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 receiver is $10^{-x} \, W$. The value of $x$ is ....... .
$\left[\right.$ Gain in $\left. dB =10 \log _{10}\left(\frac{ P _{0}}{ P _{i}}\right)\right]$
JEE MAIN 2021, Diffcult
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Sound level decreases by $5\, dB$ every $km$ so sound level decreased in $20\, km =100 \,dB$
$\beta_{2}-\beta_{1}=10 \log _{10} \frac{ I _{2}}{ I _{1}}$
$-100=10 \log _{10} \frac{ I _{2}}{ I _{1}} \Rightarrow \frac{ I _{1}}{ I _{2}}=10^{10}$
$I _{2}=10^{-10} I _{1} \Rightarrow P _{2}=10^{-10} P _{1}=10^{-8}\, W$
$\mathrm { x } = 8$
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