A diode detector has a load of 1kΩ shunted by a 10000 pF

A diode detector has a load of 1kΩ shunted by a 10000 pF capacitor. The diode has a forward resistance of 1Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping, with modulating signal frequency of 10kHz will be—

We know that

RC ≤	1	√	1 − μ²
	ω_m		μ²

given that R = 1 kΩ = 1000Ω
C = 10000 pf = 10^–8 f
ω_m = 2π f_m = 2π × 10 × 10³ radian/sec
μ =? (neglecting diode forward resistance i.e.1Ω)

1000 × 10⁻⁸ ≤	1	√	1 − μ²
	2π × 10 ×10³		μ²

0.628 ≤	√	1 − μ²
		μ²

0.394 ≤	1 − μ²
	μ²

or
0.394 μ² ≤ 1 − μ²
or
1.394 μ² ≤ 1
or

μ² ≤	1
	1.394

or
μ ≤ 0.8469
or
μ_max ≤ 0.847
Hence alternative (A) is the correct choice.