Correct Answer - Option 4 : (a) and (d)

__Concept__**:**

General equation of FM modulated signal is expressed as:

(S)_{FM} = A_{C} cos [ω_{c}t + k_{f} ∫m(t) dt]

m(t) = Modulating signal

k_{f} = Frequency sensitivity (Hz/volt)

ω_{c} = Carrier frequency (rad/sec)

A_{c} = Carrier Amplitude

Total peak to peak voltage = 2A_{c}

Also, the frequency deviation in FM is defined as:

Δf = |k_{f}.m(t)|_{max}

__Calculation__**:**

Given 2A_{C} = 1.2

So the peak voltage is:

**A**_{c} = 0.6 volts

Option (a) is therefore correct.

Given frequency deviation Δf = 10 kHz

Since m(t) is a sin wave with 0.6 V as peak amplitude, frequency deviation can be written as:

⇒ (k_{f})(0.6) = 10 kHz

\( \Rightarrow {k_f} = \frac{{10}}{{0.6}}\;\frac{{kHz}}{{Volts}}\)

The detector, however, senses voltage or ampere per Hz or rad/sec. It is therefore expressed as Volts/Hz

Taking the reciprocal of frequency sensitivity we get the detector sensitivity (D) as:

\(D = \frac{{0.6}}{{10}}\frac{{Volts}}{{kHz}}\)

**D = 60 μV/Hz**

Hence both Option (a) and (d) are therefore correct.