Question

Chandigarh radio station broadcast at $1200\;{\rm{kHz}}$. At what metre Chandigarh station would be turned in your transistor?

Answer 1

The above problem can be resolved using the mathematical formulation of the frequency of signal waves. Consider a signal wave of some specific wavelength and some definite value of frequency of the signal wave. This signal wave will then propagate in a specific medium such that the distance upto which the amplified signal reaches out becomes equal to the wavelength of the signal wave. The variables like the speed of light are constant, for which the subsequent substitution of the value of frequency is applied to obtain the final desired result.

Complete step by step solution
Given data:
The frequency of the broadcast is, $f = 1200\;{\rm{kHz}} = 1200\;{\rm{kHz}} \times \dfrac{{1000\;{\rm{Hz}}}}{{1\;{\rm{kHz}}}} = 12 \times {10^5}\;{\rm{Hz}}$.
Let d be the distance at which the Chandigarh station is turned on to the transistor.
Then the expression for the distance is given as,
$d = \dfrac{c}{f}$
Here, c is the speed of light and its value is $3 \times {10^8}\;{\rm{m/s}}$.
Solve by substituting the values in above equation as,

d = \dfrac{c}{f}\\\ \Rightarrow d = \dfrac{{3 \times {{10}^8}\;{\rm{m/s}}}}{{12 \times {{10}^5}\;{\rm{Hz}}}}\\\ \Rightarrow d = 250\;{\rm{m}} \end{array}$$ Therefore, the Chandigarh station would be turned in the transistor at the distance of $$250\;{\rm{m}}$$. **Note:** To resolve the given problem, one must go through the various concepts and applications of the wavelength formula or the frequency formula. The signal wave frequency will vary inversely with the wavelength of the wave but will show the direct relationship with the velocity of light. In many cases, this value remains constant, and other variables can be calculated accordingly. Moreover, the concept has wider applications in the field of broadcasting and amplifications.