Answer:
Option B
Explanation:
$\tau= RC= 100\times 10^{3}\times250\times 10^{-12}s$
= 2.5 x 10-5 s
The higher frequency which can be detected with tolerable distortion is
$f=\frac{1}{2\pi m_{a}RC}$
$=\frac{1}{2\pi \times0.6\times2.5\times 10^{-5}}Hz$
$=\frac{100\times 10^{4}}{25 \times1.2\pi }Hz$
$=\frac{4}{ 1.2\pi }\times 10^{4}Hz$
= 10.61 kHz