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JEE Mains · Physics · STD 11 - 14. waves and sound
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 \(\left. dB =10 \log _{10}\left(\frac{ P _{0}}{ P _{i}}\right)\right]\)
- A \(4\)
- B \(12\)
- C \(16\)
- D \(8\)
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 \(20\, km =100 \,dB\) \(\beta_{2}-\beta_{1}=10 \log _{10} \frac{ I _{2}}{ I _{1}}\) \(-100=10 \log _{10} \frac{ I _{2}}{ I _{1}} \Rightarrow \frac{ I _{1}}{ I _{2}}=10^{10}\)…
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