Generally speaking, the CRT can be viewed as a part of the CRO. If one really wants to find out, perform the experiment cautiously in the lab. However, it seemed that the CRT in this question is highly simplified and designed to function principally like a vacuum tube Diode (forget about the X-Y deflection plates and the Phosphorous screen). If that is the case and it isn't a trick question, then we can analyze such electrical circuit in theory:

As you probably already know that diodes are used in AC circuits where it is desired to convert AC voltages to DC voltages, because diode has low resistance-to-current flow in the forward-biased direction. The simplest of all the rectifier circuits is a half-wave rectifier (Click it), when the diode is connected to a load resistance R as shown in the following figure. During the positive half-cycle from 0 to π of the voltage supply Vs, the diode is forward-biased so current flows through the diode and resistor where it develops an output voltage drop, Vout = Vs − Vd, and the built-in potential of a silicon diode Vd is approximately 0.7 V. The diode is reverse-biased during the negative half-cycle from π to 2π, so no current flows, and thus Vout = 0 V. And this is the well-known half-wave ripple in the signal across the load resistor where the name of the rectifier circuit comes from.



To get rid of the undesirable large ripple, we add a “smoothing” capacitor across the resistor, and the voltage across the resistor-capacitor (RC) combination will be as shown in the next figure. When the instantaneous voltage of the AC source Vs, is higher than the instantaneous value of capacitor voltage Vc, the diode conducts, and the capacitor C charges up to the positive peak of the output voltage Vout. When the instantaneous voltage of the AC source Vs, falls below the instantaneous value of capacitor voltage Vc, the diode is reverse biased and the capacitor C is discharged through load resistance R. The resulting capacitor voltage Vc varies between a maximum value of Vout and a minimum value of Vout − Vr(pp). Theoretically, the peak-to-peak ripple voltage Vr(pp), can be made sufficiently small by choosing the time constant τ = RC much larger that the period T = 1/f of the input sinusoidal AC waveform. However, it is not very practical to use larger capacitors due to the constraints of both cost and size. Therefore, the full-wave rectifier with capacitor-input DC filter (Click it) is a much more practical approach.



Come back to the original question, I'd probably choose “A” because it is the closest image to the predicted waveform as indicated by the solid green “slight” ripple below.