Q 12-14-137JEE MainJEE Main 2021 (24 Feb, Shift 2)Easy
A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is $-5$ dB per km and cable length is 20 km. The power received at the receiver is $10^{-x}$ W. The value of $x$ is ______. [Gain in dB $= 10\log_{10}\left(\dfrac{P_o}{P_i}\right)$]
Numerical value type. Enter your answer.
Answer: 8
Total gain $= -5\times20 = -100$ dB:
$$10\log_{10}\frac{P_o}{P_i} = -100 \Rightarrow \frac{P_o}{P_i} = 10^{-10}$$
$P_o = 100\times10^{-10} = 10^{-8}$ W, so $x = 8$.
Solution by Sreeraj P, M.Sc Physics