How to detect Resonance

The resonant frequency of an LC circuit is given by

$$ f = \frac {1}{2 \pi \sqrt {LC}} $$

and plugging in the values you have chosen gives

$$ f = \frac {1}{2 \pi \sqrt {0.1 \cdot 100 \mu}} = \frac {1}{2 \pi \cdot 0.00316 } = 50.3~Hz$$

So your resonant frequency is almost exactly on the mains frequency of 50 Hz.

1. Why does resonance increases the voltage?

It's like pushing a swing or pendulum. If you do it at the natural frequency of oscillation and at the right point in the cycle then every push will boost a little bit more. In the case of the pendulum or swing air resistance will eventually limit the amplitude of the oscillation. In the LC circuit the internal resistance will cause losses which increase with increasing amplitude until an equilibrium is reached.

2. Why the voltage does not increase at a frequency of say 2 KHz?

See the formula.

3. If I made this experiment practically, Will the transformer wear out?

Transformers lifespan is mostly determined by temperature which is usually a result of too-much current or insulation breakdown which is a result of too high a voltage being applied. You would need to know the winding insulation breakdown voltage to figure that out. At 50 V you are unlikely to have a problem.

No, your source voltage doesn't connect across your tuned circuit.

You need to model the antenna as a short monopole of maybe 0.025 wavelengths. Now it will have a capacitive reactance of about 1000 ohms and the voltage generator will be in series with this impedance back to 0V: -

Picture taken from here.

This voltage source is in series with the capacitor (impedance ~1000 ohms reactive) at your operating frequency. Say you operate at 1 MHz (nominally), 1000 ohms is represented by 159 pF.

You should get good tuning with this but not brilliant because the headphones are still taking energy from the tuned circuit and lowering its Q factor.