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JEE Mains · Physics · STD 11 - Waves and sound
- JEE Mains 2021
- Hard
- Numerical
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 \(\left. dB =10 \log _{10}\left(\frac{ P _{0}}{ P _{i}}\right)\right]\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(8\)
Step-by-step Solution
Detailed explanation
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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