A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance of 100 kilo ohm. Find the maximum modulated frequency which could be detected by it.

We are given a diode envelope detector in which the shunt condenser (capacitor) has a capacitance

$$C = 250 \text{ pF}$$

and the load resistance is

$$R = 100 \text{ k}\Omega.$$

The depth of modulation is quoted as 60 %, so the modulation index is

$$m = 0.60.$$

For a diode detector to follow the envelope of an amplitude-modulated (AM) signal faithfully, the RC time constant must be small enough so that the capacitor can discharge appreciably during one half-cycle of the lowest modulation frequency. A standard design inequality for avoiding diagonal clipping is stated as

$$R\,C \;\le\; \frac{1}{2\pi\,f_m\,m},$$

where $$f_m$$ is the highest (maximum) modulating signal frequency that we want the detector to reproduce without noticeable distortion and $$m$$ is the modulation index.

We want to know that limiting frequency $$f_{m(\max)}$$, so we solve the above inequality for $$f_m$$. Rewriting, we obtain

$$f_{m(\max)} \;=\; \frac{1}{2\pi\,R\,C\,m}.$$

Now we substitute the numerical values, first converting every quantity to SI units:

$$R = 100 \text{ k}\Omega \;=\; 100 \times 10^3 \;\Omega = 1.0 \times 10^5 \;\Omega,$$

$$C = 250 \text{ pF} \;=\; 250 \times 10^{-12} \text{ F} = 2.50 \times 10^{-10} \text{ F}.$$

The RC time constant therefore is

$$R\,C \;=\; (1.0 \times 10^5)\,(2.50 \times 10^{-10}) = 2.50 \times 10^{-5} \text{ s} = 25 \;\mu\text{s}.$$

Next we insert this value, together with $$m = 0.60,$$ into the expression for $$f_{m(\max)}$$:

$$f_{m(\max)} = \frac{1}{2\pi\,R\,C\,m} = \frac{1}{2\pi \,(2.50 \times 10^{-5})\,(0.60)}.$$

Combining the numbers step by step, we first multiply the denominator:

$$2\pi = 6.2832,$$

$$6.2832 \times 0.60 = 3.76992,$$

$$3.76992 \times 2.50 \times 10^{-5} = 9.4248 \times 10^{-5} \text{ s}.$$

Now we take the reciprocal to get the frequency:

$$f_{m(\max)} = \frac{1}{9.4248 \times 10^{-5}} = 1.0617 \times 10^{4} \text{ Hz}.$$

Expressing this in kilohertz,

$$f_{m(\max)} \approx 10.62 \text{ kHz}.$$

Hence, the correct answer is Option D.