Report Issue

JEE MAIN 2021
24-02-2021 S2
Question
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 $10^{-\mathrm{x}} \mathrm{W}$. The value of x is $\_\_\_\_$ . [Gain in $\mathrm{dB}=10 \log _{10}\left(\frac{\mathrm{P}_0}{\mathrm{P}_{\mathrm{i}}}\right)$ ]
Write Your Answer
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Solution Image
Question Tags
JEE Main
Physics
Easy
Start Preparing for JEE with Competishun