The capacitance from the volume control knob through the pot is large enough to allow RF to couple through - from your body, when you touch the knob - to the volume control circuitry, where it gets rectified/detected (turned into an audio signal) and sent through the rest of the amp to the loudspeaker.
Like LvW says in his answer, note that what we call Rds is not a physical resistor present in the MOSFET but it is a phenomenon which is presented by a resistor called Rds in the small signal model of the MOSFET.
You take a MOSFET, you apply DC voltages and currents to it so that it will have a certain operating point. For example, an operating point where the drain current Ids = 1 mA and Vds is 3 V. For this imaginary NFET Vt = 1 V so this NMOS is in saturation.
Now that we know the operating point of this NMOS, we can calculate values for some small signal parameters of this NMOS at this operationg point. These parameters are all derivatives For example:
$$gm = dId / dVgs$$
and
$$Rds = dVds / dId$$
Note how Rds is the derivative of Vds/Id !
The values of gm and Rds result from the physical properties of the MOSFET. So for a different MOSFET (for example, one with a longer channel) these values will be different. In general, Rds will be larger for a MOSFET with a longer channel.
But this does not explain yet why this is so.
What does explain it is the Channel length modulation effect.
For MOSFETs with very short channels the drain is (physically close to the part of the MOSFET's channel which determines the drain current when it is in saturation. As the voltage on the drain increases the depletion layer around the drain also increases in size. Worst case this depletion region can even touch the channel. This results in a low ohmic path between drain and source and Rds will be very low.
If the drain is physically further away from the source that depletion region cannot get anywhere near the channel so the channel will determine the current without the drain and it's depletion region interfering. This results in a more ideal current source behavior of the channel. For a high Rds, this is what is needed, it means dId will be very small (only small drain current variations due to changes in Vds).
Best Answer
The power consumption of an amplifier tends to have no dependence on the gain, or input signal level. It may have a dependence on the power output. It will always have some static consumption.
In a class A amplifier, there is a high static power consumption, and virtually no dependence on the output power. Power efficiency improves from very poor at low output to poor at higher output.
In a class B amplifier, the static consumption is low, and the power supply current input increases with the load current. This means that the power efficiency is low at low output powers, and improves with higher output power.
In a class D amplifier, the static consumption is low, and the power supply power input increases with the power output at more or less constant power efficiency.
This assumes we are talking about a 'power amplifier'. Small signal amplifiers tend to be class A, though many opamps have class AB or B outputs, which are covered as above.