so by the formula V=IR the voltage across ends of the battery will be 0.
But then where does the 6 volts applied by the battery go?
The voltage is still there. You're thinking with ideal components - perfect voltage sources with no internal resistance and wires with no resistance. Ideal components can give you strange, impossible answers when you do things like short out supplies or connect them in parallel. In the real world, batteries and wires have resistances (and inductances and capacitances) since they're generally not made with superconducting materials. A standard 6V lantern battery has an internal resistance of ~1 ohm and 1' or 30cm of 20 AWG/0.5mm^2 wire has a resistance of about 10 milliohms so your short circuit would look something like this:
simulate this circuit – Schematic created using CircuitLab
And you'd end up with Isc of 5.94 A with 5.94 V dropped across the internal resistance of the battery and 0.0594 V across your wire, which is also the terminal voltage of the battery.
There is internal resistance in the battery, and a small but not zero resistance in the wire. The battery may be unable to supply the resulting high current, which causes its voltage to significantly decrease under the load. Whatever power it can sustain will be dissipated as heat. The wire, the battery, or both, will become hot. If the sustained heat is not enough to cause a failure, the whole system will effectively become a little heat-producing circuit. If the produced heat is more than one component can withstand, a failure will occur. Failure could be as dramatic as an exploding battery or melting conductor that could start a fire, or as uneventful as the battery discharging and the heat produced not being sustained. It all depends on the capacity of the battery and the size of wire forming the short.
Well in an ideal case you are correct, there is no way the voltage will dissappear, current will be infinity and a black hole is created, this is the problem of edge cases in our mathematical modeling in electrical circuits that assumes perfection, same issue is thier in an RC circuit for example (connecting a resistor in series with capacitor) if resistance is zero, current at first second is infinity and capacitor is charged in zero time; A black hole will pop out of existance, but in the real case there is always resistance of wires and even internal resistance of battery it self.
So back to your simple wires shorting the battery, their has to be some resistance that drops the voltage.
Also check something called thevinin equivalent it is a really handy way of describing any circuit and always have a resistance in series that causes the voltage to actually drop mathematically. This has really cool applications for maximum power transfer and other stuff.
A battery holds charge chemically which means there is a barrier that separates positive and negative ions similar to a capacitor. Different batteries have different types of barriers and chemistry.
When the terminals of a battery are connected to each other (shorted,) the electrons flow to the other side through the wire making the voltage difference between the terminals to 0 same as a capacitor.
The time it takes to reach the 0 volt state depends on the maximum discharge current of the battery.
In theory, if you short circuit a voltage source the current would be infinite to maintain the same supply voltage. That is not possible in practice. When an actual power supply or a battery is shorted, very high current flows for very short time. Afterwards, if it is a battery then it would be fully discharged (probably damaged too,) or if it is some sort of power supply then the fuse would be popped, or the supply would be cut-off because of short-circuit protection, or some components in the supply would be blown because it couldn't handle the large current.
