QJEEM13Ph9acc: JEE Main 2013 Physics - Class 12 JEEM13Ph9acc Solution: A diode detector is used to detect an amplitude modulated wave…

Given : Resistance $R=100$ kilo $ohm=100$ $\times {10^3}\Omega$

$\therefore$ Capacitance $C=250$ picofarad $= 250 \times {10^{ - 12}}F$

$\tau = RC = 100 \times {10^3} \times 250 \times {10^{ - 12}}\,\sec$

$= 2.5 \times {10^7} \times {10^{ - 12}}\sec$

$= 2.5 \times {10^{ - 5}}\sec$

The higher frequency which can be detected with tolerable distortion is

$f = {1 \over {2\pi {m_a}RC}}$

$= {1 \over {2\pi \times 0.6 \times 2.5 \times {{10}^{ - 5}}}}Hz$

$= {{100 \times {{10}^4}} \over {2.5 \times 1.2\pi }}Hz$

$= 10.61KHz$

This condition is obtained by applying the condition that rate of decay of capacitor voltage must be equal or less than the rate of decay modulated signal voltage for proper detection of modulated signal.