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A particular station of All India Radio, New Dehli, broadcasts on a frequency of 1,368 kHz (kilohertz), the wavelength of the electromagnetic radiation emitted by the transmitter is : [speed of light, \mathrm{e=3.0\times10^{8}ms^{-1}}]
 

Option: 1

219.3\; m


Option: 2

219.2\; m


Option: 3

2192\; m

 


Option: 4

21.92\; cm


Answers (1)

best_answer

We know that,

\nu = \frac{c}{\lambda}

\mathrm{\therefore \lambda= \frac{c}{\nu} = \frac{3 \times 10^8\ m/s}{1368 \times 10^3 \ s}=219.3\ m}

Therefore, the correct option is (1).

Posted by

Ritika Jonwal

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