When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of 220 V. This means

EXPLANATION:

→When a voltage measuring device is connected to AC mains.

The meter shows the r.m.s value (V_rms) and not the peak value (V₀)

Here, Vrms = 220 V

∴ V0 = √2× Vrms = √2×220 V ≈ 311 V

So, options (1) and (2) are incorrect. 

Again, If the input voltage is v then,

Vrms = $\sqrt{<v^2>}$

∴ The meter is calibrated as such to read $\sqrt{<v^2>}$

So, option (3) is correct and (4) is incorrect. 

So, the correct answer is option (3).