Statement (a) — correct. Industrial noise comes from switching events: motor commutators sparking, thermostats and relays opening, fluorescent tube strikes, welding gear, ignition systems. Each is a brief, high-amplitude transient rather than a continuous background, which is exactly what "impulse noise" means. Its spectrum is broad because the pulses are short, and it is worst at the lower frequencies.

Statement (b) — correct. A mixer is a deliberately non-linear device, and non-linearity is what generates noise as well as the wanted difference frequency. It also folds noise from both the signal and the image frequencies into the IF. In practice a mixer's noise figure is several decibels worse than that of a good RF amplifier at the same frequency. This is precisely why a superheterodyne receiver puts an RF amplifier ahead of the mixer: by Friis's formula,

\(F_{total}=F_1+\dfrac{F_2-1}{G_1}\)

the first stage's noise figure dominates and the mixer's contribution is divided by the RF stage's gain.

Statement (c) — correct. Thermal noise is white: its available power depends on temperature and bandwidth but not on where in the spectrum that bandwidth sits.

\(P_n=kTB\)

A 1 MHz slice at 10 MHz and a 1 MHz slice at 10 GHz carry the same noise power, because the formula contains no centre frequency at all. (The flatness holds until quantum effects intervene, far above any radio frequency.)

Statement (d) — wrong, and it is the one to discard. Impulse noise voltage is strongly dependent on bandwidth. A short pulse has a wide, flat spectrum, so a receiver admits more of its energy the wider its passband:

\(V_{impulse}\propto B\)

This is why narrowing the IF bandwidth is the standard cure for ignition interference, and why an AM broadcast receiver crackles far more than a narrowband communications set on the same signal.

Assemble. (a), (b) and (c) are correct, which is option 2.

The contrast worth keeping. Thermal noise voltage grows as \(\sqrt{B}\) since its power grows as B, whereas impulse noise voltage grows as B itself — which is why impulse noise dominates in wideband systems and thermal noise in narrowband ones.

Hence, the correct statements are (a), (b) and (c).