This figure is a crystal radio reciever and there is no active element involved in it.
LC filter is passive. Then how one can know that what output voltage LC circuit is offering to the diode and earphones?
Can the output voltage from LC circuit be found by using a multimeter/voltmeter or there is any other way to find the output voltage from LC circuit?
This discussion assumes the circuit is used to receive medium-frequency AM broadcast band.
Since there are no active elements in this circuit, the only RF power available comes from the antenna, and possibly from the earth connection wire too.
Very little power is available: matching impedances with inductance and/or capacitance is essential to delivering audible power to headphones. The diode is the element that generates audio power from the available RF power. Causing a good deal of headphone audio current to flow is the goal.
Measuring voltages at either of the two nodes in this circuit (with respect to the third node earth, which is assumed as zero volts) requires knowledge of the circuit impedance at that node.
The diode efficiency at converting RF to audio is helped by having a high RF voltage at the antenna node. This is helped by the tuned LC resonator, which maximizes its impedance at one particular RF frequency. The antenna is often shown as a simple wire, but it usually adds a capacitive reactance, and small series resistance to this resonator.
simulate this circuit – Schematic created using CircuitLab
In this simple circuit, power transfer is aided by keeping impedances at NODE1 and at NODE2 high. High-impedance headphones are rare today, but more common decades ago. Low impedance headphones (or speaker) generally give poor results because impedance-matching is not optimal.
Note that both audio current and RF current flow through L1.
TLDR:
We usually assume that NODE3 is always at zero volts for both RF and audio. Many folks omit this connection, and get disappointing results due to the whole circuit floating...little RF power is delivered to the LC resonator. NODE3 must be grounded in an effort to produce RF currents in the LC resonator.
Measuring voltages at NODE1 or NODE2 with a grounded oscilloscope might yield respectable voltages, but when the oscilloscope is disconnected, less audio is heard (because the ground is now missing).
Measuring audio voltages from NODE2 to NODE3 with a multimeter might give some indication provided a good earth ground is connected to NODE3 (very few multimeters include their own ground connection). A digital multimeter often gives about three readings-per-second, and these numbers might jump about randomly, indicating audio voltage or current.
The LC circuit exists to select the AM station in this corcuit here. These devices are intended to be connected to an antenna large enough to pick up significant AM signal (there is no amplification, the only power source is the received signal itself). Also the earpiece should be very high impedance in order to efficiently convert the tiny signal into sound.
You can't really know from the diagram what the voltage out is. It's just a filtered (frequency selected) version of the signal picked up on the antenna. Without it, you should hear wideband noise, with any transmitted signals probably lost in the noise.
The voltage across the speaker is just the voltage available on the (loaded) antenna (which depends on received signal strength and the antenna itself), half-rectified, filtered according to the characteristics of the tank circuit, and less a diode drop.
