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(B) The attenuation in decibels $(dB)$ is given by the formula: $10 \log_{10} (I/I_0) = -\alpha_{dB} x$,where $\alpha_{dB}$ is the attenuation constan

Vedclass · Accepted Answer

$0.0602$

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(B) The attenuation in decibels $(dB)$ is given by the formula: $10 \log_{10} (I/I_0) = -\alpha_{dB} x$,where $\alpha_{dB}$ is the attenuation constant in $dB/km$.Given that the intensity falls by $50$ percent,the final intensity $I = 0.5 I_0$,which means $I/I_0 = 1/2$.The distance $x = 50 \ km$.Substituting these values into the formula:$10 \log_{10} (1/2) = -\alpha_{dB} 	imes 50$$10 (\log_{10} 1 - \log_{10} 2) = -50 \alpha_{dB}$Since $\log_{10} 1 = 0$ and $\log_{10} 2 \approx 0.3010$:$10 (0 - 0.3010) = -50 \alpha_{dB}$$-3.010 = -50 \alpha_{dB}$$\alpha_{dB} = 3.010 / 50 = 0.0602 \ dB/km$.

$List-I$	$List-II$
$A.$ Facsimile	$I.$ Static Document Image
$B.$ Guided media Channel	$II.$ Local Broadcast Radio
$C.$ Frequency Modulation	$III.$ Rectangular wave
$D.$ Digital Signal	$IV.$ Optical Fiber

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$List-I$ $List-II$
$A.$ Facsimile $I.$ Static Document Image
$B.$ Guided media Channel $II.$ Local Broadcast Radio
$C.$ Frequency Modulation $III.$ Rectangular wave
$D.$ Digital Signal $IV.$ Optical Fiber
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